FEEDBACK AMPLIFIERS
Xs
+
Xi
Xo
Xf
βf
* What is negative feedback?
*
*
*
*
Adding the feedback signal to the input so as to
partially cancel the input signal to the amplifier.
What is feedback?
Taking a portion of the signal arriving at the load
and feeding it back to the input.
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.
Q. Why do you think that this type of feedback is called NEGATIVE?
1
Feedback Amplifier Analysis
Xs
+
Xi
Xo
Xf
βf
X
f
= β f Xo
where β f is called the feedback factor
X o = AX i
where A is the amplifier ' s gain , e.g . voltage gain
Xi = Xs − X
f
where X i is the net input signal to the basic amplifier ,
X s = the signal from the source
The amplifier ' s gain with feedback is given by
Af =
Xo
AX i
=
Xs
Xi + X
=
f
1+
A
X
A
=
f
Xi
1+
β f Xo
Xi
=
A
1+ β f A
<A
2
Advantages of Negative Feedback
* Gain desensitivity - less variation in amplifier gain with changes in
β (current gain) of transistors due to dc bias,
temperature, fabrication process variations, etc.
* Bandwidth extension - extends dominant high and low frequency poles to
higher and lower frequencies, respectively.
ωL
ω Hf = 1 + β f A ω H
ω Lf =
1+ β f A
(
)
(
)
* Noise reduction - improves signal-to-noise ratio
* Improves amplifier linearity - reduces distortion in signal due to gain
variations due to transistors
Ì
Ì
Disadvantages of Negative Feedback
Loss of gain, may require an added gain stage to compensate.
Added complexity in design
3
Basic Types of Feedback Amplifiers
* There are four types of feedback amplifiers. Why?
Ì Output sampled can be a current or a voltage
Ì Quantity fed back to input can be a current or a voltage
Ì 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 βf 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
Ì First term refers to nature of feedback connection at the input.
Ì Second term refers to nature of sampling connection at the output.
4
Basic Types of Feedback Amplifiers
Series - Shunt
Shunt - Series
Series - Series
Shunt - Shunt
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.
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
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 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.
=
Vf
Vo
AVf =
=βf
AV '
1 + β f AV '
Rif = Ri (1 + β f AV ')
Rof =
Ro
(1 + AV ' β f )
10
Series-Shunt Feedback Amplifier - Practical Case
Summary of Feedback Network Analysis
11
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
Ì
Find βf from βf = Vf/Io’
23
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
Ω
1 − 1.48 x10 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
β
V
1
+
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
Shunt-Shunt Feedback Amplifier - Practical Case
38
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
+
_ V o’
−1
=
= − 0.021 mA / V
47 K
Calculate gain with feedback ARfo
β f ARo = − 0 .021 mA / V ( − 350 K ) = + 7 .4
ARfo
*
− 350 K
ARo
=
=
= − 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
Shunt-Series Feedback Amplifier - Practical Case
52
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
0.0256V
VT
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 =
=
3 . 3 K + 101 (2 . 5 K
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
)
= 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
Summary of Feedback Amplifier Analysis
*
Xs
Xi
Xo
Xf
βf
*
ωL
=
(1 + β f Ao )
*
ω Hf = (1 + β f Ao )ω H
*
*
ω Lf
Identify the amplifier configuration by:
Ì Output sampling
“ Io = series configuration
“ Vo = shunt configuration
Ì Feedback to input
“ Io = shunt configuration
“ Vo = series configuration
Calculate loading effects of feedback
network
Ì On input
Ì On output
Calculate appropriate midband gain A’
(modified by loading effects of feedback
network)
Calculate feedback factor βf.
Calculate midband gain with feedback
Af.
A fo =
Ao
1 + β f Ao
63
Summary of Feedback Amplifier Analysis
64
Feedback Amplifier Stability
*
A(Š)
Feedback analysis
Ao
Ì
Midband gain with feedback A =
fo
Afo
Ì
New low and high 3dB frequencies
(
Ì
ω H ω Hf
ωi
ω
Amplifier becomes unstable (oscillates) if
at some frequency Ši we have
*
*
⎡ Im{β (ω i )A(ω i )}⎤
0
and φ (ω i ) = tan −1 ⎢
⎥ = −180 = −π radians
⎣ Re{β (ω i )A(ω i )}⎦
sin ce
β (ω i )A(ω i ) = β (ω i )A(ω i ) e jφ (ω i ) = e − jπ = −1
so
A f (ω i ) =
A (ω i )
1 + β (ω i )A(ω i )
=
A (ω i )
1−1
→∞
*
ωL
1 + β f Ao
(
)
Ri
(1 + β f AIo ')
(
Rof = Ro 1 + β f AIo '
)
Amplifier’s frequency characteristics
A( s ) = AM FH ( s ) FL ( s )
Feedback amplifier’s gain
Af (s) =
β (ω i )A(ω i ) = 1
ω Lf =
1 + β f Ao
Modified input and output resistances, e.g.
Rif =
ω Lf ω L
)
ω Hf = 1 + β f Ao ω H
Ao
A (s)
1 + β f (s) A (s)
→ Af (ω) =
A (ω)
1 + β f (ω) A (ω)
Define Loop Gain as βfA
β (ω )A(ω ) = β (ω )A(ω ) e jφ (ω )
Ì
Ì
Magnitude β (ω ) A(ω )
Phase
⎡ Im{β (ω )A(ω )}⎤
⎥
⎣ Re{β (ω )A(ω )}⎦
φ (ω ) = tan −1 ⎢
65
Feedback Amplifier Instability
βfA(dB)
General form of Magnitude plot of βfA
*
Amplifier with one pole
Ì Phase shift of - 900 is the maximum
A(ω ) =
ω1
φ (ω )
General form of Phase plot of βfA
00
- 450
- 900
- 1350
- 1800
1+ j
ω
ω HP1
Cannot get - 1800 phase shift.
Ì No instability problem
Amplifier with two poles
Ì Phase shift of - 1800 the maximum
Ì
ω
0 dB
Ao
*
A(ω ) =
ω
Ao
⎛
ω ⎞⎛
ω
⎜1 + j
⎟⎜1 + j
⎜
ω HP1 ⎟⎠⎜⎝
ω HP 2
⎝
⎞
⎟
⎟
⎠
Can get - 1800 phase shift !
For stability analysis, we define two
important frequencies
Ì Š1 is where magnitude of βfA goes to
unity (0 dB)
Ì Š180 is where phase of βfA goes to -180o
Ì Instability problem if Š1 = Š180
Ì
ω180
*
A f (ω1 = ω180 ) =
A (ω1 )
1 + β (ω1 ) A(ω1 )
=
A (ω1 )
1−1
→∞
66
Gain and Phase Margins
βfA(dB)
*
*
ω180
0 dB
φ (ω )
- 30
Gain
margin
= −1350 − (−1800 ) = +450
*
ω1
00
ω180
- 450
Example
Phase margin = φ (ω1 ) − φ (ω180 )
ω
ω1
Gain and phase margins
measure how far amplifier is
from the instability condition
Phase margin
Phase margin = φ (ω1 ) − φ (ω180 )
Gain margin
ω
Gain margin = β A(ω1 ) − β A(ω180 )
Example
Gain margin = β A(ω1 ) − β A(ω180 )
= 0 dB − (−30 dB) = +30 dB
- 900
- 1350
- 1800
Phase
margin
Φ(ω1)
Φ(ω180)
*
What are adequate margins?
Ì
Ì
Phase margin = 500 (minimum)
Gain margin = 10 dB (minimum)
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Numerical Example - Gain and Phase Margins
βfA(dB)
23.5 dB
Ao = 150
ω1
Pole 1
Ao β f = 15
Pole 2
β f = 0.1
Ao β f (dB) = 23.5 dB
15
ω ⎞⎛
ω ⎞
⎛
j
+
1
⎜1 + j
⎟
⎜
⎟
8 x105 ⎠⎝
3 x106 ⎠
⎝
From plots of magnitude and phase
Ao (ω ) β f =
Gain
margin
ω1 = 6 x106 rad / s and ω180 = 3 x107 rad / s
φ (ω1 ) = −1400 and φ (ω180 ) = −1800
Phase Shift
(degrees)
so phase margin is
φ (ω1 ) − φ (ω180 ) = −1400 − (−1800 ) = +400
Ao β f (ω1 ) = 0 dB and Ao β f (ω180 ) = −28 dB
ω180
Phase
margin
so gain margin is
Ao β f (ω1 ) − Ao β f (ω180 ) = 0 − (−28 dB) = +28 dB
Note that the gain and phase margins depend on
the feedback factor βf and so the amount of
feedback and feedback resistors.
* Directly through the value of βf.
* Indirectly since the gain Ao and pole frequencies
are influenced by the feedback resistors, e.g. the
loading effects analyzed previously.
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Amplifier Design for Adequate Gain and Phase Margins
A(dB)
General form of Magnitude plot of A
1/βf(dB)
A(ω180)
0 dB
φ (ω )
00
Gain
margin
General form of Phase plot of A
*
*
*
ω180
- 450
1
ω1 ω
180
ω1
βf
ω
ω
- 900
- 1350
- 1800
Ì
Ì
*
*
Phase
margin
How much feedback to use? What βf ?
A change of βf means redoing βf A(Š) plots!
Is there an alternate way to find the gain and
phase margins? YES
Ì Plot magnitude and phase of gain A(Š)
instead of βf A(Š) vs frequency
Ì Plot horizontal line at 1/ βf on magnitude
plot (since βf is independent of frequency)
Ì Why? At intersection of 1/ βf with A(Š), Š
= Š1
Φ(ω1)
Φ(ω180)
= A(ω
)
so
β f A(ω1 ) = 1
So this intersection point gives the value
where ω = Š1 !
Then we can find the gain and phase
margins as before.
If not adequate margins, pick another βf
value, draw another 1/ βf line and repeat
process.
Note: βf is a measure of the amount of
feedback. Larger βf means more feedback.
Ì Recall for βf = 0, we have NO feedback
βf =
Ì
Xf
Xo
so β f = 0 → X f = 0
Af =
A
→A
1+ β f A
As βf increases, we have more feedback
69
Example - Gain and Phase Margins
Alternate Method
A(dB)
Ao = 150
ω1
43.5 dB
Ao (ω ) =
1/βf(dB)
A(ω180)
Gain
margin
β f = 0 .1
so
1
βf
Ao ( dB ) = 43 .5 dB
150
ω ⎞⎛
ω ⎞
⎛
1+ j
⎟
⎜1 + j
5 ⎟⎜
8 x10 ⎠⎝
3 x10 6 ⎠
⎝
so
1
βf
=
1
= 10
0 .1
( dB ) = 20 log 10 = 20 dB
From plots of magnitude and phase
ω1 = 6 x10 6 rad / s and ω180 = 3 x10 7 rad / s
φ (ω1 ) = −140 0 and φ (ω180 ) = −180 0
φ (ω )
ω180
Phase
margin
so phase margin is
φ (ω1 ) − φ (ω180 ) = − 140 0 − ( − 180 0 ) = + 40 0
1 / β f (ω1 ) = 20 dB and
Ao (ω180 ) = −8 dB
so gain margin is
1 / β f (ω1 ) − Ao (ω180 ) = 20 − ( − 8 dB ) = +28 dB
These are the same results
as before for the gain and
70
phase margins .
Summary of Feedback Amplifier Stability
* Feedback amplifiers are potentially
unstable
Ì Can break into oscillation at a
particular frequency ωi where
A(Š)
Ao
Afo
ω Lf ω L
β (ω i )A(ω i ) = 1
ω H ω Hf
⎡ Im{β (ω i )A(ω i )}⎤
0
and φ (ω i ) = tan −1 ⎢
⎥ = −180 = −π radians
⎣ Re{β (ω i )A(ω i )}⎦
sin ce
ωi ω
β (ω i )A(ω i ) = β (ω i )A(ω i ) e jφ (ω i ) = e − jπ = −1
so
A f (ω i ) =
Oscillation instability is a design problem
*
for any amplifier with two or more
high frequency poles. This is the case for ALL
amplifiers since each bipolar or MOSFET
transistor has two capacitors and each capacitor
gives rise to a high frequency pole!
A (ω i )
1 + β (ω i )A(ω i )
=
A (ω i )
1−1
→∞
Need to design feedback amplifier
with adequate safety margin
Ì Minimum gain margin of 10 dB
Ì Minimum phase margin of 50o.
Ì Adjust βf by varying size of Rf .
“ Smaller Rf is, the larger is
the amount of feedback.
71