Feedback
Xs
Xi
+
Xo
Xf
βf
* What is feedback?
Taking a portion of the signal arriving at the load and
feeding it back to the input.
* What is negative feedback?
Adding the feedback signal to the input so as to
partially cancel the input signal to the amplifier.
* Doesn’t this reduce the gain? Yes, this is the price we pay for using feedback.
* Why use feedback?
Provides a series of benefits, such as improved bandwidth,
that outweigh the costs in lost gain and increased
complexity in amplifier design.
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
1
Feedback Amplifier Analysis
Xs
Xi
+
Xo
Xf
βf
X f   f Xo
X o  AX i
where  f is called the feedback factor
where A is the am plifier' s gain, e.g . voltage gain
Xi  X s  X f
where X i is the net input signal to the basic am plifier,
X s  the signal from the source
The am plifier' s gain with feedback is given by
Af 
Xo
AX i


Xs
Xi  X f
ECE 352 Electronics II Winter 2003
1
A

Xf
Xi
A
1
 f Xo

A
1  f A
A
Xi
Ch. 8 Feedback
2
Advantages of Negative Feedback
* Gain desensitivity - less variation in amplifier gain with changes in
 (current gain) of transistors due to dc bias,
temperature, fabrication process variations,
etc.
* Bandwidth extension - extends dominant high and low frequency poles to
L
higher and lower frequencies, respectively.
 Hf  1   f A  H
 Lf 
1  f A




* Noise reduction - improves signal-to-noise ratio
* Improves amplifier linearity - reduces distortion in signal due to gain
variations due to transistors
* Cost of these advantages:
 Loss of gain, may require an added gain stage to compensate.
 Added complexity in design
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
3
Basic Types of Feedback Amplifiers
* There are four types of feedback amplifiers. Why?
 Output sampled can be a current or a voltage
 Quantity fed back to input can be a current or a voltage
 Four possible combinations of the type of output sampling and input
feedback
* One particular type of amplifier, e.g. voltage amplifier, current amplifier,
etc. is used for each one of the four types of feedback amplifiers.
* Feedback factor f is a different type of quantity, e.g. voltage ratio,
resistance, current ratio or conductance, for each feedback configuration.
* Before analyzing the feedback amplifier’s performance, need to start by
recognizing the type or configuration.
* Terminology used to name types of feedback amplifier, e.g. Series-shunt
 First term refers to nature of feedback connection at the input.
 Second term refers to nature of sampling connection at the output.
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
4
Basic Types of Feedback Amplifiers
Series - Shunt
Shunt - Series
Series - Series
Shunt - Shunt
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
5
Method of Feedback Amplifier Analysis
* Recognize the feedback amplifier’s configuration, e.g. Series-shunt
* Calculate the appropriate gain A for the amplifier, e.g. voltage gain.
 This includes the loading effects of the feedback circuit (some
combination of resistors) on the amplifier input and output.
* Calculate the feedback factor f
* Calculate the factor f A and make sure that it is:
1) positive and 2) dimensionless
A
* Calculate the feedback amplifier’s gain with feedback Af using A f  1   f A
* Calculate the final gain of interest if different from the gain
calculated, e.g. Current gain if voltage gain originally determined.
* Determine the dominant low and high frequency poles for the
original amplifier, but taking into account the loading effects of the
feedback network.
* Determine the final dominant low and high frequency poles of the
L
amplifier with feedback using


 Hf  1   f A  H
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
 Lf 
1   f A
6
Series-Shunt Feedback Amplifier - Ideal Case
*
Assumes feedback circuit does not load down the basic
amplifier A, i.e. doesn’t change its characteristics
 Doesn’t change gain A
 Doesn’t change pole frequencies of basic
amplifier A
 Doesn’t change Ri and Ro
*
For the feedback amplifier as a whole, feedback does
change the midband voltage gain from A to Af
Basic Amplifier
Feedback Circuit
A
Af 
*
1  f A
Does change input resistance from Ri to Rif


Rif  Ri 1   f A
Equivalent Circuit for Feedback Amplifier
*
Does change output resistance from Ro to Rof
Rof 
*
Ro
1  f A
Does change low and high frequency 3dB frequencies


 Hf  1   f A  H
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
 Lf 
L
1  f A


7
Series-Shunt Feedback Amplifier - Ideal Case
Midband Gain
V
A V
AVf  o  V i 
Vs Vi  V f
AV
AV
AV


Vf
 f Vo 1   f AV
1
1
Vi
Vi
Input Resistance
Vi  V f Vi   f Vo
V
Rif  s 

 Ri 1   f AV
Ii
Ii
Vi 
 R
i



Output Resistance
It
Vt
V  AV Vi
It  t
Ro
But Vs  0 so Vi  V f
and V f   f Vo   f Vt so
It 





Ro
V
Ro
so Rof  t 
It
1  AV  f

ECE 352 Electronics II Winter 2003
Ch. 8 Feedback

Vt  AV  V f
Vt  AV  f Vt

Ro
Ro
Vt 1  AV  f
8


Series-Shunt Feedback Amplifier - Ideal Case
Low Frequency Pole
For A  
1
L
A fo 
where
then A f   
Ao
s
Ao
 Lf 
1   f Ao

 A
o


 1 L

s

A 

1   f A  

Ao
1   f


1 L

s
L
1   f Ao
then A f   
Ao
s
1
H
where
A fo 
Ao
1   f Ao


 Ao

s
 1
H

ECE 352 Electronics II Winter 2003






A 

1   f A  

Ao
1   f
s

1

H


 Hf   H 1   f Ao






Ao
 L

1  s   f Ao 




Ao

 1   f Ao






 L  1 
1 
 
 1   f Ao  s 

A fo
  Lf
1 

s





Low 3dB frequency lowered by feedback.
High Frequency Pole
For A  














Ao


s
  f Ao 
1 
 H



Ao

 1   f Ao







s
1 

  H 1   f Ao 



A fo

1  s
  Hf

Upper 3dB frequency raised by feedback.
Ch. 8 Feedback
9




Practical Feedback Networks
*
Vi
Vo
Vf
*
*
*
* How do we take these
loading effects into account?
*
ECE 352 Electronics II Winter 2003
Feedback networks consist of a set of resistors
 Simplest case (only case considered here)
 In general, can include C’s and L’s (not
considered here)
 Transistors sometimes used (gives variable
amount of feedback) (not considered here)
Feedback network needed to create Vf feedback
signal at input (desirable)
Feedback network has parasitic (loading) effects
including:
Feedback network loads down amplifier input
 Adds a finite series resistance
 Part of input signal Vs lost across this series
resistance (undesirable), so Vi reduced
Feedback network loads down amplifier output
 Adds a finite shunt resistance
 Part of output current lost through this shunt
resistance so not all output current delivered to
load RL (undesirable)
Ch. 8 Feedback
10
Equivalent Network for Feedback Network
*
*
*
*
*
*
*
*
*
ECE 352 Electronics II Winter 2003
Need to find an equivalent network for the
feedback network including feedback effect
and loading effects.
Feedback network is a two port network
(input and output ports)
Can represent with h-parameter network
(This is the best for this particular feedback
amplifier configuration)
h-parameter equivalent network has FOUR
parameters
h-parameters relate input and output
currents and voltages
Two parameters chosen as independent
variables. For h-parameter network, these
are input current I1 and output voltage V2
Two equations relate other two quantities
(output current I2 and input voltage V1) to
these independent variables
Knowing I1 and V2, can calculate I2 and V1 if
you know the h-parameter values
h-parameters can have units of ohms, 1/ohms
or no units (depends on which parameter)
Ch. 8 Feedback
11
Series-Shunt Feedback Amplifier - Practical Case
*
*
*
Feedback network consists of a set of resistors
These resistors have loading effects on the basic
amplifier, i.e they change its characteristics, such as
the gain
Can use h-parameter equivalent circuit for feedback
network
 Feedback factor f given by h12 since
Vf
V
h12  1

f
V2 I 0 Vo
1
Feedforward factor given by h21 (neglected)
 h22 gives feedback network loading on output
 h11 gives feedback network loading on input
Can incorporate loading effects in a modified basic
amplifier. Basic gain of amplifier AV becomes a new,
modified gain AV’ (incorporates loading effects).
Can then use feedback analysis from the ideal case.

*
*
AV '
AVf 
1   f AV '
Rif  Ri 1   f AV '
 Hf  1   f AV ' H
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
 Lf 
Rof 
Ro
1  AV '  f 
L
1   f AV '
12
Series-Shunt Feedback Amplifier - Practical Case
Summary of Feedback Network Analysis
*
*
*
*
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
How do we determine the h-parameters
for the feedback network?
For the input loading term h11
 Turn off the feedback signal by
setting Vo = 0.
 Then evaluate the resistance seen
looking into port 1 of the feedback
network (also called R11 here).
For the output loading term h22
 Open circuit the connection to the
input so I1 = 0.
 Find the resistance seen looking into
port 2 of the feedback network (also
called R22 here).
To obtain the feedback factor f (also
called h12 )
 Apply a test signal Vo’ to port 2 of the
feedback network and evaluate the
feedback voltage Vf (also called V1
here) for I1 = 0.
 Find f from f = Vf/Vo’
13
Series-Shunt Feedback Amplifier - Practical Case
Summary of Approach to Analysis
Basic Amplifier
*
Practical Feedback Network
*
Evaluate modified basic amplifier
(including
loading effects of feedback network)
 Including h11 at input
 Including h22 at output
 Including loading effects of source resistance
 Including load effects of load resistance
Analyze effects of idealized feedback network
using feedback amplifier equations derived
AVf 
AV '
1   f AV '
Rif  Ri '1   f AV '
Modified Basic Amplifier
 Hf  1   f AV ' H
*
Idealized Feedback Network
ECE 352 Electronics II Winter 2003
Rof 
Ro '
1  AV '  f 
 Lf 
L
1   f AV '
Note
 Av’ is the modified voltage gain including the
effects of h11 , h22 , RS and RL.
 Ri’, Ro’ are the modified input and output
resistances including the effects of h11 , h22 , RS
and RL.
Ch. 8 Feedback
14
Example - Series-Shunt Feedback Amplifier
*
*
Two stage amplifier
Each stage a CE
amplifier
* Transistor
parameters
Given: 1= 2 =50,
rx1=rx2=0
* Coupled by
capacitors, dc biased
separately
* DC analysis:
I
I C1  0.94 m A, g m1  C1  36 m A/ V ,
VT
DC analysis for each stage can be done separately
since stages are isolated (dc wise) by coupling capacitors.
r 1 
I
I C 2  1.85 m A, g m 2  C 2  71 m A/ V ,
VT
r 2 
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
1
 1.4 K
g m1
2
 0.7 K
g m2
15
Example - Series-Shunt Feedback Amplifier
*
Redraw circuit to show



Feedback circuit
Type of output sampling
(voltage in this case = Vo)
Type of feedback signal to input
(voltage in this case = Vf)
+
_
Vi
+
Vo
_
+
Vf _
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
16
Example - Series-Shunt Feedback Amplifier
Input Loading Effects
R1  h11  R f 1 R f 2
 0.1K 4.7 K
Vo=0
 0.098 K
Output Loading Effects
R2  h22  R f 1  R f 2
I1=0
 4.7 K  0.1K  4.8 K
Amplifier with Loading Effects
R2
R1
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
17
Example - Series-Shunt Feedback Amplifier
*
*
Construct ac equivalent circuit at midband frequencies including
loading effects of feedback network.
Analyze circuit to find midband gain
(voltage gain for this series-shunt configuration)
R1
R2
R2
R1
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
18
Example - Series-Shunt Feedback Amplifier
Midband Gain Analysis
AVo 
 Vo
Vo

V
Vs
 2
 V 2

 V
 o1

 Vo1

 V
  1

 V 1

 V
 i1
 Vi1

 V
 s






Vo
  g m 2 RC 2 R2  71m A/ V 4.9 K 4.8 K  172
V 2



V 2
 1 since rx 2  0
Vo1

Vo1
  g m1 RC1 R12 R22 r 2  36m A/ V 10K 47K 33K 0.7 K  22.8
V 1
V 1
I 1r 1
r 1
1. 4 K



 0.22
Vi1
I 1r 1  I 1  g m1V 1 R1
r 1  1   R1
1.4 K  510.098K 
V
I r
 I 1  g m1V 1 R1
Ri1  i1   1  1
 r 1  1   R1  1.4 K  510.098K   6.4 K
I 1
I 1
6.4 K 150K 47K
Ri1 R11 R21
Vi1
5 .4 K



 0.52
Vs
Rs  Ri1 R11 R21
10.4 K
5 K  6.4 K 150K 47K
AVo 
Vo
  1721 22.80.220.52  449
Vs
AVo ( dB)  20 log 449  53dB
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
19
Midband Gain with Feedback
*
f 
*
Determine the feedback factor f
Xf
Xo

Vf
Vo '

Rf 1
Rf 1  Rf 2

0.1K
 0.021
0.1K  4.7 K
Calculate gain with feedback Avf
 f AVo  449(0.021)  9.4
AVfo 
AVo
1   f AVo

449
449

 43.1
1  449(0.021) 10.4
AVfo (dB)  20 log 43.1  32.7dB
*
Note
 f Avo > 0 as necessary for negative feedback
 f Avo is large so there is significant feedback. For f Avo  0, there is almost
no feedback.
 Can change f and the amount of feedback by changing Rf1 and/or Rf2.
 NOTE: Since f Avo >> 0
 f AVo  449(0.021)  9.4
AVfo 
1
f

Rf1  Rf 2
Rf1
 1
Rf 2
Rf1
AVfo 
AVo
1   f AVo

AVo
1
1


 47.6
 f AVo  f 0.021
AVfo (dB)  20 log 47.6  33.6dB
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
20
Input and Output Resistances with Feedback
*
Determine input Ri and output Ro resistances with loading effects of feedback network.
Ro  R2 RC 2
Ri  RS  RB1 Ri1
 5K  38.5 K 6.4 K  10.5K
*
 4.9 K 4.8K  2.4 K
Calculate input Rif and output Rof resistances for the complete feedback amplifier.

Rif  Ri 1   f AVo

Rof 
 10.5K 1  449(0.021)  109.5K
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
Ro
1   f AVo

2.4K
 0.23K
10.4
21
Equivalent Circuit for Series-Shunt Feedback Amplifier
*
*
*
Voltage gain amplifier
Modified voltage gain, input
and output resistances
 Included loading effects of
feedback network
 Included feedback effects
of feedback network
 Include source resistance
effects
Significant feedback, i.e.
f Avo is large and positive
 f AVo  449(0.021)  9.4
AVfo 
V
AVfo   o
 VS

AVo
 
 43.1

 f 1   f AVo
AVfo (dB)  20 log 43.1  32.7dB
ECE 352 Electronics II Winter 2003


Rif  Ri 1   f AVo  109.5K
Ro
Rof 
 0.23K
1   f AVo
Ch. 8 Feedback

AVo
A
1
 Vo 
1   f AVo  f AVo  f
Rf1  Rf 2
Rf1
 1
Rf 2
Rf 1
 47.6
AVf (dB )  20 log 47.6  33.6dB
22
Low Frequency Poles and Zeros for Series-Shunt Feedback Amplifier
*
*
*
Six capacitors:
 Input and output coupling capacitors C1 and C5
 Emitter bypass capacitors C3 and C4
 Interstage coupling capacitors C2
 Feedback coupling capacitor C6
Analyze using Gray-Searle (Short Circuit) Technique one capacitor at a time
Find dominant low frequency pole (highest frequency one)
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
23
Example - Input Coupling Capacitor’s Pole Frequency
V
RxC1  x  RS  RB1 Ri1
Ix
where
V
I r  I 1  g m1V 1 R1
Ri1  i1   1  1
I 1
I 1
 r 1  1   R1  1.4 K  510.098K   6.4 K
so
RxC1  RS  RB1 Ri1  5 K  35.8 K 6.4 K  10.4 K
 PL1 
Equivalent circuit for C1
1
R xC1C1

1
 19.2 rad / s
10.4 K 5F
Note that there are some loading effects of
the feedback network on this pole
frequency. In Ri1 the feedback resistors
determine R1
R1  R f 1 R f 2
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
24
Example - Interstage Coupling Capacitor’s Pole Frequency
V
RxC 2  x  RC1  RB 2 r 2
Ix
 10K  19.4 K 0.7 K  10.7 K
so
 PL2 
Equivalent circuit for C2
1
RxC 2C2

1
 18.7 rad / s
10.7 K 5F
Note: No RE2 since
C4 shorts it out.
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
25
Example - Feedback Coupling Capacitor’s Pole Frequency
V
RxC 6  x  RC 2  R2
Ix
 4.9 K  4.8K  9.7 K
so
1
1
 PL6 

 20.6 rad / s
RxC 6C6 9.7 K 5F
Equivalent circuit for C6
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
26
Example - Emitter Bypass Capacitor’s Pole Frequency
REx 
Vx
Ix
I E1 
Vx
RE1
Vx   I 1 r 1  Rs RB1   I1R1
At node E I x  I E1  I 1  g m1V 1  0 so
Vx 

 Ix  R 
E1

I 1   
1  g m1r 1 


At node E ' I E1  I x  I1  0 so
Equivalent circuit for C3

V 
I1  I E1  I x    I x  x 
so Vx becomes
RE1 

Vx   I 1 r 1  Rs RB1   I1R1
Vx 

 Ix  R 

V 
E1
  R1  I x  x 
 r 1  Rs RB1 
RE1 
1  g m1r 1 



IE1
ECE 352 Electronics II Winter 2003
 r  Rs RB1

V 
  1
 R1  I x  x 
RE1 
 1  g m1r 1

Rearrangin g, we get
R xC3 
r 1  Rs RB1  R1 1  g m1r 1  R  0.2 K
Vx

r 1  Rs RB1  R1  RE1 1  g m1r 1  E1
Ix
 PL3 
1
1

 100 rad / s
RxE1C E1
0.2 K 50 F 
Ch. 8 Feedback
27
Example - Emitter Bypass Capacitor’s Pole Frequency
REx 
Vx
Ix
IE2 
Vx
RE 2
At node E I x  I E 2  I  2  g m 2V 2  0 so
I x  I E 2  I  2 1  g m 2 r 2  
Vx
 I  2 1  g m 2 r 2 
RE 2
But we also know that
Vx   I  2 r 2  RC1 RB 2  so I  2  
Equivalent circuit for C4
Vx
r 2  RC1 RB 2
so I x becomes

1  g m 2 r 2  
 1
I x  Vx 


R
r

R
R


E
2

2
C
1
B
2


Rearrangin g, we get
R xC4 
IE2
Vx
 RE 2
Ix

RC1 RB 2 
 r 2 


1  g m 2 r 2  


10 K 19.4 K 
  0.7 K
 4.7 K  0.7 K 

50


1
1
 PL 4 

 28.5 rad / s
RxE 2C E 2 0.7 K 50 F 
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
28
Example - Output Coupling Capacitor’s Pole Frequency

RxC 5  RC 2 R f 1  R f 2

 4.9 K 4.8K  2.4 K
so
Equivalent circuit for C5
 PL5 
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
1
RxC 5C5

1
 41.6 rad / s
2.4 K 10F
29
Zeros for Series-Shunt Feedback Amplifier Example
*
*
Coupling capacitors C1, C2 and C5 give zeros
at  = 0 since ZC = 1/sC and they are in the
signal line.
Emitter bypass capacitors C3 and C4 give a
zero when the impedance ZCE || RE.
Z C E RE 
1
 1
1


 RE Z C
E






Z C E RE   when s 
Z3 
Z 4 
Vo
  g m 2 RC 2 Z C 6  R2 
V 2
RC 2




1
1


Z C 6  R2    
RC 2  1


 R2  


 sC6
 
*
1
1  sR2C6  sRC 2C6 
 RC 2 1  sR2C6  

 

 RC 2 1  sR2C6  
1  sC6 R2  RC 2  
ECE 352 Electronics II Winter 2003
1
1
 1




sC
E
R

 E


RE
1  sRE C E
1
RE C E
1
1

 4.3 rad / s
RE1C3 4.7 K 50F 
1
RE 2C 4

1
 4.3 rad / s
4.7 K 50F 
Feedback capacitor C6 gives a zero when
[ZC6 +R2] || RC2 when
1  sC6 R2  RC 2   0 or s 
 z6 
Ch. 8 Feedback
1
C6 R2  RC 2 
1
1

 20.6 rad / s
C6 R2  RC 2  5F 4.8K  4.9 K 
30
Series-Shunt Example - Low Frequency
*
Midband Gain
Low Frequency Poles
AVo  449
AVo (dB)  53 dB
PL1  19.2 rad / s
PL4  28.5 rad / s
Low Frequency Zeros
 Z 1  0 rad / s  Z 4  4.3 rad / s
 f  0.021
1   f AVo  10.4
PL2  18.7 rad / s
PL5  41.6 rad / s
 Z 2  0 rad / s
AVfo  43.1
AVfo (dB)  32.7 dB
PL3  100 rad / s
PL6  20.6 rad / s
 Z 3  4.3 rad / s
Low 3dB Frequency
ECE 352 Electronics II Winter 2003
 PL  100 rad / s
 PLf 
Ch. 8 Feedback
 PL5  0 rad / s
 Z 6  20.6 rad / s
100 rad / s
 PL

 9.6 rad / s
1   f AVo
10.4
31
Series-Shunt Example - High Frequency
*
*
*
*
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
Substitute hybrid-pi model for
transistor with C and C
Short all coupling capacitors
and emitter bypass capacitors
Include loading effects of
feedback network R1 and R2
Find high frequency poles and
zeros using Gray-Searle (Open
Circuit) Method
32
Series-Shunt Example - High Frequency Pole - C1
Given: C1 = 15 pF
I x  I  1  I s or I s  I x  I  1  I x 
Also I1  I  1 (1  g m1r 1 )  I x 
IS
Vx
since
r 1
I 1 
V 1 Vx

r 1 r 1
Vx
(1  g m1r 1 )  I x
r 1

V

V 
Vx  I s RS RB1   I1 R1   I x  x RS RB1    x (1  g m1r 1 )  I x  R1
r 1 

 r 1

Rearrangin g
R xC  1 
I1

r 1 RS RB1  R1 
Vx

I x r 1  RS RB1  (1  g m1r 1 ) R1
1.4 K 5 K 35.8 K  0.098 K 
1.4 K  5 K 35.8 K  51(0.098 K )
1.4 K (4.49 K )
 0.95 K
1.4 K  4.49 K
1
1
 7.0 x10 7 rad / s

 PH 1 
R xC  1C 1 0.95 K 15 pF

ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
33
Series-Shunt Example - High Frequency Pole - C1
V 1  Ve1  I 1 r 1  R1 1  g m1r 1 
Given: C1 = 1.2 pF
Also V 1  Ve1  I S RB1 RS   I x  I 1 RB1 RS 


RB1 RS
so I 1  I x 

 r 1  R1 1  g m1r 1   RB1 RS 
and
Vx  Vo1  V 1  Ve1  I 1 r 1  R1 1  g m1r 1 
 R R r  R 1  g m1r 1 
 I x  B1 S  1 1

 r 1  R1 1  g m1r 1   RB1 RS 
Node C
So combining with our expression above for Vx  Vo1 we get
Vo1
Vo1
I x  g m1V 1 
 I x  g m1r 1I 1 
0
RC1 RB 2 r 2
RC1 RB 2 r 2




V
R xC 1  x
Ix



RB1 RS
Vo1
I x  g m1r 1I x 
0

 r 1  R1 1  g m1r 1   RB1 RS  RC1 RB 2 r 2




 
RB1 RS
so Vo1   I x 1  g m1r 1 
  RC1 RB 2 r 2

 r 1  R1 1  g m1r 1   RB1 RS  

ECE 352 Electronics II Winter 2003


 RB1 RS r 1  R11  g m1r 1   RC1 RB 2 r 2
so using I 1

 rr1 RR1  RRB1RRS 11ggm1rr1
1


 35.8K 5K 1.4 K  0.098K (51)  10K 19.4 K 0.7 K

B1 S
1
m1  1
K  35.8K 5K (51)
11..44KK 350.098
.8K 5K  0.098K (51)
 16.1K
 PH 2 
1
1
 5.2 x107 rad / s
RxC1C1 16.1K (1.2 pF )
Ch. 8 Feedback

34
Series-Shunt Example - High Frequency Pole - C2
Given: C2 = 12 pF
V
RxC 2  x  RC1 RB 2 r 2
Ix
 10K 19.4 K 0.7 K  0.63K
 PH 3 
1
RxC 2C 2

1
0.63K 12 pF 
 1.32x108 rad / s
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
35
Series-Shunt Example - High Frequency Pole - C2
Given: C2 = 1.4 pF
KCL at C2
Vo
RC 2 R2 
so
Vo 
Also
 g m 2V 2  I x 

RC 2 R2   g m2 Vx  Vo   I x  0
 I x  g m 2Vx  RC 2 R2
Vo

1  g m 2 r 2

Vo  Vx  I x RC1 RB 2 r 2

so com bining we get




V
RxC 2  x  RC 2 R2  RC1 RB 2 r 2 1  g m 2 RC 2 R2
Ix

 



 4.9 K 4.8 K  10K 19.4 K 0.7 K 1  71m A/ V 4.9 K 4.8 K
 111.5 K
 PH 4 
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
1
RxC 2C 2

1
 6.4 x106 rad / s
111.5 K 1.4 pF 
36

Series-Shunt Example - High Frequency Zeros - C1 & C2
I2
Node C2
For CE amplifier, a high frequency zero
occurs when ZH = gm/C
g
 ZH 1  m1 
C1
36 m A/ V
1.2 pF
 3.0 x1010 rad / s
71 m A/ V
g
 ZH 2  m2 
 5.1x1010 rad / s
C 2
1.4 pF
ECE 352 Electronics II Winter 2003
 I  2  g mV 2 
Vo
V
 o 0
RC 2 R2
V 2  Vo
V  Vo
 2
 sC  2 V 2  Vo  so substituti ng
ZC 2
 1

 sC 
2 

V
V
 sC  2 V 2  Vo   g m 2V 2  o  o  0
RC 2 R2
But I  2 



1
1
V 2 g m 2  sC  2  Vo  sC  2 

R2 RC 2

and rewriting as a voltage ratio, we get
Rearrangin g




 g m 2  sC  2
V0

V 2

1
1
 sC  2 

R2 RC 2

g
So V0  0 when s  m 2
C 2
Ch. 8 Feedback




  0





s
1 


 gm2



C 2  


  gm

1
1 
 sC  2 


R2 RC 2 

37
Series-Shunt Example - High Frequency
*
Midband Gain
AVo  449
AVo (dB)  53 dB
 f  0.021
1   f AVo  10.4
AVfo  43.1
AVfo (dB)  32.7 dB
High 3dB Frequency
High Frequency Poles
High Frequency Zeros
PH1  7.0 x107 rad / s
PH 3  1.3x108 rad / s
 ZH 1  3.0 x1010 rad / s
 ZH 2   rad / s
PH 2  5.2 x107 rad / s
PH 4  6.4 x106 rad / s
 ZH 3  5.1x1010 rad / s
 ZH 4   rad / s
 PH  6.4x106 rad / s
ECE 352 Electronics II Winter 2003


 PHf   PH 1   f AVo  6.4x106 rad / s10.4  6.7 x107 rad / s
Ch. 8 Feedback
38