The Series-Series Feedback Amplifier

13 차시

8장 FEEDBACK, 전자회로

1. Desensitize the gain.

1 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.

o i

A I

≡V

1

o f

s

I A

A ^≡V ⁼ + Aβ

The Series-Series Feedback Amplifier

The ideal case

- A is a transconductance,

while βis a transresistance.

- Closed-loop transconductance

/

s s

if

V V

R = I =V R _i V^s _i Vⁱ AVⁱ

R R

V V

β

= = +

- Input resistance with feedback

if /

i i i

R = I =V R _{i i}

i i

R R

V V

= =

(

)

if i

R = R + Aβ

The Series-Series Feedback Amplifier

The ideal case

0 of

t Vs

R V

I ₌

≡

(

)

of o

R = + Aβ R

(

)

(

)

V = I − AV R = I + A Iβ R - Output resistance with feedback

0

i f t

V = −V = −βI = −βI

3

* The negative feedback increase the output resistance.

o i

A I

≡V

1

o f

s

I A

A ^≡V ⁼ + Aβ

The Series-Series Feedback Amplifier

The ideal case

- A is a transconductance,

while βis a transresistance.

- Closed-loop transconductance

/

s s

if

V V

R = I =V R _i V^s _i Vⁱ AVⁱ

R R

V V

β

= = +

- Input resistance with feedback

if /

i i i

R = I =V R _{i i}

i i

R R

V V

= =

(

)

if i

R = R + Aβ

0 of

t Vs

R V

I ₌

≡

(

)

of o

R = + Aβ R

(

)

(

)

V = I − AV R = I + A Iβ R - Output resistance with feedback

0

i f t

V = −V = −βI = −βI

* The negative feedback increase the output resistance.

The practical case

The Series-Series Feedback Amplifier

- Find the A circuit and β circuit of the ideal amp..

1) Represent the feedback network in terms of z parameters.(a series circuit at the input and a series circuit at the output)

5

2) Omit z₂₁I₁.

21 21

( feedback basic )

amplifier network

z z

The practical case

The Series-Series Feedback Amplifier

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

1) Represent the feedback network in terms of z parameters.(a series circuit at the input and a series circuit at the output)

2) Omit z₂₁I₁.

3) Include z₁₁and z₂₂with the basic amp..

- If the basic amp. is unilateral,

12 ^basic 12 ^feedback

amplifier network

z z

21 21

( feedback basic )

amplifier network

z z

then the circuit (c) is equivalent to the ideal amp..

12 12

amplifier network

The practical case

The Series-Series Feedback Amplifier

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

1) Represent the feedback network in terms of z parameters.(a series circuit at the input and a series circuit at the output)

2) Omit z₂₁I₁.

3) Include z₁₁and z₂₂with the basic amp..

- If the basic amp. is unilateral,

12 ^basic 12 ^feedback

amplifier network

z z

21 21

( feedback basic )

amplifier network

z z

7

then the circuit (c) is equivalent to the ideal amp..

12 12

amplifier network

The Series-Series Feedback Amplifier

1

1 12

2 I 0

z V β I

=

= =

- 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.(open if series and short if shunt)

- Determination of β

*βshould be found with port 1 open.

Summary

'

out of L

R = R −R

in if s

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.

The Series-Series Feedback Amplifier

1

1 12

2 I 0

z V β I

=

= =

- 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.(open if series and short if shunt)

- Determination of β

*βshould be found with port 1 open.

9

( )

( )

1 1 2

1

1 1 2

//

//

c C

i e E F E

R r

V

V r R R R

α π

= −

 

+ + 

EXAMPLE 8.2 – MC1553

Assuming I_C1=0.6mA, I_C2=1mA, I_C3=4mA, h_fe=100, and r_o=∞, find A,β, the closed-loop gain A_f=I_o/V_s, voltage gain V_o/V_s, R_in=R_ifand R_of. If r_o3=25kΩ, R_out= ?

- To find A= I_o/V_iwe should determine the gain of the 1^ststage.

The Series-Series Feedback Amplifier

- Substituting r_e1=41.7Ω, r_π2=h_fe/g_m2=100/40=2.5kΩ, α₁=0.99, R_c1=9kΩ, R_E1=100Ω, R_F=640Ω, and R_E2=100Ω.

1 14.92 V/V

c

i

V V = −

EXAMPLE 8.2 – MC1553

The Series-Series Feedback Amplifier

( ) ( ^{( )} )

{ }

2

2 2 3 2 1

1

// 1 //

c

m C fe e E F E

c

V g R h r R R R

V = − +  + + 

2

1

131.2 V/V

c

c

V V = −

( )

( )

3 1

o e

I I

= =

- The gain of the 2^ndstage.

- Substituting g_m2=40mA/V, R_C2=5kΩ, h_fe=100, r_e3=25/4=6.25Ω, R_E2=100Ω, R_F=640Ω, and R_E1=100Ω.

- The gain of the 3^rdstage.

11

( )

( )

2 3 3 2// 1

c b e E F E

V =V = r R R R

+ +

( )

1 10.6 mA/V

6.25 100 // 740

= =

+

( )

( )

1 1 2

1

1 1 2

//

//

c C

i e E F E

R r

V

V r R R R

α π

= −

 

+ + 

EXAMPLE 8.2 – MC1553

Assuming I_C1=0.6mA, I_C2=1mA, I_C3=4mA, h_fe=100, and r_o=∞, find A,β, the closed-loop gain A_f=I_o/V_s, voltage gain V_o/V_s, R_in=R_ifand R_of. If r_o3=25kΩ, R_out= ?

- To find A= I_o/V_iwe should determine the gain of the 1^ststage.

The Series-Series Feedback Amplifier

- Substituting r_e1=41.7Ω, r_π2=h_fe/g_m2=100/40=2.5kΩ, α₁=0.99, R_c1=9kΩ, R_E1=100Ω, R_F=640Ω, and R_E2=100Ω.

1 14.92 V/V

c

i

V V = −

- The gain of the 2^ndstage.

( ) ( ^{( )} )

{ }

2

2 2 3 2 1

1

// 1 //

c

m C fe e E F E

c

V g R h r R R R

V = − +  + + 

2

1

131.2 V/V

c

c

V V = −

( )

( )

3

2 3 3 2 1

1 //

o e

c b e E F E

I I

V =V = r R R R

+ +

( )

1 10.6 mA/V

6.25 100 // 740

= =

+ - The gain of the 2^ndstage.

- Substituting g_m2=40mA/V, R_C2=5kΩ, h_fe=100, r_e3=25/4=6.25Ω, R_E2=100Ω, R_F=640Ω, and R_E1=100Ω.

- The gain of the 3^rdstage.

The Series-Series Feedback Amplifier

14.92 131.2 10.6 10 3 20.7 A/V

o

i

A I V

= = − × − × × − =

2

1

2 1

f E

E

o E F E

V R

I R R R R

β = = ×

+ +

100 100 11.9 100 640 100

= × = Ω

+ +

20.7 83.7 mA/V

= =

Io A

A = =

- Combining the gains of 3 stages

- In the circuit of figure (c)

- The closed-loop gain

13

83.7 mA/V 1 20.7 11.9

= =

+ ×

1

o f

s

A ⁼V ⁼ + Aβ

The Series-Series Feedback Amplifier

3 3

3

o c C o C

f C

s s s

V I R I R

V V V A R

− −

= = − = −83.7 10× ⁻³×600= −50.2 V/V

(

)

if i

R = R + Aβ

(

)

(

⁽

⁾ )

i fe e E F E

R = h + r + R R +R =

( )

13.65 1 20.5 11.9 3.34 MΩ

R =if + × =

- The voltage gain

- The input resistance of the feedback amp.

The Series-Series Feedback Amplifier

14.92 131.2 10.6 10 3 20.7 A/V

o

i

A I V

= = − × − × × − =

2

1

2 1

f E

E

o E F E

V R

I R R R R

β = = ×

+ +

100 100 11.9 100 640 100

= × = Ω

+ +

20.7 83.7 mA/V

= =

Io A

A = =

- Combining the gains of 3 stages

- In the circuit of figure (c)

- The closed-loop gain

15

83.7 mA/V 1 20.7 11.9

= =

+ ×

1

o f

s

A ⁼V ⁼ + Aβ

3 3

3

o c C o C

f C

s s s

V I R I R

V V V A R

− −

= = − = −83.7 10× ⁻³×600= −50.2 V/V

(

)

if i

R = R + Aβ

(

)

(

⁽

⁾ )

i fe e E F E

R = h + r + R R +R =

( )

13.65 1 20.5 11.9 3.34 MΩ

R =if + × =

- The voltage gain

- The input resistance of the feedback amp.

The Series-Series Feedback Amplifier

(

) (

)

out o m o of

R = +r + g r R r_π

( )( )

25 1 160 25 35.6 // 0.625 2.5 MΩ

= + + × =

(

)

(

)

of o

R = R + Aβ = + × =

( )

2// 1 3

1

C

o E F E e

fe

R R R R r R

  h

= + + + + =

- The output resistance of the A circuit

- The output resistance of the feedback amp.

- In the circuit of figure (d)

The Shunt-Shunt Feedback Amplifier

- A is a transresistance, and β is a transconductance.

o f

s

A V

= I

f 1 A A

Aβ

= +

1

i if

R R

Aβ

= +

- Input resistance with feedback - Closed-loop transresistance

- Output resistance with feedback

17

1

o of

R R

Aβ

= +

- Output resistance with feedback

The Shunt-Shunt Feedback Amplifier

- A is a transresistance, and β is a transconductance.

o f

s

A V

= I

f 1 A A

Aβ

= +

1

i if

R R

Aβ

= +

- Input resistance with feedback - Closed-loop transresistance

- Output resistance with feedback 1

o of

R R

Aβ

= +

- Output resistance with feedback

The Shunt-Shunt Feedback Amplifier

- Assuming that the basic amp. is unilateral and that the forward transmission through the feedback is negligibly small.

- The actual input and output R of the feedback amp. exclude R_sand R_L

12 basic 12 feedback amplifier network

y y

21 feedback 21 basic network amplifier

y y

1 1

in 1

if s

R R R

 

=  − 

19

if s

R R

 

 

1 1

out 1

of L

R R R

 

=  − 

The Shunt-Shunt Feedback Amplifier

EXAMPLE 8.3

Find the voltage gain V_o/V_s, R_inand R_of=R_out. The transistor has β=100.

- DC analysis

( )

0.7 0.07 47 3.99 47

C B B

V = + I + = + I

( )

12 1 0.07

4.7

C

B

V β I

− = + +

The Shunt-Shunt Feedback Amplifier

EXAMPLE 8.3

- I_B=0.015mA, I_C=1.5mA, and V_C=4.7V.

- In the circuit of figure (d)

(

)

i s f

V_π = I R R r_π

(

)

o m f C

V = −g V_π R R

(

)(

)

o

m f C s f

i

A V g R R R R r

I ^π

= = − = −

- The input and output resistance of the A circuit

21

// // 1.4 kΩ

i s f

R = R R r_π = // 4.27 kΩ

o f C

R = R R =

- The input and output resistance of the A circuit

- In the circuit of figure (e)

1 1

47 kΩ

f

o f

I

V R

β = = − = −

The Shunt-Shunt Feedback Amplifier

EXAMPLE 8.3

Find the voltage gain V_o/V_s, R_inand R_of=R_out. The transistor has β=100.

- DC analysis

- I_B=0.015mA, I_C=1.5mA, and V_C=4.7V.

- In the circuit of figure (d)

( )

0.7 0.07 47 3.99 47

C B B

V = + I + = + I

( )

12 1 0.07

4.7

C

B

V β I

− = + +

(

)

i s f

V_π = I R R r_π

// // 1.4 kΩ

i s f

R = R R r_π =

(

)

o m f C

V = −g V_π R R

(

)(

)

o

m f C s f

i

A V g R R R R r

I ^π

= = − = −

// 4.27 kΩ

o f C

R = R R =

- The input and output resistance of the A circuit

- In the circuit of figure (e)

1 1

47 kΩ

f

o f

I

V R

β = = − = −

The Shunt-Shunt Feedback Amplifier

- The closed-loop gain

- The voltage gain

- The input resistance of the feedback amp.

358.7 358.7

41.6 kΩ 1 1 358.7 / 47 8.63

o f

s

V A

A I Aβ

− −

= = = = = −

+ +

41.6 4.16 V/V 10

o o

s s s

V V

V I R

= = − −

1.4 162.2

1 8.63

i if

R R

Aβ

= = = Ω

+ R =165Ω

23

in 165

R = Ω

4.27 495

1 8.63

o of

R R

Aβ

= = = Ω

+

- The output resistance of the feedback amp.

An important note

* Assumption - Most of the forward transmission occurs in the basic amp., and most of reverse transmission(feedback) occurs in the feedback network.

The Shunt-Series Feedback Amplifier

1

i if

R R

Aβ

= +

- Input resistance with feedback - Closed-loop current gain

- Output resistance with feedback 1

o f

s

I A

A ^≡ I ⁼ + Aβ

(

)

of o

R = R +Aβ

The Shunt-Series Feedback Amplifier

- Assuming that the basic amp. is unilateral and that the forward transmission through the feedback is negligibly small.

- The actual input and output R of the feedback amp. exclude R_sand R_L

1 1

in 1

if s

R R R

 

=  − 

12 basic 12 feedback amplifier network

g g

21 feedback 21 basic network amplifier

g g

25

if s

R R

 

 

out of L

R = R −R

The Shunt-Series Feedback Amplifier

EXAMPLE 8.4

Find the current gain I_out/I_in, R_inand R_out. The transistor has β=100 and V_A=75V.

- DC analysis

1

12 15 1.57 V 100 15

VB =

+

1 1.57 0.7 0.87 V

VE − =

1 0.87 / 0.54 1mA

IE = =

1 12 10 1 2 V

VC − × =

2 2 0.7 1.3 V

VE − =

2 1.3 / 3.4 0.4 mA

IE =

2 12 0.4 8 8.8 V

VC − × =

The Shunt-Series Feedback Amplifier

27

( )

1 _{i s} // _E2 _f // _B// 1

V_π = I R R + R R r_π 

(

)

2 2//

b o

e E f

I V

r + R R

( ) ( )

{ }

2 1 1 1// 1// 2 1 2//

b m o C E f

V = −g V_π r R ^r_π + β + R R ^

* neglecting r_o2

The Shunt-Series Feedback Amplifier

201.45 A/A

o

i

A I

≡ I −

(

)

// // // 1.535 kΩ

i s E f B

R = R R +R R r_π = - Open-loop current gain

- Input resistance of the A circuit

(

)

o E f e

R r

R R R r

= + + + β =

+

2 3.4

0.254

f E

I R

β ≡ = − = − = −

- Output resistance of the A circuit (* neglecting r_o2)

- In the circuit of figure (d)

1+ Aβ =52.1

2

2

0.254 13.4

E

o E f

I R R

β ≡ = − = − = −

+

The Shunt-Series Feedback Amplifier

29

3.87 A/A 1

o f

s

I A

A ^≡ I ⁼ + Aβ ^{= −} 1 29.5

i if

R R

Aβ

= = Ω

+

1 29.5

1 / 1 /

in

if s

R = R R Ω

−

- Input resistance of the ideal feedback structure

- Input resistance of the practical feedback structure

- Closed-loop current gain

- Since R_in≈R_if , I_in≈I_s.

2 2

2 2

out out C c C o

in s L C s L C s

I I R I R I

I I = R R I R R I

+ +

/ 3.44 A/A

out in

I I = −

(

)

of o

R = R +Aβ

( )

2 1 2 2//

out o m of

RR_out==r18.1MΩ +g r_π R  - Output resistance

The Shunt-Series Feedback Amplifier

201.45 A/A

o

i

A I

≡ I −

(

)

// // // 1.535 kΩ

i s E f B

R = R R +R R r_π = - Open-loop current gain

- Input resistance of the A circuit

(

)

o E f e

R r

R R R r

= + + + β =

+

2 3.4

0.254

f E

I R

β ≡ = − = − = −

- Output resistance of the A circuit (* neglecting r_o2)

- In the circuit of figure (d)

1+ Aβ =52.1

3.87 A/A 1

o f

s

I A

A ^≡ I ⁼ + Aβ ^{= −}

2

2

0.254 13.4

E

o E f

I R R

β ≡ = − = − = −

+

1 29.5

i if

R R

Aβ

= = Ω

+

1 29.5

1 / 1 /

in

if s

R = R R Ω

−

- Input resistance of the ideal feedback structure

- Input resistance of the practical feedback structure

- Closed-loop current gain

- Since R_in≈R_if , I_in≈I_s.

2 2

2 2

out out C c C o

in s L C s L C s

I I R I R I

I I = R R I R R I

+ +

/ 3.44 A/A

out in

I I = −

(

)

of o

R = R +Aβ

( )

2 1 2 2//

out o m of

R = r  +g r_π R  18.1MΩ

Rout =

- Output resistance

Summary

31