electronics fundamentals - staff.uob.edu.bhstaff.uob.edu.bh/files/600435156_files/lesson_2_transistors.pdf · Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education,

Electronics Fundamentals 8th edition

Floyd/Buchla© 2010 Pearson Education, Upper Saddle

River, NJ 07458. All Rights Reserved.

Lesson 2: Transistors and Applications

electronics fundamentalscircuits, devices, and applications

THOMAS L. FLOYD

DAVID M. BUCHLA


Floyd/Buchla

Lesson 2

© 2010 Pearson Education, Upper Saddle


Introduction

A transistor is a semiconductor device that controls current between

two terminals based on the current or voltage at a third terminal.

It is used for amplification or switching of electrical signals.

The basic structure of the bipolar junction transistor, BJT, determines

its operating characteristics.

DC bias is important to the operation of transistors in terms of setting

up proper currents and voltages in a transistor circuit.

Two important parameters are αDC and βDC


Floyd/Buchla

Lesson 2



Bipolar junction transistors (BJTs)

The BJT is a transistor with three regions and two pn

junctions. The regions are named the emitter, the base, and

the collector and each is connected to a lead.

There are two types of

BJTs – npn and pnp.

n

n

pp

p

n

E (Emitter)

B (Base)

C (Collector)

E

B

C

Separating the regions

are two junctions.

Base-Collector

junction

Base-Emitter

junction


Floyd/Buchla

Lesson 2



Bipolar junction transistors (BJTs)

FIGURE 17–2 Transistor symbols.


Floyd/Buchla

Lesson 2



BJT biasing

For normal operation, the base-emitter junction is forward-

biased and the base collector junction is reverse-biased.

npn

BE forward-

biased

BC reverse-

biasedFor the npn transistor, this

condition requires that the base

is more positive than the emitter

and the collector is more

positive than the base.

+

+

For the pnp transistor, this

condition requires that the base

is more negative than the

emitter and the collector is more

negative than the base.

pnp

+

+


Floyd/Buchla

Lesson 2



BJT currents

FIGURE 17–4 Transistor currents.

E C BI I I


Floyd/Buchla

Lesson 2



BJT currents

A small base current (IB) is able to control a larger collector

current (IC). Some important current relationships for a BJT

are:

I

I

I

IB

IE

ICE C BI I I

C DC EαI I

C DC BβI I

Where αDC (dc alpha) = IC/ IE

Where βDC (dc beta) = IC/ IB


Floyd/Buchla

Lesson 2



Voltage-divider bias

R1

R2

RC

RE

VB

VE

VC

Because the base current is small, the approximation

2B CC

1 2

RV V

R R

is useful for calculating the base voltage.

After calculating VB, you can find

VE by subtracting 0.7 V for VBE.

VE = VB - VBE

Next, calculate IE by applying Ohm’s

law to RE:

C EI IThen apply the approximation

Finally, you can find the collector voltage

fromC CC C CV V I R

EE

E

VI

R

ECCE VVV


Floyd/Buchla

Lesson 2




Calculate VB, VE, and VC for the circuit.

2B CC

1 2

6.8 k15 V =

27 k + 6.8 k

RV V

R R

R1

R2

RC

RE

VE = VB 0.7 V =

C E 2.32 mAI I

C CC C C 15 V 2.32 mA 2.2 kV V I R

EE

E

2.32 V 2.32 mA

1.0 k

VI

R

27 k

6.8 k 1.0 k

2.2 k

+15 V

2N3904

3.02 V

2.32 V

9.90 V


Floyd/Buchla

Lesson 2




Determine VB, VE, VC, VCE, IB, IE, and IC in the given

Figure, The 2N3904 is a general-purpose transistor with a

typical βDC = 200.


Floyd/Buchla

Lesson 2



Collector characteristic curves

The collector characteristic curves are a family of

curves that show how collector current varies with

the collector-emittervoltage for a given IB.

IC

VCE0

IB6

IB5

IB4

IB3

IB2

IB1

IB = 0

The saturation region

occurs when the base-

emitter and the base-

collector junctions are

both forward biased.

The curves are divided into

three regions:

The active region is after

the saturation region.

This is the region for

operation of class-A

operation.

The breakdown region

is after the active region

and is is characterized by

rapid increase in collector

current. Operation in this

region may destroy the

transistor.


Floyd/Buchla

Lesson 2





Floyd/Buchla

Lesson 2



Draw the family of collector characteristics curves for the

circuit in the given figure for IB = 5 µA to 25 µA in 5 µA

increments. Assume βDC = 100



Floyd/Buchla

Lesson 2



Draw the family of collector characteristics curves for the

circuit in the given figure for IB = 5 µA to 25 µA in 5 µA

increments. Assume βDC = 100


IB IC

5 µA 0.5 mA

10 µA 1.0 mA

15 µA 1.5 mA

20 µA 2.0 mA

25 µA 2.5 mA

mAAI

II

C

BDCC

5.05100


Floyd/Buchla

Lesson 2



Load lines

A load line is an IV curve that represents the response of

a circuit that is external to a specified load. For example, the load line for the

Thevenin circuit can be found by

calculating the two end points: the

current with a shorted load, and the

output voltage with no load.

+12 V

2.0 k

00

4 8 12

2

4

6

I (mA)

V (V)

ISL = 6.0 mA

VSL = 0 V

INL = 0 mA

VNL = 12 V

Load line


Floyd/Buchla

Lesson 2



Load lines

The IV response for any load will intersect the load line

and enables you to read the load current and load

voltage directly from the graph.

+12 V

2.0 k

00

4 8 12

2

4

6

I (mA)

V (V)

Read the load current

and load voltage from the graph if

a 3.0 k resistor is the load.

3.0 k

RL =

IV curve for

3.0 k resistor

VL = 7.2 V IL = 2.4 mA

Q-point


Floyd/Buchla

Lesson 2



Load lines

The load line concept can be extended to a transistor

circuit. For example, if the transistor is connected as a

load, the transistor characteristic

+12 V

2.0 k

00

4 8 12

2

4

6

I (mA)

V (V)

curve and the base current

establish the Q-point.


Floyd/Buchla

Lesson 2



Load lines

Load lines can illustrate the operating conditions for a

transistor circuit.

00

4 8 12

2

4

6

I (mA)

V (V)

If you add a transistor load to the last

circuit, the base current will establish

the Q-point. Assume the base current

is represented by the blue line.

+12 V

2.0 k

For this base current,

the Q-point is:

Assume the IV curves are as shown:

The load voltage (VCE) and current

(IC) can be read from the graph.


Floyd/Buchla

Lesson 2



Load lines

For the transistor, assume the base current is

established at 10 A by the bias circuit. Show the Q-

point and read the value of VCE and IC.

00

4 8 12

2

4

6

IC (mA)

VCE (V)

+12 V

2.0 k

Bias circuit

IB = 5.0 A

IB = 10 A

IB = 15 A

IB = 20 A

IB = 25 A

The Q-point is the

intersection of the

load line with the

10 A base current.

VCE = 7.0 V; IC = 2.4 mA


Floyd/Buchla

Lesson 2



Signal (ac) operation

When a signal is applied to a transistor circuit, the output can have a

larger amplitude because the small base current controls a larger collector

current.

This increase is called amplification

The ratio of the ac collector current (Ic) to the ac base current (Ib) is

designated by βac (the ac beta) of hfe

FIGURE 17–13 An amplifier with voltage-divider bias with capacitively

coupled input signal. Vin and Vout are with respect to ground.

b

cac

I

I


Floyd/Buchla

Lesson 2



Signal (ac) operation

When a signal is applied to a transistor circuit, the

output can have a larger amplitude because the small

base current controls a larger collector current.

00

4 8 12

2

4

6

IC (mA)

VCE (V)

IB = 5.0 A

IB = 10 A

IB = 15 A

IB = 20 A

IB = 25 A

For the load line and

characteristic curves from the last

example (Q-point shown) assume IB

varies between 5.0 A and 15 A

due to the input signal. What is the

change in the collector current?

Reading the collector current, IC varies from 1.2 mA to 3.8 mA.

The operation along

the load line is shown in red.


Floyd/Buchla

Lesson 2



CE amplifier

In a common-emitter amplifier, the input signal is applied

to the base and the output is taken from the collector. The

signal is larger but inverted at the output.

R1

R2

RC

REInput coupling

capacitor

Output coupling

capacitor

Bypass

capacitor

C3

C2

C1

VC

C


Floyd/Buchla

Lesson 2



Summary

CE amplifier

The bypass capacitor increases voltage gain. It shorts the signal around

the emitter resistor, RE, in order to increase the voltage gain. To

understand why let us consider the amplifier without the bypass

capacitor as explained the preceding equations.

R1

R2

RC

REInput coupling

capacitor

Output coupling

capacitor

Bypass

capacitor

C3

C2

C1

VC

C


Floyd/Buchla

Lesson 2



Formulas

Av (Voltage gain) = Vout / Vin

The signal voltage at the base is approximately equal to

Vb ≡ Vin ≡ Ie (re + RE)where re is the internal emitter resistance of the transistor.

Lowercase italic subscript indicate signal (ac) voltages and signal

(alternating currents)

Vout = IcRC


Floyd/Buchla

Lesson 2



Formulas without the bypass capacitor

Av now can be expressed as

Av = Vout / Vin = IcRC / Ie (re + RE)

Since Ic ≡ Ie, the currents cancel and the gain is the ratio of the resistance.

Av = RC / (re + RE)

If RE is much greater than re, then

Av ≡ RC / RE


Floyd/Buchla

Lesson 2



Formulas with the bypass capacitor

Av = RC / re

If the bypass capacitor is connected across RE, it effectively shorts the

signal to ground leaving only re in the emitter. Thus the voltage gain of

the CE amplifier with the bypass capacitor shorting RE is:

The transistor parameter re is important because it determines the

voltage gain of a CE amplifier in conjunction with RC. A formula for

estimating re is given without derivation in the following equation:

re≡ 25mV / IE


Floyd/Buchla

Lesson 2



Summary

Voltage gain of a CE amplifier

Calculate the voltage gain of the CE amplifier. The dc

conditions were calculated earlier; IE was found to be 2.32 mA.

Coutv

in e

V RA

V r

R1

R2

RC

RE C3

C2

C1

VCC = +15 V

27 k

6.8 k 1.0 k

2.2 kE

25 mV 25 mV10.8

2.32 mAer

I

204

2.2 F

1.0 F

100 F

2.2 k

10.8

Sometimes the gain will be shown with a

negative sign to indicate phase inversion.

2N3904


Floyd/Buchla

Lesson 2



Phase Inversion

Rin = Vb / Ib

Vb = Iere

Ie ≡ βac Ib

AC Input Resistance

The output voltage at the collector is 180 degrees out of phase with

the input voltage at the base. Therefore, the CE amplifier is

characterized by a phase inversion between the input and output. The

inversion is sometimes indicated by a negative voltage gain.

Rin ≡ βac Ib re / Ib

The Ib terms cancel, leaving

Rin ≡ βac re

Total Input Resistance of a CE amplifier:

Rin(tot) = R1║R2║Rin

Rin(tot) = R1║R2║βac reRC has no effect because of the reverse-biased,

base-collector junction.


Floyd/Buchla

Lesson 2



Current Gain:

Ai= Ic / Is

Where Is is the source current and is calculated by Vin /

Rin(tot)

The signal current gain of a CE amplifier is

The power gain of a CE amplifier is the product of the voltage gain

and the current gain.

Ap≡ Av Ai

Power Gain:


Floyd/Buchla

Lesson 2



Decibel (dB) Measurement

VOLTAGE RATIO: dB = 20 log (Vout / Vin)

POWER RATIO: dB = 10 log (Pout / Pin)

The decibel (dB) is a logarithmic measurement of the ratio of one

voltage to another or one power to another, which can be used to

express the input-to-output relationship .


Floyd/Buchla

Lesson 2



R1

R2

RC

RE C3

C2

C1

VCC = +15 V

27 k

6.8 k 1.0 k

2.2 k

2.2 F

1.0 F

100 F

Input resistance of a CE amplifier

The input resistance of a CE amplifier is an ac resistance

that includes the bias resistors and the resistance of the

emitter circuit as seen by the base.

Because IB << IE, the emitter

resistance appears to be

much larger when viewed

from the base circuit. The

factor is (ac+1), which is

approximately equal to ac.

Using this approximation,

Rin(tot) = R1||R2||acre.

2N3904


Floyd/Buchla

Lesson 2




FIGURE 17–20 Total input resistance.


Floyd/Buchla

Lesson 2



R1

R2

RC

RE C3

C2

C1

VCC = +15 V

27 k

6.8 k 1.0 k

2.2 k

2.2 F

1.0 F

100 F


Rin(tot) = R1||R2||acre

Calculate the input resistance of the CE amplifier. The

transistor is a 2N3904 with an average ac of 200. The value of re

was found previously to be 10.8 . Thus, acre = 2.16 k.

2N3904= 27 k||6.8 k||2.16 k

= 1.55 k

Notice that the input resistance of

this configuration is dependent on

the value of ac, which can vary.


Floyd/Buchla

Lesson 2



CE amplifier

Determine the voltage gain, current gain, and power

gain for the CE amplifier given. DC ac = 100. Also, express the

voltage and power gains in decibels.


Floyd/Buchla

Lesson 2



CC amplifier (emitter-follower)

In a common-collector amplifier, the input signal is

applied to the base and the output is taken from the

emitter. There is no voltage gain, but there is power gain.

R1

R2RE

C1

VC

C

The output voltage is nearly

the same as the input; there is

no phase reversal as in the

CE amplifier.

The input resistance is larger

than in the equivalent CE

amplifier because the emitter

resistor is not bypassed.


Floyd/Buchla

Lesson 2



The voltage gain of a CC amplifier is approximately 1, but

the current gain is always greater than 1.

Voltage Gain

Av (Voltage Gain)

Av = Vout / Vin

Vout = IeRE

Vin= Ie (re + RE)

Av = IeRE / Ie (re + RE)

The gain expression simplifies to:

Av = RE / (re + RE)

It is important to notice here that the

gain is always less than 1. Because re is

normally much less than the RE, then a

good approximation is Av = 1


Floyd/Buchla

Lesson 2



Rin = Vb / Ib

Vb = Ie (re + RE)

Ie ≡ βac Ib

Rin ≡ βac Ib (re + RE) / Ib

AC Input Resistance

The emitter-follower is characterized by a high input resistance, which

makes it a very useful circuit. Because of the very high input

resistance , the emitter follower can be used as a buffer to minimize

loading effects when one circuit is driving another.

The Ib terms cancel, leaving

Rin ≡ βac (re + RE)

If RE is at least ten times larger than re , then the

input resistance at the base is:

Rin ≡ βac RE

Total Input Resistance of a CC amplifier:

Rin(tot) = R1║R2║Rin


Floyd/Buchla

Lesson 2



Current Gain:

Ai= Ie / Is

Where Is is the signal current and is calculated by Vin / Rin(tot)

Since Ie = Vout / RE and Is = Vin / Rin(tot) then Ai can also be expressed as

assuming Vout / Vin = 1:

Ai= (Vout / RE )/ (Vin / Rin(tot) ) = Rin(tot) / RE

The signal current gain for the emitter-follower is

The power gain of is the product of the voltage gain and the current gain. For

the emitter-follower, the power gain is approximately equal to the current gain

because the voltage gain is approximately equal to 1.

Ap≡ Ai

Power Gain:


Floyd/Buchla

Lesson 2



CC amplifier

R1

R2RE

C1

VCC = +15 V

22 k

27 k1.0 k

10 F

2N3904

Calculate re and Rin(tot) for the CC amplifier. Use = 200.

2B CC

1 2

27 k15 V = 8.26 V

22 k + 27 k

RV V

R R

E B 0.7 V = 7.76 V V V

E

25 mV 25 mV

7.76 mAer

I

EE

E

7.76 V7.76 mA

1.0 k

VI

R

Rin(tot) = R1||R2||ac(re+ RE)

= 22 k||27 k|| 200 (1.0 k) = 9.15 k

Because re is small compared to RE, it

has almost no affect on Rin(tot).

3.2


Floyd/Buchla

Lesson 2



CC amplifier

Determine the input resistance of the emitter-follower in the given

figure. Also find the voltage gain, current gain and power gain.


Floyd/Buchla

Lesson 2



The BJT as a switch

BJTs are used in switching applications when it is

necessary to provide current drive to a load.

In cutoff, the input voltage is too

small to forward-bias the

transistor. The output (collector)

voltage will be equal to VCC.

In switching applications, the transistor

is either in cutoff or in saturation.RC

VCC VCC

RC

VOUT

IIN = 0 = VCC

IIN > IC(sat)/DC

= 0 V

When IIN is sufficient to saturate

the transistor, the transistor acts

like a closed switch. The output is

near 0 V.


Floyd/Buchla

Lesson 2



The BJT as a switch

FIGURE 17–30 Ideal switching action of a transistor.


Floyd/Buchla

Lesson 2



The BJT as a switch

Conditions in Cutoff A transistor is in cutoff when the base-emitter junction is not forward biased.

Conditions in Saturation When the base-emitter junction is forward-biased and there is enough base

current to produce a maximum collector current, the transistor is saturated.

The minimum value of base current to produce saturation is:

FIGURE 17–30 Ideal switching action of a

transistor.

CCcutoffCE VV )(

C

CCsatC

R

VI )(

DC

satC

B

II

)(

(min) VVBE 7.0) VVV INRB7.0

(min)

(max)

B

R

BI

VR B


Floyd/Buchla

Lesson 2



(a) For the transistor switching circuit in the given figure, what is

VCE when VIN = 0 V?

(b) What minimum value of IB will saturate this transistor if the

βDC is 200?

(c) Calculate the maximum value of RB when VIN = 5 V.

The BJT as a switch


Floyd/Buchla

Lesson 2



The FET

The field-effect transistor (FET) is a voltage controlled

device where gate voltage controls drain current. There

are two types of FETs – the JFET and the MOSFET.

JFETs have a conductive channel

with a source and drain connection

on the ends. Channel current is

controlled by the gate voltage. p

n

S (Source)

G (Gate)

D (Drain)

S

G

D

n

n

p p

The gate is always operated with

reverse bias on the pn junction formed

between the gate and the channel. As

the reverse bias is increased, the

channel current decreases.n-channel JFET p-channel JFET


Floyd/Buchla

Lesson 2



The FETFIGURE 17–33 Effects of VGG on channel width and drain current

(VGG = VGS).


Floyd/Buchla

Lesson 2



The FET FIGURE 17–34 JFET schematic symbols.


Floyd/Buchla

Lesson 2



The FET: MOSFET

The MOSFET (Metal Oxide Semiconductor FET)

differs from the JFET in that it has an insulated gate

instead of a pn junction between the gate and channel.

Like JFETs, MOSFETs have a conductive channel with the

source and drain connections on it.

p

S (Source)

G (Gate)

D (Drain)

S

G

D

n

n

p

p-channel

MOSFET

n-channel

MOSFET

ChannelSubstrate

Channel current is

controlled by the gate

voltage. The required gate

voltage depends on the type

of MOSFET.


Floyd/Buchla

Lesson 2



The FET: MOSFET

FIGURE 17–35 Representation of the basic structure of D-MOSFETs.


Floyd/Buchla

Lesson 2



In addition to the channel designation, MOSFETs are subdivided

into two types – depletion mode (D-mode) or enhancement mode

(E-mode).

The D-MOSFET has a

physical channel which

can be enhanced or

depleted with bias. For this

reason, the D-MOSFET

can be operated with either

negative bias (D-mode) or

positive bias (E-mode).

The FET: MOSFET

FIGURE 17–36 Operation of n-channel D-MOSFET.


Floyd/Buchla

Lesson 2



The FET: MOSFET

FIGURE 17–37 D-MOSFET schematic symbols.


Floyd/Buchla

Lesson 2



The E-mode MOSFET has no physical channel. It can only be operated

with positive bias (E-mode). Positive bias induces a channel and enables

conduction as shown here with a p-channel device.

p

S

G

D

nn

p

E-MOSFET with biasE-MOSFET

n

D

G

S

induced channel

n

The FET: MOSFET

FIGURE 17–38 E-MOSFET construction and

operation.


Floyd/Buchla

Lesson 2



The FET: MOSFET

FIGURE 17–39 E-MOSFET schematic symbols.


Floyd/Buchla

Lesson 2



JFET biasing

JFETs are depletion mode devices – they must be

operated such that the gate-source junction is reverse

biased.

The simplest way to bias a JFET is

to use a small resistor is in series

with the source and a high value

resistor from the gate to ground.

The voltage drop across the source

resistor essentially reverse biases

the gate-source junction.

n-channel p-channel

RG

+VDD VDD

RG RSRS

RD RD

VG = 0 V VG = 0 V

+VS VS

Because of the reverse-biased

junction, there is almost no

current in RG. Thus, VG = 0 V.


Floyd/Buchla

Lesson 2



JFET biasing

FIGURE 17–40 Self-biased JFETs (IS = ID in

all FETs).

For the n-channel JFET, the gate-to-

source voltage is

For the p-channel JFET, the gate-to-

source voltage is

The drain voltage with respect to ground

is

Since VS=IDRS, the drain-to-source

voltage is

SDSGGS RIVVVV 0

SDGS RIV

SDGS RIV

DDDDD RIVV

SDDDDDS RRIVV


Floyd/Buchla

Lesson 2



Find VDS and VGS in the JFET circuit below, given that ID =

5mA

JFET biasing


Floyd/Buchla

Lesson 2



D-MOSFET biasing

D-MOSFETs can be operated in either depletion mode

or in enhancement-mode. For this reason, they can be

biased with various bias circuits.

The simplest bias method for a D-

MOSFET is called zero bias. In this

method, the source is connected directly

to ground and the gate is connected to

ground through a high value resistor.

n-channel D-MOSFET

with zero bias

RG

+VDD

RD

VG = 0 V

Only n-channel D-MOSFETs

are available, so this is the

only type shown.


Floyd/Buchla

Lesson 2



D-MOSFET biasing FIGURE 17–42 A zero-biased D-MOSFET.

Recall that depletion/enhancement

MOSFETs can be operated with either

positive or negative values of VGS.

A simple bias method, called zero bias, is

to set VGS = 0 V so that an ac signal at the

gate varies the gate-to-source voltage

above and below this bias point.

Since VGS = 0 V, ID = IDSS as indicated.

IDSS is defined as the drain current when

VGS = 0 V.

The drain-to-source voltage is expressed

asDDSSDDDS RIVV


Floyd/Buchla

Lesson 2



Determine the drain-to-source voltage in the circuit of the

given figure. The MOSFET data sheet give VGS(off) = -8 V and

IDSS = 12 mA

D-MOSFET biasing


Floyd/Buchla

Lesson 2



E-MOSFET biasing

E-MOSFETs can use bias circuits similar to BJTs but

larger value resistors are normally selected because of

the very high input resistance.

The bias voltage is normally

set to make the gate more

positive than the source by an

amount exceeding VGS(th).

Drain-feedback bias Voltage-divider bias

RG

+VDD

RD

RD

+VDD

R1

R2


Floyd/Buchla

Lesson 2



E-MOSFET biasingFIGURE 17–44 E-MOSFET biasing arrangements.

Recall that enhancement-only MOSFETs

must have a VGS greater than the

threshold value VGS(th) .

In either bias arrangement, the purpose is

to make the gate voltage more positive

than the source by an amount exceeding

VGS(th) .

In the drain-feedback circuit, there is

negligible gate current and, therefore, no

voltage drop across RG . As a result,

VGS = VDS.

Equation for the voltage-divider bias is

given by DDGS VRR

RV

21

2DDDDDS RIVV


Floyd/Buchla

Lesson 2



Determine the amount of drain current in the given figure. The

MOSFET has a VGS(th) of 3 V.

E-MOSFET biasing


Floyd/Buchla

Lesson 2



Bipolar

junction

transistor

(BJT)

Class A

amplifier

Saturation

Selected Key Terms

An amplifier that conducts for the entire input

cycle and produces an output signal that is a

replica of the input signal in terms of its

waveshape.

A transistor with three doped semiconductor

regions separated by two pn junctions.

The state of a transistor in which the output

current is maximum and further increases of

the input variable have no effect on the output.


Floyd/Buchla

Lesson 2



Cutoff

Q-point

Amplification

Common-

emitter (CE)

Class B

amplifier

The dc operating (bias) point of an amplifier.

Selected Key Terms

A BJT amplifier configuration in which the

emitter is the common terminal.

The non-conducting state of a transistor.

An amplifier that conducts for half the input

cycle.

The process of producing a larger voltage,

current or power using a smaller input signal

as a pattern.


Floyd/Buchla

Lesson 2



Junction field-

effect transistor

(JFET)

MOSFET

Depletion mode

Enhancement

mode

Metal-oxide semiconductor field-effect

transistor.

A type of FET that operates with a reverse-

biased junction to control current in a channel.

Selected Key Terms

The condition in a FET when the channel is

depleted of majority carriers.

The condition in a FET when the channel has

an abundance of majority carriers.


Floyd/Buchla

Lesson 2



Quiz

1. The Thevenin circuit shown has a load line that crosses

the y-axis at

a. +10 V.

b. +5 V.

c. 2 mA.

d. the origin.

+10 V

5.0 k


Floyd/Buchla

Lesson 2



Quiz

1. The Thevenin circuit shown has a load line that crosses

the y-axis at

a. +10 V.

b. +5 V.

c. 2 mA.

d. the origin.

+10 V

5.0 k


Floyd/Buchla

Lesson 2



Quiz

2. In a common-emitter amplifier, the output ac signal will

normally

a. have greater voltage than the input.

b. have greater power than the input.

c. be inverted.

d. all of the above.


Floyd/Buchla

Lesson 2



Quiz

2. In a common-emitter amplifier, the output ac signal will

normally



c. be inverted.



Floyd/Buchla

Lesson 2



Quiz

3. In a common-collector amplifier, the output ac signal

will normally



c. be inverted.

d. have all of the above.


Floyd/Buchla

Lesson 2



Quiz

3. In a common-collector amplifier, the output ac signal

will normally



c. be inverted.

d. have all of the above.


Floyd/Buchla

Lesson 2



Quiz

4. The type of amplifier shown is a

a. common-collector.

b. common-emitter.

c. common-drain.

d. none of the above.

R1

R2RE

C1

VC

C


Floyd/Buchla

Lesson 2



Quiz

4. The type of amplifier shown is a

a. common-collector.

b. common-emitter.

c. common-drain.


R1

R2RE

C1

VC

C


Floyd/Buchla

Lesson 2



Quiz

5. A major advantage of FET amplifiers over BJT amplifiers

is that generally they have

a. higher gain.

b. greater linearity.

c. higher input resistance.



Floyd/Buchla

Lesson 2



Quiz

5. A major advantage of FET amplifiers over BJT amplifiers

is that generally they have

a. higher gain.

b. greater linearity.

c. higher input resistance.



Floyd/Buchla

Lesson 2



Quiz

6. A type of field effect transistor that can operate in either

depletion or enhancement mode is an

a. D-MOSFET.

b. E-MOSFET.

c. JFET.



Floyd/Buchla

Lesson 2



Quiz

6. A type of field effect transistor that can operate in either

depletion or enhancement mode is an

a. D-MOSFET.

b. E-MOSFET.

c. JFET.



Floyd/Buchla

Lesson 2



Quiz

8. A transistor circuit shown is a

a. D-MOSFET with voltage-divider bias.

b. E-MOSFET with voltage-divider bias.

c. D-MOSFETwith self-bias.

d. E-MOSFET with self bias.RD

+VDD

R1

R2


Floyd/Buchla

Lesson 2



Quiz

8. A transistor circuit shown is a

a. D-MOSFET with voltage-divider bias.

b. E-MOSFET with voltage-divider bias.

c. D-MOSFETwith self-bias.

d. E-MOSFET with self bias.RD

+VDD

R1

R2


Floyd/Buchla

Lesson 2



Quiz

10. If you were troubleshooting the circuit shown here,

you would expect the gate voltage to be

a. more positive than the drain voltage.

b. more positive than the source voltage.

c. equal to zero volts.

d. equal to +VDD

RD

+VDD

R1

R2


Floyd/Buchla

Lesson 2



Quiz

10. If you were troubleshooting the circuit shown here,

you would expect the gate voltage to be

a. more positive than the drain voltage.

b. more positive than the source voltage.

c. equal to zero volts.

d. equal to +VDD

RD

+VDD

R1

R2

Documents

electronics fundamentals - staff.uob.edu.bhstaff.uob.edu.bh/files/600435156_files/lesson_2_transistors.pdf · Electronics Fundamentals 8th edition Floyd/Buchla © 2010 Pearson Education,