Sem5 ae sscp3323 01 ae v2

SSCP 3323 Advanced Electronics SEMESTER 1, Academic

Session 2015/2016

Week Date Topic Notes

1 6-Sep-15INTRODUCTION: Course introduction, basic requirement. Review on transistor,

capacitor’s role and frequency response.

2 13-Sep-15SMALL SIGNAL AMPLIFIER: Introduction to small signal amplifiers, analysis

using h-parameters and p-model, examples of calculation.

3 20-Sep-15 Multi-stage amplifier – analysis and examples

4 27-Sep-15 POWER AMPLIFIER: Class A, class B, class AB, class C, examples of calculation.

5 4-Oct-15DIFFERENTIAL AMPLIFIER: Introduction to differential amplifiers, analysis,

calculation and examples of differential amplifiers.

6 11-Oct-15OPERATIONAL AMPLIFIER (OPAM): ideal characteristics, inverting amplifier, non-

inverting amplifier, voltage follower.

7 18-Oct-15Summing amplifier, integrator, differentiator, active filter, low pass filter, high

pass filter.

8 25-Oct-15

DIGITAL SYSTEM & NUMBERS: Principle of digital electronics, analogue and

digital system, numbers system – decimal, binary, hexadecimal, octal. BINARY

LOGIC GATES: AND, OR, INVERTER, NAND, NOR, XOR, XNOR.

9 1-Nov-15Boolean expressions, minterm and maxterm circuits, truth tables, Karnaugh

maps, data selectors.Test1

10 8-Nov-15 SEMESTER BREAK

11 15-Nov-15

INTEGRATED CIRCUITS (IC): TTL and CMOS, IC specification and simple

interfacing. CODES: BCD code, excess-3 code, gray code, ASCII code, encoders,

seven-segment displays, decoders.

12 22-Nov-15FLIP-FLOPS: RS, JK and D flip-flops, latches, Schmitt trigger, ripple counters,

synchronous counters, Up and down counters.

13 29-Nov-15Frequency dividers, TTL and CMOS IC counters. Serial and Parallel load shift

registers, universal shift register, TTL and CMOS shift register ICs.

14 6-Dec-15

DIGITAL ARITHMETIC CIRCUITS: Binary addition, half and full adders, three-bit

adders, binary subtraction, using adders for subtraction, four-bit

adder/subtractors, IC adders, Parallel adder/subtractor systems.

Test2

15 13-Dec-15Binary multiplication, binary multipliers. 2s complement notation, addition and

subtraction. 2s complement adder/subtractors.

16 20-Dec-15 Study week

17 27-Dec-15 Final examination

Assessment

Quiz 2x5 = 10

TEST 2 X 15 = 30%

Assignments = 10%

Final Exam = 50%

References:

(1)

(2) T.F. Bogart Jr.,J. S. Beasley and G. Rico, Electronic Devices and Circuits,Prentice Hall

(3) A.P. Malvino, Electronic Principles, McGraw Hill

(4) Roger L. Tokheim, Digital Electronics, Mc Graw Hill.

Donald A. Neamen, MICROELECTRONICS - Circuit analysis and design , McGraw Hill.

LECTURE 01 SSCP3323-Advanced Electronics 1

SSCP 3323 – ADVANCED ELECTRONICS

Pre-requisites : Basic Electronics

Lecturer : Assoc. Prof. Dr. Yaacob Mat Daud

Room : C21-314

Contacts

Tel. : 07 - 5534070

Mobile : 012 - 7581247

e-mail : [email protected]


Synopsis

This course is designed to expose the students to

• small signal analysis using an equivalent circuit h-parameter

• analysis of simple circuits such as:

• Filters

• wave generators

• Oscillators

• basic digital system

At the end of the course, students should be able to

• analyze small signal amplifier and analyze simple circuits

• explain various logic principles and devices employed in digital system, as

well as the function of digital circuit in the development of a simple computer


Synopsis

This course introduces students to the basic concepts and knowledge

of the analogue and digital electronics.

For the analogue part, the transistor circuits, small signal amplifiers,

power amplifies, differential amplifiers, operational amplifiers (OPAMP)

and its application circuits are discussed.

For the digital part, basic concepts and principles in understanding

digital circuits and systems including number codes and systems,

Boolean algebra, logic gates, Karnaugh maps, IC specification and

interfacing, encoding and decoding, flip-flops, counters, shift registers

and digital arithmetic circuits are discussed.


Learning Outcomes

After completing this course, the student must have the ability to:

1. State and explain the characteristics and equation of the transistor

equivalent circuits model

2. Analyze and solve the various type of amplifier circuit problems

3. State and explain the characteristics and operations of the OPAMP

4. Design the OPAMP circuits for simple application

5. State the characteristics and specifications of the TTL and CMOS digital

ICs

6. State and explain the principle of digital systems and devices such as logic

gates, encoders, decoders, flip-flops, counters and shift registers

7. Explain and solve problems involving digital arithmetic circuits


Evaluation

The evaluation will be based on the following:

Quiz 1 : 5%

Quiz 2 : 5%

Test 1 : 15%

Test 2 : 15%

Assignments (6) : 10%

Total Course Work : 50%

Final Exam. : 50%


Text and References

Text Book:

Donald A. Neamen – MICROELECTRONICS Circuit

Analysis and Design, 3rd Edition, 2007, McGraw-Hill.

References:

1. T.F. Bogart Jr., J.S. Beasley and G. Rico, ELECTRONIC

DEVICES AND CIRCUITS, Prentice Hall

2. A.P. Malvino, ELECTRONIC PRINCIPLES, McGraw-Hill

3. Roger L. Tokheim, DIGITAL ELECTRONICS, McGraw-

Hill


Quick Review

What are DC and AC ?

DC (Direct Current)

The polarity (sign) of the voltage/current does not change

Examples:

Constant DC voltage DC wave voltage


Quick Review

AC (Alternating Current)

The polarity (sign) of the voltage/current changes periodically with

time

Examples:


Quick Review


Quick Review

Electronic Components

1. Source

Current source


Quick Review

2. Passive components


Quick Review

LfLXL 2

Reactance

Capacitor :

Inductor :

For DC (f 0)

Capacitor : XL 0

Inductor : XC

)2/(1/1 CfCXC


Quick Review

Resistors

Its nature that opposes the current is independent of

frequency

Capacitors

When frequency increases, the opposition to the current

decreases

Inductors

When the frequency decreases, the opposition to the

current increases


Quick Review

Transformers


Quick Review

3. Active Components

Semiconductor Devices

Diode


Quick Review


Quick Review


Quick Review


Quick Review


Quick Review

Circuit Analysis/Problem Solving


Quick Review


Quick Review

LECTURE 02 SSCP 3323 - Advanced

Electronics

1

BJT Amplifier Analysis Using h-Parameters

Introduction

There are three simple

configurations of BJT

amplifiers: common-emitter,

common-collector, and

common-base

We will look at the the

common-emitter because it

provides voltage gain

The figure shows a typical

CE amplifier

+VCC

RC

vs

B

C

E

R1

C1

C2

RE

R2

RL

RS


Electronics

2


The purpose of the analysis is to determine various quantities

related to the amplifier, such gain, input and output impedances

To analyze the circuit we need to use a model of the amplifier

A basic amplifier model of all amplifiers is as shown below:

RS

vin

Zin A

v v

in

Zout

vout

General Amplifier Model


Electronics

3


The amplifier model represents input conditions, the amplification

factor or gain (hFE or ), and output conditions

The signal source has some internal resistance and is shown as RS

The transistor can be shown as input impedance (Zin), a voltage

source (vin), the output voltage (vout), and some output impedans

(Zout)

The value of vout is vin times the voltage gain (Av) of the transistor

vout is the voltage developed across the load resistance (RL)

As vin changes, vout changes by a factor of Av


Electronics

4


TWO PORT NETWORKS – h-PARAMETER BJT MODEL

Depending on the application, it may be used in a number of

different ways to develop different models

We will use it to develop the h-parameter model

The h-parameter model is typically suited to transistor circuit

modeling. It is important because:

its values are used on specification sheets

it is one model that may be used to analyze circuit behavior

it may be used to form the basis of a more accurate transistor

model


Electronics

5


The h - parameter model has values that are complex numbers that

vary as a function of:

Frequency

Ambient temperature

Q-Point

A two-port network is shown below:

Two-port active device


Electronics

6


We may select two of the four quantities as the independent

variables and express the remaining two terms of the chosen

independent variables

If the current i1 and the voltage v2 are independent and if the two-

port is linear, we may write

2121111 vhihv

2221212 vhihi

The quantities h11, h12, h21, h22 are the h, or hybrid, parameters

because they are not all alike dimensionally


Electronics

7


Then, the h parameters are defined as follows :

01

111

2

iv

hi

vh

This is defined as input resistance

with output short circuited (ohm)

02

112

1

ri

hv

vh

This is defined as reverse open–

circuit voltage gain (dimensionless)

01

221

2

fv

hi

ih

02

222

1

oi

hv

ih

This is defined as forward short-

circuit current gain (dimensionless)

This is defined as output conductance

with input open circuited (mho)


Electronics

8


Rewrite:

211 vhihv ri

212 vhihi of

The Model:

Hybrid circuit


Electronics

9


TRANSISTOR HYBRID MODEL

Consider the common emitter configuration,

sticscharacteri input ),(1 CEBBE vifv

sticscharacteri output ),(2 CEBC vifi

Using Taylor’s series expansion and neglecting higher order terms,

CE

ICE

BEB

VB

BEBE v

v

vi

i

vv

BCE

CE

ICE

CB

VB

CC v

v

ii

i

ii

BCE


Electronics

10


Rewrite

ce

ICE

BEb

VB

BEbe v

v

vi

i

vv

BCE

ce

ICE

Cb

VB

Cc v

v

ii

i

ii

BCE

or

cerebiebe vhihv

ceoebfec vhihi


Electronics

11


Subscript e indicates common emitter configuration, and

0

ceCEvb

be

VB

BEie

i

v

i

vh

0

bBice

be

ICE

BEre

v

v

v

vh

0

ceCEvb

c

VB

Cfe

i

i

i

ih

0

bBice

c

ICE

Coe

v

i

v

ih


Electronics

12


Summary


Electronics

13



Electronics

1



The common-emitter (CE) h-parameter model for a BJT amplifier is

shown below :

h - parameter model for CE


Electronics

2


For the configuration, we note from KCL that

cbe iii

The circuit models and equations are valid for either an npn or a

pnp transistor and are independent of the type of load or the

method of biasing

The h-parameters can be obtained graphically from the transistor

characteristics


Electronics

3


For example, the parameters hfe and hoe can be obtained from the

output characteristics of the CE transistor

Similarly the parameters hre and hie can be obtained from the input

characteristics

The h-parameters will depend on the temperature and the

quiescent point

Manufacturers usually provide curves of h-parameters versus

temperature, VCE and IC


Electronics

4


Graphical Representation of h-Parameter for BJT Small-Signal

Model

Small-signal models for the BJT describe the operation of the

transistor as a linear amplifier of small AC signals

They make use of the relative linearity of the base and collector

curves near the Q point, and are quite useful if the transistor

voltages and currents remain within some range near the Q point

We will use the convention that each voltage and current is the sum

of a DC component (the quiescent value, denoted by uppercase)

and a small-signal AC component (Δuppercase). Subscript Q

indicates a value at the Q point:


Electronics

5


h-Parameter Small-Signal Model


Electronics

6


Parameter hie represents the forward resistance of the BE junction

and equals the slope of the IB - VBE curve at the Q point

Parameter hre indicates the small dependence of the IB-VBE curve

on the value of VCE. We can usually assume that hre = 0


Electronics

7


Parameter hfe represents the current gain of the transistor and is

approximately equal to β

Parameter hoe represents the slope of the IC-VCE curves in the

linear active region, indicating that they are not flat


Electronics

8

ANALYSIS USING h-PARAMETER

Analysis of a BJT Amplifier Using h-Parameters

Objectives of the analysis are to find:

Current gain, AI = IL/I1 = - I2/I1

Input impedance, Zi = V1/I1

Voltage gain, Av = VL/V1

Output impedance, Zo = V2/I2 with Vs = 0 and ZL=


Electronics

9


Assumption:

Sinusoidal varying voltages and currents

h-parameters remain substantially constant over the operating

range

To form a transistor amplifier, it is only necessary to connect an

external load and signal source, as shown below, and to bias the

transistor properly

A two-port network equivalent of a transistor amplifier


Electronics

10


This two-port active network represents a transistor in any one of

the three possible configurations

When the active network in the circuit is replaced by the small-

signal hybrid model of the transistor, the circuit becomes

A two-port network with small-signal hybrid model


Electronics

11


Using h-parameter model,

The current gain, or current amplification, AI

For the transistor amplifier stage, AI is defined as the ratio of output

to input currents, or

From the figure,

211 VhIhV ri

212 VhIhI of

1

2

1 I

I

I

IA L

I

LLL ZIZIV 22


Electronics

12


Substituting V2 into the previous equation gives

)( 212 Lof ZIhIhI

12 )1( IhZhI fLo

Lo

fI

Zh

h

I

IA

11

2

LECTURE 04 SSP 2323 - Advanced Electronics 1


The Input Impedance Zi

The resistance Rs in the figure represents the signal-source

resistance

The impedance seen looking into the amplifier terminals (1, 1’) is

the amplifier input impedance Zi, or

Dividing equation by I1, gives

1

1

I

VZ i

211 VhIhV ri

1

2

1

1

I

Vhh

I

VZ rii



Substituting V2 into the above equation gives

Substituting AI into the above equation gives

ILriLL

ri AZhhI

ZIhh

I

V

11

1 )(

Lo

Lfrii

Zh

Zhhh

I

VZ

11

1



The Voltage Gain, or Voltage Amplification, Av

The ratio of output voltage V2 to input voltage V1 gives the voltage

gain of the transistor, or

Substitute V2 and V1 into the above equation gives

1

2

V

VA v

i

LI

i

LLv

Z

ZA

ZI

ZIA

1



Voltage gain taking into account the source resistance

This gain is defined as

The Thevenin’s equivalent for the source is shown below

sv

ssV

V

VA

VV

VV

V

VA

s

1

1

122

Thevenin’s equivalent for the source



From the Thevenin’s equivalent for the source

Substitute V1 into the equation of gain above gives

Substitute Av into the equation gives

ssi

i VRZ

ZV

1

si

ivV

RZ

ZAA

s

si

LIV

RZ

ZAA

s



Current gain AIs taking into account the source resistance

This gain is defined as

The Norton’s equivalent for the source is shown below

sI

ssI

I

IA

II

II

I

IA

s

1

1

122

Norton’s equivalent for the source



From the Norton’s equivalent for the source

Hence

Note that, if Rs = , then AIs = AI.

ssi

s IRZ

RI

1

Isi

sI A

RZ

RA

s



Hence AI is the current gain for an ideal current source Independent of the transistor characteristics, the voltage and

current gains, taking source impedance into account are related as

The output admittance

By definition, the output impedance

s

LIsV

R

ZAA

s

oo

YI

VZ

1

2

2



This is obtained by setting the source voltage Vs to zero and the

load impedance ZL to infinity, and driving the output terminal from a

generator V2 and drawing a current of I2

with Vs = 0 and RL =

But,

Divide this equation by V2:

2

2

V

IYo

212 VhIhI of

of hV

Ih

V

I

2

1

2

2



Applying KVL at the input terminal:

Substitute the above equation in Yo gives

21)(0 VhIRh rsi

si

r

Rh

h

V

I

2

1

si

rfoo

Rh

hhh

V

IY

2

2



Summary

Table 1 – Small-signal analysis of a transistor amplifier



Equivalent Circuit

An equivalent circuit is a simplified version of the original amplifier

circuit obtained when considering the flow of DC or AC current in

the circuit.

A capacitor appears as an open circuit and an inductor appears as

a short circuit to DC current flow.

Conversely, a capacitor appears as a short circuit and an inductor

appears as an open circuit to AC current flow.

Hence, for an amplifier circuit there are two types of equivalent

circuit, i.e. the DC equivalent circuit (also known as the bias circuit)

and the AC equivalent circuit.



DC Equivalent Circuit

DC equivalent circuit of an amplifier circuit is obtained by

open circuiting all capacitors and short circuiting all

inductors found in the circuit.

For example, consider the amplifier circuit in figure (a).

After open circuiting all the capacitors in the amplifier

circuit, the DC equivalent circuit is obtained as shown in

Figure (b).



Figure (a) Figure (b)

+ VCC

R1

R2

RC

RE

+ VCC

R1

R2

RC

RE

CE

C1

C2



AC Equivalent Circuit

AC equivalent circuit of an amplifier circuit is obtained by short circuiting all capacitors and open circuiting all inductors found in the circuit.

Coupling capacitors may be regarded as short-circuits to the AC signal, and therefore need not be included, together with any resistors which themselves are short circuited by capacitors

The DC supply in the amplifier circuit is a very low resistance path to the signal frequencies and can be replaced (in the AC equivalent circuit only!) by a short circuit.



Example:

+ VCC

R1

R2

RC

RE

CE

C1

C2

+ VCC

R1

R2

RC



+ VCC

R1

R2

RC R

1

R2

RC

Sumber

unggul

Rs

= 0

VCC



R1

R2

RC R

1

R2

RC



Finally….

R1

B

E

C

E

hre

vce

hfe

ib

hoe

hie

R2

RC

R2

R1

RC

AC Equivalent Circuit Hybrid Parameter Circuit



Example

The transistor used as an amplifier in a circuit has the h-parameters as shown in the Figure below:

(a) Draw the equivalent circuit

(b) Calculate

(i) the effective input resistance,

(ii) the output resistance,

(iii) the current gain, and

(iv) the voltage gain,

if component values are as follows:

R1 = 47 k, R2 = 10 k and RL = 1.5 k



The Amplifier Circuit:



The hybrid parameter equivalent circuit:



Solution

(a) The effective input resistance is the result of R1, R2 and

hie in parallel.

hie = 8.8 k

Thus

k 3.4

1111

21

in

iein

R

hRRR

The larger the values of R1 and R2, the closer Rin will

be to the value of hie.



(b) The output resistance is the reciprocal of hoe.

hoe = 60 S

Thus

Rout = (1/ hoe)

which gives Rout = 16.7 k

(c) The current gain (Ai) is given by Ib/Ic.

where R’L is the result of RL and 1/hoe in parallel.

Lbfece

L

cec RIhV

R

VI and



The ‘minus’ signs here are the result of the assumed direction of

the output current.

Thus,

552 k 1.5

k 1.38 x 600 gain current

k 1.38 giving

k 16.71

and k 38.1

)( gain current and

L

out

oe

L

L

Lfe

b

ci

L

Lbfec

R

Rh

R

R

Rh

I

IA

R

RIhI



(d) The voltage gain (Av) is given by Vce/Vbe.

192 - k 4.3

k 38.1600

gain voltage thus

,k 3.4

and

in

Lfe

be

cev

in

inbbeLbfece

R

Rh

V

VA

R

RIVRIhV

The ‘minus’ sign represents the 180 phase shift between

output and input signal voltages for the common emitter

amplifier



Important Points

If the effect of hoe is negligible, which is sometimes the

case, then R’L becomes RL, resulting in the current and

voltage gains being as follows

in

Lfev

fei

R

RhA

hA

)( gain Voltage

)( gain Current

Further more, if Rin = hie, then

ie

Lfev

h

RhA )( gain Voltage


The Hybrid- Model of BJT

Introduction

Since long, industrial and educational institutions have heavily relied

on the hybrid parameters because they produce more accurate

results in the analysis of amplifier circuits

In fact, hybrid-parameter equivalent circuit continues to be popular

even today

But their use is beset with the following difficulties:

– The values of h-parameters are not so readily available

– Their values vary considerably with individual transistors even of the same type

number

– Their values are limited to a particular set of operating conditions for reasonably

accurate results

– The dominant model used for small-signal analysis of a BJT in the forward-active

region, the h-parameter model, does not contain frequency sensitive elements

and is therefore invariant with respect to changes in frequency



It is therefore necessary to

introduce a new BJT model or

to reinterpret an old model to

include frequency-dependent

terms using the Ebers-Moll

model as a basis for creating

the new model

Consider the base current

versus base emitter voltage

characteristics, with small

time-varying signals

superimposed at the Q-point,

as shown in the figure:



Since the sinusoidal signals are small, we can treat the slope at the

Q-point as a constant, which has units of conductance

The inverse of this conductance is the small signal resistance define

as r

The resistance r is called the diffusion resistance or base emitter

input resistance



Note that r is a function of the Q-point parametes

Another parameter is called transconductance and is written as gm

This conductance relates a current in the collector to a voltage in

the BE circuit



By considering the Early effect, there is a small signal transistor

output resistance, ro that lies between collector and emitter



We can develop a slightly different form for the output of the

equivalent circuit

We can relate the small-signal collector current to the small-signal

base current



Stray Elements in a BJT at High Frequencies

At high frequencies, there exists some transistor characteristics that

have to be considered:

(a) Bulk Resistance

The recombination of charge carriers occurs at point B’, which lies

between JE and JC

The base current flows from the base terminal, B, to point B’

through the base region and forming a voltage drop between B’ and

B

Hence, it can be figured out that there exists a resistance known as

the bulk resistance (denoted as rx or rb) of the base

At the emitter and collector there exist bulk resistance re’e and rc’c

too, but these resistance are too small and can be neglected



(b) Diffusion Capacitance The effect of forward bias at the emitter-base junction

(JE) is to inject charge carriers from a region to another

The number of charge carriers injected into the base are

changing due to the applied alternating signal

The level of injection cannot change immediately

following the change in the signal, especially for high

frequency signal

In effect, a charge storage is formed between base and

emitter – the effect of capacitance, known as diffision

capacitance (Cbe)



(c) DepletionCapacitance On both sides of the base-emitter junction (JE) and the

collector-base junction (JC), a depletion layer is formed

When the voltage change slightly, due to the small-

signal alternating voltage at the input of the transistor,

the charge stored in the depletion layer also change

Hence, the effect of capacitance, known as the depletion

capacitance formed at the junctions

At JE, the capacitance is denoted as Cje and when it is

combined with Cbe, it is denoted as C.

At JC, the depletion capacitance is denoted as Cjc or C.



(d) Reverse-Biased Collector-Base Junction Resistance

There exists a base-collector junction resistance, known

as the reverse-biased collector-base junction resistance,

denoted as r.

The resistance provide some feedback between the

output and the input, meaning that the base current is a

slight function of the collector-emitter voltage



The Hybrid- Model

In this model, proposed by L.J. Giacoletto, the circuit

elements are arranged in a -configuration known as the

hybrid- model The hybrid- BJT model includes elements that are

negligible at low frequencies and midband, but cannot

be ignored at higher frequencies of operation

In hybrid- model, some h-parameters changes in name,

i.e.



In hybrid- model, some h-parameters changes in name,

i.e.

– hie is changed to rbe or r

– 1/hoe is changed to rce or ro

– ib is written as ib = vbe/rbe = vbe/r

– vbe is denoted as v

– Hence, hfeib = hfe(vbe/rbe) = gmvbe



The hybrid- BJT model at high frequencies is as shown

below:



Hybrid- Parameters

1. rx = ohmic resistance of base region, a few tens of ohms

2. r = dynamic resistance of base region

3. ro = collector resistance of BJT

4. C = diffusion capacitance of the base-emitter junction, 100 pF to 1000 pF.

5. gm = BJT transconductance (gm = /r = ICQ/VT)

Note:

r, C represent the characteristics of the reverse-biased collector-base junction (r several Megohms, while C 1 pF to 10 pF)



Analysis of a CE Amplifier Configuration Using

Hybrid- Model in Low- and Mid-Frequency Range

In the low- and mid-frequency range, the reactance of

the capacitances are very large and can be considered

open circuit

The hybrid- model using the transconductance and

current gain parameters is shown below:



The hybrid- model of (a) NPN and (b) PNP transistor



Basic Common Emitter Amplifier

(a) Common emitter circuit with a voltage-divider biasing

(b) The small-signal equivalent circuit



Input Resistance, Ri :

rRRRi //// 21

Output Resistance, Ro :

coo RrR //

Voltage Gain (Taking source into consideration), Avs :

s

o

s

ovs

V

V

V

V

V

VA



where

)//( como RrVgV

and

ssi

i VRR

RV

therefore

si

icomvs

RR

RRrVgA ))//((



Current Gain (Taking source into consideration), Ais :

where

and

i

o

i

ois

I

V

V

I

I

IA

VgR

RrI m

c

coo

//

ii RIV

therefore

ic

comis R

R

RrgA

)//(



Circuit with Emitter Resistor

(a) CE with emitter resistor (b) The small-signal equivalent circuit



Input Resistance, Ri :

where

Output Resistance, Ro :

bi RRRR //// 21

Eb

bb Rr

I

VR )1(

co RR



Voltage Gain (Taking source into consideration), Avs :

where

Therefore

s

b

b

b

b

o

s

ovs

V

V

V

I

I

V

V

VA

s

si

ib

b

bbcbo V

RR

RV

R

VIRIV

and , ),

si

i

Ecvs

RR

R

RrRA

)1(

1)(



If Ri >> Rs and if (1 + )RE >> r then the small-signal gain

is approximately

E

c

E

cvs

R

R

R

RA

)1(

Current Gain (Taking source into consideration), Ais :

i

b

b

o

i

ois

I

I

I

I

I

IA



where

E

ii

b

iibbo

Rr

IR

R

IRIII

)1( and

Therefore

E

iis

Rr

RA

)1(

If (1 + )RE >> r then the small-signal current gain is

approximately

E

i

E

iis

R

R

R

RA

)1(



Effect of C and C

Recall the hybrid- model of BJT :



Notice the small values of C and C, especially when

compared to typical values of Cin, Cout, and CE.

At low and midband frequencies, C and C appear as

open circuits

At high frequencies, where C and C have an effect,

Cin, Cout, and CE appear as short circuits



Introduction

At high frequencies, where C and C have an effect, Cin,

Cout, and CE appear as short circuits



To focus our attention, we will assume rx 0 and r ,

and we will use Miller Effect to replace C:

Simplified hybrid- BJT model using the Miller Effect



Using Miller Effect equation,

CAACC vv )1(1

CA

CCv

112

Individually, all capacitors in the figure has a single-pole

low-pass effect

As frequency increases they become short circuits, and

vo approaches zero



Thus there are two low-pass poles with the

mathematical form:

Thevenineqb

RCf

2

1

Because (C1 + C) >> C2, the pole due to (C1 + C) will

dominate

The pole due to C2 is usually negligible, especially when

R’L is included in the circuit



Typical amplifier response in the mid-band and high

frequencies regions. fh1 is normally due to (C1 + C),

and fh2 is normally due to C2

f

A



High-Frequency Performance of CE Amplifier

We now have tools we need to analyze (actually,

estimate) the high frequency performance of an

amplifier circuit

Consider the common-emitter amplifier, as shown

Figure (a):

Using the hybrid- equivalent for the BJT, we construct

the small-signal equivalent circuit for the amplifier, as

shown in Figure (b):



Figure (a)



Figure (b):



The circuit can be simplified further by using a Thevenin

equivalent on the input side, and by assuming the effect

of r to be negligible

Modified small-signal equivalent



Note that the Thevenin resistance

])//([//' sBxs RRrrR

Recognizing that the dominant high-frequency pole

occurs on the input side, we endeavor to calculate fh1

Thus, we ignore the effect of C on the output side,

calculate the voltage gain, and apply the Miller Effect on

the input side only

Lmo

v Rgv

vA '



Final (approximate) equivalent after applying the Miller

Effect



So we have

totalsh

CRf

'2

11

where

Lmtotal RgCCC '1

and

)//(//' sBxs RRrrR



Text



Text

SSCP 3323 – ELEKTRONIK LANJUTAN SESI 2012/13 SEM 1 TUGASAN 1

Soalan 1

Lukis kan litar setara a.t. dan litar setara a.u. bagi litar-litar amplifier dalam Rajah 1(a) dan Rajah 1(b).

(a)

(b)

Rajah 1

Soalan 2 Lukiskan model hibrid bagi litar amplifier dalam Rajah 2.

+VCC

RL

vs

B

C

E

RB

vo

Rajah 2 Dengan menggunakan model itu dapatkan (i) persamaan input dan output, (ii) impedans input dan output, (iii) gandaan voltan, (iv) gandaan arus bagi litar amplifier itu.

Example 1

Calculate the small-signal voltage gain

of the bipolar transistor circuit shown in

Figure 1.

Assume the transistor circuit

parameters are = 100, VCC = 12 V,

VBE = 0.7 V, RC = 6 k, RB = 50 k,

and VBB = 1.2 V.

Figure 1

AR

VVI

B

onBEBBBQ 10

50

7.02.1)(

mA 1)10)(100( AII BQCQ

i) DC Solution

so that

Answer

k 6.21

)026.0)(100(

CQ

T

I

Vr

mA / V 5.38026.0

1

T

CQm

V

Ig

The small-signal hybrid- parameters are

and

ii) AC Solution

Figure 2

The small-signal voltage gain is

s

o

s

ovs

V

V

V

V

V

VA

Cmo R)Vg(V

sB

VRr

rV

4.11

506.2

6.2)6)(5.38(

)(

BCm

s

ov

Rr

rRg

V

VA

Therefore

Example 2:

For the network of Figure 3, find:

i) ICQ

ii) gm, r

iii) Zi

iv)Zo

v) Av

Figure 3

i) DC Solution Answer

V81.2

k56k2.8

k2.822

RR

RVVV

21

2CC2RB

V11.2

7.081.2

VVV BEBE

mA41.1

R

VII

E

EECQ

ii) AC Solution

The small-signal hybrid- parameters are

V/mA54

026.0

mA41.1

V

Ig

T

CQm

k6.1

10x54

90

gr

3m

k35.1

r//R//RZ)iii 21i k8.6RZ)iv co

V

VA)v o

v

cmo R)Vg(Vbut

367

)k8.6(m54

RgV

V,Thus cm

o

SSCP3323 Test #1 13th

Nov 2014

1. a) Draw the h-parameters equivalent circuit of an npn transistor that connected in

common-emitter configuration. Referring to the circuit drawn, write down the

Kirchoff’s equations at the input and output ports.

b) For the amplifier circuit shown in Figure 1, h parameters for the transistor are hie =

1.0k, hf e= 80, hoe = 0, hre = 0. i) Draw the ac equivalent circuit for the amplifier using h-parameter model.

ii) Find the input impedance Zi = vs/is, current gain Ai = io/is, and voltage gain

Avs = vo/vs.

iii) What is the output impedance seen by 10k load.

2. a) Sketch the hybrid- equivalent circuit of an npn bipolar transistor.

b) State the relationships of the hybrid- parameters gm, r and ro to the transistor dc

quiescent values, IBQ, ICQ and VCEQ.

c) For the circuit in Figure 2, the transistor parameter are = 180 and ro = . i) Determine the Q-point values, IBQ, ICQ and VCEQ.

ii) Find the hybrid- parameters, gm and r.

iii) Find the small signal voltage gain Av=vo/vs

vs

360k

2.2k

VCC

Figure 1

is

io 10k

vo

Figure 2

is

vs

VCC = 5V

vi

1K

1.2K

6K

1.5K

0.2K

vo

0.1K

io



Miller’s Theorem and Its Dual Certain configurations can be analyzed more simply by using

Miller’s theorem

Consider an arbitrary circuit configuration, as shown below, with N distinct nodes 1, 2, …, N

Let the node voltages be V1, V2, …, VN and the node voltage VN = 0 since N is the ground node



By definition,

The two circuits are the same if the voltage ratio AV can be found

by some independent means

KV

VAv

1

2



Proof :

where

This can be written as

where

Also, it can be shown that

'

)1(

'121

1Z

AV

Z

VVI v

1

2

V

VAv

1

111

)1/(' Z

V

AZ

VI

v

vA

ZZ

1

'1

1

'

1

'

/11

'

/11

'2

v

v

v A

AZ

K

KZ

A

Z

K

ZZ



Example on Miller’s Theorem



Dual of Miller’s Theorem

Consider the network, as shown below, with arbitrary active and

passive linear elements between nodes 1, 2, and 3



By definition,

The two circuits are the same if excited by the same voltages V1’N

and V2’N

1

2

I

IAI



Proof :

Therefore

1113

113

1

2113

2113'1

')1(

'1

')(

ZIV

ZAIV

ZI

IIV

ZIIVV

I

N

')1(1 ZAZ I



Also, it can be shown that

Therefore,

2223'2 ZIVV N

2223

223

2

1223

2123'2

'1

1

'1

')(

ZIV

ZA

IV

ZI

IIV

ZIIVV

I

N

I

I

A

AZZ

)1('2



Example on Dual of Miller’s Theorem

SSCP 3323 – ELEKTRONIK LANJUTAN SESI 2012/13 SEM 1 TUGASAN 2 Soalan 1 Lukis model hibrid bagi litar CE dengan rintangan pemancar yang ditunjukkan dalam Rajah 3. Dengan menggunakan model hybrid itu, tentukan (i) impedans input litar (ii) impedans output litar (iii) gandaan voltan litar (iv) gandaan arus litar

+VCC

RL

vs

BC

E

RB

vo

RE

Rajah 3

1

Soalan 2 Lukis model hibrid bagi litar yang ditunjukkan dalam Rajah 4. Dengan menggunakan model hibrid litar itu, tentukan (i) impedans input litar (ii) impedans output litar (iii) gandaan voltan litar (iv) gandaan arus litar

+VCC

RC

vs

BC

E

R1

C1

C2

RER2 RL

2

SSCP 3323 – Advanced Electronics

MULTISTAGE AMPLIFIER Introduction

• To obtain the gain an electronic system needs, it is often necessary to connect amplifiers in series

• When amplifiers are connected in series, they are said

to be cascaded

• In the previous analysis, the circuit consists of only one transistor

• But the circuit analysis has involved the use of

multistage concept, e.g. in the calculation of voltage gain

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛==

s

o

s

ovs v

vvv

vvA 1

1

• A multistage amplifier is an amplifier with more than

one stage of a transistor amplifier which is formed by combining several stages of such amplifier

• The objective of constructing a multistage amplifier is to

increase the overall gain or just for the purpose of impedance matching

vi1 vo1 vi2 vo2 vi3 vo3 vi4 vo4 vin von Amp #1 Amp #2 Amp #3 Amp #4 Amp #n

Figure 7.1 –An n-stage amplifier


• The overall gain of the amplifier can be written as

⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛==

in

on

i

o

i

o

i

o

i

o

i

onv v

vvv

vv

vv

vv

vvA .....

4

4

3

3

2

2

1

1

1

where , 21 io vv = 32 io vv = , …, inno vv =− )1(

• Therefore,

vnvvvv AAAAA ××××= ...321

• Current gain can also be written in the same form:

iniiii AAAAA ××××= ...321

• There are several ways to connect the amplifiers: o RC-coupled amplifiers o Direct-coupled amplifiers o LC-coupled amplifiers o Transformer-coupled amplifiers


RC-Coupled Amplifiers

• Figure 7.2 shows a three-stage cascaded amplifier

Figure 7.2 – RC-coupled amplifier

• The first two stages are common-emitter amplifiers and the third is a common collector amplifier

• The technique used for coupling the stages together in

the circuit is called RC coupling

• The term “RC” comes from the fact that capacitors are used between stages and form an RC circuit with the impedance of the next stage

• The coupling capacitors act like an open for dc currents

but are a short for signal currents


• Because of this, each stage’s bias circuitry is independent and the dc bias current and voltages can be calculated in the same manner as that used for a single-stage amplifier

DC analysis First stage:

V6 k 22 k 37

k 22 V 16 21

2 =Ω+Ω

Ω×=

+×=

bb

bccb RR

RVV

V5.3 V 0.7 - V 6 ==−= bebe VVV

mA 2.8 k 1.9 V3.5 =Ω

==e

ee R

VI

V 5mA 2.8 k 1.8 =×Ω=×= ccRc RIV

V 11 V5 - V 61 ==−= Rcccc VVV

Second stage:

V2.9 k 2.7 k 12

k 2.7 V 16 21

2 =Ω+Ω

Ω×=

+×=

bb

bccb RR

RVV

V2.2 V 0.7 - V 2.9 ==−= bebe VVV

mA 8.4 262 V.22 =Ω

==e

ee R

VI


V 9.6mA 8.4 820 =×Ω=×= ccRc RIV

V 1.9 V6.9 - V 61 ==−= Rcccc VVV Third stage:

V8.8 k 2.2 k 1.8

k 2.2 V 16 21

2 =Ω+Ω

Ω×=

+×=

bb

bccb RR

RVV

V8.1 V 0.7 - V 8.8 ==−= bebe VVV

mA 45 180 V.18 =Ω

==e

ee R

VI

V 61 == ccc VV

• The output signal from each stage will see the input

impedance of the next stage as its load

• Figure 7.3 is an equivalent circuit of each stage of the circuit and how each stage interacts with the next

Figure 7.3 - Equivalent circuit of an RC-coupled amplifier


• The signal generator’s internal impedances is in series with the input impedance of the first stage (Q1)

• The voltage out (unloaded) of the first stage drives the

second stage’s (Q2) input impedance through the output impedance of the first stage (Q1)

• The voltage out (unloaded) of the second stage (Q2)

drives the input impedances of the third stage (Q3) through the output impedance of the second stage (Q2)

• The voltage out (unloaded) of the third stage drives the

load through the output impedance of the third stage

• As far as the signal generator is concerned, the input impedance of the circuit is equal to the input impedance of the first stage

• The output inpedance of the total circuit is equal to the

output impedance of the final stage of the system

• The gain (voltage, current, and power) of the total circuit is equal to the product of the individual stages

• In RC-coupled amplifier circuits, the capacitors block dc

bias current between stages, so each transistor’s bias circuitry is independent

• Hence, RC-coupling method is suitable for amplifying

ac signal only

• The capacitor acts like a short for the signal current and the effect each stage has on an adjacent stage must be considered


• By considering the interaction between stages the signal parameters of each stage can be calculated

• RC-coupled amplifiers are popular because each stage

has its own independent bias circuit

• Fluctuations in the dc operating point in one stage are not amplified by the next stage

Transfer Function at the Interconnecting Stage

• The exists an RC combination in between stages of an RC-coupled amplifiers, as shown in Figure 7.4

Figure 7.4 – RC coupling beween stages of amplifiers

• For dc, the output voltage is

RCtco evv /−=

• For ac, the input-output relationship is

ωω

ωωβj

CRjCjR

Rvv

i

o

−=

+=

+=

1

111

11

where RC1

=βω


Example 7.1 Figure 7.5 shows a two-stage BJT cascaded amplifier. For each transistor, hie = 1 kΩ and hfe = 100; hre, hoe, and the biasing network may be neglected. At mid frequencies, (a) draw the small-signal model, (b) find the overall voltage gain.

Figure 7.5

Solution (a) The small-signal model is shown in Figure 7.6. Because at mid frequencies the coupling and bypass capacitors act as short circuits, they are shown as direct connections in the figure.

Figure 7.6


(b) Output voltage vo is 22 100- )k 1(100 bbo iiv =Ω−= Because of the current division,

11

2 50k 1)(1

)k 1(100b

bb iii −=

Ω+Ω

−=

Also,

2k 1)(11ss

bvvi =

Ω+=

Hence,

ss

o vvv 25002

50100 =⎟⎠⎞

⎜⎝⎛ −−=

2500==s

ov v

vA


Direct-Coupled Amplifiers

• Often it is necessary to have an amplifier that is capable of amplifying dc current and voltages

• To couple a dc signal from one stage to the next, it is

necessary to remove the coupling capacitor and connect the output of one stage directly to the input of the next, as shown in Figure 7.7

Figure 7.7 – Direct-coupled amplifiers

• Voltage gain at the interconnecting stage is 1==i

ov v

vA

• In the ac equivalent circuit of a multistage amplifier, the

emitter bypass capacitors are removed so that the voltage gain of the dc signal will be the same as the gain of the dynamic signals

• Figure 7.8 shows an example of a three-stage direct-

coupled amplifiers


Figure 7.8 – Direct-coupled amplifier

DC Analysis First stage:

V2.3 k 15 k 90

k 15 V 16 21

2 =Ω+Ω

Ω×=

+×=

bb

bccb RR

RVV

V1.6 V 0.7 - V 2.3 ==−= bebe VVV

mA 1.07 k 1.5 V.61 =Ω

==e

ee R

VI

V 5mA 2.8 k 1.8 =×Ω=×= ccRc RIV

V 3.5 V10.7 - V 61 ==−= Rcccc VVV


Second stage:

V5.3 stage first == cb VV

V4.6 V 0.7 - V 5.3 ==−= bebe VVV

mA 0.92 k 5 V6.4 =Ω

==e

ee R

VI

V 9.2 k 10 mA 0.92 =Ω×=×= ccRc RIV

V 8.6 V9.2 - V 61 ==−= Rcccc VVV

Third stage:

V6.8 stage second == cb VV

V6.1 V 0.7 - V 6.8 ==−= bebe VVV

V61 == ccc VV

• The first two stages are in the common-emitter

configuration and the last stage is a common collector

• The bias for the first stage is accomplished by the standard voltage divider of Rb1 and Rb2

• The last two stages also use voltage-divider biasing,

but the voltage divider is made up of the output circuitry of the previous stage


• The voltage on the base of Q2 is the same as the

voltage on the collector of Q1

• If we assume that the base current of Q2 is small enough not to upset the output circuit of Q1, the voltage on the collector of Q1 is a function of the voltage drop across Rc and the voltage drop from the collector of the transistor to ground

• If the collector current of Q1 is much larger than the

base current of Q2, the voltage on the collector of Q1 can be calculated in the same manner as used in single-stage voltage-divider biased circuits

• Once the collector voltage of Q1 is known, the base

voltage of Q2 is also known because they are directly coupled, and at the same potential

• With the base voltage of Q2 known, the collector

voltage of Q2 can be found in the same manner used in a single-stage amplifier

• This method can be continued for the following stages


Example 7.2 For the two-stage CE-CC configuration in Figure 7.9 the hybrid parameters of each stage are hie = 2 kΩ and hfe = 100. Find the input and output resistances and individual, as well as overall, voltage and current gains.

Figure 7.9

Solution

• Note that, in a cascade of stages, the collector resistance of one stage is shunted by the input resistance of the next stage

• Hence it is advantageous to start the analysis with the last stage.

• In addition, it is convenient to compute, first, the current gain, then the input resistance and the voltage gain.

• Finally, the output resistance may be calculated if desired by starting this analysis with the first stage and proceeding toward the output stage.


• For the CC output stage, we have, using standard formulas,

10112 =+= feI hA

507)5)(101(2)1(2 =+=++= Lfeii RhhR

996.0507

2112

2 =−=−=i

ieV R

hA

• Note the high input resistance of the CC stage and that

its voltage gain is close to unity

1001

11 −=−=

−= fe

b

cI h

iiA

Ω=+=++= k 1.12)1.0)(101(2)1( 11 efeieI RhhR

• The effective load on the first stage, its voltage gain, and the output resistance are

Ω==+

= k 95.4512

)507)(5(21

211

iC

iCL RR

RRR

9.401.12

)95.4)(100(1

111 −=

−==

i

Liv R

RAA

Ω== k 5' 11 Co RR

• Since R’o1 is is the effective source resistance for Q2, then,


Ω=+

=++

= 3.69101

500020001

' 12

fe

oieo h

RhR

oLo

Loo R

RRRRR ' 4.68

5069)5000)(3.69('

22

222 =Ω==

+=

• The voltage and current gains of the cascade are,

75.40)996.0)(9.40(21 −=−==≡ vvi

oV AA

vvA

6.98

50755)101)(100(

21

121

1

2

−=⎟⎠⎞

⎜⎝⎛

+−=

+=

−≡

iC

Cii

b

eI RR

RAAiiA

• Alternatively, Av may be computed from

74.401.12

)5)(6.98(1

2 −=−

==i

Liv R

RAA

• The biasing resistors R1 and R2 have had no effect

upon the above calculations

• They do influence the calculation of the overall voltage gain

97.35539.8539.7)74.40(

''

1

1 −=⎟⎠⎞

⎜⎝⎛−=

+=≡

sii

vs

ovs RR

RAvvA

where from Figure 7.10 we see that


Ω==== k 539.71.32

)1.12)(20(1.12//)40//40(//' 11 ii RRR

Figure 7.10

• Note that whereas the input resistance of the cascaded amplifier is the input resistance of the first stage Ri1, the resistance seen by the signal source (vs in series with Rs) is R’i1

SSP 2323 – Advanced Electronics LECTURE 9

LC-Coupled Amplifier

• It is desirable to design circuit without inductors or transformers, if possible, in amplifier circuits because these components are large, heavy, and expensive

• However, there are applications where these

components must be used and the designer should know how they function in amplifier circuits

• Figure 9.1 shows an LC-coupled amplifier

Figure 9.1 – LC-coupled amplifier

75


• In the circuit, resistor Rc has been replaced with an

inductor Lc

• The circuit can be evaluated in the same manner as in a circuit with a resistor in the collector leg

• First, the base voltage is found by using the voltage

divider formula

• Then the emitter voltage is found by substracting Vbe from the base voltage

• With the emitter voltage known, the emitter current can

be found by dividing the emitter voltage by the dc resistance in the emitter leg

• The collector current is assumed to be the same as the

emitter current, so the collector current times the dc resistance of Lc will give the dc voltage drop across Lc

• To find the voltage on the collector we substract this

voltage from Vcc

• The dc resistance of Lc is usually small, however, and the drop across Lc very small. For this reason the voltage on the collector is approximately equal to Vcc

• Because the dc collector voltage is always equal to Vcc,

the collector current cannot be determined by measuring the collector voltage

76


• However, by measuring the emitter voltage and knowing the value of Re, the current flowing through the transistor can be calculated

• Signal parameters are calculated in the same way as

for the circuit with a resistor in the collector leg

• When calculating voltage gain, however, RL(ac) is equal to the XL of the collector inductor in parallel with the load

• Recall that XL = 2πfL and is dependent on frequency,

which in turn makes the voltage gain of the circuit dependent on frequency

• Because gain is a function of frequency, this circuit is

not usable in circuits such as audio amplifiers where the input varies over a range of frequencies

• Another important difference of the LC amplifier

compared to the RC amplifier is the output voltage swing

• The maximum output voltage swing an RC amplifier

can have is equal to the value of Vcc

• In the LC amplifier, however, the collector is sitting at the value of Vcc with no signal inserted

• If an input signal causes the collector current to

increase the inductor will develop a voltage across it opposing the increase in current

77


• This voltage is in series opposing with Vcc; therefore, the voltage on the collector goes down

• If the collector current increases enough, the voltage

drop developed across Lc will equal to Vcc and the voltage on the collector will drop to zero

• Now, if the input signal suddenly caused the collector

current to increase, the collector inductor would again develop a voltage across it to oppose the change in current

• However, this time it is opposing a decreasing current

rather than an increasing current, so the voltage developed is reversed

• This voltage is in series aiding with Vcc and the voltage

on the collector rises above the value of Vcc

• If the collector current decreases enough, the voltage developed across the collector inductor will equal the value of Vcc

• Because this voltage is in series aiding with Vcc, the

voltage on the collector will rise to twice the value of Vcc

• Three important things to note about LC amplifiers are: o The dc voltage on the collector is equal to

approximately Vcc o The voltage gain is dependent on frequency, and o The maximum output voltage swing can be

approximately twice the value of Vcc

78


Transformer-Coupled Amplifier

Figure 9.2 – (a) Transformer-coupled amplifier (b) Dot convention on transformer windings

• The sign convention for a transformer is that if i1 and i2

both enter the dot, then the equation of current becomes i1 +i2 = 0

• However, if one of the current (i1 or i2) enter the dot and

the other one leaves the dot, then 021 =− ii or 01 2 =− ii

• For a transformer with the number of turns n1 in the

primary and n2 in the secondary,

02211 =+ inin or 02211 =− inin or 01122 =− inin

• The second equation is the relationship between the primary and secondary voltages and their number of turns, i.e.

1

2

1

2nn

vv

= or 2

2

1

1nv

nv

=

• By using the above equations we can obtain the

relationship between input impedance Zi and ZL

79


• Define 1

1ivZi = , but 2

1

21 v

nnv ⎟⎟

⎠

⎞⎜⎜⎝

⎛=

• Hence ⎟⎟⎠

⎞⎜⎜⎝

⎛⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

2

2

1iv

nnZi and ZL22 iv −=

• Thus Li Zii

nnZ ⎟⎟

⎠

⎞⎜⎜⎝

⎛ −⎟⎟⎠

⎞⎜⎜⎝

⎛=

1

2

2

1

• But 2

1

1

2nn

ii

−=

• Therefore Li ZnnZ

2

2

1⎟⎟⎠

⎞⎜⎜⎝

⎛=

• ZL is the load which is connected at the output of the

transformer; if there is nothing connected at the output, ZL = ∞

• An example of transformer-coupled amplifier is shown

in Figure 9.3.

• Assume that the h-parameter of the transistor are

0 50 k 2 ===Ω= oerefeie hhhh

80


Figure 9.3 An example of transformer-coupled amplifier

• For small-signal analysis, all capacitors and voltage sources are shorted

Figure 9.4 – Circuit in Figure 9.3 when all capacitors and voltage sources are shorted

• Then, all transistors are replaced by their h-parameter equivalent circuit

• To calculate voltage gain Av = vo/vi, split this formula

into several intermediate stages

81


i

o

i

ov v

vvv

vv

vv

vv

vvA 1

1

2

2

3

3

4

4××××=≡

• Although the amplifier consists of two cascaded

transistors, the calculation of Av involves five stages

51

4=

vvo

• V4/V3 is the voltage gain of transistor Q2

Lrefeoeieie

LfeZhhhhh

Zhvv

)(3

4−+

−=

• Here, ZL is the load seen by transistor Q2, i.e. Z4, the

output impedance of transistor Q2

• Since 0= then = oere hhie

feh

Zhvv 4

3

4 −=

• Z4 can be calculated using the formula for input

impedance of a transformer, that is

Li ZnnZ

2

2

1⎟⎟⎠

⎞⎜⎜⎝

⎛=

• Hence,

20008015 2

4 =×⎟⎠⎞

⎜⎝⎛=Z

82


• Thus,

50k 2

)2000)(50(3

4 −=−=vv

51

2

3 =vv

dan ie

feh

Zhvv 2

1

2 −=

• Calculate Z2 from

3

2

2 15 ZZ ⎟⎠⎞

⎜⎝⎛=

• Calculate Z3 from

Ω==+

−== k 213 ie

Loe

Lfereiei h

ZhZhhhZZ

• Therefore,

125015 2

1

2 −=⎟⎠⎞

⎜⎝⎛−= ie

ie

fe hhh

vv

• But 141 =

ivv

• Thus,

3

1

1

2

2

3

3

4

4

1010

14)1250(

51)50(

51

×=

⎟⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛−⎟

⎠⎞

⎜⎝⎛=××××=≡

i

o

i

ov v

vvv

vv

vv

vv

vvA

83

1

SSCP 3323 – ELEKTRONIK LANJUTAN SESI 2013/14 SEM 1 ASSIGNMENT 3

Question 1

Figure 1 shows an RC-coupled two-stage cascade amplifier. For each stage, hie

= 1 k and hfe = 100. The values of hre, hoe and the biasing network can be neglected. If this amplifier is operated in its mid-band range, (a) draw the small signal model, and (b) calculate the voltage gain, of the amplifier.

+ VCC

R11

R12

RC1

RE

CE

C1

C2100 k 1 k

10 k

RS

1 k

R21

R22

100 k

10 k

RC2

RE

1 k

vO

vs

CE

Figure 1

2

Question 2 a) Describe the following parameter in a differential amplifier

(i) Common-mode gain (ii) Differential-mode gain (iii) Common mode rejection ratio (CMRR)

(b) For the circuit in Figure 2, the transistor parameters are =100, VA = and IQ = 2 mA. Determine the CMRR value of the diff-amp.

+15V

10K 10K

2K 2K V1 V2

5K

-15V

Figure 2

To obtain the gain an electronic system needs, it is often

necessary to connect amplifiers in series.

When amplifiers are connected in series, they are said

to be cascaded.

In the previous analysis, the circuit consists of only one

transistor.

But the circuit analysis has involved the use of

multistage concept, e.g. in the calculation of voltage gain

s

o

s

ovs

v

v

v

v

v

vA 1

1

A multistage amplifier is an amplifier with more than one

stage of a transistor amplifier which is formed by

combining several stages of such amplifier.

The objective of constructing a multistage amplifier is to

increase the overall gain or just for the purpose of

impedance matching.

vi1 vo1 vi2 vo2 vi3 vo3 vi4 vo4 vin von

Amp #1 Amp #2 Amp #3 Amp #4 Amp #n

Figure 1: An n-stage amplifier

in

on

i

o

i

o

i

o

i

o

i

onv

v

v

v

v

v

v

v

v

v

v

v

vA .....

4

4

3

3

2

2

1

1

1

21 io vv 3i2o vv inno vv )1(

vnvvvv AAAAA ...321

The overall gain of the amplifier can be written as

where

Therefore,

, ,

Current gain can also be written in the same form:

iniiii AAAAA ...321

There are several ways to connect the amplifiers:

RC-coupled amplifiers

Direct-coupled amplifiers

LC-coupled amplifiers

Transformer-coupled amplifiers

•Figure 2 shows a three-stage cascaded amplifier

Figure 2: RC-coupled amplifier

The first two stages are common-emitter amplifiers and the

third is a common collector amplifier.

The technique used for coupling the stages together in the

circuit is called RC coupling.

The term “RC” comes from the fact that capacitors are

used between stages and form an RC circuit with the

impedance of the next stage.

The coupling capacitors act like an open for dc

currents but are a short for signal currents.

Because of this, each stage’s bias circuitry is

independent and the dc bias current and voltages can

be calculated in the same manner as that used for a

single-stage amplifier.

In RC-coupled amplifier circuits, the capacitors block

dc bias current between stages, so each transistor’s

bias circuitry is independent

Hence, RC-coupling method is suitable for amplifying

ac signal only.

V6 k 22 k 37

k 22 V 16

21

2

bb

bccb

RR

RVV

V5.3 V 0.7 - V 6 bebe VVV

mA 2.8 k 1.9

V3.5

e

ee

R

VI

V 5mA 2.8 k 1.8 ccRc RIV

V 11 V5 - V 61 Rcccc VVV

DC analysis

First stage:

V2.9 k 2.7 k 12

k 2.7 V 16

21

2

bb

bccb

RR

RVV

V2.2 V 0.7 - V 2.9 bebe VVV

mA 8.4 262

V.22

e

ee

R

VI

V 9.6mA 8.4 820 ccRc RIV

V 1.9 V6.9 - V 61 Rcccc VVV

Second stage:

V8.8 k 2.2 k 1.8

k 2.2 V 16

21

2

bb

bccb

RR

RVV

V8.1 V 0.7 - V 8.8 bebe VVV

mA 45 180

V.18

e

ee

R

VI

V 61 ccc VV

Third stage:

CC amplifier- Suitable for impedance

matching purposes.

Example

Figure 3 shows a two-stage BJT cascaded amplifier. For

each transistor, hie = 1 k and hfe = 100; hre, hoe, and the

biasing network may be neglected. At mid frequencies,

(a) draw the small-signal model,

(b) find the overall voltage gain.

Figure 3

Solution

(a) The small-signal model using hybrid- is:

s

ov

v

vA,gainVoltage)b

Output voltage vo is

22 100- )k 1(100 bbo iiv

Because of the current division,

11

2 50k 1)(1

)k 1(100b

bb i

ii

Also,

2k 1)(11

ssb

vvi

Hence,

ss

o vv

v 25002

50100

2500s

ov

v

vA


84

POWER AMPLIFIER Introduction

The amplifier circuits we have studied so far deal with small signal

These form the early stages of signal processing systems which are designed to give good voltage gain, hence the name voltage amplifiers

Consider the block diagram of a simple audio amplifier, shown in figure 10.1

Figure 10.1 – Audio amplifier stages

The microphone produces a very small signal, in the millivolt range

The first two stages amplify this audio signal and it becomes larger in voltage level but can provide only low current (low power)

The last stage produces a much larger signal which also has to provide sufficient power to the drive the loud speakers coil


85

Thus the power content of the output stage of this amplifier must be much larger than the small power of the input

The output stage of the power amplifier must be designed to meet the power requirement

In this course, we are interested only in power amplifiers using BJTs, and will not consider other types of power electronics that, for example, use thyristors

Two important functions of the output stage are to provide a low output resistance so that it can deliver the signal power to the load without loss of gain and to maintain linearity in the output signal

A low output resistance implies the use of emitter-follower (for BJT) or source-follower (for MOSFET) circuit configurations

A power amplifier has to handle large voltage and current swings, and so it must have large power gain and high efficiency in converting the power drawn from the dc supply to signal form

A measure of linearity of the output signal is the total harmonic distortion (THD)

This figure of merit is the rms value of the harmonic components of the output signal, excluding the fundamental, expressed as a percentage of the fundamental


86

A particular concern in the design of the output stage is to deliver the required signal power to the load efficiency

This specification implies that the power dissipated in the transistors of the output stage should be as small as possible

The output transistors must be capable of delivering the required current to the load, and must be capable of sustaining the required output voltage

Classes of Power Amplifier

Power amplifiers can be divided into four classes, i.e. o Class A o Class B o Class AB o Class C

The amplifier circuits covered in previous lectures are known as class A, which are linear as these operate in a small region in the middle of the load line, as shown in Figure 10.2


87

Figure 10.2 – Class A operating point

This gives the best possible output swing without clipping and so the output signal is a faithful replica of the input signal

In other words the signal distortion is kept low – the greatest advantage of class A operation

Class B amplifier operation is shown in Figure 10.3 wherein the operating point is at cut-off end of the load line


88

Figure 10.3 – Class B operation

This is achieved by operating the base-emitter junction of the transistor with zero bias, so that only the positive halves of the input signal will be amplified, while the negative halves are cut-off

Thus in the class B, the amplifier has a conduction

angle of 180 in contrast to class A where the

conduction angle is 360 , i.e. the entire input cycle

Biasing an amplifier at cut-off end of the load line saves power as there is no drain in the supply during the non-conduction period as in this period the transistor dc current is zero

In class A, however, the operating point is in the middle of the load line so that about half the supply voltage drops across the transistor


89

Hence, the current flowing through the transistor is half of its voltage drops across the transistor is half of its saturation current.

These values of voltage drop and current produce a power loss in the transistor which is constant in class A

Also there is a drain on power supply even if no signal is being amplified

There are also class AB and class C amplifiers – again, it is a question of bias which controls the operating point, the conduction angle and the class of operation

Figure 10.4 shows the relative location of class AB operating point which is mid between class A and class B operating points

Figure 10.4 – Location of operating points

In a class C amplifier the operating point is chosen that output current (or voltage) is zero for more than one half of the signal cycle

SSP 3323 – Advanced Electronics

90

Class-A Operation

• In class-A operation, an output transistor is biased at a quiescent current ICQ and conducts for the entire cycle of the input signal

Figure 11.1 – Collector current versus time characteristics

for class-A amplifier

• A basic common-emitter configuration is shown in Figure 11.2, where the bias circuitry has been omitted, for convenience

Figure 11.2 – A basic CE configuration


91

• Also, in this standard class-A amplifier configuration,

no inductors or transformers are used

• The dc load line is shown in Figure 11.3, where the Q-point is assumed to be in the center of the load line, so that VCEQ = VCC/2

Figure 11.3 - Dc load line for a basic CE amplifier

• If a sinusoidal input signal is applied, sinusoidal variations are induced in the collector current and collector-emitter voltage

• The instantaneous power dissipation in the transistor,

neglecting the base current is

CCEQ ivP =


92

• For a sinusoidal input signal, the collector current and collector-emitter voltage can be written as

tIIi pCQC ωsin+=

and

tVVv pCC

CE ωsin2

−=

• If we consider the absolute possible variations, then

Ip=ICQ and Vp=VCC/2

• Therefore, the instantaneous power dissipation in the transistor is

( )tIVivP CQCCCCEQ ω2sin1

2−==

• Figure 11.4 is a plot of the instantaneous transistor

power dissipation

Figure 11.4 - Instantaneous power dissipation versus time in

the CE transistor amplifier


93

• Since the maximum power dissipation corresponds to

the quiescent value, the transistor must be capable of handling a continuous power dissipation of VCCICQ/2 when the input signal is zero

• The power conversion efficiency is defined as

( )

( )SL

PP

powersupply power load signal

=η

where LP is the average ac power delivered to the load

and SP is the average power supplied by the VCC power source

• For the standard class-A amplifier and sinusoidal input

signals, the average ac power delivered to the load is pp IV2

1

• Using the absolute possible variations, we have

4)(

221(max) CQCC

CQCC

L

IVIVP =⎟

⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛=

• The average power supplied by the VCC source is

CQCCS IVP =

• The maximum attainable conversion efficiency is therefore


94

%25(max) 41

⇒=CQCC

CQCC

IVIV

η

• Keep in mind that the maximum possible conversion

efficiency may change when a load is connected to the output of the amplifier

• This efficiency is relatively low; therefore, standard

class-A amplifiers are normally not used when signal powers greater than approximately 1 W are required


95

Class-B Operation Idealized Class-B Operation

• For class-B operation, an output transistor conducts for only one-half of each sine wave input cycle, as shown by the collector current versus time characteristics in Figure 11.5

Figure 11.5 – C collector current versus time

characteristics for class-B operation

• An idealized class-B output stage that consists of a complementary pair of electronic devices is shown in Figure 11.6

Figure 11.6 – Idealized class-B output stage with

complementary pair A and B, of electronic devices


96

• When vI = 0, both devices are off, the bias currents are zero, and vO = 0

• For vI > 0, device A turns on and supplies current to the

load, as shown in Figure 11.7

Figure 11.7 – Device A turns on for vI > 0,

• For vI < 0, device B turns on and sinks current from the

load, as shown in Figure 11.8

Figure 11.8 – Device B turns on for vI < 0,


97

• The voltage transfer characteristics of the circuit is shown in Figure 11.9

Figure 11.9 – Voltage transfer characteristics for an idealized class-B output stage with complementary pair A and B, of electronic devices

Approximate Class-B Circuit

• Figure 11.10 shows an output stage that consists of a complementary pair of bipolar transistors

Figure 11.10 – Basic complementary push-pull output stage


98

• When vI = 0, both transistors are cut off and the output voltage is vO = 0

• If we assumed a B-E cut-in voltage of 0.6 V, then the

output voltage vO remains zero as long as the input voltage is in the range -0.6 ≤ vI ≤ +0.6 V

• If vI becomes positive and is greater than 0.6 V, then Qn

turns on and operates as an emitter follower

• The load current iL is positive and is supplied through Qn, and the B-E junction of Qp is reverse biased

• If vI becomes negative by more than 0.6 V, then Qp

turns on and operates as an emitter follower

• Transistor Qp is a sink for the load current, which means that iL is negative

• The circuit is called a complementary push-pull

output stage

• Transistor Qn conducts during the positive half of the input cycle, and Qp conducts during the negative half-cycle

• The transistors do not both conduct at the same time

• The transfer characteristics for the complementary

push-pull output stage is shown in Figure 11.11


99

Figure 11.11 – Voltage transfer characteristics of basic complementary push-pull output stage

• When either transistor is conducting, the voltage gain, which is the slope of the curve, is essentially unity as a result of the emitter follower

• The output voltage for a sinusoidal input signal is

shown in Figure 11.12

• When the output voltage is positive, the npn transistor is conducting, and when the output voltage is negative, the pnp transistor is conducting

• It can be seen from the figure that each transistor

actually conducts for slightly less than half the time

• The the bipolar push-pull circuit shown in Figure 11.10 is not exactly a class-B circuit


100

Figure 11.12 – Crossover distortotion of basic

complementary push-pull output stage Crossover Distortion

• From Figure 11.11, it can be seen that there is a range of input voltage around zero volts where both transistors are cut off and vO is zero

• This portion of the curve is called the dead band, and it

produces a crossover distortion, as shown in Figure 11.12 for a sinusoidal input signal

• Crossover distortion can be virtually eliminated by

biasing both Qn and Qp with a small quiescent collector current when vI is zero


101

Idealized Power Efficiency

• If the complementary push-pull output stage circuit is considered as an idealized version in which the base-emitter turn-on voltages are zero, then each transistor would conduct for exactly one-half cycle of the sinusoidal input signal

• The circuit would be a class-B output stage, and the

output voltage and load current would be replicas of the input signal

• The collector-emitter voltages would also show the

same sinusoidal variation

• Figure 11.13 shows the effective load line of the ideal class-B output stage

Figure 11.13 - Effective load line of the ideal class-B output stage


102

• The Q-point is at zero collector current, or at cutoff for both transistor – thus, the quiescent power dissipation in each transistor is zero

• The output voltage for this idealized class-B output

stage can be written

tVv po ωsin= where the maximum possible value of Vp is VCC

• The instantaneous power dissipation in Qn is

CnCEnQn ivp =

• The collector current is

πωπ

πωω

2for 0

0for sin

≤≤=

≤≤=

t

ttRV

iL

pCn

where Vp is the peak output voltage

• From Figure 11.13, it can be seen that the collector-emitter voltage can be written as

tVVv pCCCEn ωsin−=

• Therefore, the total instantaneous power dissipation in

Qn is


103

( )πωπ

πωωω

2tfor 0

t0for sinsin

≤≤=

≤≤⎟⎟⎠

⎞⎜⎜⎝

⎛−= t

RV

tVVpL

ppCCQn

• The average power dissipation is therefore

L

p

L

pCCQn R

VRVV

P4

2

−=π

• The average power dissipation in transistor Qp is

exactly the same as that for Qn because of symmetry

• A plot of the average power dissipation as a function of Vp is shown in Figure 11.14

Figure 11.14 – Average power dissipation in each transistor versus peak output voltage for class-B output stage


104

• The power dissipation first increases with increasing output voltage, reaches a maximum, and finally decreases with increasing Vp

• The maximum average power dissipation is determined

by setting the derivative of QnP with respect to Vp equal to zero, producing

L

CCQn R

VP 2

2

(max)π

=

which occurs when πCC

PpVV

Qn

2(max)

=

• The average power delivered to the load is

L

pL R

VP

2

21⋅=

• Since the current supplied by each power supply is half

a sinewave, the average current is )/( Lp RV π

• The average power supplied by each source is therefore

⎟⎟⎠

⎞⎜⎜⎝

⎛== −+

L

pCCSS R

VVPP

π


105

• And the total power supplied by the two sources is

⎟⎟⎠

⎞⎜⎜⎝

⎛=

L

pCCS R

VVP

π2

• The conversion efficiency is

( )( ) CC

p

L

pCC

L

p

S

L

VV

RV

V

RV

PP

⋅=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅==

42

21

power supply power load signal

2

π

π

η

• The maximum possible efficiency, which occurs when

Vp = VCC, is

%5.784

(max) ⇒=πη

• This maximum efficiency value is substantially larger

than that of the standard class-A amplifier


106

Class-AB Operation

• In class-AB operation, an output transistor is biased at a small quiescent current IQ and conducts for slightly more than half a cycle

• The collector current versus time characteristics for a

class-AB amplifier is shown in Figure 12.1

Figure 12.1 - The collector current versus time characteristics for a class-AB amplifier

• Up to this point, we have used voltage divider bias for

all our class-B amplifiers

• Problems can develop with the class-B or class-AB amplifier when voltage divider bias is used

o Crossover distortion can occur o Thermal runaway can occur o The amplifier is not stable over a wide range of

temperatures

• The class-AB amplifier with diode bias eliminates the problem of crossover distortion, thermal runaway, and poor temperature stability

• By using diodes that have characteristics similar to the

emitter-base junctions within the transistors, the


107

amplifier will be correctly biased for class-AB operation regardless of circuit temperature

• This eliminates the problem of crossover distortion

caused by temperature changes

• A biasing circuit that can be used to eliminate the problems of crossover distortion and thermal runaway is shown in Figure 12.2

Figure 12.2 – Diode biasing

• The circuit shown, called diode bias, uses two diodes in place of the resistor(s) between the base of Q1 and the base of Q2


108

• The diodes in the bias circuit are called compensating diodes and are chosen to match the characteristic values of VBE for the two transistors

• In class-B amplifiers, the transistors are biased at

cutoff, causing the value of ICQ for the amplifier to be approximately equal to zero

• When diode biased is used, the transistors are actually

biased just above cutoff, i.e. there will be some measurable amount of ICQ when diode biased is used

• The dc voltages at various point in the circuit is shown

in Figure 12.3

Figure 12.3 – DC voltages at various point in diode biased class-AB amplifier


109

• In order to understand the operation of the circuit, the following assumptions are made: 1. VCEQ is approximately one-half of the value of VCC 2. The current through R2 causes 5.3 V to be developed

across the resistor R2

• If the above conditions are met, the base of Q2 will be at 5.3 V and VE of Q2 will be 6 V

• Since VB for this pnp transistor is 0.7 V more negative

than VE, Q2 will conduct

• With 1.4 V being developed across the biasing diodes, VB of Q1 will be 5.3 V + 1.4 V = 6.7 V

• Since VB for this npn transistor is 0.7 V more positive

than VE, Q1 will also conduct

• Thus both transistors in the diode biased amplifier will conduct, and some measurable amount of ICQ will be present, as shown in Figure 12.4


110

Figure 12.4 – DC currents in diode biased class-AB amplifier

Class-AB Operation

• To simplify the discussion, an assumption is made: a transistor will conduct until its base and emitter voltages are equal, at which time it will turn off

• The circuit response to the waveform is shown in Figure

12.5


111

Figure 12.5 - Circuit response to the waveform

• Q1 will conduct as long as its base voltage is more positive than 0.7 V

• The value of VB drops to 6 V at t1 and t3 since the -0.7 V

value of Vin subtracts from the 6.7 V value of VB(Q1)

• Thus, it can be assumed that Q1 conducts for the entire time between t1 and t3

• The same principle applies to Q2: At t2 and t4, Q2 will

turn off, since the +0.7 V value of Vin adds to the value of VB(Q2), causing VB and VE to be equal

• Thus, Q2 conducts for the entire time between t2 and t4

• The transistors in the diode bias circuit will conduct for

slightly more than 180°, i.e. for a portion of the input cycle that is greater than 180° and less than 360°

• Both transistors will be conducting at the same time for

a small portion of the wave


112

• Since both transistors in class-AB amplifier are conducting when the input signal is at zero volts, the amplifier does not have the crossover distortion problems that the class-B amplifier may have

• Crossover distortion only occurs when both transistors

are in cutoff, so the problem will not normally occur in class-AB amplifiers

Class-AB Amplifier Analysis

• Consider a class-AB amplifier, shown in Figure 12.6

Figure 12.6 – Class-AB amplifier and its load lines

• For this circuit, IC(sat) is found as

mA 750)8(2

122(sat) ===

L

CCC R

VI


113

• VCE(off) is found as

V 62

122(off) === CC

CEVV

• Using these two values, the dc and ac load lines are

plotted, as shown in Figure 12.7

Figure 12.7 – Class-AB amplifier load lines

• Next, the value of I1 is determined as

mA 4.10 1020

V 1.4 - V 124.1

211 =

Ω=

+−

=RR

VI CC


114

• The average current in the collector circuit of the amplifier is found as

mA 5.238 8

V) (0.159)(12159.0(ave) 1 =

Ω==

L

CCC R

VI

• Using this value and the value of I1 calculated earlier,

the value of ICC is found as

mA 248.9 mA 10.4 mA 5.2381(ave) 1 =+=+= III CCC

• The total power being drawn from the supply is

W2.99 mA) (248.9 V) 12( === CCCCS IVP

• The load power is

W2.25 )8(8

V) 12(8

22

=Ω

==L

CCL R

VP

• The efficiency of the amplifier is

%25.75%100 =×=S

L

PPη


115

Inductively Coupled Class-A Amplifier

• Delivering a large power to a load generally requires both a large voltage and a high current

• In a common-emitter circuit, this requirement can be

met by replacing the collector resistor with an inductor, as shown in Figure 12.8

• The inductor is a short circuit to a dc current, but acts

as an open circuit to an ac signal operating at a sufficiently high frequency

• The entire ac current is therefore coupled to the load

Rajah 12.8 – Inductively coupled class-A amplifier


116

• The dc and ac load lines are shown in Figure 12.9

Figure 12.9 – DC and AC load lines for inductively coupled class-A amplifier

• The resistance of the inductor is assumed to be negligible, and the emitter resistor value is small

• The quiescent collector-emitter voltage is then

approximately VCEQ ≈ VCC

• The ac collector current is

L

cec R

vi −=


117

• To obtain the maximum symmetrical output-signal swing, which will in turn produce the maximum power, we want

L

CCCQ R

VI ≅

• For this condition, the ac load line intersects the vCE

axis at 2VCC

• The use of an inductor or storage device results in an output ac voltage swing that is larger than VCC

• The polarity of the induced voltage across the inductor

may be such that the voltage adds to VCC, producing an output voltage that is larger than VCC

• The absolute maximum amplitude of the signal current

in the load is ICQ; therefore, the maximum possible average signal power delivered to the load is

21

21(max)

22

L

CCLCQL R

VRIP ⋅==

• If the power dissipation in the bias resistors R1 and R2

is neglected, the average power supplied by the VCC source is

2

L

CCCQCCS R

VIVP ==


118

• The maximum possible power conversion efficiency is then

50% 21(max)(max) 2

2

21

⇒=⋅

==L

CC

L

CC

RV

RV

S

L

PPη

• This demonstrates that, in a standard class-A amplifier,

replacing the collector resistance with an inductor doubles the maximum possible power conversion efficiency

Transformer-Coupled CE Amplifier

• The design of an inductively coupled amplifier to achieve high power conversion efficiency may be difficult, depending on the relationship between the supply voltage VCC and the load resistance RL

• The effective load resistance can be optimized by using

a transformer with proper turns ratio

• A CE amplifier with a transformer-coupled load in the collector circuit is shown in Figure 12.10


119

Figure 12.10 – Transformer-coupled CE amplifier

• The dc and ac load lines are shown in Figure 12.11

Figure 12.11 – Dc and ac load lines of a transformer-coupled CE amplifier


120

• If any resistance in the transformer is neglected and RE

is assumed small, the quiescent collector-emitter voltage is

CCCEQ VV ≅

• Assuming an ideal transformer, the currents and

voltages in Figure 12.9 are related

CL iai = and 12 a

vv =

where a is the ratio of primary to secondary turns, or simply the turns ratio

• Dividing voltages by currents, gives

2112 1/v

aiv

iaa

iv

CCL

⋅==

• The load resistance is RL ≡ v2/iL, and we can define a

transformed load resistance as

' 2221L

LCL Ra

iva

ivR =⋅==

• The turns ratio is designed to produce the maximum

symmetrical swing in the output current and voltage; therefore


121

22' 2

LCQ

CC

CQ

CCL Ra

IV

IVR ===

• The maximum power delivered to the load is equal to

the maximum average power delivered to the primary of the ideal transformer, as follows:

CQCCL IVP21(max) =

where VCC and ICQ are the maximum possible amplitudes of the sinusoidal signals

• If the power dissipation in the bias resistors R1 and R2

is neglected, the average power supplied by the VCC source is

CQCCS IVP =

• The maximum possible power conversion efficiency is

%5021(max)(max) ⇒==

S

L

PPη


122

Class-C Operation

• In class-C operation an output transistor conducts for less than half a cycle

• The collector current versus time characteristics of a

class-C amplifier is shown in Figure 12.12

Figure 12.12 - The collector current versus time characteristics of a class-C amplifier

• The transistor circuit ac load line, including an extension beyond cutoff, is shown in Figure 12.13

Figure 12.13 – Effective ac load line of a class-C amplifier


123

• For class-C operation. The transistor has a reverse-biased B-E voltage at the Q-point

• The effect of this class of operation is illustrated in

Figure 12.13

• Note that the collector current is not negative, but is zero at the quiescent point

• The transistor conducts only when the input signal

becomes sufficiently positive during its positive half-cycle

• The transistor therefore conducts for less than a half-

cycle, which defines class-C operation

• Class-C amplifiers are capable of providing large amounts of power, with conversion efficiencies larger than 78.5%

• These amplifiers are normally used for radio-frequency

(RF) circuits, with tuned RLC loads that are commonly used in radio and television transmitters

• The RLC circuits convert drive current pulses into

sinusoidal signals

• Since this is a specialized area, these circuit will not be analyzed in this course

SSCP3323 Exercise 1 Sem I 2014/15

1. For the amplifier circuit shown in Figure 1,

h parameters for the transistor are hie =

11.4K, hfe = 200, hoe = 0, hre = 0. Draw the

ac equivalent circuit for the amplifier and

determine;

i) Ai = io/ii and Ais = io/is

ii) Av = vo/vi and Avs = vo/vs

iii) Zib = vi/ii and Zi = vi/is

iv) Zo = vo/io when vs = 0.

2. Draw the ac equivalent circuit for the amplifier shown in figure 2. With hie=1.0K,

hfe = 50 while hoe and hre can be ignored, determine the voltage gain, vo/vs and

input impedance, vi/is.

3. The parameters of the transistor in the circuit in Figure 3 are = 100, and VA =

.

a) Determine R1 and R2 to obtain a bias-stable circuit with the Q-point in the

center of the load line.

b) Determine the voltage gain, Av = vo/vs.

4. Consider the circuit shown in Figure 4. The transistor parameters are =100 and

VA=100V. Determine the input impedance Zi = vi/is, voltage gain Av = vo/vs and

current gain Ai = io/is.

vs

27K

15K

2.2K

1.2K

VCC=9V

Fig. 4

10K

is

vi

i0

2K

vo

R1

R2

RC=2K

RE=1K

+12V

Fig. 3

vs

vo

vs

100k

680

1k

VCC

Fig. 2

200

is

vi

ii

2k vo

195k

35k

6k

1k

VCC

Fig. 1

200

vs

is

vi

ii i0

10k vo

5. Referring to the circuit in Figure 5, parameters for transistor Q1 are hie = 1.43K

and hfe =100 and parameters for transistor Q2 are hie = 0.95K and hfe =100.

With hoe and hre can be ignored, find vo/vs and io/is.

6. For the circuit in Figure 6, the transistor has parameters of hie = 1.5K, hfe =

200, hoe = 25x10-6

S and hre = 2.5x10-5

. Estimate the current gain, io/ii, input

impedance, vi/is and overall voltage gain, vo/vs.

Fig.6

46K

9K

1.5K

470

VCC

150

vs

is

vi

ii

80

i0 vo

5 : 1

vs

620K

3.3K

VCC

1K

is Q1

io

2.2K

Q2

68K

220

50K

Fig. 5

10K

vo

2

Introduction • Power amplifiers are used to deliver a relatively high amount of

power, usually to a low resistance load.

• Typical load values range from 300W (for transmission antennas) to 8W (for audio speaker).

• Ideal power amplifier will deliver 100% of the power it draws from the supply to load. In practice, this can never occur.

• The reason for this is the fact that the components in the amplifier will all dissipate some of the power that is being drawn form the supply.

3

Amplifier Efficiency h • A figure of merit for the power amplifier is its efficiency, h .

• Efficiency ( h ) of an amplifier is defined as the ratio of ac output power (power delivered to load) to dc input power .

• By formula :

• .

%100)(

)(%100

dcP

acP

powerinputdc

poweroutputac

i

oh

-the average power from the dc supply is

Pdc = VCC IQ watts

- The average signal power in load resistor R is

Watts2

RIP

L2p

ac

Classification of Power Amplifiers

4

Class A operation

-In Class A operation, IC is flowing the whole cycle (2p). Q point is at the center of the operating range. IBQ>0

-Here R1 and R2 provide required IBQ for class A operation.

-poor efficiency (normally less than 25%) -low distortion

5

Class B operation

In Class B operation, IC is flowing half cycle (p) Q point is at cutoff (IC=0) of the operating range. IBQ=0

Here R1 provides required IBQ =0 for class B operation.

-good efficiency (up to 78%) -considerable distortion

• Conducts only half of the cycle

Vcc/2

6

Class AB operation

In Class AB operation, IC is flowing more than half cycle (>p) Q point is at small amount above cutoff (IC>0) of the operating range. IBQ>0.

Here R2 and diode D provides required small IBQ for class AB operation.

-efficiency depends on bias -distortion depends on bias

• Conduction more than half cycle

7

Class C operation

In Class C operation, IC should be flowing less than half cycle (< p) Q point is at below cutoff (IC< 0) of the operating range. IBQ< 0

-high efficiency (approaching 100%)

The class C operation is achieved by reverse biasing BE junction, which sets the dc operating point below cutoff and allows only the portion of the input signal that overcomes the reverse bias to cause collector current flow.

Vce

Ic

Class A

Class AB

Class B

Class C

9

Efficiency Ratings

• The maximum theoretical efficiency ratings of class-A, B, and C amplifiers are:

Amplifier Maximum Theoretical

Efficiency, hmax

Class A 25%

Class B 78.5%

Class C 99%

%2525.0V

R2

R8

V

P

P(max)

R2

V

R2

V

VIV(max)P

R8

V

R22

Vcc

(max)P

V)pp(V,level)pp(VimummaxAt

2CC

L

L

2CC

dc

ac

L

2CC

L

CC

CCCQCCdc

L

2CC

L

2

ac

CCoo

h

Class A Power Amplifiers

dc

ac

CQCCdc

L

2o

L

2

o

ac

o

P

P

IVP

R8

)pp(V

R2

2)pp(V

P

level)pp(VanyAt

h

11

2

I)2/V(

2

IVP Pccpp

ac

p

)(VIP ccP

dc

(Half-wave rectifier →

p P

avg

II

%5.78785.04P

Pη

dc

acmax

p

Class B Power Amplifiers

12


124

DIFFERENTIAL AMPLIFIERS Introduction

• Differential amplifier, or diff-amp, is a special multitransistor circuit configuration

• Diff-amp is a fundamental building block of analog

circuits

• It is the input stage of virtually every opamp, and it is the basis of a high-speed digital logic circuit family, called emitter-coupled logic

• A block diagram of the diff-amp is shown in Figure 13.1

Figure 13.1 – Diff-amp block diagram

• There are two input terminals and one output terminal

• Ideally, the output signal is proportional to only the difference between the two input signals

• The ideal output voltage can be written as

)( 21 vvAv volo −=

where Avol is called open-loop voltage gain


125

• In the ideal case, if v1 = v2, the output voltage is zero

• A non-zero output voltage is obtained if v1 and v2 are

not equal

• The differential-mode input voltage is defined as

21 vvvd −=

• The common-mode input voltage is defined as

221 vvvcm

+=

• This equations shows that if v1 = v2, the differential-

mode signal is zero and the common-mode input signal is vcm = v1 = v2

• If each pair of input voltages were applied to the ideal

difference amplifier, the output voltage in each case would be exactly the same

• However, amplifiers are not ideal, and the common-

mode input signal does not affect the output

• One goal of the design of diff-amps is to minimize the effect of the common-mode input signal


126

Basic BJT Differential Pair

• Figure 13.2 shows the basic BJT differential-pair configuration

Figure 13.2 – Basic BJT differential-pair configuration

• Two identical transistors Q1 and Q2, whose emitters are connected together, are biased by a constant-current source IQ, which is connected to a negative supply voltage V-

• The collectors of Q1 and Q2 are connected through

resistors RC to a positive supply voltage V+

• By design, transistors Q1 and Q2 are to remain biased in the forward-active region


127

• Here, it is assumed that the two collector resistors RC are equal, and that vB1 and vB2 are ideal sources, meaning that the output resistances of these sources are negligibly small

• Since both positive and negative bias voltages are used

in the circuit, the need for coupling capacitors and voltage divider biasing resistors at the inputs of Q1 and Q2 has been eliminated

• If the input signal voltages vB1 and vB2 in the circuit

shown in Figure 13.2 are both zero, Q1 and Q2 are are still biased in the active region by the current source IQ

• The common-emitter voltage vE would be on the order

of -0.7 V

• This circuit, then, is referred to as a dc-coupled diff-amp, so differences in dc input voltages can be amplified

• Although the diff-amp contains two transistors, it is

considered a single-stage amplifier – the analysis shows that it has characteristics similar to those of the common-emitter amplifier

Basic Diff-Amp With Common-Mode Input

• Consider the circuit in which the two base terminals are connected together and a common-mode voltage vcm is applied as shown in Figure 13.3


128

Figure 13.3 - Basic Diff-Amp With Common-Mode Input

• The transistors are biased “on” by the constant-current source, and the voltage at the common emitters is

)(onVvv BEcmE −=

• Since , Q1 and Q2 are matched or identical, current IQ

splits evenly between the two transistors, and


129

221Q

EEI

ii ==

• If base currents are negligible, then 11 EC ii ≅ and

22 EC ii ≅ , and

21 2 CCQ

C vRI

Vv =−= +

• This equation shows that for an applied common-mode

voltage, IQ splits evenly between Q1 and Q2 and the difference between vC1 and vC2 is zero

Basic Diff-Amp With Differential-Mode Input

• A basic diff-amp with differential-mode input is shown in Figure 13.4

• If vB1 increases by a few millivolts and vB2 decreases by

the same amount, or 2/1 dB vv = and 2/2 dB vv −= , the voltages at the bases of Q1 and Q2 are no longer equal

• Since the emitters are common, this means that the B-

E voltages on Q1 and Q2 are no longer equal

• Since vB1 increases and vB2 decreases, then vBE1>vBE2, which means that iC1 increases by ΔI above its quiescent value and iC2 decreases by ΔI below its quiescent value


130

Figure 13.4 - A basic diff-amp with differential-mode input

• A potential difference now exists between the two collector terminals

• The differential output is

C

CCQ

CCQ

CC

RI

RII

VRII

Vvv

Δ=

⎥⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛Δ+−−⎥

⎦

⎤⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛Δ−−=− ++

2

2212

• A voltage difference is created between vC1 and vC2

when a differential-mode input voltage is applied


131

Small-Signal Equivalent Circuit Analysis

• If the diff-amp is assumed to be operating in the linear range, the gain and other characteristics of the amplifier can be derived using the small-signal equivalent circuit

• The small-signal equivalent circuit of the bipolar

differential-pair configuration is shown in Figure 14.1

Figure 14.1 - Small-signal equivalent circuit of the bipolar differential-pair configuration

• Here, it is assumed that the Early voltage is infinite for the two emitter-pair transistors, and that the constant-current source is not ideal but can be represented by a finite output impedance, Ro

• Resistances RB are also included, representing the

output resistance of the signal voltage sources


132

• All voltages are represented by their phasor components

• Since the two transistors are biased at the same

quiescent current, then πππ rrr == 21

• Writing a KCL equation at node Ve, using phasor notation,

o

emm R

Vr

VVgVgr

V=+++

π

πππ

π

π 221

1 (14.1)

or

o

e

RV

rV

rV =⎟⎟

⎠

⎞⎜⎜⎝

⎛ ++⎟⎟

⎠

⎞⎜⎜⎝

⎛ +

ππ

ππ

ββ 1121 (14.2)

where βπ =rgm

• From the circuit, it can be seen that

B

eb

RrVV

rV

+−

=ππ

π 11 and

B

eb

RrVV

rV

+−

=ππ

π 22

• Solving for Vπ1 and Vπ2 and substituting in the above

equation,

o

e

Bebb R

VRr

VVV =⎟⎟⎠

⎞⎜⎜⎝

⎛++

−+π

β1)2( 21 (14.3)

• Solving for Ve, we obtain


133

o

B

bbe

RRr

VVV

)1(2

21

βπ

++

+

+=

(14.4)

One-Sided Output

• If we consider a one-sided output at the collector of Q2, then

B

ebCCmCo Rr

VVRRVgVV+

−−=−==

ππ

β )()( 222 (14.5)

• Substituting Equation (14.4) into (14.5) gives

⎪⎪⎭

⎪⎪⎬

⎫

⎪⎪⎩

⎪⎪⎨

⎧

++

+

−⎥⎦

⎤⎢⎣

⎡++

+−=

o

B

bo

B

B

Co

RRr

VR

RrV

RrRV

)1(2

)1( 12

β

ββπ

π

π (14.6)

• In an ideal constant-current source, the output

resistance is ∞=oR , and Equation (14.6) reduces to

)(2)( 12

B

bbCo Rr

VVRV+−

−=π

β (14.7)

• The differential-mode input is

21 bbd VVV −= (14.8)


134

• The differential-mode gain is

)(2 B

C

d

od Rr

RVVA

+==

π

β (14.9)

• It can be seen that when a common-mode signal

21 bbcm VVV == is applied, the output voltage is no longer zero

• 1bV and 2bV can be written in terms of dV and cmV as

21d

cmbVVV += (14.10)

and

22d

cmbVVV −= (14.11)

• Equation 14.10 and 14.11 simply state that the two

input signals can be written as the sum of a differential-mode input signal component and a common-mode input signal component

• Substituting Equation 14.10 and 14.11 into Equation

14.6 and rearranging terms results in the following:

cm

B

o

Cmd

B

Co V

RrR

RgVRr

RV ⋅

++

+−⋅

+=

π

πβ

β)1(21)(2 (14.12)


135

• The output voltage in Equation 14.12 can be written in the general form

cmcmddo VAVAV += (14.13)

where Ad is the differential-mode gain and Acm is the common-mode gain

• Comparing Equations 14.13 and 14.13, it can be seen

that

)(2 B

Cd Rr

RA+

=π

β (14.14)

and

B

o

Cmcm

RrR

RgA

++

+

−=

π

β )1(21 (14.15)

• It can be observed that the common-mode gain goes to

zero for an ideal current source in which ∞=oR

• For an ideal current source, oR is finite and the common-mode gain is not zero for this case of a one-sided output

• A nonzero common-mode gain implies that the diff-amp

is not ideal


136

Common-Mode Rejection Ratio

• The ability of a differential amplifier to reject a common-mode signal is described in terms of the common-mode rejection ratio (CMRR)

• The CMRR is a figure of merit for the diff-amp and is

defined as

cm

d

AA

=CMRR

• For an ideal diff-amp, 0=cmA and CMRR=∞

• Usually, the CMRR is expressed in decibels, as follows:

cm

d

AA

10log20CMRR =


137

Differential- and Common-Mode Gains

• For greater insight into the mechanism that causes differential- and common-mode gains, reconsider the diff-amp when pure differential- and common-mode signals are applied

Diff-amp with applied differential-mode input signal

• Figure 15.1 shows the ac equivalent circuit of the diff-amp with two sinusoidal input signals

Figure 15.1 – Equivalent ac circuit of thediff-amp with

applied sinusoidal differential-mode input signal

• The two input voltages are 180° out of phase, so a pure differential-mode signal is being applied to the diff-amp

• For this these inputs, vb1 + vb2 = 0, so, from the

equation


138

o

B

bbe

RRr

VVV

)1(2

21

βπ

++

+

+=

it can be seen that the common emitters of Q1 and Q2 remain at signal ground

• In essence, the circuit behaves like a balanced seesaw

• As the base voltage of Q1 goes into its positive-half

cycle, the base voltage of Q2 is in its negative half-cycle

• Then, as the base voltage of Q1 goes into its negative half-cycle, the base voltage of Q2 is in its positive half-cycle

• The signal current directions shown in the figure are

valid for vb1 in its positive half-cycle

• Since ve is always at ground potential, each half of the diff-amp can be treated as a common-emitter circuit

• Figure 15.2 shows the differential half-circuits, clearly

depicting the common-emitter configuration

• The differential-mode characteristics of the diff-amp can be determined by analyzing the half-circuit

• In evaluating the small-signal hybrid-π parameters, it

should be kept in mind that the half-circuit is biased at

2QI


139

Figure 15.2 – Differential-mode half-circuits of a diff-amp

applied with differential-mode input signal Diff-amp with applied common-mode input signal

• The ac equivalent circuit of the diff-amp with a pure common-mode sinusoidal input signal is shown in Figure 15.3

• In this case, the two input voltages are in phase

• The current source is represented as an ideal source IQ

in parallel with its output resistance Ro

• Current iq is the time varying component of the source current


140

• As the two input signals increase, voltage ve increases

and current iq increases

• Since this current splits evenly between Q1 and Q2, each collector current also increases

• The output voltage vo then decreases below its

quiescent value

• As the two input voltages go through the negative half-cycle, all signal currents shown in the figure reverse direction, and vo increases above its quiescent value

• Consequently, a common-mode sinusoidal input signal

produces a sinusoidal output voltage, which means that the diff-amp has a nonzero common-mode voltage gain

• If the value of Ro increases, the magnitude of iq

decreases for a given common-mode input signal, producing a smaller output voltage and hence a smaller common-mode gain

• With an applied common-mode voltage, the circuit

shown in Figure 15.3 is perfectly symmetrical


141

Figure 15.3 – Equivalent ac circuit of diff-amp with common-mode input signal, and resulting signal current directions

• The circuit can therefore be split into the identical common-mode half-circuit shown in Figure 15.4

• The common-mode characteristics of the diff-amp can

then be determined by analyzing the half-circuit, which is a common-emitter configuration with an emitter resistor

• Each half-circuit is biased at 2QI


142

Figure 15.4 – Common-mode half-circuits


143

Differential- and Common-Mode Input Impedances • The input impedance, or resistance, of an amplifier is

as important a property as the voltage gain

• The input resistance determines the loading effect of the circuit on the signal source

Differential-Mode Input Resistance

• The differential-mode input resistance is the resistance seen by a differential-mode signal source

• It is the effective resistance between the two input base

terminals when a differential-mode signal is applied

• A diff-amp with a pure differential input signal is shown in Figure 15.5

Figure 15.5 – BJT diff-amp with differential-mode input signal, showing differential input resistance


144

• The applicable differential-mode half-circuits are shown

in Figure 15.6

Figure 15.6 – Differential-mode half-circuits of a diff-amp applied with differential-mode input signal

• For this circuit, we have

πriv

b

d =2


145

• The differential-mode input resistance is therefore

πrivR

b

did 2==

• Another common diff-amp configuration uses emitter

resistors, as shown in Figure 15.7

Figure 15.7 – BJT diff-amp with emitter resistors

• With a pure applied differential-mode voltage, similar differential-mode half-circuits are applicable to this configuration

• To find the differential-mode input resistance, the

resistance reflection rule is used;


146

Eb

d Rri

v )1(2 βπ ++=

• Therefore,

[ ]Eb

did Rr

ivR )1(2 βπ ++==

• The above equation implies that differential-mode input

resistance increases significantly when emitter resistors are included

• Although the differential-mode gain decreases when

emitter resistors are included, a larger differential-mode voltage (greater than 18 mV) may be applied and the amplifier remains linear

Common-Mode Input Resistance

• A diff-amp with an applied common-mode voltage is shown in Figure 15.8

• The small-signal output resistance Ro of the constant-

current source is shown in Figure 15.8(a), and the equivalent common-mode half-circuits are shown in Figure 15.8(b)

• Since the half-circuits are in parallel, we can write

)2)(1()2)(1(2 ooicm RRrR ββπ +≅++=


147

Figure 15.8 – (a) BJT diff-amp with common-mode input signal, including finite current source resistance and (b) equivalent common-mode half-circuit

• The above equation is a first approximation for

determining the common-mode input resistance

• Normally, Ro is large, and Ricm is typically in the megohm range

• Therefore, the transistor output resistance ro and the

base-collector resistance rμ may need to be included in the calculation


148

• The more complete equivalent half-circuit model is shown in Figure 15.8(b)

• For this model, we have

])1[(||)]2)(1[(||2 ooicm rRrR ββμ ++=

• Therefore,

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛++⎟⎟

⎠

⎞⎜⎜⎝

⎛=

2)1(||)])(1[(||

2o

oicmrR

rR ββμ


149

OPERATIONAL AMPLIFIER Introduction

An operational amplifier (op-amp) is an electronic device that amplifies the difference voltage between the two inputs

A typical op-amp is made up of three types of amplifier circuits: a differential amplifier, a voltage amplifier, and a push-pull amplifier, as shown in Figure 16.1

Figure 16.1 – Basic internal arrangement of an op-amp

A differential amplifier is the input stage for the op-amp; it has two inputs and provides amplification of the difference voltage between the two inputs

The voltage amplifier is usually a class-A amplifier that provides additional gain


150

Some op-amps may have more than one voltage amplifier stage

A push-pull class-B amplifier is generally used for the output stage

Symbol and Terminals

The standard op-amp symbol is shown in Figure 16.2

(a) Symbol (b) Symbol with dc supply connection

Figure 16.2 – Standard op-amp symbol

It has two input terminals, the inverting input (-) and the noninverting input (+), and one output terminal

The typical op-amp operates with two dc supply voltages, one positive and the other negative, as shown in Figure 16.2(b)

Usually these dc voltage terminals are left off the schematic symbol for simplicity but are always understood to be there


151

Some typical op-amp IC packages are shown in Figure 16.3

Figure 16.3 - Some typical op-amp IC packages Ideal Op-Amp

The ideal op-amp has infinite voltage gain and an infinite input resistance (open), so that it does not load the driving source

Also, it has a zero output resistance

These characteristics are illustrated in Figure 16.4

Figure 16.4 – An ideal op-amp representation


152

The input voltage Vin appears between the two input terminals, and the output voltage is AvVin, as indicated by the internal voltage source symbol

The concept of infinite input resistance is a particularly valuable analysis tool for the various op-amp configurations

A practical op-amp, of course, falls short of these ideal standard, but it is much easier to understand and analyze the device from an ideal point of view

The Practical Op-Amp

Although modern integrated circuit (IC) op-amps approach parameter values that can be treated as ideal in many cases, no practical op-amp can be ideal

Any device has limitations, and the IC op-amp is no exception

Op-amps have both voltage and current limitations

For example, its peak-to-peak output voltage is usually limited to slightly less than the difference between the two supply voltages

Output current is also limited by internal restrictions such as power dissipation and component ratings

Characteristics of a practical op-amp are high voltage gain, high input resistance and low output resistance


153

Some of these characteristics are illustrated in Figure 16.5

Figure 16.5 – Characteristics of a practical op-amp

Op-Amp Parameters

Op-amp parameters are used to specify performance and provide for comparison of different op-amps

Three important parameters of an op-amp are o Open-loop voltage gain o Common-mode rejection ratio (CMRR) o Slew rate

Other parameters include o Input offset voltage o Input bias current o Input resistance o Output resistance o Common-mode input voltage range


154

Input Offset Voltage

The ideal op-amp produces zero volts out for zero volts in

However, in practical op-amp a small dc voltage VOUT(error) appears at the output when no differential input voltage is applied

Its primary cause is a slightly mismatch of the transistors in the differential input stage of an op-amp, as illustrated in Figure 16.6

(a) (b)

Figure 16.6 – Illustration of input offset voltage

As specified on an op-amp data sheet, the input offset voltage (Vos) is the differential dc voltage required between the inputs to force the output to zero volts

Vos is demonstrated in Figure 16.6(b)

Typical values of input offset voltage are in the range of 2 mV or less; in ideal case, it is zero


155

Input Bias Current

The input terminals of a bipolar differential amplifier are the transistor bases and, therefore, the input currents are the base current

The input bias current is the dc current required by the inputs of the amplifier to properly operate the first stage

By definition, the input bias current is the average of both input currents

The input bias current is so small in most practical applications that it can be considered to be zero

The concept of input bias current is illustrated in Figure 16.7

Figure 16.7 – Input bias current is the average of the two op-amp input currents


156

Input Resistance

Two basic ways of specifying the input resistance of an op-amp are the differential and the common-mode

The differential input resistance is the total resistance between the inverting and the non-inverting inputs, as illustrated in Figure 16.8(a)

(a) (b)

Figure 16.8 – Op-amp input resistance (a) Differential input resistance (b) Common-mode input resistance

Differential input impedance is measured by determining the change in bias current for a given change in differential input voltage

The common-mode input resistance is the resistance between each input and ground and is measured by determining the change in bias current for a given change in common-mode input voltage, as illustrated in Figure 16.8(b)

The input resistance is always very high


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Output Resistance

Output resistance is the resistance viewed from the output terminal of the op-amp, as indicated in Figure 16.9

Figure 16.9 – Op-amp output resistance

The output resistance is always very small Common-Mode Input Voltage Range

All op-amps have limitations on the range of voltages over which they will operate

The common-mode input voltage range is the range is the range of input voltages which, when applied to both inputs, will not cause clipping or other output distortion

Many op-amps have common-mode ranges of no more

than 10 V with dc supply voltages of 15 V, while in others the output can go as high as the supply voltages (this is called rail-to-rail)


158

Open-Loop Voltage Gain

The open-loop voltage gain, Aol, of an op-amp is the internal voltage gain of the device measured in the differential mode

It represent the ratio of output voltage to input voltage when there is no external components, as shown in Figure 16.10

Figure 16.10 – Open-loop op-amp

The open-loop voltage gain is set entirely by the internal design

Open-loop voltage gain can range up to 200,000 or more and is not a well-controlled parameter

Data sheets often refer to the open-loop voltage gain as the large-signal voltage gain


159

Common-Mode Rejection Ratio

The common-mode rejection ratio (CMRR) for an op-amp is a measure of an op-amp’s ability to reject common-mode signals

An infinite value of CMRR means that the output is zero when the same signal is applied to both inputs (common-mode)

An infinite CMRR is never achieve in practice, but a good op-amp does have a very high value of CMRR.

Common-mode signals are undesired interference voltages such as 50 Hz power supply ripple and noise voltages due to pick-up of radiated energy

A high CMRR enables the op-amp to virtually eliminate these interference signals from the output

The accepted definition of CMRR for an op-amp is the open-loop gain (Aol) divided by the common-mode gain

cm

ol

A

ACMRR

This is equivalent to the CMRR for a differential amplifier


160

Slew Rate

The slew rate of an op-amp is the maximum rate of change of the output voltage in response to a step input voltage

The slew rate is dependent upon the high frequency response of the amplifier stages within the op-amp

Slew rate is measured with an op-amp connected as shown in Figure 16.11

Figure 16.11 – Slew rate measurement

This particular op-amp connection is a unity-gain non-inverting configuration, which gives a worst case (slowest) slew rate

The high-frequency components of a voltage step are contained in the rising edge, and the upper critical frequency of an amplifier limits its response to a step input

The lower the upper critical frequency is, the more gradual the slope on the output for a step input


161

A pulse is applied to the input as shown in Figure 11.12, and the ideal output voltage is measured as indicated

Figure 16.12 – Step input voltage and the resulting output voltage

The width of the input pulse must be sufficient to allow the output to “slew” from its lower limit to its upper limit, as shown

It can be seen that a certain time interval, t, is required for the output voltage to go from its lower limit –Vmax to its upper limit +Vmax, once the input step is applied

The slew rate is expressed as

s)(V/ dsmicroseconper voltsrate Slew t

Vout

where )( maxmax VVVout


162

Op-Amp Configurations

An op-amp can be connected in three basic ways using negative feedback to stabilize and reduce the gain and increase frequency response

The extremely high open-loop gain of an op-amp creates an unstable condition in which an op-amp can be driven out of its linear region or it can oscillate

In addition, the open-loop gain parameter of an op-amp can vary greatly from one device to the next

Closed-Loop Voltage Gain

The closed-loop gain, Acl is the voltage gain of an op-amp with negative feedback

The amplifier configuration consists of the op-amp and an external feedback circuit that connects the output to the inverting input

The closed-loop voltage gain is then determined by the component values in the feedback circuit and can be precisely controlled by them

Noninverting Amplifier

A non inverting amplifier is an op-amp connected in a closed-loop configuration in which the input signal is applied to the noninverting input (+), as shown in Figure 17.1


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Figure 17.1 – Noninverting amplifier

A proportion of the output is applied back to the inverting input (-) through the feedback circuit, which constitutes negative feedback

The differential voltage, Vdiff, between the op-amp’s input terminals is illustrated in Figure 17.2 and can be expressed as

findiff VVV

Figure 17.2 – Differential input


164

This input differential voltage is forced to be very small as a result of the negative feedback and the high open-loop gain, Aol

Therefore, a close approximation is V0diffV

It can be assumed that fin VV

Resistor Ri and Rf form a voltage-divider network

The fraction of the output voltage, Vout, that is returned to the inverting input is found by applying the voltage-divider rule to the feedback circuit

out

fi

if V

RR

RV

The closed-loop voltage gain is inout VV

Since fin VV , the previous equation can be written as

i

f

i

fi

in

outv

R

R

R

RR

V

VA

1

This equation shows that the closed-loop voltage gain of the noninverting amplifier is not dependent on the op-amp’s open-loop gain but can be set by selecting values of Ri and Rf


165

The equation is based on the assumption that the open-loop gain is very high compared to the ratio of the feedback resistors, causing the input differential voltage, Vdiff, to be zero

Voltage-Follower

The voltage-follower is a special case of the noninverting amplifier where all of the output voltage is fed back to the inverting input by a straight connection, as shown in Figure 17.3

Figure 17.3 – Op-amp voltage follower

The straight feedback connection produces a voltage gain of approximately 1, so the closed-loop gain of the voltage-follower is unity

The most important features of the voltage-follower configuration are its very high input resistance and its very low output resistance

These features make it a nearly ideal buffer amplifier for interfacing the high-resistance sources and the low-resistance loads


166

Inverting Amplifier

An op-amp connected in a closed-loop configuration in which the input signal is applied through a series resistors to the inverting input (-) is an inverting amplifier, as shown in Figure 17.4

Figure 17.4 – Inverting amplifier

The output is fed back through Rf to the inverting input, and the non inverting input is grounded

The infinite input resistance in an ideal amplifier implies that there is no current in or out of the inverting input

If there is no current through the input resistance, then there must be no voltage drop between the inverting and noninverting inputs

This means that the voltage at the inverting input (-) is zero because the noninverting input (+) is grounded


167

This zero-voltage at the inverting input terminal is referred to as virtual ground, as illustrated in Figure 17.5

Figure 17.5 – Virtual ground

Since there is no current at the inverting input, the current through Ri and the current through Rf are equal, as shown in Figure 17.6

Figure 17.6 – Iin = If and current at the inverting input (-) is 0

The voltage across Ri equals Vin because of virtual ground on the other side of the resistor


168

Also, the voltage across Rf equals –Vout because of virtual ground

Since If = Iin,

i

in

f

out

R

V

R

V

Rearranging the terms,

i

f

in

out

R

R

V

V

Of course, inout VV is the closed-loop gain of the

inverting amplifier, i.e.

i

f

in

outv

R

R

V

VA

The above equation shows that the closed-loop voltage gain of the inverting amplifier is the ratio of the feedback resistance Rf to the resistance Ri

The closed-loop gain is independent of the op-amp’s internal open-loop gain

Thus, the negative feedback stabilizes the voltage gain

The negative sign indicates inversion


169

Effect of Negative Feedback on Bandwidth

Figure 17.7 graphically illustrates the concept of closed-loop frequency response for an op-amp with negative feedback

Figure 17.7 – Closed-loop gain compared to open-loop gain

When the open-loop gain of an op-amp is reduced by negative feedback, the bandwidth is increased

The closed-loop gain in independent of the open-loop gain up to the point of intersection of the two gain curves

This point of intersection is the critical frequency, fc(cl), for the closed-loop response, which equals the closed-loop bandwidth, BWcl


170

Notice that beyond the closed-loop critical frequency the closed-loop gain decreases at the same rate (called the roll-off rate) as the open-loop gain

Gain-Bandwidth Product

An increas in closed-loop gain causes a decrease in the bandwidth and vice versa, such that the product of gain and bandwidth is a constant

This is true as long as the roll-off rate is a fixed -20db/decade

If Acl represents the gain of any of the noninverting closed-loop configurations and fc(cl) represents the closed-loop critical frequency (same as the bandwidth), then

ololclcl fAfA

The gain-bandwidth product is always equal to the frequency at which the op-amp’s open-loop gain is unity (unity-gain bandwidth)

bandwidthgain -unityclcl fA


171

Basic Op-Amp Circuits

Op-amps are used in such a wide variety of applications

In this lecture, we will discuss only a few of the commonly used circuits, especially in the field of signal processing and instrumentation

Comparators

Operational amplifiers are often used as nonlinear devices to compare the amplitude of one voltage with another

In this application, the op-amp is used in the open-loop configuration, with the input voltage on one input and a reference voltage on the other

Zero-Level Detector

A comparator is a circuit that compares two input voltages and produces an output in either of two states indicating the greater than or less than relationships of the inputs

One application of an op-amp used as a comparator is to determine when an input voltage exceed a certain level

A zero-level detector circuit is shown in Figure 18.1


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Figure 18.1 – The op-amp as a zero-level detector

Notice that the inverting input (-) is grounded to produce a zero level and that the input signal voltage is applied to the noninverting input (+)

Because of the high open-loop voltage gain, a very small difference voltage between the two inputs drives the amplifier into saturation, causing the output voltage to go to its limit

For example, consider an op-amp having Aol = 100,000; a voltage difference of only 0.25 mV between inputs could produce an output of (0.25 mV)(100,000) = 25 V if the op-amp were capable

However, since most op-amps have output voltage

limitations of 15 V or less, the device would be driven into saturation

For many comparison applications, special op-amp comparators are used, such as the LM311 and LM711


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These ICs are generally maximize switching speed

In less stringent applications, a general-purpose op-amp works nicely as a comparator

Figure 18.2 shows the result of a sinusoidal input voltage applied to the noninverting input of the zero-level detector

Figure 18.2 - result of a sinusoidal input voltage applied to the noninverting input of the zero-level detector

When the sine wave is negative, the output is at its maximum negative level

When the sine wave crosses 0, the amplifier is driven to its opposite state and the output goes to its maximum positive level, as shown


174

It can be seen that the zero-level detector can be used as a squaring circuit to produce a square wave from a sine wave

Nonzero-Level Detector

The zero-level detector can be modified to detect positive and negative voltages by connecting a fixed reference voltage to the inverting input (-), as shown in Figure 18.3(a)

Figure 18.3 – Nonzero-level detectors

A more practical arrangement is shown in Figure 18.3(b) using a voltage divider to set the reference voltage as follows:

)(21

2 VRR

RVREF

where +V is the positive op-amp supply voltage

The circuit in Figure 18(c) uses a zener diode to set the reference voltage (VREF = VZ)


175

As long as the input voltage Vin is less than VREF, the output remains at the maximum negative level

When the input voltage exceeds the reference voltage, the output goes to its maximum positive state, as shown in Figure 18.4 with a sinusoidal input voltage

Figure 18.4 – Nonzero voltage comparator with sinusoidal input

Effects of Input Noise on Comparator Operation

In many practical situations, noise (unwanted voltage or current fluctuations) may appear on the input line

This noise voltage becomes superimposed on the input voltage, as shown in Figure 18.5, and can cause a comparator to erratically switch output states


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Figure 18.5 – Sine wave with superimposed noise

In order to understand the potential effects of noise voltage, consider a low-frequency sinusoidal voltage applied to the noninverting input (+) of an op-amp comparator used as a zero-level detector, as shown in Figure 18.6

Figure 18.6 – Effects of noise on comparator circuit

Figure 18.7 shows the input sine wave plus noise and the resulting output

It can be seen that when the sine wave approaches 0, the fluctuations due to noise cause the total input to vary above and below 0 several times, thus producing an erratic output voltage


177

Figure 18.7 – The input sine wave plus noise and the resulting output of a comparator

Reducing Noise Effects with Hysteresis

An erratic output caused by noise on the input occurs because the op-amp comparator switches from its negative output state to its positive output state at the same input voltage level that causes it to switch in the opposite direction, from positive to negative

This unstable condition occurs when the input voltage hovers around the reference voltage, and any small noise fluctuations cause the comparator to switch first one way and then the other


178

In order to make the comparator less sensitive to noise, a technique incorporating positive feedback, called hysteresis, can be used

Hysteresis uses two reference levels

The two reference levels are referred to as the upper trigger point (UTP) and the lower trigger point (LTP)

This two-level hysteresis is established with a positive feedback arrangement, as shown in Figure 18.8

Figure 18.8 – Comparator with positive feedback for hysteresis

Notice that the noninverting input (+) is connected to a resistive voltage divider such that a portion of the output voltage is fed back to the input

The input signal is applied to the inverting input (-) in this case


179

The basic operation of the comparator with hysteresis is as follows, with reference to Figure 18.7

Assume that the output voltage is at its positive maximum, +Vout(max)

The voltage fed back to the inverting input is VUTP and is expressed as

)( (max)

21

2UTP outV

RR

RV

When the input voltage Vin exceed VUTP, the output voltage drops to its negative maximum, -Vout(max)

Now the voltage fed back to the noninverting input is VLTP and is expressed as

)( (max)

21

2LTP outV

RR

RV

The input voltage must now fall below VLTP before the device will switch back to its other voltage level

This means that a small amount of noise voltage has no effect on the output, as illustrated by Figure 18.9

A comparator with hysteresis is sometimes known as a Schmitt trigger


180

The amount of hysteresis is defined by the difference of the two trigger levels

LTPUTPHYS VVV

Figure 18.9 – Operation of a comparator with histeresis. The device triggers only once when the UTP and LTP is reached; thus, there is immunity to noise that is riding on the input signal

A Comparator Application: Over Temperature Sensing Circuit

Figure 18.10 shows an op-amp comparator used in a precision over-temperature sensing circuit to determine when the temperature reaches a certain critical value

The circuit consists of a Wheatstone bridge with the op-amp used to detect when the bridge is balanced


181

Figure 18.10 – An over-temperature sensing circuit

One leg of the bridge contains a thermistor (R1), which is a temperature-sensing resistor with a negative temperature coefficient (its resistance decreases as temperature increases and vice versa).

The potentiometer (R2) is set at a value equal to the resistance of the thermistor at the critical temperature

At normal temperatures (below critical), R1 is greater than R2, thus creating an unbalanced condition that drives the op-amp to its low saturated output level and keeps transistor Q1 off

As the temperature increases, the resistance of the thermistor decreases


182

When the temperature reaches the critical value, R1 becomes equal to R2, and the bridge becomes balanced (since R3 = R4)

At this point the op-amp switches to its high saturated output level, turning Q1 on

This energizes the relay, which can be used to activate an alarm or initiate an appropriate response to the over-temperature condition

Summing Amplifier

The summing amplifier is a variation of the inverting op-amp configuration

The summing amplifier has two or more inputs, and its output voltage is proportional to the negative of algebraic sum of its input voltages

A two-input summing amplifier is shown in Figure 18.11, but any number of inputs can be used

Figure 18.11 – Two-input inverting summing amplifier


183

The operation of the circuit and derivation of the output expression are as follows.

Two stages, VIN1 and VIN2, are applied to the inputs and produce current I1 and I2 as shown

From the concepts of infinite input resistance and virtual ground, the voltage at the inverting input (-) of the op-amp is approximately 0 V, and therefore there is no current at the inverting input

This means that both input currents I1 and I2 combine at this summing point and form the total current, which is through Rf, as indicated (IT = I1 + I2)

Since fT RIV OUT , the following steps apply

ff RR

V

R

VRIIV

2

IN2

1

IN121OUT )(

If all three of the resistors are equal in value

)( 21 RRRR f , then

)( IN2IN1IN2IN1

OUT VVRR

V

R

VV

The previous equation shows that the output voltage has the same magnitude as the sum of the two input voltages but with a negative sign


184

A general expression for a summing amplifier shown in Figure 18.12 with n inputs, where all resistors are equal in value is

)...( ININ2IN1OUT nVVVV

Figure 18.12 – Summing amplifier with n inputs

Integrator

An integrator is a circuit that produces an inverted output that approximates the area under the curve of the input function

An ideal integrator is shown in Figure 18.13

Notice that the feedback element is a capacitor that forms an RC circuit with the input resistor


185

Figure 18.13 – An ideal op-amp integrator

How a Capacitor Charges

To understand how the integrator works, it is important to review how a capacitor charges

Recall that the charge Q on a capacitor is proportional to the charging current (IC) and the time (t)

tIQ C

Also, in terms of the voltage, the charge on a capacitor is

CVCQ

From these two relationships, the capacitor voltage can be expressed as

tC

IV C

C


186

This expression is an equation for a straight line which begins at zero with a constant slope of IC/C

The capacitor voltage in a simple RC circuit is not linear but is exponential

This is because the charging current continuously decreases as the capacitor charges and causes the rate of change of the voltage to continuously decrease

The advantage of using an op-amp with an RC circuit to form an integrator is that the capacitor’s charging current is made constant, thus producing a straight-line (linear) voltage rather than an exponential voltage

In Figure 18.14 the inverting input of the op-amp is at virtual ground (0 V), so the voltage across Ri equals Vin

Figure 18.14 – Currents in an integrator


187

Therefore, the input current is

i

inin

R

VI

If Vin is a constant voltage, then Iin is also a constant because the inverting input always remains at 0 V, keeping a constant voltage across Ri

Because of the very high input impedance of the op-amp, there is negligible current at the inverting input

This makes all of the input current charge the capacitor, so

inC II

The Capacitor Voltage

Since Iin is constant, so is IC – The constant IC charges the capacitor linearly and produces a linear voltage across C

The positive side of the capacitor is held at 0 V by the virtual ground of the op-amp

The voltage on the negative side of the capacitor decreases linearly from zero as the capacitor charges, as shown in Figure 18.15

This voltage is called a negative ramp and is the consequence of a constant positive input


188

Figure 18.15 – A linear ramp voltage is produced across C by the constant charging current

The Output Voltage

Vout is the same as the voltage on the negative side of the capacitor

When a constant positive input voltage in the form of a step or pulse ( a pulse has a constant amplitude when high) is applied, the output ramp decreases negatively until the op-amp saturates at its maximum negative level, as indicated in Figure 18.16

Figure 18.16 – A constant input voltage produces a ramp on the output of the integrator


189

The Rate of Change of the Output

The rate at which the capacitor charges, and therefore the slope of the output ramp, is set by the ratio IC/C.

Since iinC RVI , the rate of change or slope of the

integrator’s output voltage is

CR

V

i

inchange of rateOutput

Integrators are especially useful in triangular-wave generators

Differentiator

A differentiator is a circuit that produces an inverted output that approximates the rate of change of the input function

An ideal differentiator is shown in Figure 18.17, where the capacitor is now the input element of the circuit

Figure 18.17 – An ideal op-amp differentiator


190

A differentiator produces an output that is proportional to the rate of change of the input voltage

Although a small-value resistor is normally used in series with the capacitor to limit the gain, it does not affect the basic operation and is not shown for the purpose of this analysis

Consider a positive-going ramp voltage as an input to the differentiator circuit, as shown in Figure 18.18

Figure 18.18 – A differentiator with a ramp input

In this case, IC = Iin and the voltage across the capacitor is equal to Vin at all times (VC = Vin) because of virtual ground on the inverting input

From the basic formula, which is tCIV CC )( ,

Ct

VI C

C

Since the current at the inverting input is negligible,

CR II


191

Both currents are constant because the slope of the

capacitor voltage tVC is constant

The output voltage is also constant and equal to the voltage across Rf because one side of the feedback resistor is always 0 V (virtual ground)

fCfRout RIRIV

Substituting CtVC )( for IC

CRt

VV f

Cout

The output is negative when the input is a positive-going ramp and positive when the input is a negative-going ramp, as illustrated in Figure 18.19

Figure 18.19 – Output of a differentiator with a series of positive and negative ramps (triangle wave) on the input

During the positive slope of the input, the capacitor is charging from the input source with constant current through the feedback resistor


192

During the negative slope of the input, the constant current is in the opposite direction because the capacitor is discharging

Notice in the above equation that the term tVC is the

slope of the input

If the slope increases, Vout becomes more negative

If the slope decreases, Vout becomes more positive

So, the output voltage is proportional to the negative slope (rate of change) of the input

The constant of the proportionality is the time constant, RfC

SSP3323 Integrator and Differentiator

1

1) If the square wave, Vi is the input to the OPAM, draw the shape the output

voltage, Vo.

2) If the input voltage to the differential amplifier is shown in Figure 2, draw the

shape of the output voltage. Vi

5V 0 1 2 3 4 5 6 7 time, t (sec) -5V

1


Question 1

(a) Explain the term loading effect which occurs when two circuits or electronic equipments are connected together.

(b) State the condition for impedance matching between the two circuits or electronic equipments.

(c) Suggest a method to solve the problem of impedance mismatch using an operational amplifier.

Question 2 Figure 1 shows an operational amplifier circuit built to operate as an inverting amplifier.

+ V

- V

R2

R1

vi v

o

Figure 1

(a) If R1 = 1k , and R2 = 2 k , calculate the output voltage when (i) the input voltage is + 0.5 V (ii) the input voltage is – 1.0 V

(b) Given R1 = 1 k . Calculate the necessary value of R2 in order for the output to be ten times the input voltage.

(c) The gain is required to be fully variable between a minimum of -1 and a

maximum of -11. Suggest how this might be achieved, sketching a suitable circuit and calculating component values.

What restrictions are there for the input signal voltage when the amplifier is at maximum gain?


193

7. DIGITAL ELECTRONICS 7.1 Introduction

Engineers generally classify electronic circuits as being either analog or digital in nature

Historically, most electronics products contained analog circuitry

Most newly designed electronic devices contain at least some digital circuitry

Analogue electronics are designed and used to process analogue signals.

An analogue signal is a fluctuating voltage which can have any numerical value. i.e it may be tiny fractions of a volt or it may be hundreds of volts. It may be a constant voltage or rapidly changing.

The key feature that separates it from digital electronics is this ability to assume any value within a continuous range

In many ways this is a more true reflection of the real world than digital signals.

A digital signal can only have one of two possible values

The exact value of these voltages depends on the particular type of digital circuit but one of the most common systems uses +5 volts and 0 volts

In this system the +5V is referred to as the digital High (or simply HI) and the 0V as digital Low (or LO)

7.2 Analog Circuits versus Digital Circuits

Analog electronic systems has been more popular in the past


194

Real-world information dealing with time, speed, weight, pressure, light intensity, and position measurements are all analog in nature

A simple analog electronic system for measuring the amount of liquid in a tank is illustrated in Figure 19.1

Figure 19.1 – Analog system used to interpret float level in water tank

The input to the system is a varying resistance

The processing proceeds according to the Ohm’s law formula, I = V/R

The output indicator is an ammeter which is calibrated as a water tank gage

As the water rises, the input resistance drops

Decreasing the resistance R causes an increase in current I

Increased current causes the ammeter (water tank gage) to read higher

This analog system is simple and efficient – the gage gives an indication of the water level in the tank

If more information is required about the water level, then a digital system such as the one shown in Figure 19.2 might be used


195

Figure 19.2 – Digital system used to interpret float level in water tank

Digital systems are required when data must be stored, used for calculations, or displayed as numbers and/or letters

In the system shown in Figure 19.2, the input is still a variable resistance as it was in the analog system

The resistance is converted into numbers by the analog-to-digital (A/D) converter

The central processing unit (CPU) of a computer can manipulate the input data, output the information, store the information, calculate things such as flow rates in and out, calculate the time until the tank is full (or empty) based on flow rates, and so forth

Digital systems are valuable when calculations, data manipulations, data storage, and alphanumeric outputs are required

Some of the advantages given for using digital circuitry instead of analog are as follows: o Inexpensive ICs can be used with few external

components o Information can be stored for short periods or

indefinitely o Data can be used for precise calculations


196

o Systems can be designed more easily using compatible digital logic families

o Systems can be programmed and show some manner of “intelligence”

o Alphanumeric information can be viewed using a variety of electronic displays

o Digital circuits are less affected by unwanted electrical interference called noise

The limitations of digital circuitry are as follows: o Most “real-world” event are analog in nature o Analog processing is usually simpler and faster

Digital circuits are appearing in more and more products primarily because of low-cost, reliable digital ICs

Other reason for their growing popularity are accuracy, added stability, computer compatibility, memory, ease of use, simplicity of design, and compatibility with alphanumeric displays

7.3 Digital Signal

Digital signals are composed of two well-defined states

The two states of a wire are usually represented by some measurement of electric current: Voltage is the most common, but current is used in some logic families

A threshold is designed for each logic family

When below that threshold, the wire is "low," when above "high." Digital circuits establish a "no man's area" or "exclusion zone" that is wider than the tolerances of the components.

The circuits avoid that area, in order to avoid indeterminate results.


197

It is usual to allow some tolerance in the voltage levels used; for example, 0 to 2 volts might represent logic 0, and 3 to 5 volts logic 1

A voltage of 2 to 3 volts would be invalid and would occur only in a fault condition or during a logic level transition, as most circuits are not purely resistive, and therefore cannot instantly change voltage levels

However, few logic circuits can detect such a fault, and most will just choose to interpret the signal randomly as either a 0 or a 1.

The levels represent the binary integers or logic levels of 0 and 1

In active-high logic, "low" represents binary 0 and "high" represents binary 1

Active-low logic uses the reverse representation.

Examples of binary logic levels:

Technology L voltage H voltage Notes

CMOS 0V to VCC/2 VCC/2 to VCC VCC = supply voltage

TTL 0V to 0.8V 2V to VCC VCC is 4.75V to 5.25V

ECL -1.175V to -VEE .75V to 0V VEE is about -5.2V VCC=Ground

A TTL digital signal could be made manually by using a mechanical switch, as shown in Figure 19.3

As the blade of the single-pole, double-throw (SPDT) switch is moved up and down, it produces the digital waveform shown at the right


198

At time period t1 the voltage is 0V or LOW

At time t2 the voltage is +5V or HIGH

At t3 the voltage is again 0V or LOW, and at t4 it is again +5V or HIGH

Figure 19.3 Generating a digital signal with a switch

The action of the switch causing the LOW, HIGH, LOW, HIGH waveform is called toggling

By definition, to toggle means to switch over to an opposite state, e.g. from LOW to HIGH or from HIGH to LOW

One problem with mechanical switch is contact bounce which occurs when it toggle from LOW to HIGH, as shown in Figure 19.4

Although this happens in a very short time, digital circuits are fast enough to see this as a LOW, HIGH, LOW, HIGH waveform

Figure 19.4 – Waveform of contact bounce caused by a mechanical switch


199

To cure the contact bounce problem, a debounced logic switch is used, as shown in Figure 19.5

Figure 19.5 – Adding a debouncing latch to a simple switch to condition the digital signal

Another way of generating a digital signal is by using a push-button switch, as shown in Figure 19.6

Figure 19.6 – Push button will not generate a digital signal

However, this simple arrangement will not work because there exists an undefined state

When the push button is pressed, a HIGH of about _5V is generated at the output


200

When the push button is released, however, the voltage at the output is undefined, because there is an open circuit between the power supply and the output

This will not work properly as a logic switch

A normally open push-button switch can be used with a special circuit to generate a digital pulse, for example, a one-shot multivibrator circuit, as shown in Figure 19.7

Figure 19.7 – Push button used to trigger a one-shot multivibrator for a single-pulse digital signal

By using this circuit, for each press of the push button, a single short, positive pulse is output from the one-shot circuit

The pulse width of the output is determined by the design of the multivibrator and not by how long the push button is pressed

A series of pulses can be generated by using multivibrator circuits, as shown in Figure 19.8 or by using a free-running clock using a 555 timer, as shown in Figure 19.9


201

Figure 19.8 – Free-running multivibrator generates a string of digital pulses

Figure 19.9 – Free-running clock using a 555 timer IC


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7.4 Number Systems

A number system is a code that uses symbols to refer to a number of items

The decimal number system uses the symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9

The decimal number system contains 10 symbols and is sometimes called the base 10 system

The binary number system uses only the two symbols 0 and 1 and is sometimes called the base 2 system

Two digital states can be assigned various names, for example ON/OFF, true/false, high/low, 1/0, etc.

The notation 1 and 0 has established the binary number system (base 2)

Octal number (base 8) and hexadecimal number (base 16) also widely used in digital systems

However, decimal number (base 10) is seldom used in digital systems, although this number system is widely used in our daily life.

7.4.1 Binary, Octal and Hexadecimal Number

Consider a three digit decimal number, e.g. abc

We can write abc base 10 (abc10) as abc10 = a x 10

2 + b x 10

1 + c x 10

0

Similarly, the binary number abc base 2 (abc2) can be written as

abc2 = a x 2

2 + b x 2

1 + c x 2

0


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Every digit is known as a bit and can only have two values, either 0 or 1

The left most bit is the highest bit and is known as the most significant bit (MSB), while the right most bit is the lowest bit, known as the least significant bit (LSB)

Conversion from binary number system to decimal system can be done using some rules

However, using calculator or table is the easiest

Table 19.1 – Number Systems

Decimal Binary Hex Octal

00 0000 00 00

01 0001 01 01

02 0010 02 02

03 0011 03 03

04 0100 04 04

05 0101 05 05

06 0110 06 06

07 0111 07 07

08 1000 08 10

09 1001 09 11

10 1010 0A 12

11 1011 0B 13

12 1100 0C 14

13 1101 0D 15

14 1110 0E 16

15 1111 0F 17

Eight octal numbers are represented by 0, 1, 2, 3, 4, 5. 6, and 7, while sixteen hexadecimal (Hex) numbers are represented by 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F


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In octal number system, the number abc base 8 (abc8) can be written as

abc8 = a x 8

2 + b x 8

1 + c x 8

0

In hexadecimal number system, the number abc base 16 can be written as

abc16 = a x 16

2 + b x 16

1 + c x 16

0

Any binary number can be converted to octal number by grouping bits in 3-bit groups

To convert into Hex number, the bits is grouped into 4-bits groups

Conversion from octal to Hex will be easier by first converting the number into binary form

________________________________________________ Example 19.1 Convert the binary number 1001 1110 to Hex and

decimal Solution 1001 11102 = 9E16 = 2

7 + 2

4 + 2

3 + 2

2 + 2

1

= 15810 ________________________________________________ Example 19.2 Convert the octal number 1758 to Hex Solution 1758 = 001 111 1012 = 07D16


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________________________________________________ Example 19.3 Convert the decimal number 146 to binary number by

subtracting repeatedly the highest value of the power of two (2

7, 2

6, …) from its remainder

Solution 14610 = 128 + 16 + 2 = 2

7 + 2

4 + 2

1

= 1001 00102 ________________________________________________ ________________________________________________ Example 19.4 Use the method in Example 19.3 to convert the number

78510 to Hex by subtracting power of 16 Solution 78510 = 3 x 16

2 + 16 + 1

= 31116 ________________________________________________ 7.4.2 Number Representation

We define the following terms: word : a binary number which consists of certain number of bits, e.g. 16, 32, 64,…, bits nibble : a 4-bit word byte : an 8-bit word

Sometimes the 16-bit words are called short words and the 32-bit words are long words


206

The n-bit word can represents 2n different numbers in

the range from 0 to 2n-1

Negative numbers are represented by 2’s-complement notation

To obtain 2’s-complement value from any values, first complement the number (by inverting every bit) and then add 1 to it.

All negative numbers contains the highest bits of value 1

The range of negative number is from -2n-1

to 2n-1

-1

The advantage of 2’s-complement notation is in adding two numbers electronically

7.4.3 Binary Codes

Binary Coded Decimal (BCD) is decimal number coded using binary method

Excess-3 (XS3) Code is Binary Coded Decimal plus 3

Gray Code is ordinary binary number which is arranged such that only one bit change state for each increament of the number


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Decimal 8421 BCD XS3 BCD

10s 1s 10s 1s

0 0000 0011 0011

1 0001 0011 0100

2 0010 0011 0101

3 0011 0011 0110

4 0100 0011 0111

5 0101 0011 1000

6 0110 0011 1001

7 0111 0011 1010

8 1000 0011 1011

9 1001 0011 1100

10 0001 0000 0100 0011

11 0001 0001 0100 0100

Decimal Binary Gray Code

0 0000 0000

1 0001 0001

2 0010 0011

3 0011 0010

4 0100 0110

5 0101 0111

6 0110 0101

7 0111 0100

8 1000 1100

9 1001 1101

10 1010 1111

11 1011 1110

12 1100 1010

13 1101 1011

14 1110 1001

15 1111 1000


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7.4.4 Boolean Algebra

The binary 0 and 1 states are related to the logic variables TRUE or FALSE

If A and B are two logic variables and Q is a result of a logical operation, then we define

1. Q ≡ A AND B ≡ A∙B

such that Q is TRUE if and only if A and B are TRUE

2. Q ≡ A OR B ≡ A∙+ B such that Q is TRUE if A or B is TRUE

3. Q ≡ NOT A ≡ A such that Q is true if A is FALSE

A good way to display the results of Boolean algebra operation is by using a truth table

The followings are some of the Boolean rules

1

0

11

1

0

00

AA

AA

AAA

A

AA

AA

A


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BABAA

BABAA

BABAA

BABAA

ABAA

ABAA

CBACBA

CBACBA

CABACBA

ABBA

ABBA

AA

)(

)(

)()(

)()(

)(


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7.5 Logic Gates

The basic building block of any digital circuit is a logic gate

The term “logic” is usually used to refer to a decision-making process

A logic gate, then, is a circuit that can decide to say yes or no at the output based upon the inputs

Positive logic is an electronic representation such that the TRUE state is when the voltage is HIGH

Negative logic is an electronic representation such that the TRUE state is when the voltage is LOW

In digital circuits, all inputs must be connected

Logic gates can be constructed by using simple switches, relays, vacuum tubes, transistors and diodes, or ICs

Because of their availability, wide use, and low cost, ICs will be used to construct digital circuits

Logic devices are grouped in several families, each having its own rules

Commonly used logic families are o Resistor-Transistor Logic (RTL) o Diode-Transistor Logic (DTL) o Transistor-Transistor Logic (TTL) o N-Channel Metal-Oxide Silicon (NMOS) o Complementary Metal-Oxide Silicon (CMOS) o Emitter-Coupled Logic (ECL)

ECL is the fastest logic family

MOS is for low power and usually used for Large Scale Integrated (LSI) technology

TTL is usually used for small size integrated circuit


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7.5.1 The AND Gate

The AND gate is sometimes called the “all or nothing gate”

Figure 21.1 shows the basic idea of the AND gate using simple switches

In the circuit, both switches A and B must be closed to get the lamp to light

Figure 21.1 – AND circuit using switches

The AND gates are usually constructed using diodes and transistors and packaged inside an IC

In digital circuit, the AND gate is represented as a symbol shown in Figure 21.2

Figure 21.2 – AND gate logic symbol


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The practical AND gate circuit is shown in Figure 21.3

Figure 21.3 - Practical AND gate circuit

The input states and the corresponding output states of the logic gates can be organized in tabular form, known as truth table

The truth table for the AND circuit is shown in Figure 21.4

Figure 21.4 – AND truth table


213

There are four ways to express the function of AND gate, as shown in Figure 21.5

Figure 21.5 – Four ways to express the logical ANDing of A and B

The truth table shows that the AND function’s unique output is a HIGH output only when all inputs are HIGH


214

7.5.2 The OR Gate

The OR gate is sometimes called the “any or all gate”

Figure 21.6 illustrate the basic idea of the OR gate using simple switches

Figure 21.6 – OR circuit using switches

The circuit shows that the output lamp will light when either or both of the input switches are closed but not when both are open

A truth table for the OR circuit is shown in Figure 21.7

Figure 21.7 – OR gate truth table


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The truth table is said to describe the inclusive OR function

The unique output from the OR gate is a LOW only when all inputs are LOW

The logic symbol for the OR gate is shown in Figure 21.8

Figure 21.8 – OR gate symbol and Boolean expression

A brief summary of the OR function is shown in Figure 21.9


216

Figure 21.9 – Four ways to express the logical ORing of inputs A and B


217

7.5.3 The Inverter and Buffer

An inverter is a NOT circuit which has only one input and one output

The function of the NOT gate is to give an output that is not the same as the input

The logic symbol for the inverter is shown in Figure 21.10

Figure 21.10 – Logic symbol and Boolean expression for an inverter

The truth table for the inverter is shown in Figure 21.11

The truth table shows that if the voltage at the input of the inverter is LOW, the the output voltage is HIGH

However, if the input voltage is HIGH, the the output is LOW


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INPUT OUTPUT

A Y=A’

0 1

1 0

Figure 21.11 – Truth table for an inverter

The effect of double inverting is the output which is equal to the original input, as shown in Figure 21.12

Figure 21.12 Effect of double inverting

Two symbols found in logic diagrams that look something like an inverter symbol are shown in Figure 21.13


219

Figure 21.13 – (a) Alternative inverter logic symbol (note bubble at input) (b) Non-inverting buffer/driver logic symbol

The logic symbol in Figure 21.13(a) is an alternative symbol for an inverter and performs the NOT function

The placement of the “invert bubble” on the left side of the inverter symbol in Figure 21.13(a) might suggest that this is an active LOW input

The symbol in Figure 21.13(b) is that of a noninverting buffer/driver

The noninverting buffer serves no logical purpose (it does not invert), but it is used to supply greater drive current at its output than is normal for a regular gate


220

7.5.4 The NAND Gate

The NAND gate is a NOT AND, or an inverted AND function

The standard logic symbol for the NAND gate is shown in Figure 21.14(a)

The little invert bubble (small circle) on the right end of the symbol means to invert the output of AND

Figure 21.14(b) shows a separate AND gate and inverter being used to produce the NAND logic function

Figure 21.14 – (a) NAND gate logic symbol (b) A Boolean expression for the output of a NAND gate

The truth table for NAND gate is shown in Figure 21.15

The unique output from the NAND gate is a LOW only when all inputs are HIGH

A brief summary of the NAND function is shown in Figure 21.16


221

Figure 21.15 – Truth tables for AND and NAND gates

Figure 21.16 – Four ways to express the logical NANDing of inputs A and B


222

7.5.5 The NOR Gate

The NOR gate is actually a NOT OR gate

In other words, the output of an OR gate is inverted to form a NOR gate

The logic symbol for the NOR gate is shown in Figure 21.17(a)

Note that the NOR symbol is an OR symbol with a small invert bubble (small circle) on the right side

The NOR function is being performed by an OR gate and an inverter in Figure 21.17(b)

Figure 21.17 – (a) NOR gate logic symbol (b) Boolean expression for the output of a NOR gate

The truth table for the NOR gate is shown in Figure 21.18

Notice that the NOR gate truth table is just the complement of the output of the OR gate

The unique output of the NOR function is a HIGH output only when all inputs are LOW


223

Figure 21.18 – Truth table for OR and NOR gates

A brief summary of the NOR function is shown in Figure 21.19

Figure 21.19 – Four ways to express the logical NORing of inputs A and B


224

7.5.6 The Exclusive OR Gate

The exclusive OR (XOR) gate is sometimes referred to as the “odd but not even gate”

The logic symbol for the XOR gate is shown in Figure 21.20(a)

The Boolean expression for the XOR function is illustrated in Figure 21.20(b)

Figure 21.20 – (a) XOR gate logic symbol (b) Boolean expression for the output of an XOR gate

Not that if any but not all of the inputs are 1, then the output will be a binary, or logical, 1

The OR and XOR gate truth table is shown in Figure 21.21, for comparison

The unique characteristic of the XOR gate is that it produces a HIGH output only when an odd number of HIGH inputs are present


225

Figure 21.21 – Truth table for OR and XOR gates

A brief summary of the XOR function is shown in Figure 21.22

Figure 21.22 – Four ways to express the XORing of inputs A, B and C


226

7.5.7 The Exclusive NOR Gate

The logic symbol for the exclusive NOR (XNOR) gate” is shown in Figure 21.23(a)

The Boolean expression used to illustrate the XNOR function is shown in Figure 21.23(b)

Figure 21.23 – (a) XNOR logic symbol (b) Boolean expression for the output of an XNOR gate

The truth table for the XNOR gate is shown in Figure 21.24

Notice that the output of the XNOR gate is the complement of the XOR truth table

Figure 21.24 – Truth table for XOR and XNOR gates


227

A brief summary of the XNOR function is given in Figure 21.25

Figure 21.25 – Four ways to express the logical XNORing of inputs A, B and C


228

7.6 The NAND Gate as a Universal Gate

NAND gates seem to be more widely available than many other types of gates

NAND gates can be used to make other types of gates

The chart in Figure 21.26 NAND gates would be wired to create any of the other basic logic functions

Figure 21.26 – Substituting NAND gates

SSP3323 Advanced Electronic Logic Circuit

Question 1

a) Convert the binary number 1110 1011 10012 to hexadecimal number.

b) Hexadecimal BA216 = __________2.

c) Convert the hexadecimal number AFC16 to decimal number.

d) By using Boolean algebra, prove that ABABA )()(

e) Draw the pulse train at output Y in Figure 1(a) when C = 1. f) Write the Boolean expression for the logic

circuit shown in Figure 1(b). Draw a three variable truth table that would describe the logic function.

Figure 1(a) Figure1(b)

Question 2

a) i) Write a Boolean expression for the logic diagram shown in Figure 2.

ii) Draw a three variable truth table that would describe the logic function.

Figure 2

b) A 4-input circuit has an output, Y, given by the truth table below

A B C D Y A B C D Y

0 0 0 0 0 1 0 0 0 1

0 0 0 1 1 1 0 0 1 1

0 0 1 0 1 1 0 1 0 0

0 0 1 1 1 1 0 1 1 0

0 1 0 0 1 1 1 0 0 0

0 1 0 1 1 1 1 0 1 0

0 1 1 0 1 1 1 1 0 0

0 1 1 1 1 1 1 1 1 0

i) Write down the unsimplified Boolean expression ii) Use a Karnaugh Map to minimise the Boolean expression iii) Draw a circuit to solve this problem using a 1-of -8 data selector.


229

7.8 Combining Logic Gates

• Combinational logic is defined as an interconnection of logic gates to generate a specific logic function where the inputs result in an immediate output; having no memory or storage capabilities

• This is also sometimes called combinatorial logic • Digital circuits that have a memory or storage capability

are called sequential logic circuits 7.8.1 Constructing Circuits from Boolean Expressions

• Boolean expression is used as a guide in building logic circuit

• For example, consider the Boolean expression

YCBA =++

• The expression is read as “A or B or C equals output Y” and is realized using logic gate as shown in Figure 22.1

Figure 22.1 – Logic diagram for Boolean expression YCBA =++

• Next, consider a Boolean expression

YCBBABA =•+•+•


230

• The expression is read as “not A and B, or A and not B, or not B and C equals output Y”

• The steps in constructing the Boolean expression is illustrated in Figure 22.2

Figure 22.2 – Steps 1 in constructing the Boolean expression YCBBABA =•+•+•


231

Figure 22.3 – Steps 2 in constructing the Boolean expression YCBBABA =•+•+•

• Boolean expression comes in two forms: the sum-of-products (SOP), for example YCBBABA =•+•+• and the product-of-sum (POS), for example

YFEED =+•+ )()(


232

• The sum-of-product form is called the minterm form in engineering text

• The product-of-sum is called the maxterm form by engineers, technicians and scientists

7.8.2 Drawing a Circuit from a Maxterm Boolean

Expression • Suppose the maxterm Boolean expression is

YBACBA =+•++ )()(

• The steps in constructing a logic circuit for this Boolean

expression is as follows:

Step 1: (Refer to Figure 22.4)

Figure 22.4 – Step 1 in constructing a product-of sums logic circuit


233

Step 2: (Refer to Figure 22.5)

Figure 22.5 – Step 2 in constructing a product-of sums logic circuit

• In summary, we work from right to left (from output to input) when converting a Boolean expression to a logic circuit

• Maxterm and minterm Boolean expression both can be converted to logic circuits

• Minterm expressions create AND-OR logic circuits • Maxterm expressions create OR-AND logic circuits


234

7.9 Truth Tables and Boolean Expressions

• Boolean expressions are a convenient method of describing how a logic circuit operates

• The truth table is another precise method of describing how a logic circuit works

• The works in digital electronics involve converting information from truth table form to a Boolean expression

7.9.1 Truth Table to Boolean Expression

• Consider the truth table shown in Figure 22.6

Figure 22.6 – Forming a mintern Boolean expression from a truth table


235

• Notice that only two of the eight possible combinations

of inputs A, B, and C generate a logical 1 at the output • The two combinations that generate a 1 output are

ABC •• and ABC •• • The two combinations are ORed together to form the

Boolean expression for the truth table 7.9.1 Boolean Expression to Truth Table

• Consider the Boolean expression in Figure 22.7(a)

Figure 22.7 – Constructing a truth table from a minterm Boolean expression


236

• It appears that two combinations of inputs A, B, and C generate a logical 1 at the output

• Figure 22.7(b) shows the correct combinations of A, B, and C that are given in Boolean expression and mark 1 in the output column

• All other outputs in the truth table are 0 • The Boolean expression and the truth table both

accurately describe the operation of the same logic circuit

• The procedure for producing maxterm Boolean expressions from a truth table is quite different

7.10 Simplifying Boolean Expressions

• Consider the Boolean expression, as shown in Figure 22.8(a)

YBABABA =•+•+•

• Constructing a logic circuit for this Boolean expression,

requires three AND gates, two inverters, and one 3-input OR gate, as shown in Figure 22.8(b)

• Figure 22.8(c) details the truth table for the Boolean expression

• The truth table look similiar to the truth table for a 2-input OR gate, with expression YBA =+

• This shows that the Boolean expression YBABABA =•+•+• can be simplified to just the

expression YBA =+ • This saves a lot in terms of the number of gates to be

used to realize the circuit


237

Figure 22.8 – Simplifying Boolean expression (a) Unsimplified Boolean expression (b) Complex logic diagram (c) Truth table (d) Simplified Boolean expression: 2-input OR by inspection (e) Simple logic diagram


238

7.11 De Morgan’s Theorems

• Boolean algebra, the algebra of logic circuits, has many laws or theorems

• One of the very useful theorem is De Morgan’s theorems

• The theorems allow us to convert back and forth minterm to maxterm forms of Boolean expressions

• The theorems also allow us to eliminate long overbars that cover several variables

• De Morgan’s theorems can be stated in the form shown in Figure 22.9

Figure 22.9 – De Morgan’s theorems and practical examples


239

7.12 Minterms-to-Maxterms or Maxterms-to-Minterms

• Four steps are need to convert a maxterm to minterm Boolean expression or from minterm to maxterm form

• The four steps, which are based on De Morgan’s theorems, are as follows:

Step 1. Change all ORs to ANDs and all ANDs to Ors Step 2. Complement each individual variable (add short overbars to each) Step 3. Complement the entire function (add long overbar to entire function) Step 4. Eliminate all groups of double overbars

• Examples of the steps in the conversion are shown in

Figure 22.10 and Figure 22.11


240

Figure 22.10 – Four-step process using De Morgan’s second theorem to convert conventional NAND to alternative NAND. Note that the long overbar is eliminated


241

Figure 22.11 – Four-step process using De Morgan’s theorems to convert from maxterm-to-minterm form. Note that the long overbar is eliminated


242

7.13 Karnaugh Maps

• In 1953 Maurice Karnaugh publish an article about his system of mapping and thus simplifying Boolean expressions

• Figure 23.1 illustrate a Karnaugh map

Figure 23.1 – The meaning of squares in a Karnaugh map

• The four squares (1, 2, 3, 4) represent the four possible

combinations of A and B in a two variable truth table • Square 1 in the Karnaugh map, then, stands for BA • ,

square 2 for BA • , and so forth • For an expression YBABABA =•+•+• , the map is

shown in Figure 23.2


243

Figure 23.2 – Marking 1s on a Karnaugh map

• The filled-in Karnaugh map is now ready for looping • The looping technique is shown in Figure 23.3

Figure 23.3 – Looping 1s together on a Karnaugh map

• Adjacent 1s are looped together in groups of two, four, or eight

• Looping continues until all 1s are included inside a loop • Each loop represents a new term in the simplified

Boolean expression


244

• Since there are two loops in the table, it means that there are two terms ORed together in the new simplified Boolean expression, as shown in Figure 23.4

Figure 23.4 – Simplifying a Boolean expression from a Karnaugh map

• The summary of the steps to apply the Karnaugh map to simplify a Boolean expressions are as follows:

1. Start with a minterm Boolean expression 2. Record 1s on a Karnaugh map 3. Loop adjacent 1s (loops of two, or four, or eight

squares) 4. Simplify by dropping terms that contain a term and

its complement within a loop 5. OR the remaining terms (one term per loop) 6. Write the simplified minterm Boolean expression


245

7.14 Grouping techniques on Karnaugh Map

Figure 23.5 – Grouping techniques

• In Figure 23.5(a), there is no adjacent square of 1s that can be grouped together

• Hence the Boolean expression for this Karnaugh map cannot be further simplified

• In Figure 23.5(b), there are two squares that can be grouped

• These two squares can be considered as adjacent squares when the map is folded along x-axis

• Similarly, the squares of 1s in Figure 23.5(c) and Figure 23.5(d)

• The squares in Figure 23.5(e) are also considered adjacent if the map is folded twice, along x-axis and y-axis

• The more squares in a group the more simple the Boolean expression

• However, it should be noted that the number of squares in a group should be 2, 4, or 8 only


246

7.15 Karnaugh Maps with Three Variables

• Consider the unsimplified Boolean expression

YCBACBACBACBA =••+••+••+•• shown in Figure 23.6(a)

Figure 23.6 – Simplifying a Boolean expression using a Karnaugh map (a) unsimplified expression (b) mapping 1s (c) Looping 1s and eliminating variables (d) forming simplified minterm expression


247

• A three-variable Karnaugh map is illustrated in Figure

23.6(b) • Notice the eight possible combinations of A, B, and C,

which are represented by the eight squares in the map • The Karnaugh map with loops is redrawn in Figure

23.6(c) • Adjacent groups of two is are looped • The bottom loop contains both a B and B terms are

eliminated • The bottom loop still contains the A and C , giving

CA • term • The upper loop contains both a C and a C and these

are eliminated, leaving the BA • term • A minterm Boolean expression is formed by adding the

OR symbol • The simplified Boolean expression is written in Figure

23.6(d) as

YBACA =•+• 7.16 Karnaugh Maps with Four Variables

• The truth table for four variables has 16 (i.e. 24) possible combinations

• Consider the Boolean expression

YDCBADCBADCBADCBADCBADCBA

=•••+•••+•••+

•••+•••+•••


248

• The four-variable Karnaugh map in Figure 23.7(b) gives the 16 possible combinations of A, B, C, and D

• Following the same procedure as before, the simplified Boolean expression is written in Figure 23.7(d) as

YDACBA =•+••

Figure 23.7 – Simplifying a four-variable Boolean expression using a Karnaugh map


249

7.17 Don’t Cares on Karnough Map

• The following truth tables are for BCD numbers

(a) (b)

Figure 23.8 – Truth table for BCD numbers

• BCD numbers starting from 0 to 9 is represented by binary numbers 0000 to 1001

• Therefore, binary numbers 1010 to 1111 are not used • These unused combination numbers are called “don’t

cares”


250

• Let’s say we want to build a circuit for detecting number 9 in BCD

• For example, when number 9 is met then an LED is lighted

• Thus, the truth table for the circuit is as shown in Figure 23.8(b)

• The unsimplified Boolean expression for the circuit is

YDCBA =•••

• The Karnaugh map for the truth table is as shown in Figure 23.9

• Notice that the “don’t cares” value ar marked as “X” on the Karnaugh map

• For the squares which contain “X” we substitute with 0 or 1 so that we can form group with maximum number of squares (2, 4 or 8 only)

Figure 23.9 – Karnaugh map for the BCD 9 truth table

• Hence, the simplified Boolean expression is

YDA =•


251

7.18 – Data Selectors

• Manufacturers of ICs have simplified the job of solving many combinational logic problems by producing data selectors

• A data selector is often a one-package solution to a complicated logic problem

• The data selector actually contains a rather large number of gates packaged inside a single IC

• A 1-of-8 data selector is illustrated in Figure 24.1 • The device has eight data inputs, three data selector

inputs labeled A, B, and C, and one output, labeled W

Figure 24.1 – Logic symbol for a 1-of-8 data selector


252

• The basic job the data selector performs is transferring data from a given data input (0 to 7) to the output W

• Which data input is selected is determined by which binary number placed on the data selector inputs

• For example, by giving A = 1, B = 1, and C = 0, data input line 3 is selected and is passed to the output

• Data selector can be used to solve logic problems • Consider the simplified Boolean expression shown in

Figure 24.2(a)

Figure 24.2 – (a) Simplified Boolean expression (b) Logic circuit for Boolean expression


253

• A logic circuit for this complicated Boolean expression

is shown in Figure 24.2(b) • Using standard ICs, we probably would have to use

from six to nine IC packages to solve this problem • This would be quite expensive because of the cost of

the ICs and PC board space • A less costly solution to the logic problem is to use a

data selector • The Boolean expression from Figure 24.2(a) is

repeated in truth table form in Figure 24.3

Figure 24.3 – Solving logic problem with a data selector IC


254

• A 1-of-16 data selector is added in Figure 24.3 • Notice that logical 0s and 1s are placed at the 16 data

inputs of the data selector corresponding to the truth-table output column Y

• These are permanently connected for this truth table • Data selector inputs (D, C, B, and A) are switched to

the binary numbers on the input side of the truth table • For example, if data selector inputs D, C, B, and A are

at binary 0000, then a logical 1 is transferred to output W of the data selector

• Data selector is used to solve a complicated logic problem

• In this example, we found we needed at least six ICs to solve this logic problem

• Using data selector, we solve this problem by using only one IC

• Data selector seems to be an easy-to-use and efficient way to solve combinational logic problems

• Commonly available data selectors can solve logic problems with three, four, or five variables

• Data selectors are also called multiplexers • The data selector (multiplexer) can be used as a

universal logic element • It is a simple, low-cost solution to many logic problems

with from three to five input variables • Simplified gate circuits and data selector ICs have been

used to implement logic problems • More complex logic problems are created when there

are more variables or when the logic circuit has several outputs

• For these problems, designers can use a programmable array of logic gates within a single IC

• This device is called programmable array logic (PAL)


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• The PAL is based on programmable AND/OR architecture

• These programmable logic devices are available in both TTL and CMOS

• These devices are user programmable • A typical PAL may have 16 inputs and 8 outputs • The generic name for a PAL is programmable logic

device (PLD) 7.19 Programmable Logic Devices (PLDs)

• A programmable logic devices (PLD) is an IC that can be programmed by the user to excecute a complex logic function

• Simple PLDs are used to implement combinational logic • Other more complex PLDs have memory

characteristics (registers) and can be used in the design of sequential logic circuits (such as counters)

• The PLD has many inputs and multiple outputs • The PLD can implement minterm (sum-of-products)

Boolean expressions using AND-OR logic • A simplified version of a programmable logic device is

detailed in Figure 24.4


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Figure 24.4 – Simplified PLD with fuses intact (as from manufacturer)

• The simplified PLD has intact (not blown) fuses used for programming the AND gates

• The OR gate is not programmable in the device • The PLD needs to be programmed by burning open

selected fuses to implement the Boolean expression, e.g. as shown in Figure 24.5


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Figure 24.5 – Simplified PLD with selected fuses burned open to solve logic problem

Name: I/C:

From Boolean expression :

YDCBADCBADCBA

DCBADCBADCBADCBADCBADCBA

=⋅⋅⋅+⋅⋅⋅+⋅⋅⋅

+⋅⋅⋅+⋅⋅⋅+⋅⋅⋅+⋅⋅⋅+⋅⋅⋅+⋅⋅⋅

(i) ii)

Draw logic circuit for the Boolean expression. Construct a truth table for the expression

D C B A Y D C B A Y 0 0 0 0 1 0 0 0 0 0 0 1 1 0 0 1 0 1 1 1 1 1 1 1 1

Name: I/C:

(iii) Simplify the expression using Karnaugh map 00 01 11 10 00 01 11 10

(iv) (iv)

Draw a circuit of the simplified Boolean expression using logic gates Draw circuit to solve this problem using 1-of-8 data selector

1

SSCP 3323 – ELEKTRONIK LANJUTAN SESI 2013/14 SEM 1 TUGASAN 5

1. Construct logic circuits using AND, OR and NOT gates for the following

minterm Boolean expressions:

(a) YBABA

(b) YCBACA

(c) YDCDBDA

2. Construct logic circuits using AND, OR and NOT gates for the following

Boolean expressions:

(a) YBABA )()(

(b) YCBA )(

(c) YCADCBA )()()(

3. (a) Using the truth table in Figure 1

for an electronic lock, write the minterm Boolean expression for this truth table

(b) From the Boolean expression developed in part (a), draw a logic symbol diagram for the electronic lock problem

Figure 1 4. Simplify the Boolean expression

YCBACBACBACBA

by (i) Plotting 1s on a three-variable Karnaugh map (ii) Looping groups of two or four 1s (iii) Eliminating variables whose complement appears within the loop(s) (iv) Writing the simplified minterm Boolean expression 5. Simplify the Boolean expression

YCBACBACBACBA

by (i) Plotting 1s on a three-variable Karnaugh map (ii) Looping groups of two or four 1s (iii) Eliminating variables whose complement appears within the loop(s) (iv) Writing the simplified minterm Boolean expression

INPUT SWITCHES OUTPUT

C B A Y

0 0 0 0

0 0 1 0

0 1 0 1

0 1 1 0

1 0 0 0

1 0 1 1

1 1 0 0

1 1 1 0


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7.20 Integrated Circuit (IC) TTL and CMOS

Digital ICs are grouped into various families

These families are group of devices that can be used together

The ICs in a family are said to be compatible and can be easily connected to one another

One group of families is manufactured using bipolar technology

These ICs contains parts comparable to discrete bipolar transistors, diodes, and resistors

Another group of digital IC families uses metal oxide semiconductor (MOS) technology

The CMOS family is a very low power and widely used family using MOS technology

The CMOS ICs contains parts comparable to insulated-gate field-effect transistors (IGFETs)

A traditional type of IC is illustrated in Figure 25.1(a)

This case style is referred to as a dual-in-line package (DIP) by IC manufacturers

This particular IC is called a 14-pin DIP IC

Just counterclockwise from the notch on the IC in Figure 25.1(a) is pin 1

The pins are numbered counterclockwise from 1 to 14 when viewed from the top of the IC

A dot on the top of the DIP IC as in Figure 25.1(b) is another method used to locate pin 1

The smaller micropackages in Figure 25.1(c) and (d) are commonly called surface-mount technology (SMT) packages


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Figure 25.1 -

Figure 25.2 shows an example of the 7400 series of TTL ICs, indicating its pin diagram

This IC contains four 2-input AND gates, and is called a quadruple two-input AND gate

Figure 25.2 – Pin diagram for 7408 digital IC


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Figure 25.3 is a logic diagram using one of the gate in the 7408 IC

Figure 25.3 – Logic diagram for 2-input AND gate circuit

The wiring diagram to implement the 2-input AND function is shown in Figure 25.4

Figure 25.4 - Wiring diagram to implement the 2-input AND function is shown in Figure 25.3


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7.21 Understanding the Marking on Digital TTL ICs

The top of a typical TTL digital IC is shown in Figure 25.5(a)

The block form of the letters “NS” on the top of the IC shows the manufacturer as National Semiconductor

The DM7408N part number can be divided into sections as shown in Figure 25.5(b)

The prefix “DM” is the manufacturer’s code (National Semiconductor uses the letters “DM” as a prefix

The core part number is 7408, which is a quadruple 2-input AND gate TTL IC

This core part number is the same from manufacturer to manufacturer

The trailing letter “N” (the suffix) is a code used by several manufacturers to designate the DIP

Figure 25.5 – (a) Marking on a typical digital IC. (b) Decoding the part number on a typical IC


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The top of another digital IC is shown in Figure 25.6(a)

The letters “SN” on this IC stand for the manufacturer, Texas Instruments

Figure 25.6 – (a) Markings on a Texas Instruments digital IC (b) Decoding the part number of a typical low-power Schottky IC

On this unit, the suffix “J” stands for a ceramic DIP packaging, typically referred to as the commercial grade

The core part number of the IC in Figure 25.6 is 74LS08

This is similar to the 7408 quadruple 2-input AND gate IC manufactured by the National Semiconductor


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The letters “LS” in the center of the core number designate the type of TTL circuitry used in the IC – in this case “LS” stands for low-power Schottky

The internal letter(s) in a core part number of a 7400 series IC tell something about the logic family or subfamily

Typical internal letters used are: AC = FACT Fairchild Advanced CMOS Technology

logic ( a newer advanced family of CMOS) ACT = FACT Fairchild Advanced CMOS Technology

logic ( a newer advanced family of CMOS with TTL logic levels)

ALS = advanced low-power Schottky TTL logic (a sub-family of TTL)

AS = advanced Schottky TTL logic (a sub-family of TTL)

C = CMOS logic (an early family of CMOS) F = FAST Fairchild Advanced Schottky TTL logic

( a new sub-family of TTL) FCT = FACT Fairchild Advanced CMOS Technology

logic (a family of CMOS with TTL logic levels) H = high-speed TTL logic (a sub-family of TTL) HC = high-speed CMOS logic (a family of CMOS) HCT = high-speed CMOS logic (a family of CMOS

with TTL inputs) L = low-power TTL logic (a sub-family of TTL) LS = low-power Schottky TTL logic (a sub-family of

TTL) S = Schottky TTL logic (a sub-family of TTL)


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7.22 Practical CMOS Logic Gates

The older 7400 series of TTL logic devices has been extremely popular for many decades

One of its disadvantages is its higher power consumption

In late 1960s, manufacturers developed CMOS digital ICs which consume little power and were perfect for battery operated electronic devices

CMOS stands for complementary metal oxide semiconductor

Several families of compatible CMOS ICs have been developed

The first was the 4000 series

Next came the 74C00 series and more recently the 74HC00 series of CMOS digital ICs

In 1985, the FACT (Fairchild Advanced CMOS Technology) 74AC00 series, 74ACT00 series, and 74FCT00 series of extremely fast, low-power CMOS digital ICs were introduced by Fairchild

Many large-scale integrated (LSI) circuits such as digital wristwatch and calculator chips are also manufactured using the CMOS technology

7.23 Understanding the Marking on Digital CMOS ICs

A typical 4000 series CMOS IC is shown in Figure 25.7(a)

Note that pin 1 is marked as such on the top of the IC immediately counterclockwise from the notch

The CD4081BE part number can be divided into sections as shown in Figure 25.7(b)


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Figure 25.7 - (a) Marking on a typical CMOS digital IC (b) Decoding the part number on a typical CMOS digital IC

The prefix “CD” is the manufacturer’s code for CMOS digital ICs

The core part number is 4081B, which stands for a CMOS quadruple 2-input AND gate IC

The trailing letter “E” is the manufacturer’s packaging code for plastic DIP IC

The letter “B” is a “buffered version” of the original 4000A series

The buffering provides the 4000B series devices with greater output drive and some protection from static electricity


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Figure 25.8 shows pin diagram for the CD4081BE CMOS quad 2-input AND gate IC

Figure 25.8 – Pin diagram for the 4081B CMOS IC

Figure 25.9 shows an example of logic diagram for a 2-input AND gate circuit

Figure 25.9 – An example of logic diagram for a 2-input AND gate circuit


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Figure 25.10 shows the wiring diagram using the 4081B CMOS IC

Figure 25.10 - Wiring diagram using the 4081B CMOS IC

A 5 V dc power supply is shown but the 4000 series CMOS IC can use voltages from 3 to 18V dc

Care is taken in removing the 4081 from its conductive foam storage because CMOS ICs can be damaged by static charges

Do not touch the pins when inserting the 4081 CMOS IC in a socket or mounting board

VDD and VSS power connection should be made with the power off

When using CMOS, all unused inputs are tied to GND or VDD

In this example, unused inputs (C, D, E, F, H, G) are grounded


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7.24 Integrated Circuit (IC) Specifications

Integrated circuits within a logic family are designed to interface easily with one another

For example, in the TTL logic family, an output of a TTL device can be connected directly to the input of several other TTL inputs with no extra parts

ICs from the same logic family will interface properly

Interfacing between logic families and between digital ICs and the outside world is a bit more complicated

Interfacing can be defined as the design of the interconnections between circuits that shift the levels of voltage and current to make them compatible

7.24.1 TTL Logic Levels

Figure 26.1 defines the input and output voltage levels for a typical TTL inverter

Figure 26.1 – Defining TTL input and output voltage levels


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For TTL ICs, a LOW input must range from GND to 0.8V

Also a HIGH input must be in the range from 2.0 to 5.5V

The unshaded section from 0.8V to 2.0V on the input side is the undefined area, or undeterminate region

Inputs in the undefined region yield unpredictable results at the output

An output in the range of GND and 0.4V is defined as LOW, while in the range 2.4V to 5.5V is HIGH

The HIGH output depends on the the resistance value of the load placed at the output

The greater the load current, the lower the HIGH output voltage

The unshaded section of the output voltage side in Figure 26.1 is the undefined region

Suspect trouble if the output voltage is in the undefined region (0.4V to 2.4V)

7.24.2 CMOS Logic Levels

The 4000 and 74C00 series CMOS logic families of ICs operate on a wide range of power supply voltages (from +3V to +15V)

The definition of a HIGH and LOW logic level for a typical CMOS inverter from the 4000 and 74C00 series is illustrated in Figure 26.2(a)

A 10-V power supply is being used in this voltage profile diagram


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Figure 26.2 – Defining CMOS input and output voltage levels (a) 4000 and 74C00 series with voltage profile (b) 74HC00, 74AC00 and 74ACQ00 series voltage profile (c) 74HCT00, 74ACT00, 74ACTQ00, 74FCT00, 74FCTA00 series voltage profile


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The CMOS inverter shown in Figure 26.2(a) will respond to any input voltage within 70 – 100 percent of VDD (+10V in this example) as a HIGH

Likewise, any voltage within 0 to 30 percent of VDD is regarded as a LOW input to ICs in the 4000 and 74C00 series

Output voltages are normally almost at the voltage rails of the power supply

In this example, a HIGH output would be about +10V while a LOW output would be about 0V and GND

The 74HC00 series and the newer 74AC00 and 74ACQ00 series operate on a lower voltage power supply (from +2V to +6V) than the older 4000 and 74C00 series CMOS ICs

The input and output voltage characteristics are summarized in the voltage profile diagram in Figure 26.2(b)

The definition of HIGH and LOW for both input and output on the 74HC00, 74AC00, and 74ACQ00 series is approximately the same as for the 4000 and 74C00 series CMOS ICs

The 74HCT00 series and the newer 74ACT00, 74ACTQ00, 74FCT00 and 74FCTA00 series of CMOS ICs are designed to operate on a 5-V power supply like TTL ICs.

The function of the 74HCT00, 74ACT00, 74ACTQ00, 74FCT00 and 74FCTA00 series of CMOS ICs is to interface between TTL and CMOS devices

These CMOS ICs with a “T” designator can serve as direct replacements for many TTL ICs

The voltage profile diagram for the 74HCT00, 74ACT00, 74ACTQ00, 74FCT00 and 74FCTA00 series of CMOS ICs is shown in Figure 26.2(c)


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7.24.3 Noise Margin

Noise immunity is a circuit’s insensitivity or resistance to undesired voltages or noise

In digital circuits it is called noise margin

The noise margins for typical TTL and CMOS families are compared in Figure 26.3

Figure 26.3 – Defining and comparing TTL and CMOS noise margin


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The noise margin is much better for the CMOS than for the TTL family

Noise in a digital system is unwanted voltages induced in the connecting wires and printed circuit board traces that might affect the input logic levels, thereby causing faulty output indications

Figure 26.4 defines the LOW, HIGH, and the undefined regions for TTL inputs

Figure 26.4 – TTL input logic levels showing noise margin

If the actual input voltage is 0.2V, then the margin of safety between it and the undefined region is 0.6V (0.8V – 0.2V = 0.6V), i.e. the noise margin


274

In other words, it would take more than +0.6V added to the LOW voltage (0.2V in this example) to move the input into the undefined region

In actual practice, the noise margin is even greater because the voltage must increase to the switching threshold, which is shown as 1.2V in Figure 26.4

With the actual LOW input +0.2V and the switching threshold at about +1.2V, the actual noise margin is 1V (i.e. 1.2V – 0.2V = 1.0V)

7.24.4 Drive Capabilities – Fan-in and Fan-out

A bipolar transistor has its maximum wattage and collector current ratings

These ratings determine its drive capabilities

One indication of output drive capability of a digital IC is called its fan-out

The fan-out of a digital IC is the number of “standard” inputs that can be driven by the gate’s output

If the fan-out for standard TTL gates is 10, this means that the output of a single gate can drive up to 10 inputs of the gates in the same subfamily

A typical fan-out value for standard TTL ICs is 10. The fan-out for low-power Schottky TTL (LS-TTL) is 20 and for the 4000 series CMOS is considered to be about 50

Another way to look at the current characteristics of gates is to examine their output drive and input loading parameters

Figure 26.5 shows a simplified view of the output drive capabilities and input load characteristics of a standard TTL gate


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Figure 26.5 – Standard TTL voltage and current profiles

A standard gate is capable of handling 16 mA when the

output is LOW (IOL) and 400 A when the output is HIGH (IOH)

The input loading (worst-case condition) is only 40 A with the input HIGH (IIH) and 1.6mA when the input is LOW (IIL)

This means that the output of a standard TTL gate can drive 10 inputs (16mA/1.6mA = 10)

A summary of the output drive and input loading characteristics of several popular families of digital ICs is detailed in Figure 26.6

The load represented by a single gate is called the fan-in of that family of ICs


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Figure 26.6 – Output drive and input loading characteristics for selected TTL and CMOS logic families

Suppose it is required to know if the 74LS04 inverter has enough fan-out to drive the four standard TTL NAND gates on the right as shown in Figure 26.7

The voltage and current profiles for LS-TTL and standard TTL gates are sketched in Figure 26.8

The voltage characteristics of all TTL families are compatible

The LS-TTL gate can drive 10 standard TTL gates

when its output is HIGH (400 A/40 A = 10)

However, the LS-TTL gate can drive only five standard TTL gates when it is LOW (8mA/1.6mA = 5)


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Thus, the fan-out of LS-TTL gates is only 5 when driving standard TTL gates

It is true that the LS-TTL inverter can drive four standard TTL inputs in Figure 26.7

Figure 26.7 – Interfacing LS-TTL to standard TTL problem. Logic diagram of interfacing problem

Figure 26.8 – Voltage and current profiles for LS-TTL and standard TTL gates


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7.24.5 Propagation Delay

Speed, or quickness of response to a change at the inputs, is an important consideration in high-speed applications of digital ICs

Consider the waveforms in Figure 26.9

Figure 26.9 – Propagation delays for a standard TTL inverter

The top waveform shows the input to an inverter going from LOW to HIGH and then from HIGH to LOW

The bottom waveform shows the output response to the changes at the input

The slight delay between the time the input changes and the time the output changes is called the propagation delay of the inverter (measured in seconds)

The propagation delay delay for the LOW-to-HIGH transition of the input to the inverter is different from the HIGH-to-LOW delay

Propagation delays for selected TTL and CMOS families are shown in Figure 26.10


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Figure 26.10 – Graph of propagation delays for selected TTL and CMOS

7.24.6 Power Dissipation

Generally, as propagation delays decrease (increased speed), the power consumption and related heat generation increase

Historically, a standard TTL IC might have a propagation delay of about 10 ns compared with a propagation delay of about 30 to 50 ns for a 4000 series CMOS IC.

The 4000 CMOS IC, however, would consume only 0.001 mW, while standard TTL gate might consume 10 mW of power

The power consumption of CMOS increases with frequency

At 100 kHz, the 4000 series gate may consume 0.1 mW of power


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Figure 26.11 shows a graph of speed versus power of several modern TTL and CMOS families

Figure 26.11 – Speed versus power for selected TTL and CMOS families


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7.25 Interfacing Digital ICs 7.25.1 Interfacing TTL and CMOS with Switches

One of the most common means of entering information into a digital system is the use of switches or a keyboard

Figure 27.1 shows simple active-LOW switch-to-TTL interfaces

Figure 27.1 - Simple active-LOW switch-to-TTL interfaces

Pressing the push-button switch in this circuit will drop the input of the TTL inverter to ground level or LOW

Releasing the push-button switch opens the switch

The input to the TTL inverter now is allowed to “float”

In TTL, inputs usually float at a HIGH logic level

Floating inputs on TTL are not dependable

Figure 27.2 is a slight refinement of the switch input circuit


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Figure 27.2 – Active-LOW switch interface using pull-up resistor

The 10 k resistor, called pull-up resistor, has been added to make sure the input to the TTL inverter goes HIGH when the switch is open

Its purpose is to pull the input voltage up to +5V

Circuits in Figure 27.1 and Figure 27.2 illustrate active-LOW switches

They are called active-LOW switches because the inputs go LOW only when the switch is activated

Figure 27.3 illustrate an active-HIGH input switch


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Figure 27.3 – Active-HIGH switch interface using pull-down resistor

In the active-HIGH switch circuit, when the input switch is activated, the +5V is connected directly to the input of the TTL inverter

When the switch is released (opened) the input is pulled LOW by the pull-down resistor

The value of the pull-down resistor is relatively low because the input current required by a standard TTL gate may be as high as 1.6 mA

Figure 27.4 illustrates an active-LOW input switch-to-CMOS interface circuit

The 100 k pull-up resistor pulls the voltage to +5V when the input switch is open


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Figure 27.4 – An active-LOW switch-to-CMOS interface with pull-up resistor

Figure 27.5 illustrates an active-HIGH switch feeding a CMOS inverter

The 100 k pull-down resistor makes sure the input to the CMOS inverter is near ground when the input switch is open

Figure 27.5 – An active-HIGH switch-to-CMOS interface with pull-down resistor


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The resistance value of the pull-up and pull-down resistors is much greater than those in TTL interface circuits because the input loading currents are much greater in TTL than in CMOS

7.25.2 Switch Debouncing

When a switch is pressed, it will bounce several times

If the switch is interfaced to a decimal counter system, as shown in Figure 27.6, each press of the input switch should cause the decade counter to increase by 1

Figure 27.6 – Switch without debouncing circuit being interfaced to a decimal counter system

However, in practice each press of the switch increases the count by 1, 2, 3, or sometimes more

This means that several pulses are being fed into the clock (CLK) input of the counter each time the switch is pressed


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Figure 27.7 shows the switch that has been added with a debouncing circuit to make the decimal counter work properly

Figure 27.7 – Adding a debouncing circuit to make the decimal counter work properly

The decade counter will now count each HIGH-LOW cycle of the input switch

The cross-coupled NAND gates in the debouncing circuit are sometimes called the RS flip-flop or latch

Several other switch debouncing circuits are illustrated in Figure 27.8 to Figure 27.10

Figure 27.8 – A 4000 series switch debouncing circuit


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The simple debouncing circuit in Figure 27.8 will work on the slower 4000 series CMOS IC

The 40106 CMOS IC is a special inverter called Schmitt trigger inverter, which means it has a “snap action” when changing to either HIGH or LOW

A Schmitt trigger can also change a slow-rising signal (such as a sine wave) into a square wave

The switch debouncing circuit in Figure 27.9 will drive 4000, 74HC00, or FACT series CMOS or TTL ICs

Figure 27.9 – General purpose switch debouncing circuit that will drive CMOS or TTL inputs

Another general purpose switch debouncing circuit is illustrated in Figure 27.10


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Figure 27.10 – Another general purpose switch debouncing circuit that will drive CMOS or TTL inputs

This debouncing circuit can drive either CMOS or TTL inputs

The 7403 is an open-collector NAND TTL IC and needs pull-up resistors as shown in Figure 27.10

The external pull-up resistors make it possible to have an output voltage of just about +5V for HIGH

Open-collector TTL gates with external pull-up resistors are useful when driving CMOS with TTL

7.25.3 Interfacing TTL and CMOS with LEDs

LED can be used as an input indicator because it operates at low currents and voltages

The maximum current required by many LEDs is about 20 to 30 mA with about 2V applied

An LED will light dimly on only 1.7V to 1.8V and 2 mA


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CMOS-to-LED Interfacing

Interfacing 4000 series CMOS devices with simple LED indicator lamps is easy

Figure 27.11(a) and (b) show the CMOS supply voltage at +5V, and thus no limiting resistors are needed in series with the LEDs

Figure 27.11 – Simple CMOS-to-LED interfacing (a) CMOS active-HIGH (b) CMOS active-LOW

In Figure 27.11(a), when the output of the CMOS inverter goes HIGH, the LED output indicator lights

The opposite is true in Figure 27.11(b): when the CMOS output goes LOW, the LED indicator lights


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Figure 22.12(a) and (b) show the 4000 series CMOS ICs being operated on a higher supply voltage (+10V to +15V)

Figure 27.12 – Simple CMOS-to-LED interfacing (a) CMOS active-HIGH, supply voltage = 10V to 15V (b) CMOS active-LOW, supply voltage = 10V to 15V

Because of the higher voltage, a 1 k limiting resistor is placed in series with the LED output indicator lights

When the output of the CMOS inverter in Figure 27.12(a) goes HIGH, the LED output indicator lights

In Figure 27.12(b), however, the LED indicator is activated by a LOW at the CMOS output

Figure 27.13(a) and (b) show CMOS buffers being used to drive LED indicators

The circuit may operate on voltages from +5V to +15V


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Figure 27.13 – Simple CMOS-to-LED interfacing (a) CMOS inverting buffer to LED interfacing (b) CMOS non-inverting buffer to LED interfacing

Figure 27.13(a) shows the use of an inverting CMOS buffer (like the 4049 IC), while Figure 27.13(b) uses the non-inverting buffer (like the 4050 IC)

In both cases, a 1 k limiting resistor must be placed in series with the LED output indicator

TTL-to-LED Interfacing

Standard TTL gates are sometimes used to drive LEDs directly, as shown in Figure 27.14(a) and (b)

Figure 27.14 – Simple TTL-to-LED interfacing (a) TTL active-HIGH (b) TTL active-LOW


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When the output of the inverter in Figure 27.14(a) goes HIGH, current will flow through the LED causing it to light

The indicator light in Figure 27.14(b) only lights when the output of the 7404 inverter goes LOW

Circuits in Figure 27.11 to Figure 27.14 are not recommended for critical uses because they exceed the output current ratings of the ICs

However, the circuits have been tested and work properly as simple output indicators

Current Sourcing and Current Sinking

The idea of current sourcing and current sinking can be illustrated by using the circuits shown in Figure 27.15

Figure 27.15 (a) Current sourcing (b) Current sinking


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In Figure 27.15(a) the output of the TTL AND gate is HIGH and lights the LED

In this case the IC is said to be the source of current (conventional current flow from + to -)

The sourcing current is sketched on the schematic diagram in Figure 27.15(a)

The source current appears to “flow from the IC” through the external circuit (LED and limiting resistor) to ground

In Figure 27.15(b) the output of the TTL NAND gate is LOW

This LOW at the output of the NAND gate lights the LED

In this case, the IC is referred to as sinking the current

The sinking current is sketched on the schematic diagram in Figure 27.15(b)

The sinking current appears to start with +5V above the external circuit (limiting resistor and LED) and “sink to ground” through the external circuit (limiting resistor and LED) and the output pin of the NAND IC

Interfacing to LED Using Transistor Drive Circuit

The LED output indicator circuits can be improved by using transistor drivers

The improved circuit can be used with either CMOS or TTL

The LED in Figure 27.16 lights when the output of the inverter goes HIGH because this will turn on the transistor and the collector current flows from the supply through LED and the limiting resistor


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Figure 27.16 – Interfacing to LED using a transistor driver circuit. Active-HIGH output using a NPN transistor driver

The LED in Figure 27.17 lights when the output of the inverter goes LOW because this will turn on the transistor and the emitter current flows from the supply through LED and the limiting resistor

Figure 27.17 – Interfacing to LED using a transistor driver circuit. Active-LOW output using a PNP transistor driver


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Figure 27.18 is a combination of the active-HIGH and active-LOW LED indicator circuits

Figure 27.18 – Interfacing to LED using a transistor driver circuit. HIGH-LOW indicator circuit

The red LED will light when the inverter’s output is HIGH (during this time the green LED will be off)

When the output of the inverter goes LOW, transistor Q1 turns off while transistor Q2 turns on

This causes the green LED light up and the red LED goes off

This circuit is a very basic logic probe, but the accuracy is less than most logic probes

The indicator light shown in Figure 27.19 uses an incandescent lamp

When the output of the inverter goes HIGH, the transistor is turned on and the lamp lights


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When the inverter’s output is LOW, the lamp does not light

Figure 27.19 – Interfacing to an incandescent lamp using a transistor driver circuit

7.25.4 Interfacing with Buzzers, Relays, Motors and

Solenoids

The objective of many electromechanical systems is to control a simple output device

This device might be as simple as a light, buzzer, relay, electric motor, stepper motor, or solenoid

Interfacing with Buzzers

The piezo buzzer is a modern signaling device drawing much less current than older buzzers and bells

A standard TTL or FACT CMOS inverter is shown in Figure 27.20 driving a piezo buzzer directly


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Figure 27.20 – Logic device to buzzer interfacing. Standard TTL or FACT CMOS inverter driving a piezo buzzer directly

Standard TTL output can sink up to 16 mA while a FACT output has 24 mA of drive current

The piezo buzzer draws about 3 to 5 mA when sounding

Notice that the piezo buzzer has polarity markings

The diode across the buzzer is to suppress any transient voltages that might be induced in the system by the buzzer

Most logic families do not have the current capacity to drive a buzzer directly

A transistor has been added to the output of the inverter in Figure 27.21 to drive the piezo buzzer

When the output of the inverter goes HIGH the transistor is turned on and the buzzer sounds

A LOW at the output of the inverter turns the transistor off, switching the buzzer off

The diode protects against transient voltages


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The interface circuit shown in Figure 27.21 will work for both TTL and CMOS

Figure 27.21 – Logic device to buzzer interfacing. TTL or CMOS interfaced with buzzer using a transistor driver

Interfacing Using Relays

A relay is an excellent method of isolating a logic device from a high-voltage circuit

Figure 27.22 shows how a TTL or CMOS inverter could be interfaced with a relay

When the output of the inverter goes HIGH, the transistor is turned on and the relay is activated

When activated, the normally open (NO) contact of the relay close as the armature clicks downward

When the output of the inverter goes LOW, the transistor stops conducting and the relay is deactivated

The armature springs upward to its normally closed (NC) position


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The clamp diode across the relay coil prevents voltage spikes which might be induced in the system

Figure 27.22 – TTL or CMOS interfaced with a relay using a transistor driver circuit

The circuit in Figure 23.24 uses a relay to isolate an electric motor from the logic devices

Notice that the logic circuit and dc motor have separate power supplies

When the output of the inverter goes HIGH, the transistor is turned on and the NO contacts of the relay snap closed, causing the motor to operate

When the output of the inverter goes LOW, the transistor stops conducting and the relay contacts spring back to their NC position, turning the motor off


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Figure 27.23 – Using a relay to isolate higher voltage/current circuits from digital circuits. Interfacing TTL or CMOS with an electric motor

A solenoid is an electrical device that can produce linear motion

Figure 22.24 shows a solenoid being driven by a logic gate

Note the separate power supplies

This circuit works the same as the motor interface in Figure 27.23


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Figure 27.24 – Using a relay to isolate higher voltage/current circuits from digital circuits. Interfacing TTL or CMOS with a solenoid

In summary, voltage and current characteristics of most buzzers, relays, electric motors, and solenoids are radically different from those of logic circuits

Most of these electrical devices need special interfacing circuits to drive and isolate the devices from the logic circuits


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8 SEQUENTIAL LOGIC CIRCUITS 8.0 Introduction

Logic circuits can be classified into two groups, i.e. combinational logic circuits and sequential logic circuits

So far we have learnt combinational logic circuits using AND, OR and NOT gates

The basic building block for combinational logic circuits is the logic gates

Sequential logic circuits involved timing and memory devices

The basic building block for sequential logic circuits is the flip-flop (FF)

There are several types of FF, i.e. RS flip-flops, D flip-flop-flops and JK flip-flops

Flip-flops can be wired to form counters, shift registers and various memory devices

8.1 The RS Flip-Flop

The logic symbol for RS FF is shown in Figure 28.1

Figure 28.1 – Logic symbol for an RS FF


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Notice that the RS FF has two inputs, labeled S and R

The two output are labeled Q and Q

In FF the outputs are always opposite, or complementary

In other words, if 1Q , then 0Q , and so on

The two letters “S” and “R” at the inputs of the RS FF are often reffered to as the set and reset inputs

Figure 28.2 is the truth table that details the operation of the RS FF

Figure 28.2 – Truth table for RS FF

Notice that when the S and R inputs are both 0, both outputs go to a logical 1

This is called a prohibited state for FF and is not used

The second line of the truth table shows that when input S is 0 and R is 1, the Q output is set to logical 1 – this is called the set condition

The third line shows that when R is 0 and S is 1, output Q is reset (cleared) to 0 – this is called the reset condition

Line 4 in the truth table shows both inputs (R and S) at 1


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This is the idle or at rest condition and leaves Q and Q at their previous complementary states – this is called the hold condition

Notice that the FF in Figure 28.1 has invert bubbles at the R and S inputs

This invert bubbles indicate that the set and reset inputs are activated by a logical 0

RS FFs can be purchased in an IC package, or they can be wired from logic gates such as the NAND gates, as shown in Figure 28.3

Figure 28.3 – Wiring the RS FF using NAND gates Timing Diagrams

Timing diagrams or waveforms show the voltage level and timing between inputs and outputs and are similar to what would be observed on an oscilloscope

The horizontal axis of the diagram is time and the vertical axis is voltage

Figure 28.4 shows the input waveforms (R and S) and

the output waveforms (Q and Q ) for the RS FF


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Figure 28.4 – Waveform diagrams for an RS FF 8.2 The Clocked RS Flip-Flop

The logic symbol for a clocked RS FF is shown in Figure 28.5

Figure 28.5 – Logic symbol for a clocked RS FF


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It looks almost like an RS FF except that it has one extra input labelled CLK (for clock)

The timing diagram for the operation of the clocked RS FF is shown in Figure 28.6

Figure 28.6 – Waveform diagram for a clocked SR FF

Notice that the clock pulse (1) has no effect on output Q with input S and R in the 0 position

The flip-flop is in the idle, or hold mode during clock pulse 1

At the preset S position, the S (set) input is moved to 1, but output Q is not yet set to 1

The rising edge of clock pulse 2 permits Q to go to 1

Pulse 3 and 4 have no effect on output Q

During pulse 3 the flip-flop is in its set mode, while during pulse 4 it is in its hold mode

Next, input R is preset to 1

On the rising edge of clock pulse 5 the Q output is reset (or cleared) to 0


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The flip-flop is in the reset mode during both clock pulses 5 and 6

The flip-flop is in its hold mode during clock pulse 7; therefore, the normal output (Q) remains at Q

Notice that the outputs of the clocked RS FF change only on a clock pulse

The flip-flop is said to operate synchronously; it operates in step with the clock

Synchronous operation is very important in most digital circuits, where each step must happen in an exact order

Another characteristics of the clocked RS FF is that once it is set or reset it stays that way even if some inputs is changed

This is a memory characteristic, which is extremely valuable in many digital circuits

In waveform diagram in Figure 28.6, this flip-flop is in the hold mode during clock pulses 1, 4, and 7

Figure 28.7 shows a truth table for the clocked RS FF

Figure 28.7 – Truth table for a clocked RS FF


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Notice that only the top three lines of the truth table are usable; the bottom line is prohibited and not used

Observe that the R and S inputs to the clocked RS FF are active HIGH inputs – it takes a HIGH on input S while R = 0 to cause output Q to be set to 1

Figure 28.8 shows a wiring diagram of a clocked RS FF

Notice that two NAND gates have been added to the inputs of the RS FF to add the clocked feature

Figure 28.8 – Wiring a clocked RS FF using NAND gates


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8.3 The D Flip-Flop

Figure 28.9(a) shows the logical symbol for a D FF

It has only one data input (D) and a clock input (CLK)

The outputs are labeled Q and Q

Figure 28.9 – D FF (a) Logic symbol (b) Simplified truth table

The D FF is often called a delay flip-flop

The word “delay” describes what happens to the data, or information, at input D

The data (a 0 or 1) at input D is delayed one clock pulse from getting to output Q


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A simplified truth table for the D FF is shown in Figure 28.9(b)

Notice that the output Q follows input D after one clock pulse (see Qn+1 column)

A D FF may be formed from a clocked RS FF by adding an inverter, as shown in Figure 28.10

Figure 28.10 – Wiring a D flip-flop

Figure 28.11 shows a typical commercial D FF

Two extra inputs preset, (PS) and clear (CLR) have been added to the D FF

Figure 28.11 – Logic symbol for commercial D FF


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The PS input sets output Q to 1 when enabled by a logical 0

The CLR input clears output Q to 0 when enabled by a logical 0

The PS and CLR inputs will override the D and CLK inputs

Note the addition of the triangle on the CLK input in Figure 28.11

A more detailed truth table for the commercial 7474 TTL D FF isshown in Figure 28.12

Figure 28.12 – The truth table for 7474 D FF

The asynchronous (not synchronous) inputs (PS and CLR) override the synchronous inputs

The asynchronous inputs are in control of the D FF in the first three lines of the truth table


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The synchronous inputs D and CLK are irrelevant as shown by the “X”s on the truth table

The prohibited condition, line 3 on the truth table should be avoided

With both asynchronous inputs disabled (PS = 1 and CLR = 1), the D FF can be set and reset using the D and CLK inputs

The last two lines of the truth table use a clock pulse to transfer data from input D to output Q of the flip-flop

Being in step with the clock, this is called synchronous operation

Note that this flip-flop uses the LOW-to-HIGH transition of the clock pulse to transfer data from input D to output Q

D FF are sequential logic devices which are widely used temporary memory devices

D FF are wired together to form shift registers and storage registers

Since the D FF delays data from reaching the output Q one clock pulse, it is called a delay flip-flop


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8.4 The JK Flip-Flop

The JK FF has the features of all the other types of flip-flops

The logic symbol for the JK FF is illustrated in Figure 28.13

Figure 28.13 – Logic symbol for JK flip-flop

The inputs lebelled J and K are the data inputs

The input labeled CLK is the clock input

Outputs Q and Q are the usual normal and complementary outputs on a flip-flop

A truth table for the JK FF is shown in Figure 28.14


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Figure 28.14 – Truth table for JK FF

When the J and K inputs are both 0, the flip-flop is in the hold mode

In the hold mode the data inputs have no effect on the outputs – the outputs “hold” the last data present

Line 2 and 3 of the truth table show the reset and set conditions for the Q output

Line 4 illustrates the useful toggle position of the JK FF

When both data inputs J and K are at 1, repeated clock pulses cause the output to turn off-on-off-on-off-on, and so on

This off-on action is like a toggle switch and is called toggling

Figure 28.15 shows the logic symbol for the commercial 7476 TTL JK FF


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Figure 28.15 – Logic symbol for commercial JK FF

Added to the symbol are two asynchronous inputs, preset (PS) and clear (CLR)

The synchronous inputs are the J and K data and clock inputs

A detailed truth table for the commercial 7476 JK FF is shown in Figure 28.16

Recall that asynchronous inputs, such as preset (PS) and clear (CLR) override synchronous inputs

The asynchronous inputs are activated in the first three lines of the truth table

The asynchronous inputs are activated in the first three lines of the truth table

The synchronous inputs are irrelevant (overriden) in the first three lines in Figure 28.16; therefore an “X” is placed under the J, K, and CLK inputsfor these rows

The prohibited state occurs when both asynchronous inputs are activated at the same time


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Figure 28.16 – Truth table for commercial 7476 JK FF

The prohibited state is not useful and should be avoided

When both asynchronous inputs (PS and CLR) are disabled with a 1, the synchronous inputs can be activated

The bottom four lines of the truth table in Figure 28.16 detail the hold, reset, set, and toggle modes of operation for the 7476 JK FF

Note that the 7476 JK FF uses the entire pulse to

transfer data from the J and K inputs to the Q and Q outputs

1


Question 1 Consider an oven along with the associated input and output signals, shown in Figure 1.

The operation of the oven is as follows:

(i) The heater (H) will be ON when the power switch (P) is ON, the door (D) is closed,

and the temperature (T) is below the limit.

(ii) The fans (F) will be turned ON when the heater (H) is ON or when the temperature

(T) is above the limit and the door (D) is closed

(iii) The light (L) will be turned ON if the light switch (S) is ON or whenever the door

(D) is opened

Write the Boolean expressions for the heater (H), the fans (F) and the light (L) which

govern the operation of the oven and construct the logic circuit to accomplish the

operation.

Figure 1

2

Question 2 Design a mod-10 ripple counter using NAND gates. Draw the circuit diagram of your design and explain how the counter works. Draw a table and the relevant timing diagram to illustrate the counting sequence. Simplify your design by using a 7493 IC, a 4-bit binary counter as shown in Figure 2.

Figure 2


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8.5 Latches

There are three types of multivibrators – monostable multivibrators, astable multivibrators, and bistable multivibrators

The RS FF is one of several bistable multivibrators

The RS FF is most commonly known as a latch

A latch is a fundamental binary memory device for holding data

Latches are commonly used at the output of a digital device to hold the data until the next device is ready to receive the input

Latches are commonly organized into groups of 4-bits, 8-bits or more into registers

An 8-bit register would be a group of eight latches holding a byte of information

An example of commercial RS flip-flop IC is 74LS279 Quad SR Latch IC

8.5.1 IC Latches

Figure 29.1 shows an electronic encoder/decoder system without buffer memory

Figure 29.1 - An electronic encoder/decoder system without buffer memory


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In this system, when the decimal 7 on the keyboard is pressed and hold, a 7 will be observed on the seven-segment display

If the 7 on the keyboard is released, the 7 on the display will disappear

It is obvious that a memory device is needed to hold the BCD code for 7 at the inputs to the decoder

A device that serves as a temporary buffer memory is called a latch

Figure 29.2 shows a 4-bit latch has been added to the system

Figure 29.2 - An electronic encoder/decoder system with buffer memory (latch) added

In this modified system, when the decimal 7 on the keyboard is pressed and released, the seven-segnent display continues to show a 7

The term “latch” refers toa digital storage device

The D FF is a good example of device used to latch data

However, other types of flip-flops are also used for latching function

An example of commercial IC latches is the 7475 TTL 4-bit transparent latch, as shown in Figure 29.3


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Figure 29.3 – Logic symbol for commercial 7475 4-bit transparent latch

This unit has four D FF enclosed in a single IC package

The enable input (E0-1) is similar to the clock input on the D FF

When E0-1 is enabled, both D0 and D1 are transferred to their outputs

A simplified truth table for the 7475 latch IC is shown in Figure 29.4

Figure 29.4 – Truth table for 7475 D latch


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If the enable input is at a logical 1, data is transferred, without a separate clock pulse, from D input to the Q

and Q outputs

As an example, if E0-1 = 1 and D1 =1, then without a

clock pulse output Q1 will be set to 1 while 1Q would be

reset to 0

In the data-enabled mode of operation the Q outputs follow their respective D inputs on the 7475 latch

When the enable input drops to 0, the 7475 IC enters the data-latched mode

In this mode, the data that was at Q remains the same even if the D inputs change – the data is said to be latched

The 7475 IC is called a transparent latch because when enable input is HIGH, the normal outputs follow the data at the D inputs

Note that the D0 and D1 flip-flops in the 7475 IC are controlled by the E0-1 enable input whereas the E2-3 controls the D2 and D3 pairs of flip-flops

Other than its use as a latch, flip-flops can also be used as counters, shift registers, delay units, and frequency dividers


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8.6 Triggering Flip-Flops

Flip-flops can be classified as synchronous or asynchronous in their operation

Synchronous flip-flops are those that have a clock input

The clocked RS, the D, and the JK flip-flop operate in step with the clock

Synchronous flip-flops are also classified as either edge-triggered or master/slave

Figure 29.5 shows two edge-triggered flip-flops in the toggle position

Figure 29.5 – Waveforms for positive- and negative-triggered flip-flops

On clock pulse 1 the positive edge (positive-going edge) of the pulse is identified

The second waveform shows how the positive-edge triggered flip-flop toggles each time a positive-going pulse comes along (see pulses 1 to 4)

On pulse 1 in Figure 29.5 the negative edge(negative-going edge) of the pulse is also labeled

The bottom waveform shows how the negative-edge-triggered flip-flop toggles

Notice that it changes state, or toggles, each time a negative-going pulse comes along (see pulses 1 to 4)


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Especially notice the difference in timing between the positive- and negative-edge triggered flip-flops

Figure 29.6(a) shows the logic symbol of a D FF with positive-edge triggering

The symbol uses the small > inside the flip-flop near the clock input

This > symbol says data is transferred to the output on the edge of the pulse

Figure 29.6 – (a) Logic symbol for positive-edge triggered D FF (b) Logic symbol for negative-edge triggered D FF (c) Logic symbol for D latch

A logic symbol for a D FF using negative-edge triggering is shown in Figure 29.6(b)


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The added invert bubble at the clock input shows that triggering occurs on the negative-going edge of the clock pulse

A typical D latch symbol is shown in Figure 29.6(c)

The lack of a > symbol next to the enable (similar to a clock) input means that this unit is not considered an edge-triggered unit

Like the RS FF, the D latch is considered asynchronous

The JK master/slave flip-flop uses the entire pulse (positive edge and negative edge) to trigger the flip-flop

Figure 29.7 shows the triggering of a master/slave flip-flop

Figure 29.7 – Triggering the JK master/slave FF

Pulse 1 shows four positions (a to d) on the waveform

The following sequences of operation takes place in the master/slave flip-flop at each point on the clock pulse: o Point a: leading edge – isolate input from output o Point b: leading edge – enter information from J

and K inputs o Point c: trailing edge – disable J and K inputs o Point d: trailing edge – transfer information from

input to output


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8.7 Schmitt Trigger

Digital circuits prefer waveforms with fast rise and fall times

Schmitt trigger can be used to reshape or condition the input signal and make the rise and fall times of the signal very fast (almost instantaneous)

Figure 29.8 shows a Schmitt trigger inverter is being used to “square up” the input signal and make it more useful

Figure 29.8 – Schmitt trigger used for wave shaping

A voltage profile of a typical TTL inverter (7404 IC) is shown in Figure 29.9

Of special interest is the switching threshold of the 7404 IC

Figure 29.9 – TTL voltage profiles with switching threshold


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The switching threshold may vary from chip to chip, but it is always in the undefined region

Figure 29.9 shows that a typical 7404 IC has a switching threshold of +1.2 V

In other words, when the voltage rises to +1.2 V, the output changes from HIGH to LOW

However, if the voltage drops below +1.2 V, the output switches from LOW to HIGH

Most regular gates have a single switching threshold voltage whether the input voltage is rising (L to H) or falling (H to L)

A voltage profile for a 7414 Schmitt trigger inverter TTL IC is shown in Figure 20.10

Figure 29.10 – Voltage profiles for 7414 TTL Schmitt trigger IC showing switching thresholds


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Note that the switching threshold is different for positive-going (V +) and negative-going (V -) voltages

The voltage profile for the 7414 IC shows that the switching threshold is 1.7 V for a positive-going (V +) input voltage.

However, the switching threshold is 0.9 V for a negative-going (V -) input voltage

The difference between these swiching thresholds (1.7 V and 0.9 V) is called hysteresis

Hysteresis provides for excellent noise immunity and helps the Schmitt trigger square up wave form with slow rise and fall time


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9.0 Counters

Almost any complex digital system contains several counters

A counter’s job is the obvious one of counting events or periods of time or putting events into sequence

Other application of counters include dividing frequency, addressing, and serving as memory units

Flip-flops can be wired together to form circuits that counts

Because of the wide use of counters, manufacturers also make self-contained counters in IC form

Many counters are available in all TTL and CMOS families

Some counter ICs contain other devices such as signal conditioning circuitry, latches and display multiplexers

9.1 Ripple Counters

Figure 30.1 illustrate the counting in binary and decimal

With four binary places (D, C, B, and A) the count is from 0000 to 1111 (0 to 15 in decimal)

If it is required to design a counter to count from binary 0000 to 1111, a device that has 16 different output states: a modulo (mod)-16 counter is needed

The modulus of a counter is the number of different states the counter must go through to complete its counting cycle

A mod-16 counter using four JK FF is diagramed in Figure 30.2


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Figure 30.1 – Counting sequence for a 4-bit electronic counter


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(a)

(b)

Figure 30.2 – Mod-16 counter (a) Logic diagram (b) Waveform diagram

Each JK FF is in its toggle position (J and K both at 1)

Assume the outputs are cleared to 0000

As clock pulse 1 arrives at the clock (CLK) input of FF1, it toggles (on the negative edge) and the display shows 0001

Clock pulse 2 causes FF1 to toggle again, returning output Q to 0, which causes FF2 to toggle to 1

The count on the display now reads 0010

The counting continuous, with each flip-flop output triggering the next flip-flop on its negative going pulse


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Notice that FF1 must toggle for each pulse, while FF2 must toggle half as often as FF1

Each more significant bit toggles less often

FF1 triggers FF2, FF2 triggers FF3, and so on

Because one flip-flop affect the next one, it takes some time to toggle all flip-flops

The changing of states is a chain reaction that ripples through the counter

For this reason this counter is called a ripple counter

The counter in Figure 30.2 could be described as a ripple counter, a mod-16 counter, a 4-bit counter, or an asynchronous counter

The ripple and asynchronous labels mean that all the flip-flops do not trigger at one time

9.2 Mod-10 Ripple Counters

The counting sequence for mod-10 counter is from 0000 to 1001 (0 to 9 in decimal)

The mod-10 counter has four place values; 8s, 4s, 2s, and 1s

This takes four flip-flops connected as a ripple counter in Figure 30.3

Figure 30.3 – Logic diagram for a mod-10 ripple counter


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A NAND gate must be added to the ripple counter to clear all the flip-flops back to zero immediately after the 1001 (decimal 9) count

The trick is to look at Figure 30.1 and determine what the next count will be after 1001

Since the next count after 1001 is 1010 (decimal 10), the two 1s in the 1010 must be fed into a NAND gate to clear the flip-flop back to 0000

This type of counter might also be called a decade (meaning 10) counter

Ripple counters can be constructed from individual flip-flops

Manufacturers also produce ICs with all four flip-flops inside a single package

9.3 Synchronous Counters

Ripple counters are asynchronous counters because each flip-flops does not trigger exactly in step with the clock pulse

For some high-frequency operations it is necessary to have all stages of the counter trigger together

There is such a counter: a synchronous counter

Logic diagram for a synchronous counter is shown in Figure 30.4


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Figure 30.4 – A 3-bit synchronous counter

This logic diagram is for a 3-bit (mod-8) counter

Notice that the clock is connected directly to the CLK input of each flip-flop

The CLK inputs are connected in parallel

Figure 30.5 shows the counting sequence of this counter

Figure 30.5 – The counting sequence of a 3-bit synchronous counters


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The counting sequence of this counter is as follows: o Pulse 1 (row 2 in truth table)

Circuit action : Each flip-flop is pulsed by the clock Only FF1 can toggle because it is the only

one with 1s applied to both J and K input FF1 goes from 0 to 1 Output result: 001 (decimal 1)

o Pulse 2 (row 3 in truth table) Circuit action : Each flip-flop is pulsed by the clock Two flip-flops toggle because they have 1s

applied to both J and K inputs FF1 and FF2 both toggle FF1 goes from 1 to 0 FF2 goes from 0 to 1 Output result: 011 (decimal 3)

o Pulse 3 (row 4 in truth table) Circuit action : Each flip-flop is pulsed by the clock Only flip-flop toggles FF1 goes from 0 to 1 Output result: 011 (decimal 3)

o Pulse 4 (row 5 in truth table) Circuit action : Each flip-flop is pulsed by the clock All flip-flops toggle to opposite state FF1 goes from 1 to 0 FF2 goes from 1 to 0 FF3 goes from 0 to 1 Output result: 100 (decimal 4)


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o Pulse 5 (row 6 in truth table)

Circuit action : Each flip-flop is pulsed by the clock Only one flip-flop toggles FF1 goes from 0 to 1 Output result: 101 (decimal 5)

o Pulse 6 (row 7 in truth table) Circuit action : Each flip-flop is pulsed by the clock Two flip-flops toggle FF1 goes from 1 to 0 FF2 goes from 0 to 1 Output result: 110 (decimal 6)

o Pulse 7 (row 8 in truth table) Circuit action : Each flip-flop is pulsed by the clock Only one flip-flop toggles FF1 goes from 0 to 1 Output result: 111 (decimal 7)

o Pulse 8 (row 9 in truth table) Circuit action : Each flip-flop is pulsed by the clock All three flip-flops toggle All flip-flops change from 1 to 0 Output result: 000 (decimal 0)

Notice that the JK FF are used in their toggle mode (J and K at 1) or hold mode (J and K at 0)

Synchronous counters are most often purchased in IC form.

Synchronous counters are available in both TTL and CMOS


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9.4 Down Counters

A counter that counts from higher to lower numbers is called a down counter

Figure 30.6 shows a logic diagram of a mod-8 asynchronous down counter

Figure 30.6 – A 3-bit ripple down counter

Note how much the down counter in Figure 30.6 looks like the up counter in Figure 30.2

The only difference is in the “carry” from FF1 to FF2 and the carry from FF2 to FF3

The up counter carries from Q to the CLK input of the next flip-flop

The down counter carries from Q to the CLK input of

the next flip-flop

Notice that the down counter has a preset (PS) control to preset the counter to 111 (decimal 7) to start the downward count

The counting sequence for this counter is shown in Figure 30.7


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Figure 30.7 – The counting sequence of a 3-bit ripple down counter

9.5 Self-Stopping Counters

The down counter shown in Figure 30.6 recirculates

That is, when it gets to 000 it starts at 111, then 110, and so forth

However, sometimes it desired that a counter stop when a sequence is finished

Figure 30.7 shows a 3-bit down counter with selfstopping feature


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Figure 30.7 – A 3-bit down counter with self-stopping feature

An OR gate is added to place a logical 0 on the J and K inputs of FF1 when the count at outputs C, B, and A reaches 000

The preset must be enabled (PS to 0) again to start the sequence at 111 (decimal 7)

Up ordown counters can be stopped after any sequence of counts by using a logic gate or combination of gates

The output of the gate is fed back to the J and K inputs of the first flip-flop in a ripple counter

The logical 0s fed back to the J and K inputs of FF1 in Figure 30.7 place it in the hold mode

This stopsa FF1 from toggling, thereby stopping the count at 000

9.6 Counters as Frequency Dividers

An interesting and common use of counters is for frequency division

As an example of a simple system using a frequency divider is shown in Figure 30.8, which forms the basis for a digital clock


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Figure 30.8 – A 1-second timer system

The 60 Hz input frequency of sinusoidal signal formed into a square wave, is fed to a divide-by 60 circuit.

The divide-by-60 circuit can be formed by using a modulo-6 counter in series with a modulo-10 (a decade counter), as shown in Figure 30.9

Figure 30.9 – Practical divide-by-60 circuit used as a 1-second timer


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9.7 TTL IC Counters

Figure 31.1 shows the block diagram of a 7493 TTL 4-bit binary counter and Figure 31.2 shows its pin

Figure 31.1 - Block diagram of a 7493 TTL 4-bit binary counter


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Figure 30.2 – Pin configuration of 7493 IC


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The block diagram shows that the 7493 IC houses four JK FFs wired as a ripple counter

Notice that the bottom three JK FFs are prewired internally as a 3-bit ripple counter with output QB connected to the clock input of the next lower JK FF and output QC connected internally to the clock input of the bottom JK FF

Importantly, the top JK FF does not have its QA output internally connected to the next lower flip-flop

To use the 7493 IC as a 4-bit ripple counter (mod-16), the QA output has to be connected to input B which is the CLK input of the second flip-flop

A counting sequence for the 7493 IC wired as a 4-bit ripple counter is shown in figure 31.3

Figure 31.3 – Counting sequence


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The set/reset function table in Figure 31.4 shows that the 7493 counter will be reset (0000) when both R0(1) and R0(2) are HIGH

When either or both reset inputs are LOW, the 7493 IC will count

Figure 31.4 – Reset/Count function table


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9.8 CMOS IC Counters

Manufacturers of CMOS chips offer a variety of counters in IC form

Only one type of such counters, as shown in Figure 31.5, will be discussed in this section

Figure 31.5 – CMOS dual 4-bit binary counter IC (74HC393). (a) Function diagram (b) pin configuration (c) Detailed logic diagram (d) Pin diagram

Note that the IC contains two 4-bit binary ripple counters

The table in Figure 31.5(b) gives the names and functions of each input and output pins on the 74HC393 IC

Note that the clock inputs are labeled with the letters

CP instead of CLK, as used ealier


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9.9 A 3-Digit BCD Counter

Figure 31.6 shows the functional block diagram of a 4553 (MC14553) CMOS 3-Digit BCD Counter

Figure 31.6 – The functional block diagram of 4553 3-Digit BCD Counter

This IC contains three cascaded decade counters

Cascading counters means that the 1s BCD counter triggers the 10s counter as it recirculates from 1001BCD to 0000BCD

In like manner, the 10s counter triggers the 100s counter as it recirculates from 1001BCD to 0000BCD

A truth table drawn in Figure 31.7 for the 4553 3-digit BCD counter IC shows a few of the modes of operation


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Figure 31.7 – Partial truth table for 4553 3-Digit BCD counter

The BCD output from the three counters are fed through the the three 4-bit transparent latches

The BCD data is then transferred to a display multiplexer circuit

The display multiplexing circuit will drive three 7-segment displays


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Figure 31.8 shows a 3-Digit U

Figure 31.8 – A 3-Digit Up Counter circuit

Assignment - Digital SSP3323

1 a) Draw the circuit for the function YXYXF using

i. 2-input AND and OR gates, and a NOT.

ii. 2-input NAND gate only.

b) Prepare a truth table for the following Boolean expression, zyxF

c) Table 1 is a part of the truth table for a 2-bit comparator circuit. The input to

the comparator is number 2 bits, A = {A1, A0} and B = {B1, B0}. Comparator

output is G and L which will show whether the two numbers A and B is

greater or smaller. If A> B then G = 1 and if B> A then L = 1. If A = B then

G = L = 0.

i) Complete the truth table.

ii) Write the Boolean expression for G and L from the truth table.

iii) Simplify the expression for G and L using Karnaugh map

iii) Draw the logic circuit for the expression of G and L

Table 1

2). a) Explain the following flip-flops (Truth table, logic symbol and its

operation):

i) RS Flip-Flop

ii) D Flip-Flop

iii) JK Flip-Flop

A1 A0 B1 B0 G L

0 0 0 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 1 1 0 1

0 1 0 0 0

0 1 0 1 0 0

0 1 1 0 1

0 1 1 1 0

1 0 0 0 1 0

1 0 0 1 0

1 0 1 0 0

1 0 1 1 0

1 1 0 0 1 0

1 1 0 1 0

1 1 1 0 0

1 1 1 1 0 0

2-bit Comparator

A=A1A0 B=B1B0

G L

Assignment - Digital SSP3323

b) Sketch the waveform of the output Y against the input X of figure 2(a).

c) Explain the followings (Truth table, logic diagram and its operation):

i) Decade counter (Modulo 10 counter)

ii) Step down counter

d) Sketch the output waveforms, A, B and C after each clock pulse for the

shift register shown in Figure 2(b).

A

B

C

Figure 2(b)

Figure 2(a)

SSP 3323

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1. Referring to Figure 1, the transistor has the following parameters:

hie = 1 k, hfe = 100, hoe = 0.2 mS, hre = 0.

+VCC

B

C

E

C1

C2

40 k1.5 k

10 k 300 Vo

Vi

Figure 1

(a) Draw the AC equivalent circuit for the circuit given in Figure 1.

[5 marks]

(b) Draw the hybrid equivalent circuit for the circuit given in Figure 1.

[8 marks]

(c) Determine

(i) the input resistance,

(ii) the output resistance,

(iii) the voltage gain, and

(iv) the current gain

of the amplifier

[ 12 marks]

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2. (a) What is a cascade amplifier and show that the overall gain, A of an n-

stages cascade amplifier is

nAAAAA ...321

[5 marks]

(b) Two of the amplifier stages, as shown in Figure 1, are connected in

cascade. A load resistance of 1 k is connected to the output of the second stage. Draw the small signal equivalent circuit of the cascade

amplifier together with the load.

[8 marks]

(c) Determine the total voltage gain of the cascade amplifier

[12 marks]

3. (a) State the condition for impedance matching between any two circuits or

electronic equipments.

[6 marks]

(b) Suggest a method to solve the problem of impedance mismatch

between any two circuits using an operational amplifier.

[7 marks]

+ V

- V

R2

R1

vi v

o

Figure 2

(c) Figure 2 shows an operational amplifier circuit built to operate as an

inverting amplifier. The gain is required to be fully variable between a

minimum of -1 and a maximum of -11.

(i) Suggest how this might be achieved, sketching a suitable circuit

and calculating component values.

SSP 3323

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[8 marks]

(ii) What restrictions are there for the input signal voltage when the

amplifier is at maximum gain?

[4 marks]

4. (a) What is the difference between combinational logic circuits and

sequential logic circuits?

[5 marks]

(b) Consider an oven along with the associated input and output signals,

shown in Figure 3. The operation of the oven is as follows:

(i) The heater (H) will be ON when the power switch (P) is ON, the

door (D) is closed, and the temperature (T) is below the limit.

(ii) The fans (F) will be turned ON when the heater (H) is ON or

when the temperature (T) is above the limit and the door (D) is

closed

(iii) The light (L) will be turned ON if the light switch (S) is ON or

whenever the door (D) is opened

Write the Boolean expressions for the heater (H), the fans (F) and the

light (L) which govern the operation of the oven and construct the logic

circuit to accomplish the operation.

Figure 3

[10 marks]

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(c) Given the following Boolean expression:

DCBADCBA

DCBADCBADCBADCBAY

Simplify the Boolean expression using Karnaugh map and construct the

logic circuit for the simplified Boolean expression.

[ 10 marks]

5. (a) Describe with the aid of device symbol and truth table, the three types

of flip-flops; RS-FF, D-FF and JK-FF..

[9 marks]

(b) Referring to the block diagram of the D flip-flop in Figure 4(a), copy

the timing diagram in Figure 4(b) and sketch the output Q and Q

correspond to the inputs CLK, PS, D and CLR of the flip-flop.

Figure 4

[8 marks]

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(c) Referring to the block diagram of the JK flip-flop in Figure 5(a), copy

the timing diagram in Figure 5(b) and sketch the output Q corresponds

to the inputs CLK, PS, J, K and CLR of the flip-flop.

Figure 5

[8 marks]

1. a) Explain two methods to create a connection to several multi-stage amplifier.

[2 marks] b) For a transformer with primary and secondary windings ratio as n1 : n2 and the

load on the secondary terminals , ZL , show that input impedance for

transformers is Li Zn

nZ

2

2

1

[4 marks]

c) Give two reasons for using a multistage amplifier over a single-stage circuit in a design.

[4 marks]

d) Figure 1 is a multi-stage amplifier circuit using coupling transformers

Figure 1

If the hybrid parameters for the transistor are hie =2K, hfe = 50, hre = hoe = 0

Calculate the voltage gain Av=vo/vi for this amplifier.

[10 marks]

2. a) Power amplifier are generally classified according to the percent of time the

output transistors are conducting or ‘turned on’. Based on the statement, how

Class-A, Class-B and Class-AB amplifier are classified? Discuss the advantage and

disadvantage between Class-A and Class-B amplifier.

[6 marks] b) Explain the following on a differential amplifier

i) common mode

ii) differential mode

iii) CMRR (common mode rejection ratio)

[6 marks]

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c) A differential amplifier with differential mode gain, Ad = 105 and CMRR = 104.

The input voltage is v1 = 105 V and v2 = 95 V. Calculate

i) common mode voltage, vc

ii) differential mode voltage, vd

iii) the magnitude of the common mode gain, AC

iv) the output voltage, vo

[8 marks]

3) a) Give four (4) main characteristics of the Operational Amplifier (Op-Amp)

[4 marks]

b) Refer to the Op-Amp circuit in Figure 3a, sketch the output voltage, vo when the

input voltage, vi = 10V sin t.

[4 marks]

c) Refer to the op-amp circuit shown in Figure 3b;

i) Which point is known as ”virtual ground” and why?

ii) Which resistor is the feedback resistor?

iii) Find the voltage gain, Av

iv) Sketch the input voltage, Vin =(2V)Sint and the output voltage, Vo for

R1=1k and R2=5k.

[8 marks]

+15V

-15V

vi

vo

7V

-

Figure 3a

+

SSCP3323

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Figure 3b

d) Design an OP-AMP summing amplifier circuit that can produce an ouput voltage,

)5.0( 321O VVVV where V1, V2 and V3 are the input voltages. Set the

feedback resistor equal to 2k and the supply voltage equal to 15V.

[4 marks]

4) a) State de Morgan theorems. [2 marks]

Convert the following Boolean expression to their minterm and maxterm form.

i) CABAABCY

ii) ))(( DCBADACBY

[6 marks]

b) Prepare a truth table for the following Boolean expression, zyxF

[4 marks]

c) Table 1 is a part of the truth table for a 2-bit comparator circuit. The input to the

comparator is number 2 bits, A = {A1, A0} and B = {B1, B0}. Comparator output

is G and L which will show whether the two numbers A and B is greater or

smaller. If A> B then G = 1 and if B> A then L = 1. If A = B then G = L = 0.

i) Complete the truth table.

ii) Write the Boolean expression for G and L from the truth table.

iii) Simplify the expression for G and L using Karnaugh map

iii) Draw the logic circuit for the expression of G and L

[8 marks]

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Table 1

5). a) Explain the following flip-flops (Truth table, logic symbol and its operation):

i) RS Flip-Flop

ii) D Flip-Flop

iii) JK Flip-Flop

[6 marks]

b) Sketch the waveform of the output Y against the input X of figure 5(a).

[4 marks]

A1 A0 B1 B0 G L

0 0 0 0 0 0

0 0 0 1 0

0 0 1 0 0

0 0 1 1 0 1

0 1 0 0 0

0 1 0 1 0 0

0 1 1 0 1

0 1 1 1 0

1 0 0 0 1 0

1 0 0 1 0

1 0 1 0 0

1 0 1 1 0

1 1 0 0 1 0

1 1 0 1 0

1 1 1 0 0

1 1 1 1 0 0

2-bit Comparator

A=A1A0 B=B1B0

G L

Figure 5(a)

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c) Explain the followings with their truth table, logic diagram and its operation:

i) Decade counter (Modulo 10 counter)

ii) Step down counter

[6 marks]

d) Sketch the output waveforms, A, B, C and D after each clock pulse for the shift

register shown in Figure 5(b).

[4 marks]

A

B

C

D

Figure 5(b)

Elektronik Lanjutan SSP2323

2

1. Bagi litar dalam Rajah 1, parameter transistor adalah = 100 dan VA = .

(a) Lukis litar setara isyarat kecil hybrid- . [4 markah]

(b) Tentukan:

(i) nilai gm dan r. [4 markah]

(ii) gandaan voltan isyarat-kecil, Avs = vo / vs. [4 markah]

(iii) rintangan masukan amplifier, Ri [4 markah]

(iv) gandaan arus isyarat-kecil, Ais = io / is [4 markah]

Rajah 1

2. Berpandukan litar pada Rajah 2,

(i) lukis litar setara a.u (parameter h), bagi amplifier ini, [5 markah]

(ii) kira gandaan voltan seluruh amplifier, Avs = vo/vs, [5 markah]

(iii) kira impedans masukan, Ris dan [5 markah]

(iv) kira impedans keluaran, Ro [5 markah]

Ambil nilai parameter h bagi kedua-dua transistor sebagai hie = 2K, hfe

= 100 dan hre = hoe = 0.


3

Rajah 2

3. (a) Berdasarkan lakaran graf ciri keluaran transistor, terangkan perbezaan di

antara amplifier kelas A, B dan AB. Nyatakan kebaikan dan kelemahan tiap-

tiap kelas tersebut. [6 markah]

(b) Rajah 3 adalah litar amplifier kuasa kelas A gandingan transformer. Nilai

komponen yang digunakan adalah VCC = 10 V, RL = 8 , n1 : n2 = 3 : 1, R1 =

0.73 k, R2 = 1.55 k dan RE = 20 . Parameter transistor pula adalah hie =

4.3 , hfe = 25, hre = hoe = 0 dan VBE(on) = 0.7 V. Amplitud voltan masukan

sinusoid adalah 17 mV.

(i) Lukis litar setara parameter h untuk amplifier ini. [4 markah]

(ii) Kira gandaan voltan, sov vvA di mana Vo adalah voltan keluaran ac

merentas beban, L. [4 markah]

(iii) Kira kuasa dari bekalan kuasa, PDC [3 markah]

(iv) Tentukan kuasa ulangalik (Pac) yang dihantar kebeban [3 markah]

(v) Kira nilai kecekapan amplifier. [3 markah]


4

Rajah 3

4. (a) Nilai perintang pada Rajah 4(a) adalah R1 = 20 k, R2 = 120 k, R3 =

15 k dan R4 = 75 k.

(i) Jika voltan masukan vI = 0.20 V, kira vo1, vo, i1, i2, i3 dan i4.

[6 markah]

(ii) Tentukan arah sebenar bagi semua arus. [2 markah]

Rajah 4(a)


5

b) Lakarkan bentuk gelombang voltan keluaran dari litar kamiran jika

voltan masukannya adalah seperti dalam Rajah 4(b). Jelaskan jawapan

anda. [5 markah]

Rajah 4b

c) Dua voltan masukan, tv sin5.21 V dan v2 = +2V digunakan pada

sebuah amplifier penjumlah untuk menghasilkan voltan keluaran

tvo sin15 (V).

Nilai perintang paling besar yang boleh digunakan adalah 200 k.

Bina litar penjumlah tersebut dan tentukan nilai perintang yang

digunakan. [7 markah]

5. (a) Rajah 5 merupakan sebuah litar logik yang mempunyai tiga nilai masukan A,

B dan C dan satu nilai keluaran, Y.

(i) Tuliskan pernyataan Boole untuk litar logik pada Rajah 5. [4 markah]

(ii) Buatkan jadual kebenaran untuk pernyataan Boole yang diperolehi.

[4 markah]


6

Rajah 5

(b) Ringkaskan rumus berikut dengan menggunakan aljabar Boole atau Peta

Karnaugh

(i) CABAABCY [3 markah]

(ii) ACBCAY [3 markah]

(iii) )()(________

BABAY [3 markah]

(iv) ))(( DCBADACBY [3 markah]

6. (a) Nyatakan tiga jenis flip-flop bersama simbol dan jadual kebenarannya.

[9 markah]

(b) Lukis bentuk gelombang output, Q pada Rajah 6(a), 6(b) dan 6(c)

Q

[3 markah]

Rajah 6(a)


7

Q

[4 markah]

Rajah 6(b)

Q

[4 markah]

Rajah 6(c)


8

7. (a) Terangkan dengan ringkas yang berikut:

(i) Pembilang riak (ripple counter), [2 markah]

(ii) Pembilang selari (Parallel counter), [2 markah]

(iii) Pembilang dekad (decade counter), [2 markah]

(iv) Pembilang menurun (Count-down counter). [2 markah]

(b) Lukis litar logik pembilang mod 6 (membilang 0 hingga 5 kemudian 0, 1, 2,

3,….). [4 markah]

(c) Lukiskan bentuk output C, B dan A setiap denyut jam (Clock pulse) pada

Rajah 7(a). [4 markah]

Clock

C

B

A

Rajah 7(a)


9

(d) Lukis bentuk gelombang pada output, Q dari FF1, FF2, FF3 dan FF3 pada

Rajah7(b). [4 markah]

Rajah 7(b)


10

8. (a) Dengan ringkas terangkan yang berikut:-

(i) Penambah Separuh, (Half Adder), [2 markah]

(ii) Penambah Penuh (Full Adder), [2 markah]

(iii) Penolakan Separuh (Half Substractor), [2 markah]

(iv) Penolakan Penuh (Full Substractor) dan [2 markah]

(v) terangkan bagaimana litar Penolakan Penuh (Full Substractor) dapat

dibuat dari litar Penambah Penuh (Full Adder). [2 markah]

(b) Dengan menggunakan litar pendarab (multiplier) jenis tambah dan anjak (add

and shift), senaraikan turutan kandungan pendaftar ‘accumulator’ dan

pendaftar pendarab (multiplier register) bagi menyelesaikan masalah berikut:-

[6 markah]

11102 X 1012 = ?

Accumulator Register Multiplier Register

(c) Lukis bentuk gelombang bagi D dan Bout pada Rajah 8. [4 markah]

D

Bout

Rajah 8

SSCP 3323

2

1. (a) What is a cascade amplifier?

Show that the overall gain, A of an n-stages cascade amplifier is

nAAAAA ...321

[4 marks]

(b) Figure 1 shows a two-stage RC-coupled amplifier. The h-parameter of

both transistors are hie = 1 k and hfe = 100. The values of hre, hoe and the biasing network can be neglected. The amplifier is operating in its

midband region. Draw the ac equivalent circuit and the parameter-h

equivalent circuit of the amplifier.

+ V

CC

R11

R12

RC1

RE

CE

C1

C2100 k 1 k

10 k

RS

1 k

R21

R22

100 k

10 k

RC2

RE

1 k

vO

vs

CE

Figure 1

[8 marks]

(c) Calculate the voltage gain of the amplifier circuit shown in Figure 1.

[ 8 marks]

2. (a) Describe the following parameter in a differential amplifier:

(i) Common - mode gain

(ii) Differential- mode gain

(iii) Common mode rejection ratio (CMRR)

[6 marks]

(b) When the inputs to a certain differential amplifier are vi1 = 0.1sint and

vi2 = - 0.1sint, it is found that the outputs are vo1 = - 5sint and vo2 =

5sint. When both inputs are 2sint, the outputs are vo1 = - 0.05sint

and vo2 = 0.05sint. Find the CMRR in dB. [8 marks]

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(c) Figure 2 shows a differential amplifier. For transistors Q1 and Q2, hfe =

90 and hie = 1 k. Calculate the DC values of IE, IC and VC.

Q1

Q2

VEE

= - 20 V

vE

RC1 R

C2

IE

VCC

= 20 V

10 k10 k

RE 10 k

IC1

vC1

vi

Figure 2

[6 marks]

3. (a) State the condition for impedance matching between two circuits or

electronic equipments and suggest a method to solve the problem of

impedance mismatch using an operational amplifier.

[6 marks]

(b) What is meant by slew rate and explain how it affects the performance

of an operational amplifier.

[6 marks]

(c) Figure 3 shows an operational amplifier circuit. Calculate the voltages

V1 and Vo.

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+

-

+

-

V i = 3 V

V 1

V o

10 k

10 k

10 k

10 k

10 k

Figure 3

[8 marks]

4. (a) Convert the following numbers

i) binary number 1100 0011 1101 to hexadecimal number.

(ii) Hexadecimal 9A2 to binary.

(iii) Hexadecimal number ABC to decimal number.

[6 marks]

(b) Write the Boolean expression for the logic circuit shown in Figure 4 and

draw a three variable truth table of the logic function.

Figure 4

[4 marks]

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(c) A 4-input circuit has an output, Y, given by the truth table below

A B C D Y A B C D Y

0 0 0 0 0 1 0 0 0 1

0 0 0 1 1 1 0 0 1 1

0 0 1 0 1 1 0 1 0 0

0 0 1 1 1 1 0 1 1 0

0 1 0 0 1 1 1 0 0 0

0 1 0 1 1 1 1 0 1 0

0 1 1 0 1 1 1 1 0 0

0 1 1 1 1 1 1 1 1 0

i) Write down the unsimplified Boolean expression

ii) Use a Karnaugh Map to minimise the Boolean expression

iii) Draw a circuit to solve this problem using a 1-of -8 data selector.

[10 marks]

5. (a) Draw a logic symbol and truth table for the following flip-flops:

(i) J-K

(ii) Clocked R-S

[4 marks]

(b) i) List and explain two types of edge triggered flip-flops.

[2 marks]

ii) Referring to the block diagram of the JK flip-flop in Figure 5(a),

sketch the output Q corresponds to the inputs CLK, PS, J, K and

CLR given in Figure 5(b).

[4 marks]

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(c) (i) What type of flip-flops are useful in wiring counters.

(ii) Draw a logic symbol diagram for a 2-bit ripple down counter.

(iii) Redesign the 2-bit counter in (ii) to count from 11 to 00 and then

stop.

[5 marks]

Figure 5(a)

J

CLK

PS

Q

CLR

K Q

Figure 5(b)

CLK

PS

CLR

K

Q

time

Q

J

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(iv) List the binary counting sequence for counter circuit in Figure 5(c).

[5 marks]

Figure 5(c)


342

10.0 Shift Registers

A shift register is a group of memory cells grouped together and considered a single unit

The register can be used to simply store information for later use or the register can be designed to act on the data as is the case of a shift register

A shift register usually modifies the contents by shifting data right or left

The term latch may be used to describe the register used to store data

A buffer register is a specific use of a storage device that holds data that is waiting to be transferred

Shift registers are constructed by wiring flip-flops together

Shift registers in IC form are also available

Registers often are used to store data momentarily

Figure 32.1 shows a typical example of where registers might be used in a digital system, e.g. a calculator

Figure 32.1 – A digital system using registers


343

Notice the use of registers to hold information from the encoder for the processing unit

A register is also being employed for temporary storage between the processing unit and the decoder

Registers are also used at other locations within a digital system

One method of describing shift register characteristics is by how data is loaded into and read from the storage units

Four categories of shift registers (in this example, each storage device is an 8-bit register) are illustrated in Figure 32.2

The registers are classified as: 1. Serial in – serial out 2. Serial in – parallel out 3. Parallel in – serial out 4. Parallel in – parallel out


344

Figure 32.2 – Shift register characteristics (a) serial in – serial out (b) Serial in – parallel out (c) Parallel in – serial out (d) Parallel in – parallel out


345

10.1 Serial Load Shift Registers

Figure 32.3 shows a basic shift register, constructed from four D FF

Figure 32.3 – A 4-bit serial load shift register using D FFs

This register is called a 4-bit shift register because it has four places to store data: A, B, C, and D

The table in Figure 32.4 shows the operation the shift register


346

Figure 32.4

First, clear (CLR input to 0) all the outputs (A, B, C, and D) to 0000 (see line 1 in the table)

The output remain 0000 while they await a clock pulse

Pulse the CLK input once; the output now shows 1000 because the 1 from the D input of FF A has been transferred to the Q output on the clock pulse

Now, enter 1s on the data input (clock pulse 2 and 3); these 1s shift across the display to the right

Next, enter 0s on the data input (clock pulses 4 to 8); it can be seen that the 0s being shifted across the display

On clock pulse 9 enter a 1 at the data input

On pulse 10 the data input is returned to 0


347

Pulses 9 to 13 show the single 1 on display being shifted to the right

Line 15 shows the 1 being shifted out the right end of the shift register and being lost

The circuit is referred to as a serial load shift register because only one bit of data at a time can be entered in the register

For example, to enter 0001 in this serial load shift register needs four steps (line 11 to line 14 in Figure 32.4)

10.2 Parallel Load Shift Register

Figure 32.5 illustrates a system that permits parallel loading of four bits at once

Figure 32.5 – Block diagram of a 4-bit parallel load recirculating shift register

These inputs are the data inputs A, B, C, and D

This system couldalso incorporate a recirculating feature that would put the output data back into the input so that it is not lost


348

A wiring diagram of the 4-bit parallel load recirculating shift register is shown in Figure 32.6

Figure 32.6 – Wiring diagram of a 4-bit parallel load recirculating shift register

In this register, the recirculating lines leading from the Q

and Q outputs of FF D back to the J and K inputs of FF

A

These feedback lines cause the data that would normally be lost out of FF D to recirculate through the shift register

The CLR input clears the outputs to 0000 when enabled by a logical 0

The parallel load data inputs A, B, C, and D are connected to the preset (PS) inputs of the flip-flops to set 1s at any output position (A, B, C, D)

If the switches attached to the parallel load data inputs are even temporarily switched to a 0, that output will be preset to a logical 1


349

The clock pulsing the CLK inputs of the JK FF will cause data to be shifted to the right

The data from FF D will be recirculated back to FF A

Figure 32.7 illustrates the operation of the parallel load shift register

Figure 32.7 – Operation of a 4-bit parallel load recirculating shift register

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1) i) By using suitable flip-flops, draw a logic diagram of a mod-6 ripple up

counter.

ii) Figure 1 shows a logic diagram of a counter. Complete its counting sequence

in Table 1.

Figure 1

Number of

Pulse sequence

Binary counting

sequence

Decimal Count

C B A

0 1 1 1 7

1

2

3

4

5

6

7

9

9

10

11

Table 1.

2). (a) (i) Draw a logic symbol diagram of a 3 bits serial load shift-right register.

(ii) Explain how you would clear to 000 the 3-bit you drew in (i).

(iii) After clearing the 3-bit register, explain how you would enter 010 into

the register you drew in (i).

(iv) With the original output = 010 (A=0, B=1 and C=0), list the contents of

the register you drew in (i) after one, two and three clock pulses.

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3). a) Describe the classification of shift registers.

b) Figure 2 shows the wiring diagram of a 4-bit parallel load recirculating

shift register. Table 2 shows the operation of this shift register. Complete

the values of output A, B, C and D.

Figure 2

Table 2

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Input Output

A B S C

0 0

0 1

1 0

1 1

A + B Co

Input Output

A B D B

0 0

0 1

1 0

1 1

A - B Di Bo

4) a) Complete the truth tables for HA (Half Adder) and HS (Half Subtractor) in

Table 3 and draw their logic circuit using gates .

Table 3

b) Draw the full adder and Co outputs for each set of input pulses

shown in Figure 3.

Figure 3

Co

5. a) Figure 4 shows full-adders and XOR gates are connected to form an

arithmetic circuit. Study the circuit carefully and answer the following;

i) Complete the truth table of the full-adder below.

Input Output

Cin A B Co

0 0 0

0 0 1

0 1 0

0 1 1

1 0 0

1 0 1

1 1 0

1 1 1

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ii) Complete the table below.

M B N

0 0

0 1

1 0

1 1

iii) For each input A, B and M given in the table below find the output W,

X, Y and Z.

Input Output

A3 A2 A1 A0 B3 B2 B1 B0 M W X Y Z

0 1 0 0 0 0 1 1 0

0 1 0 1 1 1 0 0 0

0 1 1 1 0 0 1 1 1

1 0 0 0 1 1 0 1 1

1 1 0 0 0 0 1 0 1

iv) Explain briefly the arithmetic function of the circuit in Figure 4.

b) Convert the following signed decimal numbers to their 4-bit 2s compliment

form:

+1, -1, +7, -7.

c) Do the binary addition, subtraction and multiplication problems below.

(i) 1010 + 0110 =

(ii) 0100 + 1010 =

(iii) 0110 – 0101 =

(iv) 1100 – 1011 =

(v) 110 × 111 =

(vi) 1100 × 1010 =

d) (i) Show a table of 2’s complement for all 4-bit positive and negative

numbers.

(ii) Calculate the sum of the 2’s complement number 0110 and 1001. Give

the answer in 2’s complement and decimal.

SSP3323

6

FA

Cin

A

B

Co

FA

Cin

A

B

Co

FA

Cin

A

B

Co

FA

Cin

A

B

Co

A0

A1

A2

A3

B0

B1

B2

B3

M

N0

N1

N2

N3

Z Y X W

Figure 4


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11. Binary arithmetic and arithmetic circuits

11.1 Binary Addition

0 + 0 = 0 0 + 1 = 1 1 + 0 = 1 1 + 1 = 0 Carry 1 or 102 The symbol + indicates added (Add) not OR.

Example 1 1 1

1 0 02 = 410 1 0 1 510 +0 1 02 = 210 0 1 1 310

1 1 0 610 1 0 0 0 810 Truth table for add operations is as follows: - The output consists of two parts: Sum (total) and Carry out.

Input Output

A B S C

0 0 0 0

0 1 1 0

1 0 1 0

1 1 0 1

A + B Co

Block form Logic Form

Half Adder

From the truth table, we get = A B and Co = A B


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Input Output

A B Cin Co

0 0 0 0 0

0 0 1 1 0

0 1 0 1 0

0 1 1 0 1

1 0 0 1 0

1 0 1 0 1

1 1 0 0 1

1 1 1 1 1

A + B + Cin Sum Cout

Input Output

A B D B

0 0 0 0

0 1 1 1

1 0 1 0

1 1 0 0

A - B Di Bo

• This operation is called Half Adder, which can only be used for adding the first column only. • But for the next column, the Full Adder (FA) circuit’s is used to take into account the result of the addition of Carry on previous column.

Truth Table for Full Adder is as follows: -

= A + B + Cin

We use two HA, for making two additions. First operation is between A and B, both for (A + B) and Cin. Cout from FA values taken from one of the HA Cout (by OR operation).

Binary Subtraction. 0 - 0 = 0 0 - 1 = 1 Borrow 1 1 - 0 = 1 1 - 1 = 0 Example 0 10

1 0 210 - 0 1 110

0 1 110


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Input Output

A B Bin Di Bo

0 0 0 0 0

0 0 1 1 1

0 1 0 1 1

0 1 1 0 1

1 0 0 1 0

1 0 1 0 0

1 1 0 0 0

1 1 1 1 1

A - B - Bin Di Bout

From the truth table, the difference, Di = A B and borrow is

Bo = BA Logic circuit for HS binary subtraction is as follows: -

Similar to the case of additions, HS can only be used in the first column, while the next column we use Full Substractor (FS) Truth table for FS is


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Here is a logic circuit for FS built from two HSs and an OR gate.

Examples of the use of HS and FS 64s 32s 16s 8s 4s 2s 1s A 1 1 1 0 1 0 1 11710

-B 0 0 1 1 1 0 0 2810 Di 1 0 1 1 0 0 1 8910 (0) (1) (1) (0) (0) (0) Bin Bin Bin Bin Bin Bin Bin

Bo Bo Bo Bo Bo Bo Bo Bo (0) (0) (1) (1) (0) (0) (0)

• In the first column, the operation of HS was used. Di = A - B = 1 - 0 = 1 and Bo = 0

• In the second column Di = A - B - Bin = 0 - 0 - 0 = 0 and Bo = 0. Bin is Bo from HS operations in the first column

• All operations of FS were used for the next column.

Parallel Addition and Subtraction

Usually the addition and subtraction operations carried out in parallel in which only the first column only using HA or HS, whereas all subsequent column using FA or FS.

To perform the FA / FS in the second column and the next, the

value of Cin / Bin taken from Cout / Bout from operations in the

previous column.


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Use FA only for the addition • If we want to use FA only for the addition, we can do by making Cin

= 0 for the first column. Results are the same with HA operation Circuit is as follows: -

Use the adder for subtraction. • With a little modification to the circuit in parallel addition it can be used for parallel subtraction. Di = A - B = A + (-B).

• (-B) here is the negative value of a signed number or 2's

complement number of (-B) = B + 1. So Di = A – B = A +

(-B) = A + B + 1


351

• We use all FA circuit where Cin = 1 for FA in the first row • Circuit for 4 bits subtraction is as follows: -

Comparing subtraction circuit using the FA with addition circuits using FA alone, there is little difference. In the first column, the addition with Cin = 0 whereas in subtraction, Cin = 1

Then in the subtraction the values of Bin is inverted but not in the addition

So the addition and subtraction circuits can share the FA but we need some extra circuitry, so that the use of FA during addition and subtraction can be done correctly


352

Circuit is as follows: -

• for add operation, the Control Mode is set to 0. Then Cin = 0 in FA1

and B = Bi • But when the subtraction to be done, the Control Mode is set to 1,

so Cin = 1 in FA1 and B = iB

• Actually, this circuit can be used for addition and subtraction of unsigned binary (2's complement).


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13 1101

x10 x1010

00 0000

13 1101

130 0000

1101

10000010

Binary Multiplication 0 0 1 1 x0 x1 x0 x1 0 0 0 1 The traditional method is used for the multiplication of binary numbers and decimal • This method is used in the logic circuit for multiplying two binary numbers • The method used is as follows: -.

• Rows 1 and 2 are two values to be multiplied. Line 3 is the first partial product (0 x 1101 = 0000) • Line 4 shows a shift to the left of the second digit of the product of (1 x 1101 = 1101) in the fifth row • Line 6 is the sum of the product of the first and the second (which is shifted to the left), is 11010 • Shift left in 7th line, for the third partial product (0 x 1101 = 0000) in 8th row • 9th Line for the total product of 011010. Displacements in the 10th row for the fourth partial product (1 x 1101 = 1101) in the 11th row • 12th Line is the final product of the total, which is also the actual product. • Multiplication method is called add and shift method.


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There are three important facts in add and shift method.

1) the product = 0000 if bits in the multiplier is 0, and the product

= multiplicand, if bits in the multiplier is 1

2) The end product can be twice the size of the multiplicand.

3) The product of the first digit is shifted rightward compared by

the product of a second.

From the above observations, the binary multiplier circuit can be

constructed.

• binary multiplier is 5-bit parallel adder with control circuit either

added / not add

• The multiplicand and multipler filled with values from the previous

example


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Operation of a binary multiplier circuit is as follows: - • Step 1 for clear (accumulator = 0) and load the data on the Multiplier • 2nd step for add operations if control = 1. • the product of Step 3 to shift rightward. • 4th step for add operations if control = 1. Add to accumulator by the multiplicand. • Step 5 for the product of rightward shift • Step 6 for add operations if control = 1. • Step 7 for the product of rightward shift • Step 8 for add operations if control = 1. • Step 9 for shift rightward product of the actual result. There are 4 times ADD operation and 4 times SHIFT operation • This method is used in the computer when doing multiplication.

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