FEEDBACK AMPLIFIERS
Introduction to Feedback




Feedback is used in virtually all amplifier system.
Invented in 1928 by Harold Black – engineer in Western
Electric Company
 methods to stabilize the gain of amplifier for use in
telephone repeaters.
In feedback system, a signal that is proportional to the
output is fed back to the input and combined with the
input signal to produce a desired system response.
However, unintentional and undesired system response
may be produced.
Feedback Amplifier

Feedback is a technique where a portion of the output
of a system (amplifier) is fed back and recombined with
input
input
A
output
b

There are 2 types of feedback amplifier:
Positive feedback
 Negative feedback

Positive Feedback

Positive feedback is the process when the output is
added to the input, amplified again, and this process
continues.
A
input
output
+
b

Positive feedback is used in the design of oscillator
and other application.
Positive Feedback - Example

In a PA system
get feedback when you put the microphone in front
of a speaker and the sound gets uncontrollably loud
(you have probably heard this unpleasant effect).
Negative Feedback

Negative feedback is when the output is subtracted
from the input.
input
A
output
b

The use of negative feedback reduces the gain. Part of
the output signal is taken back to the input with a
negative sign.
Negative Feedback - Example

Speed control
If the car starts to speed up above the desired setpoint speed, negative feedback causes the throttle
to close, thereby reducing speed; similarly, if the car
slows, negative feedback acts to open the throttle
Feedback Amplifier - Concept
Basic structure of a single - loop feedback amplifier
Structure of a feedback system
A
ß
A
ß
A
ß
AF
Two types of feedback:
negative feedback (degenerative) NF.
 The signal fed back from the output reduces the
effect of the input signal;
 stabilizing effect
positive feedback (regenerative) PF.
 The signal fed back from the output intensifies the
effect of the input signal;
 leads to instability
Effects (ADVANTAGES) of a negative feedback
Gain desensitivity
Bandwidth Extension
Noise Reduction
Reduction of the nonlinear distortion
Reduction of the nonlinear distortion
Improvement of the input and
output impedances
Disadvantages of Negative Feedback
1.
2.
Circuit Gain – overall amplifier gain is reduced
compared to that of basic amplifier.
Stability – possibility that feedback circuit will
become unstable and oscillate at high frequencies.
Classification of Amplifiers
Classify amplifiers into 4 basic categories based on
their input (parameter to be amplified; voltage or
current) & output signal relationships:




Voltage amplifier (series-shunt)
Current amplifier (shunt-series)
Transconductance amplifier (series-series)
Transresistance amplifier (shunt-shunt)
Feedback topologies
The Four Basic Feedback Topologies
26




Voltage amplifier---series-shunt feedback
voltage mixing and voltage sampling
Current amplifier---shunt-series feedback
Current mixing and current sampling
Transconducatnce amplifier---series-series feedback
Voltage mixing and current sampling
Transresistance amplifier---shunt-shunt feedback
Current mixing and voltage sampling
Feedback Configuration
Series:
connecting the
feedback signal
in series with the
input signal
voltage.
Shunt:
connecting the
feedback
signal
in shunt
(parallel) with
an input current
source
How to determine the topology
At the input:
in accordance to the method of summing up the signals

voltage: series connected (in a loop)
 current: parallel connected (in a node)

At the output:
in accordance to the way of taking (measuring) the signals

voltage: in parallel (measurement with the voltmeter)
 current: in series (measurement with the ammeter)

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.
Ì This includes the loading effects of the feedback circuit (some
combination of resistors) on the amplifier input and output.
* Calculate the feedback factor βf
* Calculate the factor βf 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 + β 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
ωL
amplifier with feedback using
(
)
ω Hf = 1 + β f A ω H
ω Lf =
(1 + β 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
Af =
*
A
1+ β f A
Does change input resistance from Ri to Rif
(
Rif = Ri 1 + β f A
Equivalent Circuit for Feedback Amplifier
*
Does change output resistance from Ro to Rof
Rof =
*
)
Ro
1+ β f A
Does change low and high frequency 3dB frequencies
(
)
ω Hf = 1 + β f A ω H
ω Lf =
ωL
1+ β f A
(
)
7
Series-Shunt Feedback Amplifier - Ideal Case
Midband Gain
V
A V
AVf = o = V i =
Vs Vi + V f
Input Resistance
AV
AV
AV
=
=
Vf
β f Vo 1 + β f AV
1+
1+
Vi
Vi
Vi + V f Vi + β f Vo
V
Rif = s =
=
= Ri 1 + β 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 = β f Vo = β f Vt so
It =
=
(
)
(
Vt − AV − V f
Vt + AV β f Vt
=
Ro
Ro
Vt 1 + AV β f
(
)
Ro
V
Ro
so Rof = t =
It
1 + AV β f
(
8
)
)
Practical Feedback Networks
*
Vi
Vo
Vf
*
*
*
* How do we take these
loading effects into account?
*
Feedback networks consist of a set of resistors
Ì Simplest case (only case considered here)
Ì In general, can include C’s and L’s (not
considered here)
Ì 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
Ì Adds a finite series resistance
Ì Part of input signal Vs lost across this series
resistance (undesirable), so Vi reduced
Feedback network loads down amplifier output
Ì Adds a finite shunt resistance
Ì Part of output current lost through this shunt
resistance so not all output current delivered to
load RL (undesirable)
9
Example - Series-Shunt Feedback Amplifier
*
*
*
*
*
Two stage amplifier
Each stage a CE
amplifier
Transistor
parameters
Given: β1= β2 =50
Coupled by
capacitors, dc biased
separately
DC analysis:
I
IC1 = 0.94 mA, gm1 = C1 = 36 mA/ V ,
VT
DC analysis for each stage can be done separately
since stages are isolated (dc wise) by coupling capacitors.
rπ1 =
β1
gm1
= 1.4K
I
IC2 = 1.85 mA, gm2 = C2 = 71 mA/ V ,
VT
rπ 2 =
β2
gm2
= 0.7K
12
Example - Series-Shunt Feedback Amplifier
*
Redraw circuit to show
Ì
Ì
Ì
Feedback circuit
Type of output sampling
(voltage in this case = Vo)
Type of feedback signal to input
(voltage in this case = Vf)
+
_
Vi
+
Vf _
+
Vo
_
13
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 = 4.8K
Amplifier with Loading Effects
R2
R1
14
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
15
Example - Series-Shunt Feedback Amplifier
Midband Gain Analysis
A Vo =
⎛ Vo
Vo
= ⎜⎜
Vs
⎝ Vπ 2
⎞⎛ Vπ 2
⎟⎜
⎟⎜ V
⎠ ⎝ o1
(
Vo
= − g m 2 RC 2 R2
Vπ 2
(
V o1
= − g m 1 R C 1 R 12
Vπ 1
)=
⎞ ⎛ V o1
⎟⎜
⎟⎜ V
⎠⎝ π 1
⎞⎛ Vπ 1
⎟⎜
⎟⎜ V
⎠ ⎝ i1
⎞ ⎛ V i1
⎟⎜
⎟⎜ V
⎠⎝ s
⎞
⎟
⎟
⎠
(
− 71 mA / V 4 . 9 K 4 . 8 K
R 22
rπ 2
)=
)=
− 172
(
Vπ 2
= 1 since
V o1
− 36 mA / V 10 K 47 K 33 K 0 . 7 K
)=
rx 2 = 0
− 22 . 8
Vπ 1
I π 1 rπ 1
rπ 1
1 .4 K
=
=
=
= 0 . 22
V i1
I π 1 rπ 1 + (I π 1 + g m 1V π 1 )R 1
rπ 1 + (1 + β )R 1
1 . 4 K + 51 (0 . 098 K )
V i1
I
r
+ (I π 1 + g m 1V π 1 )R 1
R i1 =
= π1 π1
= rπ 1 + (1 + β )R 1 = 1 . 4 K + 51 (0 . 098 K ) = 6 . 4 K
Iπ 1
Iπ 1
6 . 4 K 150 K 47 K
R i 1 R 11 R 21
V i1
=
=
Vs
R s + R i 1 R 11 R 21
5 K + 6 . 4 K 150 K 47 K
A Vo =
Vo
=
Vs
(− 172 )(1 )(−
A Vo ( dB ) = 20 log 449
22 . 8 )(0 . 22
)(0 . 52 ) =
=
5 .4 K
10 . 4 K
= 0 . 52
+ 449
= 53 dB
16
Midband Gain with Feedback
*
βf =
*
Determine the feedback factor βf
Xf
Xo
=
Vf
Vo '
=
Rf1
Rf1 + Rf 2
=
0.1K
= 0.021
0.1K + 4.7 K
Calculate gain with feedback Avf
β f AVo = 449(0.021) = 9.4
AVfo =
AVo
449
449
=
=
= 43.1
1 + β f AVo 1 + 449(0.021) 10.4
AVfo (dB ) = 20 log 43.1 = 32.7 dB
*
Note
Ì βf Avo > 0 as necessary for negative feedback
Ì βf Avo is large so there is significant feedback. For βf Avo ¡ 0, there is almost no
feedback.
Ì Can change βf and the amount of feedback by changing Rf1 and/or Rf2.
Ì NOTE: Since βf Avo >> 0
β f AVo = 449(0.021) = 9.4
AVfo ≈
1
βf
=
Rf1 + Rf 2
Rf1
=1+
Rf 2
Rf1
AVfo =
1
1
AVo
A
≈ Vo =
=
= 47.6
1 + β f AVo β f AVo β f 0.021
AVfo (dB ) = 20 log 47.6 = 33.6dB
17
Input and Output Resistances with Feedback
*
Determine input Ri and output Ro resistances with loading effects of feedback network.
Ri = RS + RB1 Ri1
Ro = R2 RC 2
= 5K + 38.5K 6.4 K = 10.5K
*
= 4.9 K 4.8 K = 2.4 K
Calculate input Rif and output Rof resistances for the complete feedback amplifier.
(
Rif = Ri 1 + β f AVo
)
= 10.5 K [1 + 449(0.021)] = 109.5 K
Rof =
Ro
1 + β f AVo
=
2.4 K
= 0.23K
10.4
18
Equivalent Circuit for Series-Shunt Feedback Amplifier
*
*
*
Voltage gain amplifier
Modified voltage gain, input
and output resistances
Ì Included loading effects of
feedback network
Ì Included feedback effects
of feedback network
Ì Include source resistance
effects
Significant feedback, i.e.
βf Avo is large and positive
β f AVo = 449(0.021) = 9.4
AVfo =
⎛V ⎞
AVo
= 43.1
AVfo = ⎜⎜ o ⎟⎟ =
⎝ VS ⎠ f 1 + β f AVo
AVfo (dB) = 20 log 43.1 = 32.7dB
(
)
Rif = Ri 1 + β f AVo = 109.5 K
Ro
Rof =
= 0.23K
1 + β f AVo
=
AVo
1 + β f AVo
Rf1 + Rf 2
Rf1
≈
AVo
1
=
β f AVo β f
= 1+
Rf 2
Rf1
= 47.6
AVf (dB) ≈ 20 log 47.6 = 33.6dB
19
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 + β f ACo
Feedback changes input resistance from Ri to Rif
(
Rif = Ri 1 + β f ACo
)
*
Feedback changes output resistance from Ro to Rof
*
Feedback changes low and high frequency 3dB
frequencies
(
Rof = Ro 1 + β f ACo
(
)
ω Hf = 1 + β f ACo ω H
ω Lf =
)
ωL
1 + β f ACo
(
)
20
Series-Series Feedback Amplifier - Ideal Case
Gain (Transconductance Gain)
ACfo =
Io
A V
A
ACo
ACo
= Co i = Co =
=
Vf
β f I o 1 + β f ACo
Vs Vi + V f
1+
1+
Vi
Vi
Input Resistance
Rif =
Vs Vi + V f Vi + β f I o
=
=
= Ri 1 + β f ACo
Ii
Ii
⎛Vi ⎞
⎜ R⎟
i⎠
⎝
(
)
Output Resistance
V
R of =
It
But V s = 0 so V i = −V f
and V f = β f I o = β f I t so V i = − β f I t
V = (I t − ACo V i )R o = (I t − ACo (− β f I t ))R o
+
V
-
= I t (1 + β f ACo )R o
so R of =
V
V
= R o (1 + β f ACo )
It
21
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
Ì Feedback factor βf given by
=
*
*
Vf
Io
=βf
Can incorporate loading effects in a modified basic
amplifier. Gain ACo becomes a new, modified gain
ACo’.
Can then use analysis from ideal case
ACfo =
ACo '
1 + β f ACo '
Rif = Ri '(1 + β f ACo ')
ω Hf = (1 + β f ACo ')ω H
ω Lf =
Rof = Ro ' (1 + β f ACo ')
ωL
(1 + β f ACo ')
22
Series-Series Feedback Amplifier - Practical Case
*
*
Modified basic amplifier
(including loading effects of feedback
network)
Can now use feedback amplifier equations
derived
ACfo =
ACo '
1 + β f ACo '
Rif = Ri '(1 + β f ACo ')
ω Hf = (1 + β f ACo ')ω H
*
ω Lf =
Rof = Ro ' (1 + β f ACo ')
ωL
(1 + β f ACo ')
Note
Ì ACo’ is the modified transconductance
gain including the loading effects of RS
and RL.
Ì Ri’ and Ro’ are modified input and
output resistances including loading
effects.
24
Example - Series-Series Feedback Amplifier
*
*
*
*
*
Three stage amplifier
Each stage a CE amplifier
Transistor parameters
Given: β1= β2 = β3 =100,
rx1=rx2=rx3=0
Coupled by capacitors, dc
biased separately
DC analysis (given):
IC1 = 0.60 mA, gm1 =
rπ1 =
β1
gm1
= 4.3K
IC2 = 1.0 mA, gm2 =
rπ 2 =
Note: Biasing resistors for each stage are
not shown for simplicity in the analysis.
β2
gm2
β3
gm3
IC 2
= 39 mA/ V ,
VT
= 2.6K
IC3 = 4.0 mA, gm3 =
rπ 3 =
IC1
= 23 mA/ V ,
VT
I C3
= 156 mA/ V ,
VT
= 0.64K
25
Example - Series-Series Feedback Amplifier
*
Redraw circuit to show:
Ì
Feedback circuit
Ì
Type of output sampling (current in this case = Io)
“ Collector resistor constitutes the load so Io ¡ Ic
“ Emitter current Ie=(β +1) Ib = {(β +1)/ β} Ic ¡Ic = Io
Ì
Type of feedback signal to input (voltage in this case = Vf)
Ic3 ≈ Io
Voltage fedback
to input
Io
Output current
sampling
26
Example - Series-Series Feedback Amplifier
Io
Input Loading Effects
R1
Output Loading Effects
I2=0
R1 = z11 = RE1 [ RF + RE 2 ]
= 0.1K [0.64 K + 0.1K ] = 0.088 K
R2
I1=0
R2 = z 22 = RE 2 [ RF + RE1 ]
= 0.1K [0.64 K + 0.1K ] = 0.088 K
27
Example - Series-Series Feedback Amplifier
Voltage fedback
to input
Io
Output current
sampling
Redrawn basic amplifier
with loading effects,
but not feedback.
R1
R2
28
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
series-series configuration)
ACo =
Io
Vs
Io
VS
R1
R2
29
Example - Series-Series Feedback Amplifier
Midband Gain Analysis
I‡1
I‡2
I‡3
Io
VS
Vi1
Vi3
Ri1
Io
Vπ 3
Ri3
⎛ I ⎞⎛ V ⎞⎛ V ⎞⎛ V
Io
= ⎜⎜ o ⎟⎟ ⎜⎜ π 3 ⎟⎟ ⎜⎜ i 3 ⎟⎟ ⎜⎜ π 2
Vs
⎝ Vπ 3 ⎠⎝ V i3 ⎠⎝ Vπ 2 ⎠⎝ Vπ 1
g V
= m 3 π 3 = g m 3 = 156 mA / V
Vπ 3
A Co =
Note convention on Io is
into the output of the
last stage of the amplifier.
⎞⎛ Vπ 1
⎟⎟ ⎜⎜
⎠ ⎝ V i1
⎞ ⎛ V i1
⎟⎟ ⎜⎜
⎠⎝ V s
⎞
⎟⎟
⎠
0 . 64 K
Vπ 3
I π 3 rπ 3
rπ 3
=
=
=
= 0 . 067
V i3
I π 3 rπ 3 + R 2 (I π 3 + g m 3V π 3 ) rπ 3 + R 2 (1 + g m 3 rπ 3 ) 0 . 64 K + 0 . 088 K (101 )
− g m 2 V π 2 (R C 2 [rπ 3 + R 2 (1 + g m 3 rπ 3 )])
V i3
=
= − 39 mA / V (5 K
Vπ 2
Vπ 2
− g m 1V π 1 (R C 1 rπ 2 )
Vπ 2
=
= − 23 mA / V (9 K 2 . 6 K
Vπ 1
Vπ 1
)=
Vπ 1
I π 1 rπ 1
4 .3 K
=
=
V i1
I π 1 rπ 1 + R 1 (1 + g m 1 rπ 1 ) 4 . 3 K + 101 (0 . 088 K
[0 . 64
K + 101 (0 . 088 K
)]) =
− 128
− 46 . 4
)
= 0 . 33
V i1
=1
VS
A Co =
Io
= (156 mA / V
Vs
)(0 . 067 )(− 128 )(−
46 . 4 )(0 . 33 )(1 ) = + 2 . 05 x 10 4 mA / V = 20 . 5 A / V
30
Feedback Factor and Midband Gain with Feedback
*
Determine the feedback factor βf
βf =
Xf
Xo
=
Vf '
Io '
=
RE1 I f 1
Io '
RE 2
= RE 1
RE1 + RE 2 + RF
Io ’
If1
VE2
0.1K
= 0.1K
= 0.012 K = 12Ω
0.1K + 0.1K + 0.64 K
*
Calculate gain with feedback ACfo
β f ACo = 20.5 A / V (12Ω) = +246
ACfo =
*
ACo
20.5 A / V
20.5 A / V
=
=
= 0.083 A / V = 83 mA / V
1 + β f ACo 1 + 20.5 A / V (12Ω)
247
VE 2 = I f 1 ( RF + RE1 ) = ( I o '− I f 1 ) RE 2
I f 1 ( RF + RE1 + RE 2 ) = I o ' RE 2
I f1
Io '
=
RE 2
RF + RE1 + RE 2
Note
Ì βf ACo > 0 as necessary for negative feedback
and dimensionless
Ì βf ACo is large so there is significant feedback.
Ì βf has units of resistance (ohms); ACo has units
of conductance (1/ohms)
Ì Can change βf and the amount of feedback by
changing RE1 , RF and/or RE2.
Ì Gain is largely determined by ratio of feedback
resistances
1
R + RE 2 + RF 0.1K + 0.1K + 0.64 K
1
ACfo ≈
= E1
=
= 84 = 84 mA / V
βf
RE1RE 2
0.1K (0.1K )
K
31
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
= rπ 1 + (1 + g m1rπ 1 )R1
Iπ 1
Ro = RC 3 + ∞ = ∞
= 4.3K + 101(0.088 K ) = 13.2 K
Calculate input Rif and output Rof resistances for the complete feedback amplifier.
(
Rif = Ri 1 + β f ACo
)
= 13.2 K [1 + 20.5 A / V (12Ω)]
Rof = Ro (1 + β f ACo ) = ∞
= 13.2 K (247 ) = 3.26 MΩ
32
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 = −600Ω(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
Correct voltage gain
for the amplifier
with feedback!
1 + β f AVo
Vo − I o RC 3
=
= − RC 3 ACo = −600Ω(20.5 A / V ) = −1.23 x10 4 V / V
Vs
Vs
(
)
Calculate β f AVo = 12Ω − 1.23 x10 4 V / V = −1.48 x105 Ω (Note this has units (it should not!) and a negative sign )
Calculate voltage gain with feedback from
AVfo =
AVo
1 + β f AVo
=
− 1.23 x10 4 V / V
1
= 0.083
5
1 − 1.48 x10 Ω
Ω
Magnitude is off by orders of magnitude and units are wrong!
Wrong
voltage
gain!
33
Equivalent Circuit for Series-Series Feedback Amplifier
*
*
*
Transconductance gain
amplifier A = Io/Vs
Feedback modified gain, input
and output resistances
Ì Included loading effects of
feedback network
Ì Included feedback effects
of feedback network
Significant feedback, i.e.
βf ACo is large and positive
β f ACo = 20.5 A / V (12Ω) = 246
ACfo =
=
⎛I ⎞
ACo
ACfo = ⎜⎜ o ⎟⎟ =
= 83 mA / V
V
1
+
β
A
f Co
⎝ S ⎠f
AVfo = RC 3 ACfo = −49.8V / V
(
)
ACo
ACo
1
≈
=
1 + β f ACo β f ACo β f
1
= 84 mA / V
0.012 K
Rif = Ri 1 + β f ACo = 3.3 MΩ
(
)
Rof = Ro 1 + β f ACo = ∞
34
Shunt-Shunt Feedback Amplifier - Ideal Case
*
*
*
*
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 + β f ARo
Feedback changes input resistance from Ri to Rif
Rif =
*
(1 + β f ARo )
Feedback changes output resistance from Ro to Rof
Rof =
*
Ri
Ro
(1 + β f ARo )
Feedback changes low and high frequency 3dB
frequencies
ω Hf = (1 + β f ARo )ω H
ω Lf =
ωL
(1 + β f ARo )
35
Shunt-Shunt Feedback Amplifier - Ideal Case
Gain
ARfo =
Vo
A I
A
ARo
ARo
= Ro i = Ro =
=
If
β f Vo 1 + β 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 + β f Vo
Vs
⎛
V ⎞
I i ⎜⎜1 + β f o ⎟⎟
Ii ⎠
⎝
=
Ri
(1 + β f ARo )
Output Resistance
Io ’
_
+
R of =
Vo’
V o ' I o ' R o + A Ro I i
I
=
= R o + A Ro i
Io '
Io '
Io '
But I s = 0 so I i = − I f
and I f = β f V o ' so I i = − β f V o '
− β f Vo '
Ii
=
= − β f R of
Io '
Io '
(
R of = R o + A Ro − β f R of
so R of =
)
Ro
(1 + β f ARo )
36
Shunt-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 y-parameter equivalent circuit for feedback
network
Ì Feedback factor βf given by
=
*
*
If
Vo
=βf
Can incorporate loading effects in a modified basic
amplifier. Gain ARo becomes a new, modified gain
ARo’.
Can then use analysis from ideal case
ARfo =
Rif =
ARo '
1 + β f ARo '
Ri
Rof =
(1 + β f ARo ')
ω Hf = (1 + β f ARo ')ω H
Ro
1 + β f ARo '
(
ω Lf =
)
ωL
1 + β f ARo '
(
)
37
Example - Shunt-Shunt Feedback Amplifier
*
*
*
*
Single stage CE amplifier
Transistor parameters. Given: β =100, rx= 0
No coupling or emitter bypass capacitors
DC analysis:
VBE ,active = 0.7V
0.7V
= 0.07 mA = 70µA I 47 K = I B + 0.07 mA
10K
I C = βI B I 4.7 K = I C + ( I B + 0.07 mA) = (β + 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 = ((β + 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 µA I C = βI B = 100(0.016 mA) = 1.6 mA
522K
1.6 mA
β
I
100
= 63 mA / V rπ =
=
= 1.6 K
gm = C =
VT 0.0256 V
g m 63 mA / V
IB =
39
Example - Shunt-Shunt Feedback Amplifier
*
Redraw circuit to show
Ì
Ì
Ì
Feedback circuit
Type of output sampling (voltage in this case = Vo)
Type of feedback signal to input (current in this case = If)
40
Example - Shunt-Shunt Feedback Amplifier
Input Loading Effects
R1 = RF = 47 K
Output Loading Effects
R2 = RF = 47 K
41
Example - Shunt-Shunt Feedback Amplifier
Original Feedback Amplifier
Modified Amplifier with Loading Effects,
but Without Feedback
R2
R1
Note: We converted the signal source to a
Norton equivalent current source
since we need to calculate the gain
ARfo =
Vo
ARo
=
I s 1 + β f ARo
42
Example - Shunt-Shunt Feedback Amplifier
*
*
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
43
Example - Shunt-Shunt Feedback Amplifier
Midband Gain Analysis
A Ro =
Vo
=
Vπ
⎛ V ⎞⎛ V
Vo
= ⎜⎜ o ⎟⎟ ⎜⎜ π
Is
⎝ Vπ ⎠ ⎝ I s
− g m Vπ R C R F
(
Vπ
⎞
⎟
⎟
⎠
) = − g (R
m
C
)
(
)
R F = − 63 mA / V 4 . 7 K 47 K = − 269 V / V
(
)
Vπ
= ⎛⎜ R S R F rπ ⎞⎟ = 10 K 47 K 1 . 6 K = 1 . 3 K
⎝
⎠
IS
A Ro =
Vo
= (− 269 )(1 . 3 K ) = − 350 K
Is
44
Midband Gain with Feedback
*
*
Determine the feedback factor βf
X
I '
If '
−1
=
βf = f = f =
X o Vo ' (− I f R f ) RF
+
_ Vo’
−1
=
= − 0.021 mA / V
47 K
Calculate gain with feedback ARfo
β f ARo = − 0 .021 mA / V ( − 350 K ) = + 7 .4
ARfo
*
ARo
− 350 K
=
=
= − 42 K
1 + β f ARo
1 + 7 .4
Note: The direction of If is
always into the
feedback network!
Note
Ì βf < 0 and has units of mA/V, ARo < 0 and has units of K>
Ì βf ACo > 0 as necessary for negative feedback and dimensionless
Ì βf ACo is large so there is significant feedback.
Ì Can change βf and the amount of feedback by changing RF.
Ì Gain is determined primarily by feedback resistance
ARfo
1
1
≈
=
= − RF = − 47 K
β f (−1 / RF )
45
Input and Output Resistances with Feedback
Ro
Ri
*
Determine input Ri and output Ro resistances with loading effects of feedback network.
Ri = RS RF rπ = 10 K 47 K 1.6 K = 1.3K
*
Ro = RC RF = 4.7 K 47 K = 4.3K
Calculate input Rif and output Rof resistances for the complete feedback amplifier.
Rif =
=
Ri
(1 + β f ARo )
Rof =
Ro
(1 + β f ARo )
=
4.3K
= 0.5 K
8.4
1.3K
= 0.15K
8.4
46
Voltage Gain for Transresistance Feedback Amplifier
*
*
Can calculate voltage gain after we calculate the transresistance gain!
ARfo − 42 K
⎛ V ⎞
⎛V ⎞
1 ⎛V ⎞
=
= − 4.2 V / V
AVfo = ⎜⎜ o ⎟⎟ = ⎜⎜ o ⎟⎟ = ⎜⎜ o ⎟⎟ =
V
I
R
R
I
R
10
K
s ⎝ s ⎠f
s
⎝ s ⎠f ⎝ s s ⎠f
Correct voltage gain
AVfo (dB) = 20 log(4.2 ) = 12.5 dB
Note - can’t calculate the voltage gain as follows:
Assume
Find
AVfo =
AVo =
Calculate
AVo
1 + β f AVo
Vo
V
A
− 350 K
= o = Ro =
= − 35 V / V
Vs I s Rs
Rs
10 K
β 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 + β f AVo
=
− 35 V / V
1 + 0 .74 mA / V
Magnitude is off by nearly a factor of five and units are wrong!
= − 20 V / V (?)
Wrong
voltage
gain!
47
Equivalent Circuit for Shunt-Shunt Feedback Amplifier
*
*
Rof
Rif
ARfoI i
*
⎛V ⎞
ARo
= − 42 K
ARfo = ⎜⎜ o ⎟⎟ =
⎝ I S ⎠ f 1 + β f ARo
Rif =
Rof =
Ri
(1 + β
f
ARo )
Ro
=
(1 + β f ARo )
1. 3 K
= 0.15 K
8.4
=
Transresistance gain amplifier A = Vo/Is
Feedback modified gain, input and
output resistances
Ì Included loading effects of feedback
network
Ì Included feedback effects of
feedback network
Significant feedback, i.e.
βf ARo is large and positive
β f ARo = −0.021mA / V (−350 K ) = +7.4
ARfo =
AVfo =
ARo
1
≈
= − 47 K
1 + β f ARo β f
ARfo
RS
= − 4.2 V / V
4 .3 K
= 0.5 K
8.4
48
Shunt-Series Feedback Amplifier - Ideal Case
*
Current feedback
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
CURRENT GAIN A = AIo = Io/Ii
For the feedback amplifier as a whole, feedback changes
midband current gain from AIo to AIfo
AIo
AIfo =
1 + β f AIo
Feedback changes input resistance from Ri to Rif
Rif =
*
Ri
(1 + β f AIo )
Feedback changes output resistance from Ro to Rof
(
Rof = Ro 1 + β f AIo
*
)
Feedback changes low and high frequency 3dB
frequencies
(
)
ω Hf = 1 + β f AIo ω H
ω Lf =
ωL
1 + β f AIo
(
)
49
Shunt-Series Feedback Amplifier - Ideal Case
Gain
AIfo =
Io
A I
= Io i =
I s Ii + I f
Input Resistance
Rif =
=
AIo
AIo
AIo
=
=
If
β f I o 1 + β f AIo
1+
1+
Ii
Ii
Vs
Vs
Vs
=
=
I s Ii + I f
Ii + β f Io
Vs
⎛
I ⎞
I i ⎜⎜1 + β f o ⎟⎟
Ii ⎠
⎝
=
Ri
(1 + β f AIo )
Output Resistance
Is=0
Io ’
⎛
V o ' R o (I o '− A Io I i )
I ⎞
=
= R o ⎜⎜ 1 − A Io i ⎟⎟
Io '
Io '
Io ' ⎠
⎝
But I s = 0 so I i = − I f
R of =
Vo’
and I f = β f I o ' so I i = − β f I o '
− β f Io '
Ii
=
= −β
Io '
Io '
(
R of = R o 1 + β f A Io
f
)
50
Shunt-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
Ì
Feedback factor βf given by
If
Io
*
*
=βf
Can incorporate loading effects in a modified basic
amplifier. Gain AIo becomes a new, modified gain
AIo’.
Can then use analysis from ideal case
AIfo =
AIo '
1 + β f AIo '
Rif =
ω Hf = (1 + β f AIo ')ω H
Ri
(1 + β
ω Lf =
f AIo ')
Rof = Ro (1 + β f AIo ')
ωL
(1 + β f AIo ')
51
Example - Shunt-Series Feedback Amplifier
*
*
*
*
*
Two stage [CE+CE] amplifier
Transistor parameters Given: β =100, rx= 0
Input and output coupling and emitter bypass capacitors, but
direct coupling between stages
Capacitor in feedback connection removes Rf from DC bias
DC bias of two stages is coupled (bias of one affects the other)
53
DC Bias Analysis
Given :
β1 = β 2 = 100
VC1
(
I B1 =
VB2 = VC1 = VBE2 + I E 2 RE2 = 0.7V + (β + 1)I B2 RE 2
RB2
15K
12V =
12V =1.6V
RB1 + RB2
15K +100K
)
IC1 = βI B1 =1008.7 µA = 0.87 mA (neglecting IB2 )
gm1 =
IC1 0.87 mA
=
= 34 mA/V
VT 0.0256 V
rπ1 =
β
gm1
3.3V − 0.7V
2.6V
=
= 7.6 µA (<<IC1 =870 A)
(β +1)RE2 101(3.4K )
I B2 =
VTH1 −VBE1
1.6V − 0.7V
=
= 8.7 µA
RTH1 + (β +1)RE1 13K + (101)0.87K
(
)
VC1 = 12 V − IC1RC1 = 12 V − 0.87 mA10K = 3.3 V
RTH1 = RB1 RB2 =100K 15K =13K
VTH1 =
rx1 = rx 2 = 0
VB2
=
100
= 2.9K
34mA/V
(
)
IC 2 = βI B2 = 100 7.6 µA = 0.76 mA
gm2 =
rπ 2 =
IC2 0.76 mA
=
= 30 mA/ V
VT
0.0256V
β
gm2
=
100
= 3.3K
30 mA/ V
54
Example - Shunt-Series Feedback Amplifier
*
Redraw circuit to show
Ì
Feedback circuit
Ì
Type of output sampling
(current in this case = Io)
Ì
Type of feedback signal to
input (current in this case = If)
Iout’
Io ’
Iout’
55
Example - Shunt-Series Feedback Amplifier
Iout’
Io ’
Input Loading Effects
Io’= 0
Output Loading Effects
R1
R1 = RF + R1 = 10 K + 3.4 K = 13.4 K
R2
R2 = RF RE 2 = 10 K 3.4 K = 2.5 K
56
Example - Shunt-Series Feedback Amplifier
Amplifier with Loading Effects but Without Feedback
57
Example - Shunt-Series Feedback Amplifier
Midband Gain Analysis
Ri2
Is
IC’
Iout’
Vi2
R1
R2
I out ' ⎛ I out ' ⎞ ⎛ I c ' ⎞ ⎛ V π 2 ⎞ ⎛ V i 2
⎟⎟ ⎜⎜
⎟⎟ ⎜⎜
⎟⎟ ⎜⎜
= ⎜⎜
'
Is
I
V
V
⎝ c ⎠⎝ π 2 ⎠⎝ i 2 ⎠⎝ Vπ 1
8K
I out '
RC 2
=
=
= 0 . 89
8 K + 1K
Ic '
RC 2 + R L
A Io =
⎞⎛ Vπ 1
⎟⎟ ⎜⎜
⎠⎝ I s
⎞
⎟⎟
⎠
Ic '
g V
= m 2 π 2 = g m 2 = 30 mA / V
Vπ 2
Vπ 2
Vπ 2
I π 2 rπ 2
r
rπ 2
3 .3 K
=
= π2 =
=
Vi2
I π 2 rπ 2 + I π 2 (1 + g m 2 rπ 2 )R 2
Ri2
rπ 2 + (1 + g m 2 rπ 2 )R 2
3 . 3 K + 101 (2 . 5 K
)
= 0 . 013
Vi2
= − g m 1 R C 1 R i 2 = − g m 1 (R C 1 [rπ 2 + (1 + g m 2 rπ 2 )R 2 ]) = − 34 mA / V (10 K [3 . 3 K + 101 (2 . 5 K
Vπ 1
(
) (
)]) =
− 327
)
Vπ
= R S R 1 rπ 1 R B 1 = 10 K 13 . 4 K 2 . 9 K 13 K = 1 . 7 K
IS
A Io =
I out '
= (0 . 89 )(30 mA / V
Is
)(0 . 013 )(− 327 )(1 . 7 K ) = − 193
58
Midband Gain with Feedback
*
Determine the feedback factor If
βf =
*
Xf
Xo
=
If '
Io '
=
− RE 2
− 3.4 K
=
(RE 2 + R f ) 3.4 K + 10 K = − 0.25
+
VE2
-
Calculate gain with feedback AIfo
β f AIo = − 193 ( − 0 .25 ) = 48
AIfo =
AIo
1 + β f AIo
=
− 193
= − 3 .9
1 + 48
VE 2 = − I f ' R f = (I o '+ I f ')RE 2
− I f ' (R f + RE 2 ) = I o ' RE 2
AIfo (dB ) = 20 log 3 .9 = 11 .8 dB
*
If '
Io '
Note
Ì
Ì
Ì
Ì
Ì
=
− RE 2
R f '+ RE 2
βf < 0 and AIo < 0
βf AIo > 0 as necessary for negative feedback
and dimensionless
βf AIo is large so there is significant feedback.
Can change βf and the amount of feedback by
changing RF.
Gain is determined by feedback resistance
AIfo ≈
1
βf
=−
RE 2 + R f
RE 2
= − 4.0
59
Input and Output Resistances with Feedback
Ro
Ri
*
Determine input Ri and output Ro resistances with loading effects of feedback network.
Ri = RS R1 RB1 rπ 1 = 10 K 13.4 K 13K 2.9 K = 1.7 K
*
Ro = RC 2 RL + ∞ = ∞
Calculate input Rif and output Rof resistances for the complete feedback amplifier.
Rif =
=
Ri
(1 + β f AIo )
Rof = Ro (1 + β f AIo ) = ∞(49) = ∞
1.7 K
= 0.035 K
49
60
Voltage Gain for Current Gain Feedback Amplifier
*
Can calculate voltage gain
− RL AIfo − 1K
⎛V ⎞
⎛ − I 'R ⎞
− RL ⎛ I out ' ⎞
⎜⎜
⎟⎟ =
(− 4.2) = + 0.42 V / V
AVfo = ⎜⎜ o ⎟⎟ = ⎜⎜ out L ⎟⎟ =
=
V
I
R
R
I
R
10
K
s ⎝
s ⎠f
s
⎝ s ⎠f ⎝ s s ⎠f
AVfo (dB) = 20 log(0.42) = −7.5 dB
*
Note - can’t calculate the voltage gain as follows:
Assume AVfo =
Find AVo =
AVo
1 + β f AVo
Vo − I o RL − AIo RL − (−193)(1K )
=
=
=
= +19.3 V / V
Vs
I s Rs
Rs
10 K
Calculate β f AVo = (− 0.25)(19.3 V / V ) = −4.8
Calculate voltage gain with feedback from
AVfo =
AVo
1 + β f AVo
=
− 19.3 V / V
1 − 4.8
= +5.8 V / V
Magnitude is off by nearly a factor of ten!
61
Equivalent Circuit for Shunt-Series Feedback Amplifier
*
*
*
Current gain amplifier A = Io/Is
Feedback modified gain, input and
output resistances
Ì Included loading effects of feedback
network
Ì Included feedback effects of
feedback network
Significant feedback, i.e.
βf AIo is large and positive
β f AIo = − 193 ( − 0 .25 ) = 48
Rif
AIfoI i
Rof
AIfo =
AIo
1 + β f AIo
=
− 193
= − 3 .9
1 + 48
AIfo (dB ) = 20 log 3 .9 = 11 .8 dB
ARfo
Rif =
Ri
(1 + β
f
AIo )
⎛I
= ⎜⎜ o
⎝ IS
= 0.035 K
⎞
AIo
⎟⎟ =
= − 3 .9
⎠ f 1 + β f AIo
AIfo ≈
1
βf
=−
RE 2 + R f
RE 2
= − 4 .0
Rof = Ro (1 + β f AIo ) = ∞
62
Feedback Amplifier
Input and output Impedances
 Summary
1. For a series connection at input or output, the
resistance is increased by (1+bA).
2. For a shunt connection at input or output, the
resistance is lowered by (1+bA).
Feedback Amplifier