Amplifiers and Stuff About Transistor Ammps

8/14/2019 Amplifiers and Stuff About Transistor Ammps

1/13

Amplifier Load

Rin

Rout

A*Vit

Vit RLVL(Output loop)(Input loop)

Source

Ros

Vos

Input

Vit Rin

Rout

A*V itOutput

AMPLIFIERS

INTRODUCTIONIn this section, we will introduce transistors as a new circuit element and will focus on the designof transistor amplifiers. To provide appropriate background for the discussion with transistors, wewill begin with a general discussion of amplifiers.

AMPLIFIERSThe defining behavior of an amplifier is that the output is in some sense larger than the input.(This is like the behavior of a magnifying glass.) Of course, we generally want the output to be afaithful replica of the inputthat is, we want the amplification to be linear. The linear relation

between the output and input, defines A, the amplification. In the sense of this relation,Aout

in=

linearmeans that A is a real constant. The purpose of the amplifier may be to amplify voltage,current, orpower. Although our main focus will be on voltage amplifiers, we will briefly considerall three types.

An operational model for an amplifier is:

In this model, Rin indicates the input resistance, Routthe output resistance, Vitthe voltage at theinput terminals, and A*Vit is the amplified version of Vit. This model comes directly fromThevenins theorem: the output section of an amplifier should be a signal source and thus theThevenin representation is an ideal voltage source in series with a resistance. In contrast, the inputsection is nota source; thus it behaves simply as a resistance. The voltage source representedin the output section is the voltage resulting from amplifying the voltage at the input terminals.

However, an amplifier is useful only if it is connected to a source supplying the signal to beamplified, and to a load where the amplified result will be delivered. Consequently, the model for

anamplifier system is:


2/13

Transistors and Amplifiers GEORGIA STATE UNIVERSITY Page 2 of 13

init os os

osos in

in

R 1V = V = V

RR + R 1 +R

LL it it

outout L

L

R 1V = A * V = A * V

RR + R 1 +R

it os os

L it it

L

os

100 5V = V = V , and20+100 6

2 200V = 100 * V = V .

2+5 7

Thus,

V 5 200 500EA = = = = 23.8

V 6 7 21

(In the sketch, Vosis the open-circuit output voltage of the source,Ros is the output resistance of

the source, and RL is the load resistance.) With an amplifier system, the standard objective is forthe voltage across the load to be as nearly equal as possible to the amplification factor times Vos.Analysis of the input loop shows the requirement for maximizing Vit, the voltage actually amplified:

Clearly, Vit is maximum when the fraction Ros / Rin is zero. A similar analysis of the output loopyields

Also clearly, VL is maximum when Rout / RL is zero. Thus, from the perspective ofamplifier design,

the objectives are to make while making inR , outR 0.

EXAMPLE. Calculate the Effective Amplification of a system where Ros = 20k, Rin = 100k, A =100, Rout = 5S, and RL = 2S. (Effective Amplification is the ratio VL / Vos.)

By analysis of the input loop:

The clear result is that EA < 100! This is a direct consequence of the loading of the source by

Rin of the amplifier and of loading of the amplifiers output by the load.

Considerations for Current and Power Amplifiers: The analysis above focused on theobjectives in a voltage-amplifier system of maximizing delivery of voltage from the source to theinput terminals of the amplifier, and maximizing delivery of voltage from the amplifier output to theload. The obvious question is what considerations hold for current and power amplifier systems.For current amplifiers, the solution is fairly obvious: current to the input terminals of the amplifier


3/13


( )

2 2 inin in in os 2

in os

RP = I R = V ,

R + R

( ) ( ) ( )

( ) ( )

=

in in

2 3 2in in os in os in os

in os os in os

3 3

in os in os

in

R Rd 1- 2 +

dR R + R R + R R + R

R + R - 2R R - R= = = 0

R + R R + R

Thus, P i .s maximum when in osR = R

is maximized when Rin is zero, and load current is maximized when Rout is large. These are theexact opposites of the desired cases for voltage amplifiers!

For power amplifiers, the objective is to maximize delivery of the power, the product V*I. Obviously,

neither Rin = 0 nor Rin64 are optimal as V is zero in one case, I in the other, and P is zero for both!Since the input and output loops are the same type of circuit, the maximization problem is the samefor both. Specifically considering the input loop:

Thus the problem is to find Rin maximizing the quantity in braces; i.e., to find Rin such that thederivative of the quantity in braces with respect to Rin equals zero:

By the same type of analysis, PL is maximum when RL = Rout.

AMPLIFIER SUMMARYThe basic result is that the amplifier presents a load to the signal source, and the delivery of signalto the load is affected by both the load resistance and the amplifiers output resistance. For avoltage amplifier, the objective of maximum voltage transfer leads to the design targets ofhaving the input resistance as high as possible and the output resistance as low as possible.

In contrast, for a current amplifier, the objective of maximum current transfer creates the goalsthat the input resistance should be as low as possible and the output resistance should be ashigh as possible. Finally, forpower amplifiers , the objective of maximum power transfer createsthe goals that the input resistance should equal that of the source, and the output resistanceshould equal that of the load.


4/13


N P N P N PC C

B

E

B

E

B

C

E

NPN

B

C

E

PNP

TRANSISTORS AND TRANSISTOR AMPLIFIER DESIGN

As indicated above, transistors are the new circuit element introduced in this section. Transistors

are related to diodes discussed previously in that they incorporate 3 sections of P- and N-dopedsemiconductors as shown below:

Also as indicated, there are two possible configurations, NPN, and PNP. Junction bipolartransistors are therefore classified with this terminology. Moreover, the connections are designatedas collector, base, and emitter, orC, B, and E, respectively. (From the simplistic sketches

shown, it is not clear that C and E are different in any way; however, they are quite different in theactual construction of a transistor.)

Also, two PN-junctions are apparent in each arrangement: base-emitter and collector-base. At thedesign level we will use, the base-emitter junction is the important one in that it will be necessaryto remember that current flow through a PN junction requires that the forward threshold voltage beestablished (e.g., Vf= ~ 0.7V for Silicon).

Circuit symbols for each type of (junction bipolar) transistor are:

Our interest is in learning how to use the electrical characteristics of transistors to accomplish thefunction of amplification. For this reason, we will focus only on the NPN type; thus, ourtypicaltransistorwill be an NPNversion constructed of silicon. (For our purposes, the basic difference

between NPN and PNP behavior is that current flow in one is in the opposite direction from theother. Thus all statements made below regarding NPN transistors also apply to PNP versionswhen the current directions are reversed.)

Electrical Behavior of Transistors: The basic circuit sketched below indicates the mainparameters of interest to us: the base current, Ib; the collector current, Ic; and the collector-emittervoltage, VCE. The graph shown below the circuit illustrates the electrical relationships betweenthese parameters as often presented in textbooks. Apparent from the graph is the characteristicthat the collector current is virtually proportional to the base current at constant VCE, and is pretty


5/13


+

+

Ib

Ic

V1

V2R1

RC

Ie

VCE

"Typical" Transistor Characteristic Curves

VCE,sat

Ib

IC

VCE

much proportional regardless of VCE as long asit is above the value indicated as VCE,sat.

(It is important to note here that many textbooks

build their amplifier design procedure on the setof curves shown in the graph. However, thisprocedure, a carry-over from vacuum-tube-based amplifier design methods, is inappropriatefor transistor amplifier design for at least tworeasons. One is that transistor manufacturers donot provide such characteristic curves for theirproducts, and the second is that manufacturingprocesses are such that the characteristics ofthe same transistor type number may vary overa wide range.)

Based on the characteristic curves, and for practical purposes, we can take the basic propertyof transistors as the base current CONTROLS the collector current. In addition, the collectorcurrent is larger than the base current by a factor of 30 to 300; thus changing the base currentleads to a corresponding change in a much larger current. In this sense, transistors are inherentlycurrent amplifiers. (It is important to point out that the base current does not create collectorcurrent; it only exercises control over the collector current. It other words, the base is like a faucet:the faucet can control the flow of water, but it cannot create water when none is present.)


6/13


CFE

B

Cfe

B

Ih = ("large- signal" current gain; very crude approximation, but useful)

I

Ih = ("small - signal" current gain; better approximation)

I

"Realistic" Relation Between

Base and Collector Currents

Realistic Ic

vs. Ib

Ic

prop. to Ib

Ic

Ib

RC

RE

RB1

RB2

VB

VE

VC

+VS

+Power

Supply

Along with the basic behavior that Ib controls Ic, is the fact that their relationship is approximatelyproportional. (Of course, this is not exactly true although a detailed description of transistorparameters is more involved than necessary for our purposes.)

Specifically, we will make use of the following relations between base and collector currents:

Sketched to the side is a reasonablerepresentation of the actual relationship

between IC and IB.

Summary of Our Typical Transistor

Characteristics:

1. Type: NPN2. Material: Silicon

(VBE = 0.7 V for IB0)3. hfe = hFE ~100

Amplifier Design. As indicated above, thebasic nature of a transistor is that of a currentamplifier. Our objective is to use thischaracter to create a voltage amplifier leadingto the obvious question of how we can convertcurrent into voltage. The method, of course isto use resistors as current-to-voltageconverters. Therefore, our standard transistor amplifier circuit will be as sketched below:


7/13


RC

RE

VB

VE

VC

+VSIc

Ib

Ie

(The power supply and its connections will be omitted from this point on. Thus, it is

understood that amplifiers require a power supply, and that the power supply + will be

connected to the point labeled +VS and with - connected to ground, or common.)

Voltage Relations for Transistors: Before the discussion ofamplifier design, it will be helpful to think through voltagerelationships appropriate for a circuit such as that of thestandard amplifier. Specifically, consider the fragment of theamplifier circuit consisting of the transistor along with thecollector and emitter resistors. We are only interested in thecase of non-zero base current; in that case the followingrelations hold:

1. VB = VE + 0.7 (VBE = VF)2. IE = VE / RE (Ohms Law)3a. I

E= I

B+ I

C(KCL)

3b. IC = hfe*IB (Transistor Property)3c. IE = IB * (hfe + 1) (3a. and 3b.)3d. IE ~ IB * hfe = IC (hfe ~ 100; can ignore difference

between 100 & 101))4. VC = VS - IC * RC (Ohms Law)

EXAMPLE:

Given the following data for the circuit fragment sketched above, calculate the voltages at theemitter and collector and the currents indicated on the sketch. (VB = 1.1V; RE= 250SSSS; RC= 3kSSSS,

VS = 15V.)

Solution: VE = (1.1 - 0.7)V = 0.4 V

IE = 0.4V / 250S = 1.6 maVC = 15V - (1.6ma) * (3kS) = 10.2VIC ~ IE = 1.6 ma

IB~ IC / hfe = 1.6ma / 100 = 16 ::::a

Basic Point: Just as the base current controls the collector current (and emitter current),

so does the base voltage control the collector and emitter voltages.

EXAMPLE:

For the resistors and VS the same as in the example above, calculate the base voltage and currentnecessary to make the collector voltage = 7.5V.

Solution: IC = (15 - 7.5)V / 3kS = 2.5 maVE ~ 2.5ma * (0.25kS) = 0.625VVB = 0.625V + 0.7V = 1.325V

IB~ IC / hfe = 2.5ma / 100 = 25 ::::a


8/13


E BC E

E E

BC S C C S C

E

S C E

C C

B E

V V - 0.7I I = =

R R

V - 0.7V = V - I R = V - R

R

Since V , 0.7, R , and R are constants,V R

A = = - .V R

RB1

RB2

VB, IB

I1

I2

+VS

AmplifyingWe can now analyze the circuit fragment as an amplifier; in fact, this portion of the standardcircuit is responsible for the amplification. The basic idea is to establish a base voltage, andtherefore a collector voltage, as a steady value. Amplification will amount to the change in thecollector voltage (the output) caused by changing the base voltage (the input). For this circuit,

amplification will be expressed as By use of the relations developed above,

out C

in B

V VA = = .

V V

we can relate amplification to the circuit elements as follows:

Thus the amplification is controlled by only the collector and emitter resistors with no explicitdependence on the transistor parameters! This of course raises the question of why the transistoris necessary. However, the transistor plays a significant role since it makes the collector andemitter voltage different by a constant amount; it makes the emitter and collector currents virtuallyequal; and thus it makes the collector voltage respond to the base voltage as described. (The -in the expression for amplification is the result of taking the output voltage from the collector toground: as IC increases, this voltage decreases.)

Role of Base Resistors RB1 and RB2:As seen above, the amplification is controlled by only RC and RE.However, to have a working amplifier circuit, it is necessary that the basevoltage and current be suitably established, and this is the purpose ofthe base resistors. Specifically, the resistors must be suitable for settingup the electrical situation sketched to the side. However, since the basiccircuit is that of a voltage divider and there are two resistors, the electricalrequirements can be met in an infinite number of ways as they requireonly that the resistors be related appropriately. This situation, one ofunderdetermined parameters gives us the opportunity to incorporate anadditional design target into the overall amplifier design.

In fact, before continuing with the discussion of methods for selectingsuitable values for the resistors, let us note that the amplificationrelation also does not uniquely determine the collector and emitterresistors. Since neither the amplification nor the electrical requirementsat the base yield unique values for the set of 4 resistors, it will be usefulto look for other desirable amplifier characteristics we may include in and meet with the overalldesign considerations. In connection with amplifiers in general we already discussed above twosuch characteristics: the input resistance and the output resistance. To see how target valuesfor these characteristics can be included in the design, we need to analyze the standard amplifiercircuit to learn the factors on which these depend.


9/13


+

BA C D E

F G

H

Power

Supply+9V

RE

VB

VE

VC

+VS

Ib

Ie

Ic, shortRC

Rout

RinVin VVout

IoutIinR =V

Iin

in

in

R =V

Iout

out open

out short

The Concept of ac Ground:To obtain the relations necessary for the equivalent input and output resistances of our amplifier,we need to introduce the concept of ac ground. The basic ideas are : (1) the difference betweenac and dc is that ac changes; and (2) that ground is the set of connections used as the 0-voltreference for all measurements. (Ground also is one connection for applying input signals, and onefor obtaining outputs; as well it often isone connection to the power supply.)Since dc ground is the set of all points atvoltages no different from ground, acground is the set of all points which donot change relative to ground. For thediscussion consider the sketch: in it, thepoints A, B, C, D, E, and G are allconnected directly to ground and thus areat dc ground; that is, they are at avoltage equal to dc ground. Similarly,points F and H are notdc ground, but areac ground as they are connected toground by a constant voltage. Insummary, all points in the sketch are ac ground.

Input and Output Resistance for our Standard Transistor Amplifier:The approach for relating the amplifiers input and output resistance to the actual circuit is basedon Thevenins Theorem. Specifically, referring to the general amplifier model as introduced above(see sketch below), the input resistance can be calculated by developing the relation betweenand input voltage and the current flowing into the amplifier, while the output resistance can be

obtained by relating the open-circuit out put voltage to the short-circuit output current. These aresummarized below:

Output Resistance:

Referring to the amplifying fragment shown to the side andrecalling that the output is the change in the collector voltage,

we can identify that The direct approach toV = Vout open Cb g .calculating the output short-circuit current is to short thecollector to ground. However, this complicates the calculation;the alternative is to recall the ac ground concept and make theshort circuit from collector to VS as indicated by the dotted line.

From this it is clear that ; thusI = Iout short Cb g


10/13


RB1

RB2

Input

(hfe + 1)RE

+VS

(ac ground)

R =V

I=

V

I=out

out open

out short

C

C

RC

R =V

I

V = V + 0.7 = V = I R

I = (I + I = h + 1 I

V = I R = h + 1 I R

R = h + 1R

inB

B

B E E E E

E B C fe B

B E E fe B E

in fe E

b gb g

b gb g

)

In summary, therefore, the amplifiers output resistance is determined by (and is equal to!) Thecollector resistor. (Of course, the exactstory is more complicated; nevertheless, this result givesthe dominant contribution to Rout.)

Input Resistance:Obtaining the expression for the input resistance requires a two-step approach: first we willconsider the basic amplifying section sketched above, and then we will consider the effect of thebase resistors. Referring to the sketch above and the relations developed earlier for the input of)VB:

This result describes only the basic amplifying section and now must be combined with the baseresistors to calculate the effective input resistance of the amplifier under operating conditions. Thecircuit to consider is as sketched below:

From the sketch, it is clear that RB2 is parallel to the effectiveinput resistance of the basic amplifying section, (h fe + 1)RE.The effect of RB1 can be included by recalling that the inputsignal is a changing voltage, and thus is an ac signal. Forsignals, therefore, VS is ground since it is a point of ac ground.Thus, for the input signal, the effective input resistance is theequivalent of RB1 // RB2 // (hfe + 1)RE.

Summary of Rin and Rout Relations:For our transistor amplifier circuit, the effective input resistance(for signals) is:

Rin = RB1 // RB2 // (hfe + 1)RE

and the effective output resistance is:

Rout = RC.

Methods for Calculating Suitable Resistor Values for the Amplifier Circuit:For our purposes, the standard problem can be summarized as: given a typical transistor (NPNsilicon) and a power supply of voltage VS, find suitable resistors to meet the specifications:

A $$$$ specified value; Rout####specified value; Rin$$$$specified value.


11/13


RB1

RB2

VB, IB

I1

I2

+VS

Example and Illustration of (One) Procedure

Given VS = 24V and a typical transistor, find resistor values to meet the following specifications:

*A*$15, Rout#3k, and Rin$12k.

Solution:

A 15 CHOOSE A = 15 =R

RR 3k CHOOSE R = 3k

R = 3k

R = 200

C

E

out out

C

E

UV|W|

Good Idea: The input and output signals are ac voltages, and the basic operation of the circuitis that input changes VB which leads to a change in VC as the output voltage.Thus, it makes sense to aim for an initial voltage at the collector maximizing theupwards and downwards change in collector voltage. Obviously, the best startingpoint is VC = VS.

AIM for VC = VS = 12V; ICRC = VS - VC = 12V; IC = (VS - VC) / RC = 12V / 3k = 4ma;

The goal now is to find out what VB and IB are necessary for the target VC:

VE = IE RE ~ IC RE = (4ma) (0.2k) = 0.8V; VB = VE + 0.7V = (0.8 + 0.7)V = 1.5VIB ~ IC / hfe = 4ma / 100 = 40 :a;

Base Rs: To meet the Rin specification and provide the necessaryelectrical conditions, RB1 and RB2 must create VB = 1.5V,provide IB = 40 :a, and be such that RB1 // RB2 // (hfe + 1) RE.An effective way to proceed is based on educated guessing.

Specifically, (hfe + 1)RE is already determined; moreover, RB2is (usually) the smaller of the two base resistors since it hasless voltage across it. Since the equivalent of a parallelarrangement of resistors is less than the smallest member, RB2will play this role.

For the example, (hfe + 1) RE = 100 (0.2k) = 20k; to make Rin$12k, the equivalent of RB1 // RB2 must be greater than orequal to 30k. Therefore, RB2 must itself be more than 30k; soTRY RB2 = 40k.

I2 = VB / RB2 = 1.5V / 40k = 37.5:a

I1 = IB + I2 = 77.5 :aRB1 = (VS - VB) / I1 = 22.5V / 77.5:a ~ 290 k

Check: Use of a trial-and-error procedure requires that the result be checked to see if itmeets the specifications:

Rin = RB1 // RB2 // (hfe + 1)RE = 290k // 40k // 20k = 12.74K > 12k, so OK.


12/13


RC

RE

RB1

RB2

+VS

Cin Cout

Amplifier Load

Rin

Rout

A*Vit

Vit RLVL(Output loop)(Input loop)

Source

Ros

Vos

CinC

out

Input and Output Coupling Capacitors:Since the input point (the base) and the output point (the collector) are at carefully established dcvoltages, the input and output (ac voltages) need to be coupled via capacitors. These serve asDC-blocking elements, which is another way of saying they create high-pass filters in the overall

circuit. We need to consider two points in this regard:

(1) what are appropriate considerations for selecting suitable values for Cin and Cout?

(2) how do these capacitors affect the overall performance of the amplifier?

Sketched below is the standard amplifier including the coupling capacitors:

These create the effective circuit as modeled below:

The basic effect is incorporation of the capacitors in the input and output loops. A reasonable rule-of-thumb is to calculate capacitor values so that their respective reactances equal the amplifiers

input resistance and the load resistance at a chosen low frequency.

EXAMPLE:

Calculate suitable input and output coupling capacitors for an amplifier with Rin = 10k to beconnected to a load RL = 100S at the target low frequency (2Hz/B).


13/13


B

C

E

NPN Transistor

E

C

B

PNP Transistor

V = RV

R + R + - j C

V = RAV

R + R - j C

EA =V

V=

AR

R + R - j C

R

R + R - jC

EA =V

V=

AR

R + R + 1C

R

R + R + 1

it inos

os in out

L Lit

out Lin

L

os

L

out L out

in

os inin

L

os

L

out L2

out

2

in

os in2

FHG

IKJ

F

H

G

GG

I

K

J

JJ

FHG

IKJF

HGGG

I

KJJJ

FH

IK

F

H

GGGG

I

K

JJJJb g b g

C

in

2FH

IK

F

H

GGGG

I

K

JJJJ

X =1

CC =

1

XC

1

2 2 10k= 25 F

C

1

2 2 100 = 2500 F

CC

in

out

=

=

e j

e j

Finally, the complete description of the loading including the capacitors requires treating the inputand output loops as ac circuits:

GLOSSARY

Documents

Amplifiers and Stuff About Transistor Ammps