Negative Feedback System

Chapter 12 Feedback

¾ 12.1 General Considerations

¾ 12.2 Properties of Native Feedback

¾ 12.3 Types of Amplifiers

¾ 12.4 Sense and Return Techniques

¾ 12.5 Polarity of Feedback

¾ 12.6 Feedback Topologies

¾ 12.7 Effect of Finite I/O Impedances

Negative Feedback System

¾ A negative feedback system consists of four components:

¾ 1) feedforward system

¾ 2) sense mechanism

¾ 3) feedback network

¾ 4) comparison mechanism

Close-loop Transfer Function

X

= KY

1

( )

( )

Y A X X

A X KY

= −

= −

Close-loop Transfer Function

1 1

1 KA A X

Y

= +

1

1 1

( )

Y A X KY Y A KY A X

= −

+ =

Example 12.1: Feedback

¾ A₁ is the feedforward network,

R₁ and R₂ provide the sensing and feedback capabilities, 1 2

1

2 1

1 A

R R

R A X

Y

+ +

=

2

1 2

K R

R R

= +

K

Comparison Error

¾ As A₁K increases, the error between the input and feedback signal decreases. Or the feedback signal approaches a

good replica of the input.

0

E ≈

X ≈ X

1 1

E X X F

X

A K

= −

= +

X

= KY

1

1 1

Y A X

= KA

∵ +

1

1

X KA X

= KA

+

Comparison Error

2

1 1

R R X

Y ≈ +

2

1 2

F

Y R X

R R =

+ ^{∵ X ≈ X}

= K

1

= K

Loop Gain

¾ When the input is grounded, and the loop is broken at an arbitrary location, the loop gain is measured to be KA₁.

test N

V KA

= − V

= 0

X

Example 12.3: Alternative Loop Gain Measurement

test

N KA V

V = − ₁

= 0

X

Incorrect Calculation of Loop Gain

¾ Signal naturally flows from the input to the output of a feedforward/feedback system. If we apply the input the

other way around, the “output” signal we get is not a result of the loop gain, but due to poor isolation.

Gain Desensitization

¾ A large loop gain is needed to create a precise gain, one that does not depend on A₁, which can vary by ±20%.

1

K >> 1

A X K

Y 1

≈

F

Y Y

X X

∴ ≈

0 E ≈

if 1 A K

X X

∴ ≈ 1

E X X

X A K

= −

= +

X

= KY

ex: g_mR_D m ox ⁽ GS TH⁾

g C W V V

μ L

= −

Ratio of Resistors

¾ When two resistors are composed of the same unit resistor, their ratio is very accurate. Since when they vary, they will vary together and maintain a constant ratio.

2

1

R R X

Y ≈ +

Example 12.4: Gain Desensitization

1 4 K =

Nominal gain

1

1

A Y

X = A K +

3.984 Y

X = ^{Y 3.968}

X =

( A

= 1000) ( A

= 500)

¾ Determine the actual gain if A₁=1000. Determine the percentage change in the gain if A₁ drops to 500.

Merits of Negative Feedback

¾ 1) Bandwidth enhancement

¾ 2) Modification of I/O Impedances

¾ 3) Linearization

1 1

1 KA A X

Y

= +

Bandwidth Enhancement

¾ Although negative feedback lowers the gain by (1+KA₀), it also extends the bandwidth by the same amount.

( )

1

0

1 A s A

s

ω

=

+ ( )

(

)

0 0

1 1

1

ω KA

s KA A X s

Y

+ +

= +

Open Loop Closed Loop

Negative Feedback

1

1 1

A Y

X = KA

∵ +

0

: -3 dB BW

ω

0

0 0 0

0 0

0 0 0

0

1 1

1 1

1 (1 )

1 A

s A

A KA

Y

A s s

X K KA KA

s ω

ω ω

ω

+ +

= = =

+ + +

+ +

+

∵

gain

bandwidth

Bandwidth Extension Example

¾ As the loop gain increases, we can see the decrease of the overall gain and the extension of the bandwidth.

Example 12.6: Unity-gain bandwidth

¾ We can see the unity-gain bandwidth remains independent of K, if KA₀ >>1 and K²<<1

0 dB

0

0 2

2 2

0 0

( ) 1

1 (1 )

A

KA Y j

X

KA

ω ω

ω

= +

+ +

0

0 2

2 2

0 0

1 1

1 (1 )

u

A

KA KA

ω ω

= +

+ +

2 2

0 0 0

2 2 2

0 0 0

0 0

(1 )

u A KA

A K A

A

ω ω

ω ω

= − +

≈ −

≈

1<< KA0

2 1

K <<

Example12.7: Open Loop Parameters

D out

m in

D m

R R

R g

R g

A

=

=

≈

1

0

0

?

?

?

in out

A R R

=

=

=

Example12.7: Closed Loop Voltage Gain

1 2

2

Assuming ,

1

D

out m D

in

R R R

v g R

v R

g R + >>

= +

( )

D m G S

i = g v − v

i

1 2

because of

m D

D

A g R

R R R

≈

+ >>

0 0

1

out in

v A

v = KA

+

Example12.7: Closed Loop I/O Impedance

⎟⎟ ⎠

⎜⎜ ⎞

⎝

⎛ + +

=

m

in

g R

R R

R R g

2 1

1

1

D m D

out

R R g

R R R R

2 1

1

+ +

=

i

2

1 2

( i R

) R R R

−

+

D D

≈ i R

+

−

2

1 2

X D

D m GS m X

i R R

i g v g v

R R

⎛ ⎞

= = ⎜ ⎝ + − ⎟ ⎠

D X

i = − i

2

1 2

v R

R + R i

D m GS

i = g v

X

D

v

R

X D X

/

i ≈ + i v R

Example: Load Desensitization

3

D

/

m D

m

R g R

g →

D m D m

D m D m

R R g

R R

R g

R R g

R R

R g

2

2

3

1 + +

→ + +

W/O Feedback Large Difference

With Feedback Small Difference

2

1 2

( / 3)

1 ( / 3)

m D

m D

g R

R g R R R

+ +

Before feedback

Linearization

Before feedback

After feedback

1 1

1

Gain at

1 x A

= KA +

( )

1

1 1 1 1

1 / 1 1 1 1 1 1

1 1

1 / 1 1 1 /

K

KA K KA K KA K KA

⎛ ⎞− ⎛ ⎞

= = ⎜ + ⎟ ≈ ⎜ − ⎟

+ + ⎝ ⎠ ⎝ ⎠

2 2

1 1

Gain at 1

1 x A

KA K KA

⎛ ⎞

= + ≈ ⎜⎝ − ⎟⎠

Four Types of Amplifiers

Current meter Voltage meter

Voltage source

Ideal Models of the Four Amplifier Types

(a) Voltage amplifier

(c) Transconductance amplifier (d) Current amplifier (b) Transresistance amplifier infinite

input impedance

zero output

impedance

R

= 0 R

= 0

R

= ∞ R

= ∞

R

= ∞

in

0

R =

Realistic Models of the Four Amplifier Types

Examples of the Four Amplifier Types

Sensing a Voltage

¾ In order to sense a voltage across two terminals, a voltmeter with ideally infinite impedance is used.

Sensing and Returning a Voltage

¾ Similarly, for a feedback network to correctly sense the output voltage, its input impedance needs to be large.

¾ R₁ and R₂ also provide a means to return the voltage.

∞

≈ +

1

R

R

Feedback Network

Sensing a Current

¾ A current is measured by inserting a current meter with ideally zero impedance in series with the conduction path.

¾ The current meter is composed of a small resistance r in parallel with a voltmeter.

Sensing and Returning a Current

¾ Similarly for a feedback network to correctly sense the current, its input impedance has to be small.

¾ R_S has to be small so that its voltage drop will not change I_out.

≈ 0 R

Feedback Network

Addition of Two Voltage Sources

¾ In order to add or substrate two voltage sources, we place them in series. So the feedback network is placed in series with the input source.

Feedback Network

e in F

v = v − v

Practical Circuits to Subtract Two Voltage Sources

¾ Although not directly in series, V_in and V_F are being

subtracted since the resultant currents, differential and

single-ended, are proportional to the difference of V_in and V_F.

Addition of Two Current Sources

¾ In order to add two current sources, we place them in

parallel. So the feedback network is placed in parallel with Feedback

Network

Practical Circuits to Subtract Two Current Sources

¾ Since M₁ and R_F are in parallel with the input current source, their respective currents are being subtracted. Note, R_F has to be large enough to approximate a current source.

e in F

i = − i i

Example 12.10: Sense and Return

¾ R₁ and R₂ sense and serve as the feedback network.

¾ M₁ and M₂ are part of the op-amp and also act as a voltage subtractor.

1

I

↑

2

I

↓

v

↑

v

↑

2

I

⇑

Voltage Return Voltage Sense

Example 12.11: Feedback Factor

mF

F g

v

K = i =

Voltage Sense

Current Return

v

↑ X

1

v

↓

2

v

↑

F X

i ↑⇒ v ⇓

Topics in Last and Today’s Lectures

¾ 12.3 Types of Amplifiers – Simple Amplifier Models

– Examples of Amplifier Types

¾ 12.4 Sense and Return Techniques

¾ 12.5 Polarity of Feedback

¾ 12.5 Feedback Topologies – Voltage-Voltage Feedback – Voltage-Current Feedback – Current-Voltage Feedback – Current-Current Feedback

Input Impedance of an Ideal Feedback Network

¾ To sense a voltage, the input impedance of an ideal feedback network must be infinite.

¾ To sense a current, the input impedance of an ideal feedback network must be zero.

Voltmeter Amperemeter

Output Impedance of an Ideal Feedback Network

¾ To return a voltage, the output impedance of an ideal feedback network must be zero.

¾ To return a current, the output impedance of an ideal feedback network must be infinite.

Voltage source Current source

¾ 1) Assume the input goes either up or down.

¾ 2) Follow the signal through the loop.

¾ 3) Determine whether the returned quantity enhances or opposes the original change.

Determining the Polarity of Feedback

Negative feedback Positive feedback

Example 12.12: Polarity of Feedback

in

↑

V I

↑ , I

↓ _V

_{↑ , V}

_↑ I

↑ , I

↓

Example 12.13: Polarity of Feedback

in

↑

V I

↑ , V

↓ _V

_{↑ , V}

_↑ I

↓ , V

↑

Example 12.14: Polarity of Feedback

in

↑

I I

↑ , V

↑ V

↓ , I

↓ I

↑ , V

↑

Voltage-Voltage Feedback

0 0

1 KA A V

V

in out

= +

1 in F

V = V − V

A V

=

F out

V = KV

0

( )

out in out

V A V KV

∴ = −

Example 12.15: Voltage-Voltage Feedback

1 2

2

A ssu m ing ( ),

( )

1 ( )

O N O P

out m N O N O P

in

R R r r

V g r r

V R

g r r

+ >>

=

+

( )

1 2

out

mN ON OP

in in

v g r r

v v ≈

∵ −

A

Input Impedance of a V-V Feedback

) 1

( A

K I R

V

in in

in

= +

¾ The impedance modification brings the circuit closer to an ideal voltage amplifier

( )

in in in F

in in in

I R V V

V I R A K

= −

= −

Example12.16: V-V Feedback Input Impedance

1 2

2

Assuming , 1 1

D in

m D

R R R

V R

I g R R g R + >>

⎛ ⎞

= ⎜ ⎝ + + ⎟ ⎠

) 1

( A₀K I R

V

in in

in = +

m

1

= g

Output Impedance of a V-V Feedback

( ^{1 R KA}

)

I

V

X X

= +

¾ A better voltage source

( )

X 0 X

X

out

V KA V

I R

= − −

KVX

KVX

−

0 X

−KA V

×

Example 12.17: V-V Feedback Output Impedance

1 2

,

2 1 2

1

A ssum ing ( ),

1 / ( ) ( )

1 1

O N O P

O N O P

out closed

m N O N O P

R R r r

r r

R R R R g r r

R

+ >>

= + + ⋅ ⋅

⎛ ⎞

≈ ⎜ + ⎟ (

)

X

X 0

R V

I = 1 KA

∵ +

Voltage-Current Feedback

1

out O

V R

I = KR +

in F

I I

= −

0

I R

=

( )

out in out

V I KV R

∴ = −

KV

=

Example 12.18: Voltage-Current Feedback

out in

Assuming is very large, open loop gain (V /I ):

( )

R

R = R − g ⋅ R

1 in D

i R

( )

0 1 2 2

out

D m D

in

V R R g R

i = = −

Example 12.18: Voltage-Current Feedback

1

Assuming is very large,

1 / 1 /

F

out out

RF

F m F

F

R

V V

I R g R

K R

= ≈

+

= −

∞

1 / g

I

F out

I = KV

∵

Example 12.18: Voltage-Current Feedback

2 1 2

2 1 2

1

out m D D

m D D

in closed

V g R R

g R R I

R

= − +

R

KR

if

out

F in closed

V R

I

g R R R

≈

>>

Input Impedance of a V-I Feedback

1

in X

X

R V

I = R K

+

¾ A better current sensor.

Current sensor or ammeter

X in

V

≈ R

in

V R

= R

0 X in

K V R

= R

0

X X

X

in in

V V

I K R

R R

− =

By KCL

Example 12.19: V-I Feedback Input Impedance

D D

m m closed

in

g g R R

R

2 1

1 2 ,

1 . 1 1

+

=

1

in X

X

R V

I = R K

∵ +

Output Impedance of a V-I Feedback

¾ A better voltage source.

K R

R I

V

X X

1 +

=

KV

= −

Input current source

KV

=

0

KV R

= −

×

Large

Input impedance

0

X X

out

V KV R R

= +

Example12.20: V-I Feedback Output Impedance

2 ,

2 1 2

1

D out closed

m D D

R R

g R R

=

+

1

o u t X

X

V R

I = R K

∵ +

2

if is very larg e

o u t D

F

R R

R

≈

1 in D

i R

( )

0

1 2 2

out in

D m D

V R

i

R g R

=

= −

∵

Current-Voltage Feedback

m m in

out

KG G V

I

= + 1

F

/

V I

=

( )

m in F

G V V

= −

KI

=

( )

out m in out

I = G V − KI

Example12.21: Current-Voltage Feedback

( )

3 3 5 1

| ||

out

m open m O O m

G = I = g ⋅ r r ⋅ g

Laser

3

(

)

m o o in

g r r V

−

+

−

[ ]

1 1 3

(

)

D m m o o in

i = g − g r r V

1

out D

I = − i

Example12.21: Current-Voltage Feedback

F

M out

K V R

= I =

Laser

out M

I R

=

Example12.21: Current-Voltage Feedback

( )

( )

1 3 3 5

1 3 3 5

|| 1

| 1 1 ||

m m O O

out m

closed

in m m m O O M M

g g r r

I G

V = K G = g g r r R ≈ R

+ ⋅ +

Laser

Topics in Last and Today’s Lectures

¾ 12.5 Polarity of Feedback

¾ 12.5 Feedback Topologies – Voltage-Voltage Feedback – Voltage-Current Feedback – Current-Voltage Feedback – Current-Current Feedback

¾ 12.6 Effect of Finite I/O Impedances – Inclusion of I/O Effects

¾ 12.7 Stability in Feedback Systems

Input Impedance of a I-V Feedback

¾ A better voltage sensor.

(1 )

in

in m

in

V R K G

I = +

in F

in in

V V I R

= −

=

m in in

G I R

=

F out

m in in

V KI

KG I R

=

=

in in in F

in m m in

I R V V

V KG I R

= −

= −

Output Impedance of a I-V Feedback

¾ A better current source.

) 1

(

out X

X

R KG

I

V = +

+

−

m X

− KG I

By KCL

X

X m X

out

I V KG I

= R −

F X

V = KI

F X

K V

= I

Laser

Example: Current-Voltage Feedback

) 1

1 (

|

) 1

1 (

|

| 1

2 1

2

2 1

1

2 1

2 1

M D

m m

m closed

out

M D

m m

m closed

in

M D

m m

D m

m closed

in out

R R

g g g

R

R R

g g g

R

R R

g g

R g

g V

I

+

=

+

=

= +

Very small

1 in m D

V g R = g

( V g R

) for open loop circuit

1 2

m m D m

G g R g

∴ = ⋅

F

M out

K V R

= I =

∵

1 / g

1 / g

out m

I G

= +

∵ V

(1

)

R K G

= +

∵ V

(1

)

R K G

= +

∵

Wrong Technique for Measuring Output Impedance

¾ If we want to measure the output impedance of a I-V closed- loop feedback topology directly, we have to place V_X in

series with K and R_out. Otherwise, the feedback will be disturbed.

I

Output current sensing

Current-Current Feedback

I

out

A

I =

( )

I in F

A I I

= −

F

/

I I

=

in F

I I

= −

KI

= I

= A I

(

− KI

)

Input Impedance of I-I Feedback

¾ A better current sensor.

I in X

X

KA R I

V

= + 1

I X

/

A V R

=

F I X

/

I = KA V R

X

X F

in

X X

I

in in

I V I

R

V V

R KA R

= +

= +

Output Impedance of I-I Feedback

¾ A better current source.

) 1

(

out X

X

R KA

I

V = +

KI

=

I X

− KA I

By KCL

X

X I X

out

I V KA I

= R −

High R

Example 12.24: Test of Negative Feedback

in

↑

I V

↑ , I

↓ _V

_{↓ , I}

_↑ V

↓ , I

↑

Laser

Example 12.24: I-I Negative Feedback

2 2

|

1

out m X m D in

I open m D

in in in

F P M

I g V g R I

A g R

I I I

I V R

K I R I R

− −

= = = = −

= = − ⋅ = −

Laser

X in D X

V = I R

2 2

D m X

I = g V

2 2

out m X

m D in

I g V

g R I

= −

= −

P M

V

= R

Very small

P out M

V = I R

Example: I-I Negative Feedback

)]

/ (

1 [

|

) /

( 1

. 1

| 1

) /

(

| 1

2 1

2

2

F M

D m

m closed

in

F M

D m

D m

closed I

R R

R g

r R

R R

R g

R g

R R

R g

R A g

+

=

= + +

= −

Laser

1

out I

in I

I A

I = KA

∵ +

1

in X

X I

V R

I = KA

∵ +

(1

)

X

V R KA

I = +

∵

Summary

¾ Determine whether the returned quantity enhances or opposes the original change.

Negative feedback Positive feedback

Voltage-Voltage Voltage-Current

Current-Voltage Current-Current

0 0

1 KA A V

V

in out

= + R (1 A₀K) I

V

in in

in = +

(

)

I

V _out

X X

= +

1

out O

in O

V R

I = KR

+ 1 ₀

X in X

R V

I = R K +

K R R I

V _out

X X

1 + 0

=

m m in

out

KG G V

I

= +

1 ^{(1 )}

in

in m

in

V R K G

I = +

I I in

out

KA A I

I

= +

1

in X

X

KA R I

V

= +

1

How to Break a Loop

¾ The correct way of breaking a loop is such that the loop does not know it has been broken. Therefore, we need to present the feedback network to both the input and the output of the feedforward amplifier.

Rules for Breaking the Loop of Amplifier Types

V-V V-I

I-V I-I

Z_out= low

Return Duplicate

Z_in= high

Z_out= low

Z_in= low

Z_in= low Z_in= high

Z_out= high Z_out= high

Sense Duplicate

Intuitive Understanding of these Rules

¾ Since ideally, the input of the feedback network sees zero impedance (Z_out of an ideal voltage source), the return

replicate needs to be grounded. Similarly, the output of the feedback network sees an infinite impedance (Z_in of an ideal voltage sensor), the sense replicate needs to be open.

¾ Similar ideas apply to the other types.

Voltage-Voltage Feedback

Rules for Calculating Feedback Factor

V-V V-I

I-V I-I

Intuitive Understanding of these Rules

¾ Since the feedback senses voltage, the input of the

feedback is a voltage source. Moreover, since the return quantity is also voltage, the output of the feedback is left open (a short means the output is always zero).

¾ Similar ideas apply to the other types.

Voltage-Voltage Feedback

Example 12.26: Breaking the Loop

( )

[ ]

( )

1 ,

2 1

1 ,

/ 1

||

g R

R R

R g

A

m open

in

D m

open v

=

+

=

Example 12.26: Feedback Factor

) 1

/(

) 1

(

) 1

/(

) /(

, ,

,

, ,

,

, ,

,

2 1

2

open v

closed out

closed out

open v

open in

closed in

open v

open v

closed v

KA R

R

KA R

R

KA A

A

R R

R K

+

=

+

=

+

=

+

=

Example 12.27: Breaking the Loop

( )

[ ]

( )

,

2 1

,

|| ||

R

R R

r r

g A

open in

OP ON

mN open

v

+

=

∞

=

+

=

Example 12.27: Feedback Factor

) 1

/(

) 1

/(

) /(

,

, ,

,

2 1

2

closed in

open v

open v

closed v

KA R

R R

KA A

A

R R

R K

+

=

∞

=

+

=

+

=

Example 12.29: Breaking the Loop

( )

[ ]

F m

open in

F D

m

m F

D F

open in

out

g R R

R R

g R g

R R I

V

1 ||

||

1 .

|

1 ,

2 2

1 1

=

− +

=

Example 12.29: Feedback Factor

)

| 1

/(

)

| 1

/(

)

| 1

/(

|

|

/ 1

, ,

out open in

out open

in closed

in

open in

out open

in out closed

in out

F

K V R

R

I K V R

R

I K V I

V I

V

R K

+

=

+

=

+

=

−

=

1 2 F

V = − I R

2 1

K I

= V

Example 12.30: Breaking the Loop

( )

open in

M L

O

O m O

O m

open in

out

R

R R

r

r g

r r

g V

I

+

=

∞

=

+

= +

,

1

1 1 5

3

3

||

|

1

ro

RL

RM

( )

3 3 5

X

m o o

in

V g r r

V = −

1 m X

g V

( )

1 1

1

o m X

out

o L M

r g V I = − r R R

+ +

M 2 1

out in closed out in open out in open

in ,closed

out ,closed out ,open out in open

K R V / I

( I / V | ) ( I / V | ) / [ 1 K ( I / V ) | ] R

R R [ 1 K ( I / V ) | ]

= =

= +

= ∞

= +

Example 12.30: Feedback Factor

Example 12.31: Breaking the Loop

m open

in

m M

L

D m

open in

out

g R

g R

R

R g

V I

=

+

= + / 1

/

| 1

1 ,

2 1

1 X

m D

in

V g R V =

A

1 / 2

A L M

X L M m

V R R

V R R g

= +

+ +

A out

I = V

+

out open ?

R =

RM

VX

+

− IX

Example 12.31: Feedback Factor

]

| ) /

( 1

[

]

| ) /

( 1

[

]

| ) /

( 1

/[

)

| /

( )

| /

(

,

,closed in open out in open

in

open in

out open

in out

closed in

out

M

V I

K R

R

V I

K R

R

V I

K V

I V

I

R K

+

=

+

=

+

=

=