dr. eman azab microelectronics elct 703 …eee.guc.edu.eg/courses/electronics/elct703...series-shunt...

MICROELECTRONICS ELCT 703 (W19) LECTURE 2: FEEDBACK AMPLIFIERS

Dr. Eman Azab

Assistant Professor

Office: C3.315

E-mail:

[email protected]

1

INTRODUCTION

Negative Feedback technique is applied to Electronic circuitsamplifiers for the following benefits:

To Stabilize the Gain against device parameters variation and aging

To increase/decrease the amplifier’s input/output Resistance

Increase the amplifier’s Bandwidth

Reduce distortion

However, it has some drawbacks:

Gain Reduction

Tendency for Oscillation (Unstable system?)

DR. EMAN AZAB

ELECTRONICS DEPT., FACULTY OF IET

THE GERMAN UNIVERSITY IN CAIRO

2

INTRODUCTION

Feedback System Structure:

The system consists of a basic Amplifier with gain ‘A’ anda feedback network with gain ‘β’

The input, feedback and output signals ‘X’ can bevoltage or current

DR. EMAN AZAB



3

FEEDBACK SYSTEM Feedback System Equations:

DR. EMAN AZAB



4

o

i

XA

X fB

o

X

X

i S fB S oX X X X X

1

oFB

S

X AA

X A

Closed-loop Gain

T A Loop Gain

1FBA

For T>>>>1

FEEDBACK SYSTEM

Gain Sensitivity to Parameter Variations

What is the sensitivity of the closed-loop gain ‘AFB’ to any variationsin the basic amplifier’s gain ‘A’?

DR. EMAN AZAB



5

FB

FB

dASensitivity

A

2 2

1 1

1 1

FBA AdA

dA A A

2 2

11

1 1

FB

FB FB

AdA dA dA

A A AA A

1

1

FB

FB

dA dASensitivity

A A T

FEEDBACK SYSTEM

Gain Sensitivity to Parameter Variations

What is the sensitivity of the closed-loop gain ‘AFB’ to any variationsin the basic amplifier’s gain ‘A’?

Any variation in the basic amplifier’s gain A will lead to a changein the closed-loop gain divided by (1+T)

For this reason the feedback amplifiers are more immune todistortion as well

DR. EMAN AZAB



6

1

1

FB

FB

dA dASensitivity

A A T

FEEDBACK SYSTEM Bandwidth of the Feedback Amplifier

What is the value of the higher 3-dB frequency of the closed-loopgain ‘AFB’?

Feedback increases the Bandwidth of the Amplifier

DR. EMAN AZAB



7

1

MFHF

H

AA

s

Assume that the basic Amplifiers gain is:

1

1

1

MF

HFB

MF

H

A

s

AA

s

1

11

MF

FB

H

A

TAs

T

, 1H FB H T

FEEDBACK CONFIGURATIONS

DR. EMAN AZAB



8

FEEDBACK CONFIGURATIONS

There are Four basic Feedback Configurations (Connections)

They are categorized based on the type of the source and output signals (Voltageor Current)

Important Note: Source Signal must be the same type as the feedback signal

DR. EMAN AZAB



9

Source & Feedback

Signal Output Signal

Feedback

Configuration (Mixing-

Sampling)

Amplifier’s Type

Voltage Voltage Series-Shunt Voltage

Current Current Shunt-Series Current

Voltage Current Series-Series Trans-conductance

Current Voltage Shunt-Shunt Trans-resistance

SERIES-SHUNT CONNECTION

The feedback network is connected in series with the I/P node andshunt with the O/P node

The feedback voltage signal is proportional to the output voltage

This configuration is used for Voltage Amplifiers

DR. EMAN AZAB



10

𝐴𝑣,𝐹𝐵 =𝑣𝑜𝑣𝑠

=𝐴𝑣

1 + 𝐴𝑣𝛽

SERIES-SHUNT CONNECTION

What is the effect of Series-Shunt feedback on the Input and OutputResistances of the voltage amplifier?

Assume the basic amplifier has finite Rin and Rout

DR. EMAN AZAB



11

,s i fb

in FB

i i

v v vR

i i

,

1iin FB

i

v AR

i

, 1in FB inR R A

Series mixing at the input

node increases the Input

Resistance of the Amplifier

SERIES-SHUNT CONNECTION What is the effect of Series-Shunt feedback on the Input and Output

Resistances of the voltage amplifier?

Short Circuit the Source voltage and add a VT at the output

DR. EMAN AZAB



12

,T

out FB

T

VR

I T i

T

out

V AvI

R

Shunt sampling at the

output node decreases the

Output Resistance of the

Amplifier

i fb Tv v V

,1

outout FB

RR

A

SHUNT-SERIES CONNECTION

The feedback network is connected in parallel with the I/P node andin series with the O/P node

The feedback current signal is proportional to the output current

This configuration is used for Current Amplifiers

DR. EMAN AZAB



13

𝐴𝑖,𝐹𝐵 =𝑖𝑜𝑖𝑠=

𝐴𝑖1 + 𝐴𝑖𝛽


What is the effect of Shunt-Series feedback on the Input andOutput Resistances of the Current amplifier?

Assume the basic amplifier has finite Rin and Rout

DR. EMAN AZAB



14

,1

s sin FB

s i

v vR

i i A

,1

inin FB

RR

A

Shunt mixing at the input

node decreases the Input



What is the effect of Shunt-Series feedback on the Input andOutput Resistances of the Current amplifier?

Open circuit the current Source

DR. EMAN AZAB



15

,T

out FB

T

VR

I T

T T

out

VI A I

R

Series sampling at the

output increases the Output


i fb Ti i I

, 1out FB outR R A

SERIES-SERIES CONNECTION

The feedback network is connected in series with the I/P nodeand in series with the O/P node

The feedback voltage signal is proportional to the output current

This configuration is used for Trans-conductance Amplifiers

DR. EMAN AZAB



16

Note: G and β are not

dimensionless

, 1out FB outR R A

, 1in FB inR R A

𝐺𝐹𝐵 =𝑖𝑜𝑣𝑠

=𝐺

1 + 𝐺𝛽

SHUNT-SHUNT CONNECTION

The feedback network is connected shunt with the I/P node andshunt with the O/P node

The feedback current signal is proportional to the output voltage

This configuration is used for Trans-resistance Amplifiers

DR. EMAN AZAB



17

Note: R and β are not

dimensionless

,1

outout FB

RR

A

,1

inin FB

RR

A

𝑅𝐹𝐵 =𝑣𝑜𝑖𝑠=

𝑅

1 + 𝑅𝛽

FEEDBACK CIRCUITS ANALYSIS

DR. EMAN AZAB



18

FEEDBACK CIRCUIT ANALYSIS

Solution Steps:

1. Identify the Feedback Configuration type whether it isseries or shunt at the Input and Output nodes of thebasic open loop amplifier

2. Calculate the feedback network gain β The source signal contribution is neglected while calculating β

If the feedback network at the input node is shunt, then consider the inputvoltage to be zero and calculate the feedback current

If the feedback network at the input node is series, then consider the inputcurrent to be zero and calculate the feedback voltage

Irrespective of the feedback network connection at the input node, if it isconnected shunt/series to the output node, then drive the feedbacknetwork with a voltage/current Source

DR. EMAN AZAB



19


Solution Steps (Cont.):

3. Redraw the basic Open loop Amplifier to find the gain‘A’

Redraw the input loop with the output signal driving thefeedback network deactivated

If the feedback network is connected shunt to the output node, thenshort circuit the output node to ground.

If the feedback network is connected series to the output node, thenopen circuit the output node.

Redraw the output loop with the input signal driving thefeedback network deactivated

If the feedback network is connected shunt to the input node, then shortcircuit the input node to ground.

If the feedback network is connected series to the input node, then opencircuit the input node

DR. EMAN AZAB



20


Solution Steps (Cont.):

4. Calculate the open loop amplifier’s gain, inputresistance and output resistance.

5. Use the feedback rules to calculate the amplifier’sclosed loop gain, closed loop input resistance andclosed loop output resistance.

DR. EMAN AZAB



21

SOLVED EXAMPLES

DR. EMAN AZAB



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EXAMPLE ONE

Find the closed loop Gain, closed loop Input andOutput Resistances of the amplifier?

Solution:

1. Feedback Network is connected shunt to the input node and shuntto the output node

Shunt-Shunt Configuration

DR. EMAN AZAB



23

EXAMPLE ONE

Solution:

2. Calculating β

DR. EMAN AZAB



24

fb

o

i

v

i o ofb

F F

v v vi

R R

1

FR

EXAMPLE ONE

Solution:

3. Redraw the basic Amplifier

Since we had shunt-shunt, we will short circuit the input andoutput nodes

DR. EMAN AZAB



25

EXAMPLE ONE

Solution:

4. Calculate the basic amplifier’s gain, input and outputResistances

DR. EMAN AZAB



26

o

i

vA

i

/ /

/ /

F Lo v i

F L out

R Rv a V

R R R

/ /i F in iV R R i

/ /

/ // /

F Lv F in

F L out

R RA a R R

R R R

, / /in A F inR R R

𝑅𝑜𝑢𝑡,𝐴 = 𝑅𝑜𝑢𝑡//𝑅𝐹//𝑅𝐿

EXAMPLE ONE

Solution:

4. Calculate the closed loop amplifier’s gain, closed loop inputand output Resistances.

DR. EMAN AZAB



27

/ // /

/ /

/ / 11 / /

/ /

F Lv F in

F L outFB

F Lv F in

F L out F

R Ra R R

R R RA

R Ra R R

R R R R

,

/ /

1

F inin FB

R RR

A

𝑅𝑜𝑢𝑡,𝐹𝐵 =

𝑅𝑜𝑢𝑡//𝑅𝐹//𝑅𝐿1 + 𝐴𝛽

EXAMPLE TWO

Find the closed-loop Gain, Input and OutputResistances of the amplifier?

Solution:

1. Feedback Network is connected in series to the input and outputloop

Series-Series Configuration

2. Calculating β

DR. EMAN AZAB



28

2 13

1 2

fb E E

o E E F

v R R

i R R R

1 1E

fbR E

o

vi R

i

1

2

1 2E

ER o

E E F

Ri i

R R R

EXAMPLE TWO

Solution (cont.):

3. Redraw the basic Amplifier

Since we had series-series, we will open circuit the input and output node

DR. EMAN AZAB



29

EXAMPLE TWO

Solution (Cont.):

4. Calculate the basic amplifier’s gain, input and output Resistances

4. Calculate the closed-loop gain, , input and output Resistances

DR. EMAN AZAB



30

o

i

iA

v

Assignment!

dr. eman azab microelectronics elct 703 …eee.guc.edu.eg/courses/electronics/elct703...series-shunt...

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