Chapter_1_Student

Applied Electronics II

Chapter 1: Feedback Amplifiers

School of Electrical and Computer Engineering

Addis Ababa Institute of Technology

Addis Ababa University

Daniel D./Abel G.

March 7,2016

Chapter 1: Feedback Amplifiers (AAIT)

Chapter One

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Overview

Overview

1

2

3

4

5

6

Overview

Types of Feedback

The General Feedback Structure

Basic Feedback Amplifier

Feedback Topologies

Properties of Negative Feedback

Gain Desensitivity

Noise/Interference Reduction

Reduction of Nonlinear Distortion

Control of Impedance level & Bandwidth Extension

Analysis of Feedback Amplifiers

Voltage-Series (Voltage Amplifier) Feedback

Method of Analysiis of Feedback Amplifiers

Current-Series (Transconductance Amplifier) Feedback

Current-Shunt (Current Amplifier) Feedback

Voltage-Shunt (Transresistance Amplifier) Feedback

7

Exercise


Chapter One

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Type of Feeedback

Types of Feedback

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

1 Posittive Feedback

A portion of the output signal is added to the input. Positve feedback is used in the design of oscillator and a number of other applications

(will be discussed in Chapter 4 and 5).

2 Negative Feedback

A portion of the output signal is subtracted from the input signal.The

basic idea of negative feedback is to trade off gain for other desirable properties listed below

Desensitize the gain

Reduce nonlinear distortion

Reduce the effect of noise

Control the input and output resistances

Extend the bandwidth of the amplifier.


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Negeative Feedback

Types of Feedback

Example

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

Solution Apply KCL at B-E loop

R

C

V

+

I

C

V

BB

= I

B

R

B

+ V

BE

( on ) + I

E

R

E

+ V

−

Assuming active-mode of operation

V

BB

I

B

R

B

+

V

BE

-

-

+

V

CE

R

E

I

E

As I

C

I

E

= (1 + β ) I increases(due to voltage drop across R

E

B

↑ and I

C

= β I

B in T, ageing ), the increase thus opposing the base-emitter voltage.

(1)

(2)

V

-


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

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

The relationship between the quantities x is

Chapter 1: Feedback Amplifiers (AAIT) x o

= Ax i x x i f

= β x o

= x s

− x f

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March 7,2016 5 / 45


Feedback Systems

Thus x o

= A ( x s

− β x o

)

The gain with feedback , A f

A f

= x o x s

A

=

1 + β A

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

if | A f

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

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

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

1

A f u

β


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The General Feedback Structure Basic Feedback Amplifier


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

Signal

Source

Comparato r or Mixer

Network

+

V i

-

I i

I f

Basic

Amplifier, gain A

I

+

V

-

Sampling

Network

I o

= I

L

+

V o

-

R

L

Feedback

Network b

Signal Source : This block is a voltage source V s with a series resistor

R s

(Thvenin’s equivalent circuit) or a current source I s with a parallel resistor R s

(Norton’s equivalent circuit)

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


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Sampling Network : Sampling blocks are shown below

Sampler

A R

L

A

Sampler

R

L b b

Figure: (a) Voltage or node sampling Figure: (b) Current or loop sampling

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

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


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Comparator or Mixer Network : Two types a series (loop) and shunt

(node). A differential amplifier is often used as mixer.

Source

Series

Mixer

Source

Shunt

Mixer

R s

V s

+

V i

-

A I s

R s

I i

A

I f

+

V f

b b

Figure: (a) Series Mixing Figure: (b) Shunt Mixing


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Basic Amplifier : A could be used to represent

V

V i

I

I i

I

V i

= A

V

, Voltage gain

= A

I

, Current gain

= G

M

, Transconductance

V

I i

= R

M

, Transresistance

They are gain of the basic amplifier without feedback


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Feedback Topologies

Feedback Topologies

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


Chapter One

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Feedback Topologies

Feedback Topologies

Figure: (a) Series-Shunt

Figure: (c) Series-Series

Figure: (b) Shunt-Series


Chapter One

Figure: (d) Shunt-Shunt

March 7,2016 12 / 45

Properties of Negative Feedback Gain Desensitivity

Gain Desensitivity

Variation in the circuit gain as a result of change in transistor parameters is reduced by negative feedback

From the previous slides the gain with feedback , A f is given as

A f

= x o x s

A

=

1 + β A

Assuming β is constant and taking the derivative of A f with respect to A , dA dA f

=

1

1 + β A

−

A

(1 + β A ) 2

β =

1

(1 + β A ) 2 or dA f

= dA

(1 + β A ) 2

Dividing both sides the gain with feedback yields dA f

A f

= dA

(1+ β A )

2

A

1+ β A

1

=

1 + β A dA

A


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Properties of Negative Feedback Gain Desensitivity

Gain Desensitivity

Hence the percentage change in A f

(due to variations in some circuit parameter) is smaller than the percentage change in A by a factor equal to the amount of feedback. For this reason, the amount of feedback, 1 + A β , is also known as the desensitivity factor .

Example

The open-loop gain of an amplifier is A = 5 × 10 4 V / V exhibits a gain change of 25% as the operating temperature changes. Calculate the percentage change if the closed loop gain A f

= 50 V / V .

dA f

A f

1

=

1 + β A dA

A

A

=

A (1 + β A ) dA

A

=

A f

A dA

A

=

50

5 × 10 4

× 25% dA f

A f

= 0 .

025%


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Properties of Negative Feedback Noise/Interference Reduction

Noise/Interference Reduction

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

This can be achieved if a preamplifer which is (relatively) noise/interference-free precessed the noise/interference-prone amplifier

Under 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


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Properties of Negative Feedback Reduction of Nonlinear Distortion

Reduction of Nonlinear Distortion

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

Negative feedback can be implemented to reduce nonlinear distortion by a factor of 1 + A β .

Assuming that the open-loop gain A β 1 , the gain with feedback

A

A f

=

1 + A β

1 u

β

It implies that A f is independent of the nonlinear properties of the transistors 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.


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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 be increased or decreased with the proper type of negative feedback circuit.

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

The advantage of negative feedback is at the cost of gain. Under certain circumstance, a negative feedback amplifier may become unstable and break into oscillation.


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Analysis of Feedback Amplifiers

Fundamental Assumtions

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

Input is transmitted through the amplifier only, not through the feedback.

The feedback signal transmitted feedback network only, not through the amplifier.

β is independent of the load and source impedance.

I i

R o

I o

+ +

V s

R i

+

−

A vo

V i

-

V o

R

L

- V f

V i

+

-

R if

R of

R’ of

ßV o

+

−

+

-

V o

Figure:

Ideal structure of a Voltage-Series feedback amplifier


Chapter One

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


A vo represents the open circuit voltage gain taking R s into account

Input Impedance: The input impedance with feedback is

R if

=

V s

I i

Also,

V s

= I i

R i

+ V f

= I i

R i

+ β V o and V o

= A vo

V i

R o

R

L

+ R

L let the

A

R v

L

= A vo

R o

R

L

+ R

L

, where into account then

A v is the voltage gain without feedback taking

V s

= I i

R i

+ β A v

V i

= I i

R i

+ β A v

I i

R i

R if

=

V

S

I i

= R i

(1 + β A v

)


Chapter One

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V f

+

V i

-

R i

+

−

A vo

V i

R o

I x

+

V x

ßV o

Figure: Ideal structure of a Voltage-Series feedback amplifier

Output Impedance: To find R of must remove the external signal (set

V s

= 0 or I s

= 0 ), let R

L

= ∞ , impress a voltage V x across the output terminals and calculate the current I x delivered by the test voltage V x

I x

=

V x

− A vo

V i

R o

Since V i

= − β V x

R of

=

V x

I x

=

R o

1 + β A vo


Chapter One

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0

The output resistance with feedback R of amplifier is R of k R

L which includes R

L as part of the

0

R of

R of

R

L

=

R of

+ R

L

=

R o

R

L

R of

1 + β A vo

+ R

L

R

L

R o

+ R

L

Taking R o

0

= R o k R

L

0

R of

=

R o

0

1 + β A v

Voltage gain with feedback: A vf taking the load into account.

V s

= V i

+ β V o

= V o

R o

+ R

A vo

R

L

L

+ β V o


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Manipulating the equation

A vf

=

V o

V s

=

A vo

1 + β A vo

R

R

L

R o

+ R

L

R

L o

+ R

L

The voltage gain with feedback without the load A vfo is

A vfo

=

V o

V s

=

A vo

1 + β A vo

In conclusion

Input Impedance: increased by a factor 1 + β A v output Impedance: decreased by a factor 1 + β A v

Voltage Gain: decreased by a factor 1 + β A v


Chapter One

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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 , R i

, and R o

.

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 its port 2

(because it is connected in shunt with the output).

3.

The loading effect at the output is obtained by open-circuiting port 1 of the feedback network (because it is connected in series with the input)

4.

The gain without feedback A is determined

5.

The feed back gain β is determined


Chapter One

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Figure: Finding the A circuit and β for the Voltage-Series feedback amplifier.


Chapter One

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Analysis 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 externally applied signal as a voltage b) Shunt mixing : If the feedback signal subtracts from the applied excitation signal as a current.

2.

Identify the sampled signal as series or shunt a) Voltage sampling : Set V o voltage sampling.

b) Current sampling : Set current sampling.

I o

= 0( R

L

= 0 . If X f

= 0( R

L

= ∞ . If X f becomes zero, we have becomes zero, we have

3.

The amplifier without feedback but taking the feedback network loading into account

1) Find the input circuit.

a) Set V o

= 0 for voltage sampling.

b) Set I o

= 0 for current sampling.


Chapter One

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2) Find the output circuit.

a) Set V i

= 0 for shunt comparison so that no feedback current enters the amplifier input.

b) Set I o

= 0 for series comparison so that no feedback voltage reaches the amplifier input.

4.

Find the feedback network β .

5.

Calculate β , A , R i and R o

.

6.

Calculate the closed loop A f

, R if

, R of

.


Chapter One

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V s

Example

Analyze the amplifier to obtain its voltage gain V o

/ V s

, input resistance

R in

, and output resistance R out

. Find numerical values for case g m 1

= g m 2

= 4 mA / V , R

D 1

= R

D 2

= 10 k

Ω and R

2

= k

Ω

. For simplicity, neglect r o of each of Q

1 and Q

2

.

R

D1

Q

1

R

2

R

D2

Q

2

V o

R out

The next step is identifying the A and

β circuit

We identify the feedback network as the voltage divider of ( R

1

, R

2

)

R

2

+

V f

R

1

+

V o

R in

R

1 -


Chapter One

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Example (Continued)

The A circuit is

+

R i

-

V i

V d1

R

D1

Q

1

R

1

R

2

Q

R

D2

2

R

2

V o

R

1

R out


Chapter One

Calculating A

1 and A

2

V d 1

= 0 − i d 1

R

D 1

V i

= V gs 1

+ i d 1

( R

1 k R

2

)

A

1

=

V d 1

V i

=

− i d 1

R

D 1

V gs 1

+ i d 1

( R

1 k R

2

)

A

1

=

− R

D 1

1 / g m 1

+ ( R

1 k R

2

)

A

1

=

− g m 1

R

D 1

1 + g m 1

( R

1 k R

2

)

March 7,2016 28 / 45



Example (continued)

From A circuit we have

V o

= 0 − i d 2

( R

D 2 k ( R

1

+ R

2

)) and V gs 2

= V d 1

A

2

=

V o

V d 1

=

− i d 2

( R

D 2 k (

V gs 2

R

1

+ R

2

))

= − g m 2

( R

D 2 k ( R

1

+ R

2

))

The open loop gain is

A =

V o

V i

= A

1

A

2

= g m 1 g m 2

R

D 1

[ R

D 2

1 + g m 1

( R

1 k ( R

1

+ R

2

)] k R

2

)

When evaluated

A =

4 × 4 × 10[10 k (1 + 9)]

1 + 4(1 k 9)

= 173 .

913 V / V


Chapter One

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Example (continued) from β circuit we have

β =

R

1

R

1

+ R

2

1

=

1 + 9

= 0 .

1 V / V

The closed loop gain

V o

V s

= A f

A

=

1 + A β

=

173 .

913

1 + 173 .

913 × 0 .

1

= 9 .

45 V / V

The input resistance is infinite because it is the input resistance of

MOSFET.

The output resistance is

R out

= R f

=

R o

1 + A β

=

R

D 2 k ( R

1

+ R

2

)

1 + A β

=

10 k (1 + 9)

1 + 173 .

913 × 0 .

1

= 271 .

87

Ω


Chapter One

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


I i

I o

+ +

V s

- V f

+

V i

-

R i G m

V i

R o

-

V o

R

L

R if

R of

R’ of

ßI o

+

− I o

Input Impedance:

R if

=

V s

I i

; V s

= I i

R i

+ β I o

; I o

= G m

V i

R o

R o

+ R

L

R if

=

I i

R i

+ β G m

I i

R i

R o

R o

+ R

L

I i

= R i

(1 + β G m

R o

R o

+ R

L

)


Chapter One

March 7,2016 31 / 45



G m

= I o

/ V i is the short-circuit transconductance, and

G

M

= G m

R o

/ ( R o

+ R

L

) is the transconductance without feedback taking the load into account.

R if

= R i

(1 + β G

M

)

Output Impedance: calculated by short-circuiting the source and replacing the source with a voltage source V x with a current of I x

I x

=

V x

R o

− G m

V i and V i

= β I x

∴ R of

=

V x

I x

=

R o

( I x

+ G m

I x

β I x

)

= R o

(1 + β G m

)

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

0

R of

= ( R of k R

L

) = ( R o k R

L

)

1 + β G m

1 + β G

M


Chapter One

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Figure: Finding the A circuit and β for the Current-Series feedback amplifier.


Chapter One

March 7,2016 33 / 45

-

+

V s



Example

Calculate the closed loop voltage gain, output resistance and input resistance for the circuit below. The output is taken from emitter current of Q

3

. The values of

R

C 1

R

F

= 9 k Ω , R

C 2

= 5 k Ω , R

C 3

= 600 Ω , R

E 1

= 100 Ω , R

E 3

= 100 Ω and

= 640 Ω . Assume that the bias circuit, which is not shown, establishes

I

C 1

= 0 .

6 mA , I transistors, h fe

C 2

= 1 mA , and I

C 3

= 100 and r o

= ∞ .

= 4 mA . Also assume that for all three

R

C1

Q

1

R

E1

R

C2

Q

2

R

F

R

C3

Q

3

I o

R

E3

V o

The β circuit.

R

F

-

+

V f

R

E1

R

E3 I o

β =

V f

I o

=

[( R

F

+ R

E 1

) k R

E 2

] I o

R

F

R

E 1

+ R

E 1

I o

β =

R

F

R

E 1

× R

E 2

+ R

E 1

+ R

E 2

= 11 .

9 Ω


Chapter One

March 7,2016 34 / 45



Example (continued)

The A circuit.

When A β 1

R

C1

R

C2

R

C3

Q

3

A f u

1

β

=

1

11 .

9

Ω

= 84 mA / V

V o lets check by determining each transistor gain

Q

2

V i

R i

Q

1

R

E1

R

F

R

F

R

E3

R

E1

I o

R o

A

1

=

V c 1

V i

− i c

( R

C 1 k r

π 2

)

= i e

( r e 1

+ [ R

E 1 k ( R

E 3

+ R

F

)])

R

E3

A

1

=

− α ( R

C 1 k r

π 2

)

( r e 1

+ [ R

E 1 k ( R

E 3

+ R

F

)])

Since

Q

2

Q

1 is biased at 0.6mA , r is biased at 1mA; thus evaluating A

1 r

π 2 e 1

= 41 .

7

Ω

.

= h fe

/ g m 2


Chapter One

A

1

=

−

14

.

92

V

/

V

March 7,2016 35 / 45



Example (continued)

The gain of Q

2

A

2

=

V c 2

V b 2

=

− i c

[ R

C 2 k ( h fe

+ 1)[ r e 3

+ ( R

E 3 k ( R

F

+ R

E 1

))]]

V b 2

A

2

= − g m 2

[ R

C 2 k ( h fe

+ 1)[ r e 3

+ ( R

E 3 k ( R

F

+ R

E 1

))]] r e 3

= 25 / 4 = 6 .

25 Ω and substituting the other values

A

2

= − 131 .

2 V / V

The gain of Q

3

I o

A

3

=

V c 2

=

I e 3

V b 3

=

1 r e 3

+ ( R

E 3 k ( R

F

+ R

E 1

)) when evaluated

A

3

= 10 .

6 mA / V


Chapter One

March 7,2016 36 / 45



Example (continued)

The gain without feedback

A = A

1

A

2

A

3

= − 14 .

92 × − 131 .

2 × 10 .

6 × 10

− 3

= 20 .

7 A / V

The gain with feedback

A f

A

=

1 + A β

=

20 .

7

1 + 20 .

7 × 11 .

9

= 83 .

7 mA / V

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

R in

= R if

= R i

(1 + A β )

R i

= ( h fe

+ 1)[ r e 1

+ ( R

E 1 k ( R

F

+ R

E 2

))] = 13 .

65 k

Ω

∴ R if

= 13 .

65(1 + 20 .

7 × 11 .

9) = 3 .

38 M Ω


Chapter One

March 7,2016 37 / 45



Example (continued)

The output resistance

R of

= R o

(1 + A β )

R o

= [ R

E 3 k ( R

F

R

C 2

+ R

E 1

)] + r e 3

+ h fe

+ 1

When evaluated R o

= 143 .

9

Ω

∴ R of

= 143 .

9(1 + 20 .

7 × 11 .

9) = 35 .

6 k

Ω


Chapter One

March 7,2016 38 / 45

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


I s

I i

+

V i

-

R i

A i

I i

R o

I o

R of

+

-

V o

R

L

R’ of

R if

ßI o

I o

A i is the short-circuit current gain taking

Input Resistance:

R s into account

I s

= I i

+ β I o

; I o

= A i

I i

R o

R

L

+ R o taking A

I

= A i

( R o

/ ( R o

+ R

L taking the load into account.


)) , where A

I

Chapter One

is current gain without feedback

March 7,2016 39 / 45



R if

=

V i

I s

=

I i

R i

I i

+ β A

I

I i

=

R i

1 + β A

I

Output Resistance: making I s

= 0 and replacing the load with a source

I x

=

V x

R o

− A i

I i

; I i

= − I f

= − β I o

= β I x

I x

=

V x

R o

− β A i

I x

;

V x

R o

= I x

(1 +

∴ R of

=

V x

I x

The output resistance with load

= R o

(1 + β A i

)

β A i

)

0

R of

= R of k R

L

= ( R o k R

L

)

1 + β A i

1 + β A

I


Chapter One

March 7,2016 40 / 45



Figure: Finding the A circuit and β for the Current-Shunt feedback amplifier.


Chapter One

March 7,2016 41 / 45

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


I i

I o

I s

+

V i

-

R i

+

−

R m

I i

R o

+

-

V o

R

L

R of

R’ of

R if

ßV o

R m is the open-circuit transresistance gain taking R s into account

Input Resistance:

I s

= I i

+ β V o

; V o

= R m

I i

R

L

R

L

+ R o taking R

M

= R m

( R

L

/ ( R o

+ R

L

)) , where R

M is transresistance gain without feedback taking the load into account.


Chapter One

March 7,2016 42 / 45



R if

=

V i

I s

=

I i

V i

+ β R

M

I i

=

R i

1 + β R

M

Output Resistance: making I s

= 0 and replacing the load with a source

I x

=

V x

− R m

I i

R o

; I i

= − I f

= − β V o

= − β V x

I x

=

V x

+ R m

β V x

R o

;

V x

R o

I x

=

(1 + β R m

)

∴ R of

=

V x

I x

=

R o

1 + β A i

The output resistance with load

0

R of

= R of k R

L

=

R o k R

L

1 + β R

M


Chapter One

March 7,2016 43 / 45



Figure: Finding the A circuit and β for the Voltage-Shunt feedback amplifier.


Chapter One

March 7,2016 44 / 45

Exercise

Exercise

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

10.36

10.52

10.57

10.65


Chapter One

March 7,2016 45 / 45

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