General structure of the feedback amplifier

12 차시

8장 FEEDBACK, 전자회로

1. Desensitize the gain.

1

1. Desensitize the gain.

2. Reduce nonlinear distortion.

3. Reduce the effect of noise.

4. Control the input and output impedances 5. Extend the bandwidth of the amplifier.

The General Feedback Structure

General structure of the feedback amplifier

- This is a signal-flow diagram, and the quantities x represent

either voltage or current signals.

The General Feedback Structure

General structure of the feedback amplifier

- The open-loop amplifier has a gain A.

- A sample of the output is related by the feedback factorβ.

- The input to the basic amplifier

f o

x = β x

o i

x = Ax

i s f

x = x − x

The General Feedback Structure

General structure of the feedback amplifier

- This is a signal-flow diagram, and the quantities x represent either voltage or current signals.

- The gain of the feedback amplifier

, Aβ : loof gain

- Aβ=+tive : negative feedback(A_f<A), Aβ=-tive : positive feedback(A_f>A) - The closed-loop gain is almost entirely by the feedback elements.

1

o f

s

x A

A ⁼ x ⁼ + A β

The General Feedback Structure

General structure of the feedback amplifier

- This is a signal-flow diagram, and the quantities x represent either voltage or current signals.

- The open-loop amplifier has a gain A.

- A sample of the output is related by the feedback factorβ.

o i

x = Ax

- The input to the basic amplifier

- The gain of the feedback amplifier

, Aβ : loof gain

- Aβ=+tive : negative feedback(A_f<A), Aβ=-tive : positive feedback(A_f>A) - The closed-loop gain is almost entirely by the feedback elements.

f o

x = β x

i s f

x = x − x

1

o f

s

x A

A ⁼ x ⁼ + A β

f 1 s

x A x

A β

= β

+

The General Feedback Structure

- Feedback signal

1

i 1 s

x x

A β

= +

- For Ab ≫ 1, xf ≈ xs, which implies that xi is reduced to almost 0.

Some Properties of Negative Feedback

Gain desensitivity

-

Assume that β is constant.

( ¹ ) ²

f

dA dA

A β

= +

( ^{1 1} )

f f

dA dA

A ⁼ + A β A

*

1+Aβ : desensitivity factor

Some Properties of Negative Feedback

( ) 1 /

M H

A s A

s ω

= +

( ) ( )

( )

f

1 A s A s

β A s

= +

- Consider an amplifier with a single pole.

- Closed-loop gain

ω

= ω

( ¹ + A

β )

1

L Lf

A

ω ω

= β +

- Assume that β is constant.

- Open-loop gain with a dominant low-frequency pole.

1 + A

β

( ) ( )

( )

/ 1

1 / 1

M M

f

H M

A A

A s

s A

β

ω β

= +

+ +

- Midband gain

( )

M

/ 1

A + A β

* Maintaining the gain-bandwidth product at a constant value.

Some Properties of Negative Feedback

Gain desensitivity

- Assume that β is constant.

( ¹ )

f

dA dA

A β

= +

( )

1 1

f f

dA dA

A ⁼ + A β A

* 1+Aβ : desensitivity factor Bandwidth Extension

- Consider an amplifier with a single pole.

( ¹ )

Hf H

A

ω = ω + β

1

L Lf

A

ω ω

= β +

- Assume that β is constant.

- Open-loop gain with a dominant low-frequency pole.

* Maintaining the gain-bandwidth product at a constant value.

( ) 1 /

M H

A s A

s ω

= +

( ) ( )

( )

f

1 A s A s

β A s

= +

- Consider an amplifier with a single pole.

- Closed-loop gain

( ) ( )

( )

/ 1

1 / 1

M M

f

H M

A A

A s

s A

β

ω β

= +

+ +

- Midband gain

( )

M

/ 1

A + A β

Some Properties of Negative Feedback

Noise reduction

- In Fig.8.2(a), the signal-to-noise ratio

/ _s / _n

S N =V V

Vs

S = A

1 2 1 2

1 2 1 2

1 1

o s n

A A A A

V V V

A A β A A β

= +

+ +

- In Fig.8.2(b), a noise free amp. A₂ precedes the original amp. A₁.

- The signal-to-noise ratio at the output

2 s n

V

S A

N =V

Figure 8.3 Curve (a) shows the amplifier transfer characteristic without feedback. Curve (b) shows the characteristic with negative feedback (β= 0.01) applied.

Some Properties of Negative Feedback

Reduction in nonlinear distortion

- When β=0.01, the closed-loop gain of curse (b)

1

1000 90.9

1 1000 0.01

Af = =

+ ×

2

100 50

1 100 0.01

Af = =

+ ×

* The order-of-magnitude change in slope has been reduced. The price paid is a reduction in voltage gain.

Figure 8.3 Curve (a) shows the amplifier transfer characteristic without feedback. Curve (b) shows the characteristic with negative feedback (β= 0.01) applied.

Some Properties of Negative Feedback

Noise reduction

- In Fig.8.2(a), the signal-to-noise ratio

/ _s / _n

S N =V V

Vs

S = A

1 2 1 2

1 2 1 2

1 1

o s n

A A A A

V V V

A A β A A β

= +

+ +

- In Fig.8.2(b), a noise free amp. A₂ precedes the original amp. A₁.

- The signal-to-noise ratio at the output

2 s n

V

S A

N =V

Reduction in nonlinear distortion

- When β=0.01, the closed-loop gain of curse (b)

1

1000 90.9

1 1000 0.01

Af = =

+ ×

2

100 50

1 100 0.01

Af = =

+ ×

* The order-of-magnitude change in slope has been

reduced. The price paid is a reduction in voltage gain. Figure 8.3 Curve (a) shows the amplifier transfer characteristic without feedback. Curve (b) shows the characteristic with negative feedback (β= 0.01)

The Four Basic Feedback Topologies

High input resistance, low output resistance Low input resistance, high output resistance

Figure 8.4 The four basic feedback topologies: (a) voltage-mixing voltage-sampling (series–shunt) topology;

(b) current-mixing current-sampling (shunt–series) topology; (c) voltage-mixing current-sampling (series–series) topology;

(d) current-mixing voltage-sampling (shunt–shunt) topology.

Voltage amplifiers

- Voltage-mixing voltage sampling - Series-shunt feedback

- Higher input resistance, lower output resistance

The Four Basic Feedback Topologies

The Four Basic Feedback Topologies

Current amplifiers

- Current-mixing current sampling - Shunt-series feedback

- Lower input resistance, higher output resistance

- I

increases, V

increases, I

increases, V

decreases, I

decreases,

and I

increases.(negative feedback)

Voltage amplifiers

- Voltage-mixing voltage sampling - Series-shunt feedback

- Higher input resistance, lower output resistance

The Four Basic Feedback Topologies

Current amplifiers

- Current-mixing current sampling - Shunt-series feedback

- Lower input resistance, higher output resistance

- I_s increases, V_g1 increases, I_d1 increases, V_d1 decreases, I_odecreases, and I_fincreases.(negative feedback)

Transconductance amplifiers

The Four Basic Feedback Topologies

- Voltage-mixing current sampling

- Series-series feedback

The Four Basic Feedback Topologies

Transresistance amplifiers

- Current-mixing voltage sampling

- Shunt-shunt feedback

Transconductance amplifiers

The Four Basic Feedback Topologies

- Voltage-mixing current sampling - Series-series feedback

Transresistance amplifiers

- Current-mixing voltage sampling - Shunt-shunt feedback

1

o f

s

V A

A ⁼V ⁼ + Aβ

/

s s

if

i i i

V V

R = I =V R _i ^s _i ^{i i}

i i

V V AV

R R

V V

β

= = +

The ideal situation

- Closed-loop voltage gain

The Series-Shunt Feedback Amplifier

- Input resistance with feedback

(

)

if i

R = R + Aβ

( ) ( )

( ) ( )

Zif

( )

( )

( ) ( )

The ideal situation

The Series-Shunt Feedback Amplifier

- Output resistance with feedback

0 t of

Vs

R V

I ₌

=

t i

o

V AV

I R

= −

- Since V_s=0

i f o t

V_i = −V_f = −βV_o = −βV_t V = −V = −βV = −βV

1

o f

s

V A

A ⁼V ⁼ + Aβ

/

s s

if

i i i

V V

R = I =V R _i ^s _i ^{i i}

i i

V V AV

R R

V V

β

= = +

The ideal situation

- Closed-loop voltage gain

The Series-Shunt Feedback Amplifier

- Input resistance with feedback

(

)

if i

R = R + Aβ

( ) ( )

( ) ( )

Zif

( )

( )

( ) ( )

0 t of

Vs

R V

I ₌

=

t i

o

V AV

I R

= −

- Since V_s=0

i f o t

V = −V = −βV = −βV

t t o

V A V

I R

β

= +

1

o of

R R

Aβ

= +

( ) ( )

( ) ( )

1

o of

Z s

Z s

A s β s

= +

The Series-Shunt Feedback Amplifier

t t o

V A V

I R

β

= +

1

o of

R R

Aβ

= +

( ) ( )

( ) ( )

1

o of

Z s

Z s

A s β s

= +

The practical situation

- Find the A circuit andβcircuit of the ideal amp..

1) Represent the feedback network in terms of h

The Series-Shunt Feedback Amplifier

1) Represent the feedback network in terms of h parameters.(the series connection at the input and the parallel connection at the output)

2) Omit h₂₁I₁.

3) Include h₁₁and h₂₂with the basic amp..

- If the basic amp. is unilateral,

then the circuit (c) becomes the ideal amp..

12 ^basic 12 ^feedback

amplifier network

h h

21 21

(

)

amplifier network

h h

The Series-Shunt Feedback Amplifier

1

1 12

2 I 0

h V β V

=

= =

-

The loading effect is found by looking into the port of the feedback network while the other port is open circuited or short-circuited so as to destroy the feedback.

- Determination of β

* β should be found with port 1 open-circuited.

The Series-Shunt Feedback Amplifier

in if s

R = R −R

1 1

out 1

of L

R R R

 

=  −  Summary

1. R_i and R_o are the input and output R of the A circuit.

2. R_if and R_of are the input and output R of the feedback amp., including R_sand R_L. 3. The actual input and output R of the feedback amp. exclude R_sand R_L.

EXAMPLE 8.1

( )

(

)

(

)

//

// //

o L id

i L o id s

R R R

V R

A≡ V = µ R^ R R^{+ }r R R R R

+ +

 + +

 

The Series-Shunt Feedback Amplifier

Find A,β, the closed-loop gain V_o/V_s, R_inand R_out. Open-loop gain µ=10⁴, R_id=100kΩ, r_o=1kΩ, R_L=2kΩ, R₁=1kΩ, R₂=1MΩ, and R_s=10kΩ.

- This amp. samples the output voltage V_o and provide a voltage signal(across R₁) mixed in series with V_s. - In the A circuit of figure (b)

A ≈ 6000V/V

EXAMPLE 8.1

1

10 V/V

V

R

V R R

β = =

+

The Series-Shunt Feedback Amplifier

Find A,β, the closed-loop gain V_o/V_s, R_inand R_out. Open-loop gain µ=10⁴, R_id=100kΩ, r_o=1kΩ, R_L=2kΩ, R₁=1kΩ, R₂=1MΩ, and R_s=10kΩ.

- In the circuit of figure (c)

1 2

V

R + R

6000 857 V/V

1 7

o f

s

V A

A ⁼ V ⁼ + A β ^{= =}

- Voltage gain with feedback

EXAMPLE 8.1

( )

(

)

(

)

//

// //

o L id

i L o id s

R R R

V R

A≡ V = µ R^ R R^{+ }r R R R R

+ +

 + +

 

The Series-Shunt Feedback Amplifier

Find A,β, the closed-loop gain V_o/V_s, R_inand R_out. Open-loop gain µ=10⁴, R_id=100kΩ, r_o=1kΩ, R_L=2kΩ, R₁=1kΩ, R₂=1MΩ, and R_s=10kΩ.

- This amp. samples the output voltage V_o and provide a voltage signal(across R₁) mixed in series with V_s. - In the A circuit of figure (b)

1 3

1 2

10 V/V

f o

V R

V R R

β = = ⁻

+

6000 857 V/V

1 7

o f

s

V A

A ⁼V ⁼ + Aβ ^{= =} A ≈ 6000V/V

- In the circuit of figure (c)

- Voltage gain with feedback

739kΩ

in if s

R = R −R =

1

o of

R R

Aβ

= +

The Series-Shunt Feedback Amplifier

- Output resistance including R_L - Input resistance excluding R_s

(

)

if i

R = R + Aβ

(

)

i s id

R = R + R + R R

- Input resistance seen by the external source

111 7 777 kΩ R =if × =

1+ Aβ

(

)

// // 667

o o L

R =r R R + R Ω

667 95.3 kΩ

of 7

R = =

of out // L

R = R R

out 100

R Ω

- Output resistance excluding R_L