Transcript Feedback Amplifiers - City University of Hong Kong

Lecture 2 Feedback Amplifier
•
•
•
•
Introduction of Two-Port Network
Negative Feedback (Uni-lateral Case)
Feedback Topology
Analysis of feedback applications
– Close-Loop Gain
– Input/Output resistances
Ref:080130HKN
EE3110 Feedback Amplifiers
1
Two-Port Network (z-parameters)
(Open-Circuit Impedance)
I1
+
z22
z11
V1

+
V1   z11
V    z
 2   21
V2
V1  z11 I1  z12 I 2

V2  z21I1  z22 I 2
I2
z12I2 +

+ z21I1

At port 1
Open-circuit
input impedance
At port 2
V
z11  1
I1 I 2  0
Open-circuit reverse z  V1
12
I 2 I1  0
transimpedance
Ref:080130HKN
z12   I1 
z22   I 2 
Open-circuit forward z  V2
21
I1 I 2  0
transimpedance
Open-circuit
output impedance
EE3110 Feedback Amplifiers
z22 
V2
I 2 I1  0
2
Two-Port Network (y-parameters)
(Short-Circuit Admittance)
I
I
1
+
1/y
1/y
11
V
1
+
22
yV
12

 I1   y11
I    y
 2   21
2
2
yV
21
V
I1  y11V1  y12V2

I 2  y21V1  y22V2
2
1
At port 1
Short-circuit
input admittance
At port 2
I
y11  1
V1 V2  0
Short-circuit reverse y  I1
12
transadmittance
V2 V1  0
Ref:080130HKN
y12  V1 
y22  V2 
Short-circuit forward y  I 2
21
V1 V2  0
transadmittance
Short-circuit
output admittance
EE3110 Feedback Amplifiers
y22 
I2
V2 V1  0
3
Two-Port Network (h-parameters)
(hybrid)
I
I
+
1/h
h
1
hV +
12
2
+
22
11
V

V1   h11 h12   I1 
 I   h
 V 
h
 2   21 22   2 
2
1

hI
V
V1  h11 I1  h12V2

I 2  h21I1  h22V2
2
21 1
At port 1
Short-circuit
input impedance
At port 2
V
h11  1
I1 V2  0
Open-circuit reverse h  V1
12
voltage gain
V2 I1  0
Ref:080130HKN
Short-circuit forward h  I 2
21
I1 V2  0
current gain
Open-circuit
output admittance
EE3110 Feedback Amplifiers
h22 
I2
V2 I1  0
4
Two-Port Network (g-parameters)
(inverse-hybrid)
I
I
1
+
 I1   g11
V    g
 2   21
2
g
1/g
11
V
1
gI
12 2

+
22
V
2
+ g21V1


At port 1
Open-circuit
input admittance
I1  g11V1  g12 I 2
V2  g 21V1  g 22 I 2
At port 2
I
g11  1
V1 I 2  0
Short-circuit reverse g  I1
12
current gain
I 2 V1  0
Ref:080130HKN
g12  V1 
g 22   I 2 
Open-circuit forward g  V2
21
V1 I 2  0
current gain
Short-circuit
output impedance
EE3110 Feedback Amplifiers
g 22 
V2
I 2 V1  0
5
z-parameter example
I1
I1
I2
+
V1
6

+
+


V2
Z11  6 Z 22  6
V1
I2
I1
+
12
3
V2

Z11  12 Z 22  3
12
+
V1

I2
3
+
6
V2

Z11  18 Z 22  9
Z12 
V1
 6
I 2 I1  0
Z12 
V1
 0
I 2 I1  0
Z12 
V1
6I
 2  6
I 2 I1  0 I 2
Z 21 
V2
 6
I1 I 2  0
Z 21 
V2
 0
I1 I 2  0
Z 21 
V2
6I
 1  6
I1 I 2  0 I1
Z   
6 6

6 6 
Z   
12 0

 0 3
18 6
Z   

 6 9
Note: (1) z-matrix in the last circuit = sum of two former z-matrices
(2) z-parameters is normally used in analysis of series-series circuits
(3) Z12 = Z21 (reciprocal circuit)
(4) Z12 = Z21 and Z11 = Z22 (symmetrical and reciprocal circuit)
Ref:080130HKN
EE3110 Feedback Amplifiers
6
y-parameter example
I1
+
V1

I1
I2
0.05S
0.1S
+
+


V2
I2
0.2S
+
V2
V1

0.025S
1
y11  0.05S y22  0.05S
y12 
I1
 0.05V2

 0.05S
V2 V1  0
V2
y21 
I2
 0.05V1

 0.05S
V1 V2  0
V1
0.05  0.05
 y   

 0.05 0.05 
1
 1

y11  

  0.0692S
 0.1 0.2  0.025
1
1
 1

y22  

  0.0769S
 0.2 0.1  0.025
y12 
I1
V2 V1  0
But I 2  y22V2  0.0769V2
 I1 I 2  I1

0.1 0.025
 I1  0.8 I 2  0.0615V2
y12  0.0615S
By reciprocal, y21  y12  0.0615S
 0.0615

 0.0615 0.0769 
EE3110 Feedback Amplifiers
 y   
Ref:080130HKN
0.0692
7
y-parameter example (Cont’)
0.05S
y11  0.05  0.692  0.1192
y22  0.05  0.769  0.1269
y21  y12 
I1
V2 V1  0
I1
0.1S
+
V1
I 2  y22V2  0.1269V2
0.05
I 0.05 S 
I 2  0.05V2
0.1269
I 0.2 S  I 2  I 0.05 S  0.0769V2

0.025S
I2
0.2S
+
V2

 y12  0.1115 y21
Note: the y-matrix is equal to
the sum of two former ones.
Therefore, y-parameters is
normally used in analysis of
shunt-shunt circuits
 0.1192  0.1115
Y   

 0.1115 0.1269 
What connection should be for
h- or g- parameters?
0 .1
0.0769V2  0.0615V2
0.1  0.025
I1  0.0615V2  0.05V2  0.1115V2
I 0.1S 
Ref:080130HKN
EE3110 Feedback Amplifiers
8
General Feedback Structure
Source
Vs +
V

-
A
V
Load
A : Open Loop Gain
A = Vo / V
 : feedback factor
 = Vf / Vo
Vf

V  Vs  V f
Close loopgain : ACL 
V f    Vo
LoopGain : T  A  
V  VS    Vo
Amountof feedback: 1  A  
Vo  A  V
Note: ACL
Ref:080130HKN
A 

Vo
A
1 T

 (
)
Vs 1  A  1  T
1

EE3110 Feedback Amplifiers
9
Negative Feedback Properties
• Negative feedback takes a sample of the output signal and applies
it to the input to get several desirable properties. In amplifiers,
negative feedback can be applied to get the following properties
– Desensitized gain : gain less sensitive to circuit component
variations
– Reduce nonlinear distortion : output proportional to input
(constant gain independent of signal level)
– Reduce effect of noise
– Control input and output impedances by applying appropriate
feedback topologies
– Extend bandwidth of amplifier
• All of these properties can be achieved by trading off gain
Ref:080130HKN
EE3110 Feedback Amplifiers
10
Gain De-sensitivity
• Feedback can be used to desensitize the closed-loop gain to variations in the
basic amplifiler.
• Assume  is constant. Take differentials of the closed loop gain equation
gives,
ACL 
A
1  A
Differential respected with A
dACL
1
dA

or
dA

CL
dA (1  A ) 2
(1  A ) 2
• Divided by Av, the close loop gain sensitivity is equal to,
dACL
dA (1  A )
1 dA


ACL (1  A ) 2
A
1  A A
• This result shows the effects of variations in A on ACL is mitigated by the
feedback amount.
• (1+A) is also called the desensitivity amount.
Ref:080130HKN
EE3110 Feedback Amplifiers
11
Basic Feedback Topologies
Depending on the input signal (voltage or current) to be amplified
and form of the output (voltage or current), amplifiers can be
classified into four categories. Depending on the amplifier
category, one of four types of feedback structures should be used.
(Type of Feedback)
(Type of Sensing)
(1) Series (Voltage)
Shunt (Voltage)
(2) Series (Voltage)
Series (Current)
(3) Shunt (Current)
Shunt (Voltage)
(4) Shunt (Current)
Series (Current)
Ref:080130HKN
EE3110 Feedback Amplifiers
12
Feedback Structure (Series-Shunt)
Basic amplifier
Voltage Gain Calculation:
Ii +
V

Vi
ri
+
 A V
ro
+
Vo

Vo  A  V
V f    Vo
Vo
   Vo
A
(Close LoopVoltage Gain)
+

Vi  V  V f 
+
 Vf=Vo
 ACL 
Vo 1 T
 (
)
Vi  1  T
where T  A
Feedback network



Voltage amplifier voltage-controlled
voltage source
Requires high input impedance, low
output impedance
Voltage-voltage feedback
Ref:080130HKN
And, we get
V A
Vo  i
1 A 
Vi  V (1  A   )
EE3110 Feedback Amplifiers
13
Input/Output Resistance (Series-Shunt)
Input Resistance:
V
Rin  i
Ii
Vi  (1  T )  V
V
Vi
Ii   
ri (1  T )  ri
Rin 
Vi
 (1  T )  ri
Ii
Output Resistance
(Closed loop output resistance with zero input voltage)
Rout |Vi 0 
Vo  A  V
Io 
ro
ro
+ AV

Io
+

Vo
V    Vo  Vi  0
V     Vo
Vo  A    Vo
Io 
ro
 Rout 
Ref:080130HKN
Vo
Io
Vo
ro
r

 o
Io 1 A   1 T
EE3110 Feedback Amplifiers
14
h-parameter Modeling
I1
z
s
+
h
11a
h V2 + h I1

12a
V
1
1/h
22a
1/y
L
 V2
21a
+

h
h I1
11f
21f
h V2 +
12f
Ref:080130HKN
1/h
22f

EE3110 Feedback Amplifiers
Only uni-lateral case
will be considered :
(1) NO reverse
dependent signal
found in the
amplifier network.
|h12a| = 0
(2) NO reverse
dependent signal
found in the
feedback network.
|h21f| = 0
15
I1
z
s
h
V
1/h
1/y
22a
11a
1/h
h I1
L
V
2
Uni-lateral
22f
21a
1
From Input port: V1  (h11a  h11 f  Z s ) I1  h12 f V2
+

h
11f
h V2 +

From output port: (h22 a  h22 f  YL )V2  h21a I1  0
P ut V2  
12f
h21a I1
back tofirst equat ion,
(h22 a  h22 f  YL )
For h12 f  0, theOpen LoopGain,
A
 h21a
h
V2

 21a
V1 (h11a  h11 f  Z s )(h22 a  h22 f  YL ) zi yo
where zi  h11a  h11 f  Z s and yo  h22 a  h22 f  YL
Wit h thefeedback f  h12 f , theClosed LoopGain is
Ref:080130HKN
 h21a
zi yo
 h21a
A
ACL 


 h h
1  A
zi yo  h21a  h12 f
1  21a 12 f
z i yo
EE3110 Feedback Amplifiers
16
Series-Shunt Example
Given : A  105 , R1  1k, R2  9k, r  10M and ro  40
Vi
V
+

Vo
Amplifier
Vi
Vf
+
V  r  r
Vo
+

 AV
R2
R1
Vf
It is observed that:
(1) Series connection in input ports
(2) Shunt connection in output ports
 Series-Shunt connection
h-parameter should be used.
Ref:080130HKN
EE3110 Feedback Amplifiers
R1
Vo
R2
Feedback
Equivalent circuit
17
h-parameter analysis
I1
V1
R2
V2 h11 f  V1
I1 V2  0
R1
h12 f   
h22 f 
h
12f
Ref:080130HKN
2
R1 R2
 0.9
R1  R2
V1
I 2 R1
R1


 0.1
V2 I1  0 I 2 ( R1  R2 ) R1  R2
I2
I2
1
1



V2 I1  0 I 2 ( R1  R2 ) R1  R2 10k
1
11f
h V
 R1 // R2 
+

h
22f
EE3110 Feedback Amplifiers
18
Firstly,set h12 f  0 for open loop circuit and
by put t ingV1  Vε and h22 f
V
1

, we get
R1  R2
And theclose loop volt agegain,
AOP
105


~ 10
1  AOP  1  (105 )(0.1)
Ref:080130HKN
1

+
r  V

1
1 
AV1
 
V2 
ro
 ro R1  R2 
T herefore,theopen loop volt agegain is
V
A( R1  R2 )
A OP  2 
~ A  105
V1 r0  R1  R2
A CL
+
+
 r

 ~V
V  V1 
r h  1
  11 f 
from out put port ,
V2  AV
 h22 f  V2  0
ro
r
I1
from theinput port ,
h
AV
+

1/h
22f
V

11f
h V2 +

12f
Th e in pu tim pe dan ce ,
R in  (1  AOP  )(h11 f  r )  
Th e ou tpu tim pe dan ce ,
ro //(1 / h22 f )
R out 
0
1  AOP 
EE3110 Feedback Amplifiers
19
2
Feedback Structure (Series-Series)
Basic amplifier
Ii
I
I o  A  V
AV
+
V
Vi
Gain Calculation :
ri

V f    Io
ro
V
+

Vf=Io +

Feedback network
Ref:080130HKN
Io
   Io
A
(Close LoopT ransadmittanceGain)
I
1 T
 ACL  o  (
)
Vi  1  T
Vi  V  V f 
where T  A
And, we get
Vi  A
1 A 
Vi  V (1  A   )
Io 
EE3110 Feedback Amplifiers
20
Input/Output Resistance (Series-Series)
Input Resistance:
Rin 
Vi
Ii
(1  T )  V

Ii
 (1  T )  ri
Output Resistance
(Closed loop output resistance with zero input voltage)
Rout |Vi 0 
frominput port ,
V  V f     I o
fromout put port ,
V
V
I o  AV  o  T  I o  o
ro
ro
 Rout 
Ref:080130HKN
Vo
Io
Vo
 (1  T )ro
Io
EE3110 Feedback Amplifiers
21
Series-Series Example
+VCC
B
+
R1
RC
vo
vs
vs
R2
r
r
E
R1//R2
RC
RE

RE
C
+
vo

Feedback network
CE amplifier with an un-bypassed emitter
Ref:080130HKN
ac small signal equivalent circuit
EE3110 Feedback Amplifiers
22
Feedback Network with z-parameter
I1
I2
ii
V1
RE
+
V2
r  v
+
 vs
z11 f 
v1
 RE
i1 i2  0
  Z12 f 
Z 22 f
v1
 RE
i2 i1  0
v
 2
 RE
i2 i1  0
Ref:080130HKN
io

z11f
gv
ro
vo
z22f
z12fio +

+

Reduce equivalent circuit
EE3110 Feedback Amplifiers
23
Close loop analysis
 r
v  
r Z
11 f
 

vs and io  gv


T henopen loop transadmittancegain is Aop 
io
rg
 
vs r  RE
T herefore,
r g
Aop
r  RE
r g
T heclose loop transadmittancegain is ACL 
 

1  Aop  1  r gRE
r  RE  r gRE
r  RE
Input impedanceis :

r gR 
R in  (r  z11 f )(1  AOL  )  (r  RE )1   E 
 (r  RE ) 
 (r  RE )  gr RE
Out put impedanceis :
R out  [(z 22 f )(1  AOL  )]
Ref:080130HKN
EE3110 Feedback Amplifiers
24
Final Rin and Rout
i
i
+
i
o
r v


vs

+

z
gv

r
11f
RC
R1//R2
z i +

+
o
z
22f
v
o

12f o
R'in
Rin
Rout
R'out
R'in  R in // R1 // R2
R'out  R out // RC
 [(r  RE )  gr RE ] // R1 // R2
 [(z22 f )(1  AOP  )] // RC
Ref:080130HKN
EE3110 Feedback Amplifiers
25
Feedback Structure (Shunt-Shunt)
Gain Calculat ion :
Basic amplifier
+
Ii
Vi

I
ri
Vo  A  I   A( I i  I f )
ro
+
 AI
+
I f    Vo
Vo
A( I i   Vo )  Vo

AIi  (1  T )Vo
(Close LoopT ransimpedance Gain)
I f=  V o
Vo 1 T
 ACL 
 (
)
Ii  1  T
where T  A
And, we get
Feedback network
Ref:080130HKN
Ii  A
Vo 
1 A 
I i  I  (1  A   )
EE3110 Feedback Amplifiers
26
Input/Output Resistance (Shunt-Shunt)
Input Resistance:
V
Rin  i
Ii

I   ri
I  (1  T )

ri
(1  T )
Output Resistance
(Closed loop output resistance with zero input voltage)
Rout |Vi 0 
frominput port ,
I    I f    Vo
from out put port ,
V  AI Vo  TVo
Io  o

ro
ro
 Rout 
Ref:080130HKN
Vo
Io
Vo
ro

I o (1  T )
EE3110 Feedback Amplifiers
27
Shunt-Shunt Example
Vcc
Rs
RC
RF
Vs +

+
V r

gV
Rc
RL
C2
RS
C1
RL
vS +

RF
CE amplifier
ac small signal equivalent circuit
Shunt-Shunt connection found!  y-parameter
Ref:080130HKN
EE3110 Feedback Amplifiers
28
Vo
I1
I1  y11V1  y12V2
I2
RF
V1
V2
I 2  y21V1  y22V2
Feedback Network
I1
1
y11 

V1 V2  0 RF
I
I
1
y12  1
 2 
V2 V1  0 V 2
RF
I2
1
y22 

V2 V1  0 RF
I2
I1
V1 RF
1/RFVo
RF
y-parameter modeling
Ref:080130HKN
EE3110 Feedback Amplifiers
29
From input port,
V  I S ( RF // r )
V
 IS 
( RF // r )
RF
+
V
Vo
gV
RF
r
Is
RC//RL

And fromoutput port,
1/RF¡ EVo
Vo
 gV  0
RF // RC // RL
Rin 
Vo   gV ( RF // RC // RL )
Open loop tranimpedance gain :
Vo
IS
AOP   gV ( RF // RC // RL )(RF // r )
Withfeedback factor  
1
,
RF
theclose loop transimpedance gain :
ACL 
AOP
1  AOP 
Ref:080130HKN

( RF // r )
(1  AOP  )
Rout 

ri
(1  AOP  )
ro
(1  AOP  )
( RF // RC // RL )
(1  AOP  )
Volt age Gain :
EE3110 Feedback Amplifiers
Vo
Vo

Vs I s ( Rs  Rin )
30
Feedback Structure (Shunt-Series)
Gain Calculat ion :
Basic amplifier
I
+
Ii
V
i

I
ri
I o  A  I   A( I i  I f )
I f    Io
AI
ro
A( I i   I o )  I o
AIi  (1  T ) I o
(Close LoopCurrent Gain)
If=  I o
Io 1 T
 ACL   (
)
Ii  1  T
where T  A
And, we get
Feedback network
Ref:080130HKN
Ii  A
Io 
1 A 
I i  I  (1  A   )
EE3110 Feedback Amplifiers
31
Input/Output Resistance (Shunt-Series)
Input Resistance:
Output Resistance
V I r
Rin  i   i
Ii
Ii
(Closed loop output resistance with zero input voltage)
Ii
 ri
(1  T )

Ii
ri

(1  T )
Rout |Vi 0 
Vo
Io
frominput port,
I    I f   I o
from output port,I o  Vo / ro  AI
Vo  ( I o  AI )ro
Vo  ( I o  T  I o )ro
 Rout 
Ref:080130HKN
Vo
 (1  T )ro
Io
EE3110 Feedback Amplifiers
32
Summary
Feedback
Structure
Close loop
gain
Input
Output
Parameter
impedance impedance
used
SeriesShunt
Vo 1 T
 (
) Rin  (1  T )  ri R  ro
out
Vi  1  T
1 T
SeriesSeries
Io 1 T
 (
) Rin  (1  T )  ri Rout  (1  T )  ro z-parameter
Vi  1  T
ShuntShun
Vo 1 T
 (
)
Ii  1  T
ShuntSeries
Io 1 T
r
 (
) Rin  i
Ii  1  T
1 T
Ref:080130HKN
Rin 
ri
1 T
Rout
ro

1 T
h-parameter
y-parameter
Rout  (1  T )  ro g-parameter
EE3110 Feedback Amplifiers
33
Supplementary
+VCC
R1
RC
R1  100
vo
vs
RE
RC  1k
R E  2 k
  200
r  10k
Find the input and output resistance from
- Two port network, and
- Circuit theory
Ref:080130HKN
EE3110 Feedback Amplifiers
34
Circuit Theory
Rin  R1 // R
but ib 
where R 
vs
ib
vs  v E
v
and E  ib (   1)
r
RE
 vs  ib r  vE  ib [r  (   1) RE ]
Rin  R1 //[r  (   1) RE ] ~ 100
ib
 ib
vs
r
R1
vE
vo
RC
RE
T o find Rout , vs  0  ib  0
 Rout  RC  1k
Ref:080130HKN
EE3110 Feedback Amplifiers
35
Two Port Network
z11 f 
v1
 RE
i1 i2  0
Z12 f 
v1
 RE
i2 i1  0
Z 22 f 
v2
 RE
i2 i1  0
ib
vo
r
RE
RE
RE
 ib
vs
RE

+
 RE i 2
Ref:080130HKN
EE3110 Feedback Amplifiers
36
T heOpen LoopT ransadmittanceGain is found by settingfeedback signal  0
vs  ib (r  RE )
io  ib
AOL 
io
ib



vs ib (r  RE ) (r  RE )



Rin  (r  RE ) 1 
RE 
 (r  RE ) 



Rout  RE 1 
RE 
 (r  RE ) 
vs
ib
r
R1
RE
 ib
RC
RE
+
 RE io
Rin  Rin // R1 ~ R1  100
  Rout // RC ~ RC  1k
Rout
R'in
Ref:080130HKN
vo
Rin
EE3110 Feedback Amplifiers
Rout
R'out
37