Applied Electronics II · Current-Shunt (Current Ampli er) Feedback Voltage-Shunt (Transresistance Ampli er) Feedback 6 Exercise Chapter 1: Feedback Ampli ers SECE March 2017 2

Applied Electronics II

Chapter 1: Feedback Amplifiers

School of Electrical and Computer EngineeringAddis Ababa Institute of Technology

Addis Ababa University

Daniel D./Getachew T./Abel G.

March 2017

Chapter 1: Feedback Amplifiers () SECE March 2017 1 / 45

Overview

Overview

1 Types of Feedback

2 The General Feedback StructureBasic Feedback Amplifier

3 Feedback Topologies

4 Properties of Negative FeedbackGain DesensitivityNoise/Interference ReductionReduction of Nonlinear DistortionControl of Impedance level & Bandwidth Extension

5 Analysis of Feedback AmplifiersVoltage-Series (Voltage Amplifier) FeedbackMethod of Analysiis of Feedback AmplifiersCurrent-Series (Transconductance Amplifier) FeedbackCurrent-Shunt (Current Amplifier) FeedbackVoltage-Shunt (Transresistance Amplifier) Feedback

6 ExerciseChapter 1: Feedback Amplifiers () SECE March 2017 2 / 45

Types of Feedback

Type of Feeedback

Most physical systems incorporate some form of feedback. Feedbackcan be broadly classified as:

1 Posittive FeedbackA portion of the output signal is added to the input. Positvefeedback is used in the design of oscillator and a number of otherapplications (will be discussed in Chapter 4 and 5).

2 Negative FeedbackA portion of the output signal is subtracted from the inputsignal.The basic idea of negative feedback is to trade off gain forother desirable properties listed below

Desensitize the gainReduce nonlinear distortionReduce the effect of noiseControl the input and output resistancesExtend the bandwidth of the amplifier.


Types of Feedback

Negeative Feedback

Example

Introducing resistor at the emitter of BJT common-emitter circuitsstabilizes the Q-point against variation transistor parameters.

V+

IC

IE

V-

VBB

IB +

-

VCE+

-VBE

RC

RB

RE

Solution Apply KCL at B-E loop

VBB = IBRB + VBE(on) + IERE + V − (1)

Assuming active-mode of operation

IE = (1 + β)IB and IC = βIB (2)

As IC increases(due to ↑ in T,aging ) thevoltage drop across RE increase thusopposing the base-emitter voltage.


The General Feedback Structure


Figure 1 show the basic structure of a feedback amplifier, where each ofthe quantities x can represent either a voltage or a current signal.

Figure: 1 General structure of the feedback amplifier, the quantities xrepresent either voltage or current signals.

The relationship between the quantities x is

xo = Axi

xf = βxo

xi = xs − xfChapter 1: Feedback Amplifiers () SECE March 2017 5 / 45


Feedback Systems

Thus xo = A(xs − βxo)The gain with feedback ,Af

Af =xoxs

=A

1 + βA

The open-loop gain, A represents the transfer gain of the basicamplifier without feedback. Implicit in the description is that thesource, the load, and the feedback network do not load the basicamplifier. That is, the gain A does not depend on any of these threenetworks.In practice this will not be the case.

if |Af | < |A| the feed back is negative or degenerative

if |Af | > |A| the feed back is positive or regenerative

If, as is the case in many circuits, the loop gain Aβ is large, Aβ 1,then it follows that

Af u1

β


The General Feedback Structure Basic Feedback Amplifier

Basic Feedback Amplifier

A basic representation of feedback amplifier is show in the Figure below

Signal

Source

Sampling

Network

Basic

Amplifier,

gain A

Comparato

r or Mixer

Network

Feedback

Network b

Ii I

If

Io = IL

RL

+

-

+

-

+

-Vi V Vo

Signal Source : This block is a voltage source Vs with a series resistorRs(Thvenin’s equivalent circuit) or a current source Iswith a parallel resistor Rs (Norton’s equivalent circuit)

Feedback Network : Usually a passive two-port network with reversetransmission β




Sampling Network : Sampling blocks are shown below

A

b

RL

Sampler

Figure: (a) Voltage or node sampling

A

b

Sampler

RL

Figure: (b) Current or loop sampling

(a) Output voltage is sampled by connecting the feedbacknetwork in shunt across the output.

(b) Output current is sampled by connecting the feedbacknetwork in series with the output.




Comparator or Mixer Network : Two types a series (loop) and shunt(node). A differential amplifier is often used as mixer.

A

b

Rs

Series

MixerSource

Vs

+

-

+

-

Vf

Vi

Figure: (a) Series Mixing

A

b

Rs

Shunt

MixerSource

Is

Ii

If

Figure: (b) Shunt Mixing




Basic Amplifier : A could be used to represent

VVi

= AV , Voltage gain

IIi

= AI , Current gain

IVi

= GM , Transconductance

VIi

= RM , Transresistance

They are gain of the basic amplifier without feedback


Feedback Topologies

Feedback Topologies

There are four basic feedback topologies, based on the parameters tobe amplified (voltage or current)and the output parameter (voltage orcurrent). They are described by the type of connection at the inputand output of the circuit.

(a) Voltage-Series (Series-Shunt) or Voltage Amplifier

(b) Current-Shunt (Shunt-Series) or Current Amplifier

(c) Current-Series (Series-Series) or Transconductance Amplifier

(d) Voltage-Shunt (Shunt-Shunt) or Transeresistance Amplifier


Feedback Topologies

Feedback Topologies

Figure: (a) Series-Shunt

Figure: (b) Shunt-Series

Figure: (c) Series-Series

Figure: (d) Shunt-Shunt


Properties of Negative Feedback Gain Desensitivity

Gain Desensitivity

Variation in the circuit gain as a result of change in transistorparameters is reduced by negative feedbackFrom the previous slides the gain with feedback,Af is given as

Af =xoxs

=A

1 + βA

Assuming β is constant and taking the derivative of Af with respect toA,

dAfdA

=1

1 + βA− A

(1 + βA)2β =

1

(1 + βA)2or dAf =

dA

(1 + βA)2

Dividing both sides the gain with feedback yields

dAfAf

=

dA(1+βA)2

A1+βA

=1

1 + βA

dA

A


Properties of Negative Feedback Gain Desensitivity

Gain Desensitivity

Hence the percentage change in Af (due to variations in some circuitparameter) is smaller than the percentage change in A by a factorequal to the amount of feedback. For this reason, the amount offeedback, 1 +Aβ, is also known as the desensitivity factor.

Example

The open-loop gain of an amplifier is A = 5× 104V/V exhibits a gainchange of 25% as the operating temperature changes. Calculate thepercentage change if the closed loop gain Af = 50V/V .

dAfAf

=1

1 + βA

dA

A=

A

A(1 + βA)

dA

A=AfA

dA

A=

50

5× 104× 25%

dAfAf

= 0.025%


Properties of Negative Feedback Noise/Interference Reduction

Noise/Interference Reduction

Under certain condition feedback amplifiers can be used to reducenoise/interference.

This can be achieved if a preamplifer which is (relatively)noise/interference-free precessed the noise/interference-proneamplifierUnder such conditions the Signal-to-Noise ratio can be improved (compare to noise/interference-prone amplifier without feedback)by the factor of the preamplifier gain


Properties of Negative Feedback Reduction of Nonlinear Distortion

Reduction of Nonlinear Distortion

Distortion in the output is due to application of large amplitudeinput signal applied beyond the linear region of operation.

Negative feedback can be implemented to reduce nonlineardistortion by a factor of 1 +Aβ.

Assuming that the open-loop gain Aβ 1, the gain with feedback

Af =A

1 +Aβu

1

β

It implies that Af is independent of the nonlinear properties of thetransistors used in the basic amplifier.

Since the feedback network usually consists of passive components,which usually can be chosen to be as accurate as one wishes.


Properties of Negative Feedback Control of Impedance level & Bandwidth Extension

Control of Impedance level & Bandwidth Extension

Control of Impedance level: The input and output impedance can beincreased or decreased with the proper type of negativefeedback circuit.

Bandwidth Extension : The improvement in frequency response andbandwidth extension (Chapter 3)

The advantage of negative feedback is at the cost of gain. Undercertain circumstance, a negative feedback amplifier may becomeunstable and break into oscillation.


Analysis of Feedback Amplifiers

Fundamental Assumtions

Some fundamental assumptions are taken in order to analyze the fourfeedback configurations.

Input is transmitted through the amplifier only, not through thefeedback.The feedback signal transmitted feedback network only, notthrough the amplifier.β is independent of the load and source impedance.

+−

Ii

Vs

Rif Rof R’of

Io+

-

Vi Ri

Ro

RL

+

-Vo

+

-

Vo

+- Vf

ßVo

AvoVi

+

−

+

−

Figure: Ideal structure of a Voltage-Series feedback amplifierChapter 1: Feedback Amplifiers () SECE March 2017 18 / 45

Analysis of Feedback Amplifiers Voltage-Series (Voltage Amplifier) Feedback

Voltage-Series (Voltage Amplifier) Feedback

Avo represents the open circuit voltage gain taking Rs into accountInput Impedance: The input impedance with feedback is

Rif =VsIi

Also,

Vs = IiRi + Vf = IiRi + βVo and Vo = AvoViRL

Ro +RL

let Av = AvoRL

Ro+RL, where Av is the voltage gain without feedback

taking the RL into account then

Vs = IiRi + βAvVi = IiRi + βAvIiRi

Rif =VSIi

= Ri(1 + βAv)




+

-

Vi Ri

Ro

Vf

ßVo

AvoVi

+

−

+−

+− Vx

Ix

Figure: Ideal structure of a Voltage-Series feedback amplifier

Output Impedance: To find Rof must remove the external signal (setVs = 0 or Is = 0), let RL =∞, impress a voltage Vx across the outputterminals and calculate the current Ix delivered by the test voltage Vx

Ix =Vx −AvoVi

Ro

Since Vi = −βVxRof =

VxIx

=Ro

1 + βAvo




The output resistance with feedback R′of which includes RL as part of

the amplifier is Rof ‖ RL

R′of =

RofRLRof +RL

=

RoRLRof +RL

1 + βAvoRL

Ro +RL

Taking R′o = Ro ‖ RL

R′of =

R′o

1 + βAv

Voltage gain with feedback: Avf taking the load into account.

Vs = Vi + βVo = VoRo +RLAvoRL

+ βVo




Manipulating the equation

Avf =VoVs

=Avo

RLRo +RL

1 + βAvoRL

Ro +RL

The voltage gain with feedback without the load Avfo is

Avfo =VoVs

=Avo

1 + βAvo

In conclusion

Input Impedance: increased by a factor 1 + βAv

output Impedance: decreased by a factor 1 + βAv

Voltage Gain: decreased by a factor 1 + βAv




In practical case

In practical case, feedback network will not be ideal VCVS.

Actually, it is resistive and will load the amplifier.

Source and load resistances will affect A, Ri, and Ro.

Source and load resistances should be lumped with basic amplifier.

Expressed as two-port network.

How To Solve

1. Identify the feedback network

2. Its loading effect at the input is obtained by short circuiting itsport 2 (because it is connected in shunt with the output).

3. The loading effect at the output is obtained by open-circuitingport 1 of the feedback network (because it is connected in serieswith the input)

4. The gain without feedback A is determined

5. The feed back gain β is determinedChapter 1: Feedback Amplifiers () SECE March 2017 23 / 45


Voltage-Series (Voltage Amplifier) FeedbackFigure: Finding the A circuit and β for the Voltage-Series feedback amplifier.


Analysis of Feedback Amplifiers Method of Analysiis of Feedback Amplifiers

Method of Analysiis of Feedback Amplifiers

Steps

1. Identify if the mixing or comparison is series or shunt

a) Series mixing : If the feedback signal subtracts from the externallyapplied signal as a voltage

b) Shunt mixing : If the feedback signal subtracts from the appliedexcitation signal as a current.

2. Identify the sampled signal as series or shunt

a) Voltage sampling : Set Vo = 0(RL = 0. If Xf becomes zero, we havevoltage sampling.

b) Current sampling : Set Io = 0(RL =∞. If Xf becomes zero, wehave current sampling.

3. The amplifier without feedback but taking the feedback networkloading into account

1) Find the input circuit.

a) Set Vo = 0 for voltage sampling.b) Set Io = 0 for current sampling.



Method of Analysiis of Feedback Amplifiers

2) Find the output circuit.

a) Set Vi = 0 for shunt comparison so that no feedback current entersthe amplifier input.

b) Set Io = 0 for series comparison so that no feedback voltage reachesthe amplifier input.

4. Find the feedback network β.

5. Calculate β,A,Ri and Ro.

6. Calculate the closed loop Af , Rif , Rof .




Example

Analyze the amplifier to obtain its voltage gain Vo/Vs, input resistanceRin, and output resistance Rout. Find numerical values for casegm1 = gm2 = 4mA/V , RD1 = RD2 = 10kΩ and R2 = kΩ. Forsimplicity, neglect ro of each of Q1 and Q2.

+−Vs

Q1

Q2

RD1

Vo

R1

R2

Rin

Rout

RD2

The next step is identifying the A andβ circuitWe identify the feedback network asthe voltage divider of (R1, R2)

+− Vo

R2

R1

+

-

Vf




Example (Continued)

The A circuit is

Q1

Q2

RD1

Vo

R1 R2

Rout

RD2

+

-Ri

R2

R1

Vi

Vd1

Calculating A1 and A2

Vd1 = 0− id1RD1

Vi = Vgs1 + id1(R1 ‖ R2)

A1 =Vd1Vi

=−id1RD1

Vgs1 + id1(R1 ‖ R2)

A1 =−RD1

1/gm1 + (R1 ‖ R2)

A1 =−gm1RD1

1 + gm1(R1 ‖ R2)




Example (continued)

From A circuit we have

Vo = 0− id2(RD2 ‖ (R1 +R2)) and Vgs2 = Vd1

A2 =VoVd1

=−id2(RD2 ‖ (R1 +R2))

Vgs2= −gm2(RD2 ‖ (R1 +R2))

The open loop gain is

A =VoVi

= A1A2 =gm1gm2RD1[RD2 ‖ (R1 +R2)]

1 + gm1(R1 ‖ R2)

When evaluated

A =4× 4× 10[10 ‖ (1 + 9)]

1 + 4(1 ‖ 9)= 173.913 V/V




Example (continued)

from β circuit we have

β =R1

R1 +R2=

1

1 + 9= 0.1 V/V

The closed loop gain

VoVs

= Af =A

1 +Aβ=

173.913

1 + 173.913× 0.1= 9.45 V/V

The input resistance is infinite because it is the input resistance ofMOSFET.The output resistance is

Rout = Rf =Ro

1 +Aβ=RD2 ‖ (R1 +R2)

1 +Aβ=

10 ‖ (1 + 9)

1 + 173.913× 0.1= 271.87Ω


Analysis of Feedback Amplifiers Current-Series (Transconductance Amplifier) Feedback

Current-Series (Transconductance Amplifier) Feedback

+−

Ii

Vs

Rif Rof R’of

Io+

-

Vi Ri Ro RL

+

-

Vo

+- Vf

+

−

GmVi

IoßIo

Input Impedance:

Rif =VsIi

; Vs = IiRi + βIo ; Io = GmViRo

Ro +RL

Rif =IiRi + βGmIiRi

RoRo+RL

Ii= Ri(1 + βGm

RoRo +RL

)




Gm = Io/Vi is the short-circuit transconductance, andGM = GmRo/(Ro +RL) is the transconductance without feedbacktaking the load into account.

Rif = Ri(1 + βGM )

Output Impedance: calculated by short-circuiting the source andreplacing the source with a voltage source Vx with a current of Ix

Ix =VxRo−GmVi and Vi = βIx

∴ Rof =VxIx

=Ro(Ix +GmβIx)

Ix= Ro(1 + βGm)

The output impedance taking the load as part of the amplifier is:

R′of = (Rof ‖ RL) = (Ro ‖ RL)

1 + βGm1 + βGM



Current-Series (Transconductance Amplifier) FeedbackFigure: Finding the A circuit and β for the Current-Series feedback amplifier.



Current-Series (Transconductance Amplifier) FeedbackExampleCalculate the closed loop voltage gain, output resistance and input resistancefor the circuit below. The output is taken from emitter current of Q3. Thevalues of RC1 = 9kΩ, RC2 = 5kΩ, RC3 = 600Ω, RE1 = 100Ω, RE3 = 100Ωand RF = 640Ω. Assume that the bias circuit, which is not shown, establishesIC1 = 0.6mA, IC2 = 1mA, and IC3 = 4mA. Also assume that for all threetransistors, hfe = 100 and ro =∞.

Q1

Q2

Q3

+

-

Vs

Vo

RE1 RE3

RF

RC1

RC2

RC3

Io

The β circuit.

+

-

RE1 RE3

RF

IoVf

β =VfIo

=[(RF +RE1) ‖ RE2]Io

RE1

RF+RE1

Io

β =RE1 ×RE2

RF +RE1 +RE2= 11.9Ω




Example (continued)

The A circuit.

Q1

Q2

Q3

Vo

RE1 RE3

RF

RC1

RC2

RC3

Io

RE3 RE1

RF+−

Vi

Ri

Ro

When Aβ 1

Af u1

β=

1

11.9Ω= 84mA/V

lets check by determining eachtransistor gain

A1 =Vc1Vi

=−ic(RC1 ‖ rπ2)

ie(re1 + [RE1 ‖ (RE3 +RF )])

A1 =−α(RC1 ‖ rπ2)

(re1 + [RE1 ‖ (RE3 +RF )])

Since Q1 is biased at 0.6mA,re1 = 41.7Ω. Q2 is biased at 1mA; thusrπ2 = hfe/gm2 evaluating A1

A1 = −14.92V/VChapter 1: Feedback Amplifiers () SECE March 2017 35 / 45



Example (continued)

The gain of Q2

A2 =Vc2Vb2

=−ic[RC2 ‖ (hfe + 1)[re3 + (RE3 ‖ (RF +RE1))]]

Vb2

A2 = −gm2[RC2 ‖ (hfe + 1)[re3 + (RE3 ‖ (RF +RE1))]]

re3 = 25/4 = 6.25Ω and substituting the other values

A2 = −131.2V/V

The gain of Q3

A3 =IoVc2

=Ie3Vb3

=1

re3 + (RE3 ‖ (RF +RE1))

when evaluatedA3 = 10.6mA/V




Example (continued)

The gain without feedback

A = A1A2A3 = −14.92×−131.2× 10.6× 10−3 = 20.7A/V

The gain with feedback

Af =A

1 +Aβ=

20.7

1 + 20.7× 11.9= 83.7mA/V

We can note that it is very close to approximate value. The input resistance

Rin = Rif = Ri(1 +Aβ)

Ri = (hfe + 1)[re1 + (RE1 ‖ (RF +RE2))] = 13.65kΩ

∴ Rif = 13.65(1 + 20.7× 11.9) = 3.38MΩ




Example (continued)

The output resistanceRof = Ro(1 +Aβ)

Ro = [RE3 ‖ (RF +RE1)] + re3 +RC2

hfe + 1

When evaluated Ro = 143.9Ω

∴ Rof = 143.9(1 + 20.7× 11.9) = 35.6kΩ


Analysis of Feedback Amplifiers Current-Shunt (Current Amplifier) Feedback

Current-Shunt (Current Amplifier) Feedback

Ii

Rif

Rof R’of

Io+

-

Vi Ri Ro RL

+

-

Vo

IoßIo

IsAiIi

Ai is the short-circuit current gain taking Rs into accountInput Resistance:

Is = Ii + βIo ; Io = AiIiRo

RL +Ro

taking AI = Ai(Ro/(Ro +RL)), where AI is current gain withoutfeedback taking the load into account.



Current-Shunt (Current Amplifier) Feedback

Rif =ViIs

=RiIi

Ii + βAIIi=

Ri1 + βAI

Output Resistance: making Is = 0 and replacing the load with asource

Ix =VxRo−AiIi ; Ii = −If = −βIo = βIx

Ix =VxRo− βAiIx ;

VxRo

= Ix(1 + βAi)

∴ Rof =VxIx

= Ro(1 + βAi)

The output resistance with load

R′of = Rof ‖ RL = (Ro ‖ RL)

1 + βAi1 + βAI



Current-Shunt (Current Amplifier) FeedbackFigure: Finding the A circuit and β for the Current-Shunt feedback amplifier.


Analysis of Feedback Amplifiers Voltage-Shunt (Transresistance Amplifier) Feedback

Voltage-Shunt (Transresistance Amplifier) Feedback

Ii

Rif

Rof R’of

Io+

-

Vi Ri

Ro

RL

+

-

VoIs

+

− RmIi

ßVo

Rm is the open-circuit transresistance gain taking Rs into accountInput Resistance:

Is = Ii + βVo ; Vo = RmIiRL

RL +Ro

taking RM = Rm(RL/(Ro +RL)), where RM is transresistance gainwithout feedback taking the load into account.



Voltage-Shunt (Transresistance Amplifier) Feedback

Rif =ViIs

=Vi

Ii + βRMIi=

Ri1 + βRM

Output Resistance: making Is = 0 and replacing the load with asource

Ix =Vx −RmIi

Ro; Ii = −If = −βVo = −βVx

Ix =Vx +RmβVx

Ro;

VxRo

=Ix

(1 + βRm)

∴ Rof =VxIx

=Ro

1 + βAi

The output resistance with load

R′of = Rof ‖ RL =

Ro ‖ RL1 + βRM



Voltage-Shunt (Transresistance Amplifier) FeedbackFigure: Finding the A circuit and β for the Voltage-Shunt feedback amplifier.


Exercise

Exercise

The following questions in the text book are exercises to be done forthe tutorial session.

10.36

10.52

10.57

10.65


Documents

Applied Electronics II · Current-Shunt (Current Ampli er) Feedback Voltage-Shunt (Transresistance Ampli er) Feedback 6 Exercise Chapter 1: Feedback Ampli ers SECE March 2017 2