Download ECE 352 Electronics II

Shunt-Shunt Feedback Amplifier - Ideal Case
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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
TRANSRESISTANCE GAIN A = ARo = Vo/Ii
For the feedback amplifier as a whole, feedback changes
midband transresistance gain from ARo to ARfo
ARo
ARfo 
1   f ARo
Feedback changes input resistance from Ri to Rif
Rif 
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1   f ARo 
Feedback changes output resistance from Ro to Rof
Rof 
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Ri
1   f ARo 
Feedback changes low and high frequency 3dB
frequencies
 Hf  1   f ARo  H
ECE 352 Electronics II Winter 2003
Ro
Ch. 8 Feedback
 Lf 
L
1   f ARo 
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Shunt-Shunt Feedback Amplifier - Ideal Case
Gain
ARfo 
Vo
A I
A
ARo
ARo
 Ro i  Ro 

If
 f Vo 1   f ARo
I s Ii  I f
1
1
Ii
Ii
Input Resistance
Rif 

Is = 0
Vs
Vs
Vs


I s Ii  I f
I i   f Vo
Vs

V 
I i 1   f o 
Ii 


Ri
1   f ARo 
Output Resistance
Io ’
_
+
Rof 
Vo’
Vo ' I o ' Ro  ARo I i
I

 Ro  ARo i
Io '
Io '
Io '
But I s  0 so I i   I f
and I f   f Vo ' so I i    f Vo '
  f Vo '
Ii

   f Rof
Io '
Io '

Rof  Ro  ARo   f Rof
so Rof 
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback

Ro
1   f ARo 
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Equivalent Network for Feedback Network
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ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
Feedback network is a two port network
(input and output ports)
Can represent with Y-parameter network
(This is the best for this feedback
amplifier configuration)
Y-parameter equivalent network has
FOUR parameters
Y-parameters relate input and output
currents and voltages
Two parameters chosen as independent
variables. For Y-parameter network,
these are input and output voltages V1
and V2
Two equations relate other two quantities
(input and output currents I1 and I2) to
these independent variables
Knowing V1 and V2, can calculate I1 and
I2 if you know the Y-parameter values
Y-parameters have units of conductance
(1/ohms=siemens) !
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Shunt-Shunt Feedback Amplifier - Practical Case
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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 y-parameter equivalent circuit for feedback
network
 Feedback factor f given by y12 since
y12 
I1
V2

V1 0
If
f
Vo
Feedforward factor given by y21 (neglected)
 y22 gives feedback network loading on output
 y11 gives feedback network loading on input
Can incorporate loading effects in a modified basic
amplifier. Gain ARo becomes a new, modified gain
ARo’.
Can then use analysis from ideal case

I2
I1
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y21V1
V1 y11
y12V2
V2
y22
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ARfo 
Rif 
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
ARo '
1   f ARo '
Ri
Rof 
1   f ARo '
 Hf  1   f ARo ' H
Ro
1   f ARo '

 Lf 

L
1   f ARo '


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Shunt-Shunt Feedback Amplifier - Practical Case
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I2
I1
V1
y21V1
y11
y12V2
ECE 352 Electronics II Winter 2003
y22
V2
Ch. 8 Feedback
How do we determine the y-parameters
for the feedback network?
For the input loading term y11
 We turn off the feedback signal by
setting Vo = 0 (V2 =0).
 We then evaluate the resistance
seen looking into port 1 of the
feedback network (R11 = y11).
For the output loading term y22
 We short circuit the connection to
the input so V1 = 0.
 We find the resistance seen looking
into port 2 of the feedback
network.
To obtain the feedback factor f
(also called y12 )
 We apply a test signal Vo’ to port 2
of the feedback network and
evaluate the feedback current If
(also called I1 here) for V1 = 0.
 Find f from f = If/Vo’
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Example - Shunt-Shunt Feedback Amplifier
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Single stage CE amplifier
Transistor parameters. Given:  =100, rx= 0
No coupling or emitter bypass capacitors
DC analysis:
VBE ,active  0.7V
0.7V
 0.07 mA  70A I 47 K  I B  0.07 mA
10 K
I C  I B I 4.7 K  I C  ( I B  0.07 mA)    1I B  0.07 mA
I10 K 
12V  I 4.7 K 4.7 K  I B  0.07 mA47 K  0.7V
12V    1I B  0.07 mA4.7 K  I B  0.07 mA47 K  0.7V
12V  0.33V  3.3V  1014.7 K   47 K I B
8.37V
 0.016 mA  16 A I C  I B  1000.016 mA  1.6 mA
522 K
1.6 mA
I

100
gm  C 
 63 mA / V r 

 1.6 K
VT 0.0256 V
g m 63 mA / V
IB 
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
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Example - Shunt-Shunt Feedback Amplifier
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Redraw circuit to show



Feedback circuit
Type of output sampling (voltage in this case = Vo)
Type of feedback signal to input (current in this case = If)
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
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Example - Shunt-Shunt Feedback Amplifier
Equivalent circuit for feedback network
I1
I2
V1 y11
Input Loading Effects
y21V1
y12V2
y22
V2
Output Loading Effects
R1= y11
R2= y22
R1  RF  47 K
ECE 352 Electronics II Winter 2003
R2  RF  47 K
Ch. 8 Feedback
8
Example - Shunt-Shunt Feedback Amplifier
Modified Amplifier with Loading Effects,
but Without Feedback
Original Feedback Amplifier
R2
R1
Note: We converted the signal source to a
Norton equivalent current source
ARo
since we need to calculate the gain A  Vo 
Rfo
I s 1   f ARo
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
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Example - Shunt-Shunt Feedback Amplifier
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Construct ac equivalent circuit at
midband frequencies including loading
effects of feedback network.
Analyze circuit to find midband gain
(transresistance gain ARo for this shuntshunt configuration)
ARo 
Vo
Is
s
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
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Example - Shunt-Shunt Feedback Amplifier
Midband Gain Analysis
ARo 
Vo

V
Vo  Vo  V

 

Is
V
   I s
 g mV RC RF

V




   g R
m
C



RF  63 m A/ V 4.7 K 47K  269V / V


V
  RS RF r   10K 47K 1.6 K  1.3K


IS
ARo 
Vo
  2691.3K   350K
Is
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
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Midband Gain with Feedback
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Determine the feedback factor f
X
I '
If '
1
f  f  f 

X o Vo '  I f R f  RF
+
_ Vo’
1

  0.021 mA / V
47 K
Calculate gain with feedback ARfo
 f ARo  0.021mA / V (350 K )  7.4
ARfo
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ARo
 350 K


  42 K
1   f ARo
1  7.4
Note: The direction of If is
always into the
feedback network!
Note
 f < 0 and has units of mA/V, ARo < 0 and has units of K
 f ACo > 0 as necessary for negative feedback and dimensionless
 f ACo is large so there is significant feedback.
 Can change f and the amount of feedback by changing RF.
 Gain is determined primarily by feedback resistance
1
1
ARfo 

  RF   47 K
 f (1 / RF )
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
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Input and Output Resistances with Feedback
Ro
Ri
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Determine input Ri and output Ro resistances with loading effects of feedback network.
Ri  RS RF r  10K 47K 1.6K  1.3K
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Ro  RC RF  4.7 K 47K  4.3K
Calculate input Rif and output Rof resistances for the complete feedback amplifier.
Rif 

Ri
Rof 
1   f ARo 
Ro
(1   f ARo )

4.3K
 0.5K
8.4
1.3K
 0.15K
8.4
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
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Voltage Gain for Transresistance Feedback Amplifier
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Can calculate voltage gain after we calculate the transresistance gain!
ARfo  42 K
V 
 V 
1 V 
AVfo   o    o    o  

  4.2 V / V
V
I
R
R
I
R
10
K
s  s f
s
 s f  s s f
AVfo (dB)  20 log 4.2  12.5 dB
Note - can’t calculate the voltage gain as follows:
Assume AVfo 
Find AVo 
Correct voltage gain
AVo
1   f AVo
Vo
V
A
 350 K
 o  Ro 
 35 V / V
Vs I s Rs
Rs
10 K



Calculate  f AVo   0.021 mA / V  35 V / V  0.74 mA / V Note this has units; it should not!
Calculate voltage gain with feedback from
AVfo 
AVo
1   f AVo

 35 V / V
1  0.74mA / V
Magnitude is off by nearly a factor of five and units are wrong!
ECE 352 Electronics II Winter 2003
Ch. 8 Feedback
 20 V / V (?)
Wrong
voltage
gain!
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Equivalent Circuit for Shunt-Shunt Feedback Amplifier
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Rof
Rif
ARfoI i
*
V 
ARo
ARfo   o  
  42 K
I
1


A
f Ro
 S f
Rif 
Rof
Ri
1  
f
ARo 

 f ARo  0.021mA / V (350 K )  7.4
ARfo 
ARo
1

  47 K
1   f ARo  f
AVfo 
ARfo
1.3K
 0.15K
8.4
Ro
4.3K


 0.5K
(1   f ARo )
8.4
ECE 352 Electronics II Winter 2003
Transresistance gain amplifier A = Vo/Is
Feedback modified gain, input and
output resistances
 Included loading effects of feedback
network
 Included feedback effects of
feedback network
Significant feedback, i.e.
f ARo is large and positive
Ch. 8 Feedback
RS
  4.2 V / V
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Frequency Analysis
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Hf  1   f ACo 'H
Lf 
ECE 352 Electronics II Winter 2003
L
1   f ACo '
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For completeness, need to add coupling
capacitors at the input and output.
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.
Ch. 8 Feedback
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