Series-Series Feedback Amplifier - Ideal Case
Voltage fedback
to input
*
Output current
sampling
*
*
*
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 this configuration, the appropriate gain is the
TRANSCONDUCTANCE GAIN A = ACo = Io/Vi
For the feedback amplifier as a whole, feedback changes
midband transconductance gain from ACo to ACfo
ACo
ACfo 
1   f ACo
Feedback changes input resistance from Ri to Rif
Rif  Ri 1   f ACo




*
Feedback changes output resistance from Ro to Rof
*
Feedback changes low and high frequency 3dB
frequencies
Rof  Ro 1   f ACo


 Hf  1   f ACo  H
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
 Lf 
L
1   f ACo


1
Series-Series Feedback Amplifier - Ideal Case
Gain (Transconductance Gain)
ACfo 
Io
A V
A
ACo
ACo
 Co i  Co 

Vf
 f I o 1   f ACo
Vs Vi  V f
1
1
Vi
Vi
Input Resistance
Rif 
Vs Vi  V f Vi   f I o


 Ri 1   f ACo
Ii
Ii
Vi 
 R
i



Output Resistance
V
Rof 
It
But Vs  0 so Vi  V f
and V f   f I o   f I t so Vi    f I t
V  I t  ACoVi Ro  I t  ACo   f I t Ro
+
V
-
 I t 1   f ACo Ro
so Rof 
V
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
V
 Ro 1   f ACo 
It
2
Equivalent Network for Feedback Network
*
*
*
*
*
*
*
*
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
Feedback network is a two port network
(input and output ports)
Can represent with Z-parameter network
(This is the best for this feedback
amplifier configuration)
Z-parameter equivalent network has
FOUR parameters
Z-parameters relate input and output
currents and voltages
Two parameters chosen as independent
variables. For Z-parameter network,
these are input and output currents I1
and I2
Two equations relate other two quantities
(input and output voltages V1 and V2) to
these independent variables
Knowing I1 and I2, can calculate V1 and
V2 if you know the Z-parameter values
Z-parameters have units of ohms !
3
Series-Series 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 z-parameter equivalent circuit for feedback
network
 Feedback factor f given by z12 since
z12 
V1
I2

I1 0
Vf
Io
f
Feedforward factor given by z21 (neglected)
 z22 gives feedback network loading on output
 z11 gives feedback network loading on input
Can incorporate loading effects in a modified basic
amplifier. Gain ACo becomes a new, modified gain
ACo’.
Can then use analysis from ideal case

*
*
ACo '
ACfo 
1   f ACo '
Rif  Ri '1   f ACo '
 Hf  1   f ACo ' H
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
 Lf 
Rof  Ro ' 1   f ACo '
L
1   f ACo '
4
Series-Series Feedback Amplifier - Practical Case
*
*
ECE 352 Electronics II Winter 2003
How do we determine the z-parameters
for the feedback network?
For the input loading term z11
 We turn off the feedback signal by
setting Io = 0 (I2 = 0 ).
 We then evaluate the resistance seen
looking into port 1 of the feedback
network (R11 =z11).
*
For the output loading term z22
 We open circuit the connection to the
input so I1 = 0.
 We find the resistance seen looking
into port 2 of the feedback network
(R22 =z22).
*
To obtain the feedback factor f (also
called z12 )
 We apply a test signal Io’ 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/Io’
Ch. 8 Feedback
5
Series-Series Feedback Amplifier - Practical Case
Original Amplifier
*
Feedback Network
*
*
Modified basic amplifier
(including loading effects of feedback network)
 Including z11 at input
 Including z22 at output
 Including loading effects of source resistance
 Including load effects of load resistance
Now have an idealized feedback network, i.e. produces
feedback effect, but without loading effects
Can now use feedback amplifier equations derived
ACo '
ACfo 
1   f ACo '
 Hf  1   f ACo ' H
Modified Amplifier
*
ECE 352 Electronics II Winter 2003
 Lf 
Rof  Ro ' 1   f ACo '
L
1   f ACo '
Note

Idealized Feedback Network
Rif  Ri '1   f ACo '

Ch. 8 Feedback
ACo’ is the modified transconductance gain
including the loading effects of z11 , z22 , RS and RL.
Ri’ and Ro’ are modified input and output
resistances including loading effects.
6
Example - Series-Series Feedback Amplifier
*
*
*
*
*
Three stage amplifier
Each stage a CE amplifier
Transistor parameters
Given: 1= 2 = 3 =100,
rx1=rx2=rx3=0
Coupled by capacitors, dc
biased separately
DC analysis (given):
I C1  0.60 m A, g m1 
r 1 
1
 4.3K
g m1
I C 2  1.0 m A, g m 2 
r 2 
Note: Biasing resistors for each stage are
not shown for simplicity in the analysis.
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
IC 2
 39 m A/ V ,
VT
2
 2.6 K
g m2
I C 3  4.0 m A, g m3 
r 3 
I C1
 23 m A/ V ,
VT
IC3
 156 m A/ V ,
VT
3
 0.64K
g m3
7
Example - Series-Series Feedback Amplifier
*
Redraw circuit to show:
 Feedback circuit

Type of output sampling (current in this case = Io)
 Collector resistor constitutes the load so Io  Ic
 Emitter current Ie=( +1) Ib = {( +1)/ } Ic Ic = Io

Type of feedback signal to input (voltage in this case = Vf)
Ic3 ≈ Io
Voltage fedback
to input
Io
Output current
sampling
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
8
Example - Series-Series Feedback Amplifier
Z-parameter equivalent circuit
for feedback circuit
Io
Input Loading Effects
R1
Output Loading Effects
I2=0
R1  z11  RE1 [ RF  RE 2 ]
R2  z22  RE 2 [ RF  RE1 ]
 0.1K [0.64 K  0.1K ]  0.088K
ECE 352 Electronics II Winter 2003
R2
I1=0
 0.1K [0.64 K  0.1K ]  0.088K
Ch. 8 Feedback
9
Example - Series-Series Feedback Amplifier
Voltage fedback
to input
Io
Output current
sampling
Redrawn basic amplifier
with loading effects,
but not feedback.
R1
ECE 352 Electronics II Winter 2003
R2
Ch. 8 Feedback
10
Example - Series-Series Feedback Amplifier
IC3
*
*
Io= IE3 ≈ IC3
Construct ac equivalent circuit at
midband frequencies including loading
effects of feedback network.
Analyze circuit to find MIDBAND GAIN
(transconductance gain ACo for this seriesseries configuration)
ACo 
Io
Vs
Io
VS
R1
ECE 352 Electronics II Winter 2003
R2
Ch. 8 Feedback
11
Example - Series-Series Feedback Amplifier
Midband Gain Analysis
I1
I2
I3
Io
VS
Vi1
Vi3
Ri1
 I o  V 3  Vi 3  V 2
Io



V 






Vs
  3  Vi 3  V 2  V 1
g V
 m 3  3  g m 3  156mA / V
V 3
ACo 
Io
V 3
Note convention on Io is
into the output of the
last stage of the amplifier.
Ri3
 V 1



 Vi1
 Vi1



 Vs




V 3
I  3 r 3
r 3
0.64 K



 0.067
Vi 3
I  3 r 3  R2 I  3  g m 3V 3  r 3  R2 1  g m 3 r 3  0.64 K  0.088 K (101)
 g m 2V 2 RC 2 r 3  R2 1  g m 3 r 3 
Vi 3

 39mA / V 5 K 0.64 K  1010.088 K   128
V 2
V 2
 g m1V 1 RC1 r 2 
V 2

 23mA / V 9 K 2.6 K   46.4
V 1
V 1
V 1
I  1r 1
4. 3 K


 0.33
Vi1
I  1r 1  R1 1  g m1r 1  4.3K  1010.088 K 
Vi1
1
VS
ACo 
Io
 156mA / V 0.067  128 46.4 0.331  2.05 x10 4 mA / V  20.5 A / V
Vs
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
12
Feedback Factor and Midband Gain with Feedback
*
Determine the feedback factor f
f 
Xf
Xo

Vf '
Io '

RE1 I f 1
Io '
VE2
0.1K
 0.1K
 0.012 K  12
0.1K  0.1K  0.64 K
*
VE 2  I f 1 ( RF  RE1 )  ( I o ' I f 1 ) RE 2
Calculate gain with feedback ACfo
I f 1 ( RF  RE1  RE 2 )  I o ' RE 2
 f ACo  20.5 A / V (12)  246
ACfo
*
Io ’
If1
RE 2
 RE1
RE1  RE 2  RF
ACo
20.5 A / V
20.5 A / V



 0.083 A / V  83 m A/ V
1   f ACo 1  20.5 A / V (12)
247
I f1
Io '

RE 2
RF  RE1  RE 2
Note
 f ACo > 0 as necessary for negative feedback
and dimensionless
 f ACo is large so there is significant feedback.
 f has units of resistance (ohms); ACo has units
of conductance (1/ohms)
 Can change f and the amount of feedback by
changing RE1 , RF and/or RE2.
 Gain is largely determined by ratio of feedback
resistances
1
R  RE 2  RF 0.1K  0.1K  0.64K
1
ACfo 
 E1

 84  84 mA/ V
f
RE1RE 2
0.1K (0.1K )
K
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
13
Input and Output Resistances with Feedback
I1
Io
Vi1
I1(1+gm1r1)
Ro
Ri = Ri1
*
Determine input Ri and output Ro resistances with loading effects of feedback network.
Ri  Ri1 
*
Vi1
 r 1  1  g m1r 1 R1
I 1
Ro  RC 3    
 4.3K  1010.088K   13.2 K
Calculate input Rif and output Rof resistances for the complete feedback amplifier.

Rif  Ri 1   f ACo

Rof  Ro (1   f ACo )  
 13.2 K 1  20.5 A / V (12)
 13.2 K 247  3.26M
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
14
Voltage Gain for Transconductance Feedback Amplifier
Io
*
Can calculate voltage gain after we calculate the transconductance gain!
V 
I R 
I 
AVfo   o    o C 3    RC 3  o    RC 3 ACfo  600(0.083 A / V )  49.8V / V
 Vs  f  Vs  f
 Vs  f
AVfo (dB)  20 log 49.8  34dB
*
Note - can’t calculate the voltage gain as follows:
Assume AVfo 
Find AVo 
AVo
1   f AVo
Correct voltage gain
for the amplifier
with feedback!
Vo  I o RC 3

  RC 3 ACo  60020.5 A / V   1.23x10 4 V / V
Vs
Vs


Calculate  f AVo  12  1.23x10 4 V / V  1.48 x105  Note this has units (it should not! and a negative sign )
Calculate voltage gain with feedback from
AVfo 
AVo
 1.23 x10 4 V / V
1

 0.083
5
1   f AVo
1  1.48 x10 

Magnitude is off by orders of magnitude and units are wrong!
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
Wrong
voltage
gain!
15
Equivalent Circuit for Series-Series Feedback Amplifier
*
*
*
Transconductance gain
amplifier A = Io/Vs
Feedback modified gain, input
and output resistances
 Included loading effects of
feedback network
 Included feedback effects
of feedback network
Significant feedback, i.e.
f ACo is large and positive
 f ACo  20.5 A / V (12)  246
ACfo 

I 
ACo
ACfo   o  
 83 mA / V
V
1


A
f Co
 S f
AVfo  RC 3 ACfo  49.8V / V
ECE 352 Electronics II Winter 2003


ACo
ACo
1


1   f ACo  f ACo  f
1
 84 m A/ V
0.012K
Rif  Ri 1   f ACo  3.3 M


Rof  Ro 1   f ACo  
Ch. 8 Feedback
16
Frequency Analysis
*
*
*
*
*
*
Hf  1   f ACo 'H
Lf 
ECE 352 Electronics II Winter 2003
L
1   f ACo '
Ch. 8 Feedback
*
*
Simplified amplifier analyzed had biasing
resistors omitted for simplicity.
For completeness, need to add biasing
resistors.
 Coupling capacitors then need to be
added to simplify biasing by isolating
each stage.
Low frequency analysis of poles for feedback
amplifier follows Gray-Searle (short circuit)
technique as before.
Low frequency zeroes found as before.
Dominant pole used to find new low 3dB
frequency.
For high frequency poles and zeroes,
substitute hybrid-pi model with C and C
(transistor’s capacitors).
 Follow Gray-Searle (open circuit)
technique to find poles
High frequency zeroes found as before.
Dominant pole used to find new high 3dB
frequency.
17