Feedback: 8.8 The Stability Problem
1
8.8.1 Transfer Function of the
Feedback Amplifier
A s 
Af  s  
1  A s    s 
A  j 
Af  j  
1  A  j    j 
Figure 8.1 General structure of the feedback amplifier. This is a signal-flow diagram, and the quantities x represent either voltage or current signals.
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8.8.1 Transfer Function of the
Feedback Amplifier
Loop Gain:
L  j    A  j     j    A  j     j   e j   
Suppose that there is a frequency, 180, for which
j  180 
 180   180  e
 e j180  1
Then
A  j 
Af  j  
1  A  j    j 
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8.8.1 Transfer Function of the
Feedback Amplifier
Af  j  
If
A  j 
1  A  j    j 
A  j    j   1 , then Af  j   A  j 
(Armstrong’s “regeneration”)
If A  j    j   1 , then Af  j   
(Armstrong’s “oscillation”)
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8.9.1 Stability and Pole Location
A quick review
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v  t   e o t e jn t  e  jn t   2 e o t cos  n t 
Figure 8.29 Relationship between pole location and transient response.
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8.9.2 Poles of the Feedback
Amplifier
Recall the closed loop transfer function
(transfer function with feedback):
A s 
Af  s  
1  A s    s 
Poles:
1  A s    s   0
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8.9.3 Amplifier with a SinglePole Response
Simple example: suppose an amplifier without
feedback has a single pole at s    P :
A0
A s 
1  s P
With negative feedback, we saw earlier that,
assuming  constant, the amplifier gain is:
A0 1  A0  
Af  s  
1  s P 1  A0  
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8.9.3 Amplifier with a SinglePole Response
A0 1  A0  
Af  s  
1  s P 1  A0  
The feedback has moved the pole along the
axis from   P to P 1  A0   .
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8.9.3 Amplifier with a SinglePole Response
Let s  j  . When  P 1  A0   ,
A0
A0
A0  P
A j   


1  j  P
j  P
j
A0 1  A0  
A0 1  A0  
Af  j   

1  j  P 1  A0   j  P 1  A0  
A0  P
Af  j   
j
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8.9.3 Amplifier with a SinglePole Response
Summary: when  P 1  A0   ,
A0  P
Af  j    A  j   
j
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8.9.3 Amplifier with a SinglePole Response
Figure 8.30 Effect of feedback on (a) the pole location and (b) the frequency response of an amplifier having a single-pole open-loop response.
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8.10.1 Gain and Phase Margins
A
dB
 20 log10 A  j    j 
Instability:
  180
AND
A   1; A  dB  0
Figure 8.36 Bode plot for the loop gain A illustrating the definitions of the gain and phase margins.
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8.10.2 Effect of Phase Margin on
Closed-Loop Response
Typical phase margin design value: 45
Value of phase margin affects shape of closedloop gain versus frequency plot.
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8.10.2 Effect of Phase Margin on
Closed-Loop Response
Example: suppose
 0
A
j


 A0 ; A0 


0;
 
real
1
Closed loop gain at low frequencies:
A  j 
A0
A0
1
 0
Af  j  




1  A  j  
1  A0 
A0  
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8.10.2 Effect of Phase Margin on
Closed-Loop Response
Suppose A  j 1    1 for some frequency  1 .
A  j 1    A  j 1   e
 
j  1
e
 
j  1
e
 j
where   0 . Thus, we can write
A  j 1  
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
e
 j
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8.10.2 Effect of Phase Margin on
Closed-Loop Response
phase margin =     180
phase margin = 180  
  180  phase margin
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8.10.2 Effect of Phase Margin on
Closed-Loop Response
Closed loop gain at frequency 1 :
Af  j 1  
A  j 1 
1  A  j 1  
1
A f  j 1  
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
1
e  j
1

e  j
; A  j 1  
1
1

e  j
 j
e

 j

1

e

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8.10.2 Effect of Phase Margin on
Closed-Loop Response
Magnitude of the closed loop gain at
frequency 1 :
Af  j 1 
1
e j
 j
e
1
1
1



 j
 j
 1 e
 1 e
 1  e j
For a phase margin of 45
  180  phase margin = 180  45  135
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8.10.2 Effect of Phase Margin on
Closed-Loop Response
Magnitude of the closed loop gain at
frequency 1 :
Af  j 1 
1
1
1
1 1



 j
 j135
 1 e
 1 e
 0.7654
Af  j 1  
1
1.31

Thus the gain peaks at 131% of its value at low
frequencies with a phase margin of 45.
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8.10.2 Effect of Phase Margin on
Closed-Loop Response
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8.10.3 An Alternative Approach
for Investigating Stability (Bode)
Construct a Bode plot for A rather than for A  .
A  dB  20log10 A   20log10 A 
A  dB  20log10 A  20log10 
A
dB
 20log10 A  20log10
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
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8.10.3 An Alternative Approach
for Investigating Stability (Bode)
Example:
105
A
5
6
7
1

j
f
10
1

j
f
10
1

j
f
10




    tan 1  f 105   tan 1  f 106   tan 1  f 107 
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8.10.3 An Alternative Approach
for Investigating Stability (Bode)
Af dB  20log10 1   
50
3.15 103
60
1.00  103
85
5.62  105
Figure 8.37 Stability analysis using Bode plot of |A|.
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8.10.3 An Alternative Approach
for Investigating Stability (Bode)
• Note that instability occurs for smaller
values of Af .
• Result that we will not prove:
– If the 20log10 1   horizontal line intersects
the 20log10 A curve on a segment for which
the slope is –20 dB/decade, then the phase
margin will be a minimum of 45º.
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8.11 Frequency Compensation
• Frequency compensation modifies the openA an
s  amplifier
loop transfer function
of
(with three or more poles) so that the
Af iss stable
closed-loop transfer function
for any value chosen for the closed-loop
gain.
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8.11 Frequency Compensation:
Theory
• The simplest method of frequency
compensation modifies the original openloop transfer function, A j   , by
introducing a new low frequency pole at  D
to form a new open-loop transfer function,
A '  j   , which has a slope of –20 dB/decade
at the intersection of the 20 log10 A '  j  
curve and the 20log10 1   curve.
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8.11 Frequency Compensation:
Example:
Stability for
closed-loop gains
of 40 dB or higher
by adding a new
pole at  D .
Figure 8.38 Frequency compensation for  = 102. The response labeled A is obtained by introducing an additional pole at fD.
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8.11 Frequency Compensation:
Difficulty:
Stability achieved,
but high open-loop
gain, and hence the
benefits of negative
feedback, only for:
f  10 Hz
2
Figure 8.38 Frequency compensation for  = 102. The response labeled A is obtained by introducing an additional pole at fD.
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8.11.3 Miller Compensation and
Pole Splitting
The Miller effect
allows shifting f P1
to f D ' and shifting
f P 2 to a higher
frequency.
Stability and high
open-loop gain for
3
f  10 Hz
Figure 8.38 Frequency compensation for  = 102. The response labeled A is obtained by introducing an additional pole at fD. The A response is
obtained by moving the original low-frequency pole to f D.
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8.11.3 Miller Compensation and
Pole Splitting
Figure 8.40 (a) A gain stage in a multistage amplifier with a compensating capacitor connected in the feedback path and (b) an equivalent circuit. Note
that although a BJT is shown, the analysis applies equally well to the MOSFET case.
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8.11.3 Miller Compensation and
Pole Splitting
Transresistance
amplifier:
Node equations:
VB
 s C1 VB  s C f VB  VC   0
B:  Ii 
R1
VC
 s C2 VC  0
C: s C f VC  VB   gm VB 
R2
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8.11.3 Miller Compensation and
Pole Splitting
B:
VB
 Ii 
 s C1 VB  s C f VB  VC   0
R1
VC
s C f VC  VB   g m VB 
 s C2 VC  0
R2
C:
Collect terms:
 1

VB 
 s C1  s C f   VC   s C f   I i
 R1



1
VB   s C f  g m   VC  s C f 
 s C2   0
R2


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8.11.3 Miller Compensation and
Pole Splitting
MATLAB:
%Declare symbolic variables
syms s R1 R2 C1 C2 Cf gm Ii Vb Vc Vo A b x
%Define equations: Ax = b
A=[1/R1+s*(C1+Cf) -s*Cf; gm-s*Cf
1/R2+s*(Cf+C2)];
b=[Ii; 0];
%Solve the equations:
x=A\b;
Vo=x(2);
Vo=simplify(Vo)
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8.11.3 Miller Compensation and
Pole Splitting
MATLAB:
Vo = R1*Ii*R2*(-gm+s*Cf)/
(s*R1*Cf*R2*gm+1+s*R1*C1+s*R1*Cf+s*R2*Cf+s^2*R2*Cf*R1*C1
+s*R2*C2+s^2*R2*C2*R1*C1+s^2*R2*C2*R1*Cf)


sC f  gm R1 R2
Vo

Ii 1  s C R  C R  C g R R  R  R   s 2 C C  C C  C  R R
2 2
f m 1 2
1
2 
f 1
2  1 2
 1 1
 1 2








D s   1  s C1 R1  C2 R2  C f gm R1 R2  R1  R2   s2 C1 C2  C f C1  C2  R1 R2




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8.11.3 Miller Compensation and
Pole Splitting



D s   1  s C1 R1  C2 R2  C f gm R1 R2  R1  R2   s2 C1 C2  C f C1  C2



 R1 R2
(3 capacitors, 2 poles? Miller’s Theorem.)



s 
s 

D s   1 
1
1



'
'

 P1   P 2 



1
1 
s2

s


 '

'
' '



 P1
P2 
P1 P 2
'
P1
Suppose one pole is dominant: 
P' 2
s
s2
D s   1 

'
' '
P1 P
1 P2
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8.11.3 Miller Compensation and
Pole Splitting



D s   1  s C1 R1  C2 R2  C f gm R1 R2  R1  R2   s2 C1 C2  C f C1  C2



 R1 R2
s
s2
D s   1 

'
' '
P1 P
1 P2
By comparison:
1
'
 P1 
C1 R1  C2 R2  C f gm R1 R2  R1  R2


Prepare for logarithmic differentiation:
 


'
ln P
1   ln C1 R1  C2 R2  C f gm R1 R2  R1  R2
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8.11.3 Miller Compensation and
Pole Splitting
Logarithmic differentiation:
 


'
ln P
1   ln C1 R1  C2 R2  C f gm R1 R2  R1  R2

'
g m R1 R2  R1  R2
1 d P1

0
'
dC
C1 R1  C2 R2  C f g m R1 R2  R1  R2
 P1
f


'
d P
1
Thus, dC f  0 , so the Miller capacitor
lowers the frequency of the lower
frequency pole.
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8.11.3 Miller Compensation and
Pole Splitting
'

What about P2 ?



D s   1  s C1 R1  C2 R2  C f gm R1 R2  R1  R2   s2 C1 C2  C f C1  C2



s
s2
D s   1 

'
' '
P1 P
1 P2
 R1 R2
By comparison:
1
' '
P
1 P2
' 
P
2
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

 C1 C2  C f C1  C2  R1 R2


1


' C C  C C  C  R R
P
1  1 2
f 1
2  1 2
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8.11.3 Miller Compensation and
Pole Splitting
'
P2 
1


' C C  C C  C  R R
P
1  1 2
f 1
2  1 2
Prepare for logarithmic differentiation:
 
  

'
'
ln P


ln

2
P1  ln C1 C2  C f C1  C2
  ln R1 R2 
'
'
C1  C2
1 d P 2
1 d P1


'
'
dC
dC
C1 C2  C f C1  C2
P 2
 P1
f
f
'

gm R1 R2  R1  R2
1 d P 2
   
'
dC f
 C1 R1  C2 R2  C f gm R1 R2  R1  R2
P
2



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


C1  C2

 C C C C C

1 2
f 1
2

Copyright  2004 by Oxford University Press, Inc.


40
8.11.3 Miller Compensation and
Pole Splitting
'

gm R1 R2  R1  R2
1 d P 2

  
'
dC
 C1 R1  C2 R2  C f gm R1 R2  R1  R2
P 2
f






C1  C2

 C C C C C

1 2
f 1
2




C f d '
C f gm R1 R2  R1  R2
C f C1  C2
P2 

'
dC
C1 C2  C f C1  C2
C1 R1  C2 R2  C f g m R1 R2  R1  R2
P 2
f
Note:


C f g m R1 R2  R1  R2


C1 R1  C2 R2  C f g m R1 R2  R1  R2

C f C1  C2


C1 C2  C f C1  C2
Microelectronic Circuits - Fifth Edition
Sedra/Smith


1
C1 C2
1
1
C f C1  C2





1
1
Copyright  2004 by Oxford University Press, Inc.
41
8.11.3 Miller Compensation and
Pole Splitting




C f d '
C f gm R1 R2  R1  R2
C f C1  C2
P2 

'
dC
C1 C2  C f C1  C2
C1 R1  C2 R2  C f g m R1 R2  R1  R2
P 2
f

For the Miller effect to be large,



gm R2  


:
C f d '
C f C1  C2
P2  1 
0
'
dC
C1 C2  C f C1  C2
P 2
f
'
d P
2
dC f


0
Microelectronic Circuits - Fifth Edition
Sedra/Smith
Copyright  2004 by Oxford University Press, Inc.
42
8.11.3 Miller Compensation and
Pole Splitting
The Miller effect can split the two poles, pushing
the frequency of the lower one lower and
pushing the higher frequency higher.
Shifting the lower frequency pole gives stability
without the addition of an extra pole.
Microelectronic Circuits - Fifth Edition
Sedra/Smith
Copyright  2004 by Oxford University Press, Inc.
43
8.11.3 Miller Compensation and
Pole Splitting
Shifting the
higher frequency
pole shifts the
– 20 dB/decade
segment to the
right and thus
gives broader
bandwidth.
Microelectronic Circuits - Fifth Edition
Sedra/Smith
Copyright  2004 by Oxford University Press, Inc.
44