Lecture 280 – Series-Shunt and Shunt-Series Feedback (3/15/02)
Page 280-1
LECTURE 280 – SERIES-SHUNT AND SHUNT-SERIES FEEDBACK
(READING: GHLM – 579-587)
Objective
The objective of this presentation is:
1.) Illustrate the analysis of series-shunt and shunt-series feedback circuits
Outline
• Series-shunt feedback with nonideal source and load
• Series-shunt example
• Shunt-series feedback with nonideal source and load
• Shunt-series example
• Summary
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 280 – Series-Shunt and Shunt-Series Feedback (3/15/02)
Page 280-2
Series-Shunt Feedback including Source and Load Resistances
Configuration:
RS
ii
i1
Basic Amplifier
h11a
h12avo
vs
h22a
RL
h21aii
h12fvo
ii
RS
h22f
h21fii
New Basic Amplifier
+
vo
-
h11a
h11f
h21aii
h22a
h22f
RL
New Feedback Network
h12fvo
Fig. 280-01
vo v2
(-h21a/h11h22)
a
=
=
A
=
=
vs v1
1+af 1+(-h21a/h11h22) h12f
ECE 6412 - Analog Integrated Circuit Design - II
+
v1
-
Series-Shunt
Feedback
Network
+
v2
Fig. 280-015
Feedback Network
h11f
vs
+
vo
-
i2
v1 = h11i1 + h12v2
i2 = h21i1 + h22v2
where for the new basic amplifier,
v1 |
h11 = i v2=0 = RS + h11a + h11f
1
v1 |
h12 = v2 i1=0 = 0
i2 |
h21 = i v2=0 = h21a
1
i2 |
h22 = v2 i1=0 = GL + h22a + h22f
-h21a
⇒ a = h11h22 and f = h12f
© P.E. Allen - 2002
Lecture 280 – Series-Shunt and Shunt-Series Feedback (3/15/02)
Page 280-3
Example 1 – Series-Shunt Feedback Amplifier
For the amplifier shown, find v2/v1,
v1/i1, and v2/i2. Assume that βF1 =
R3 =
βF2 = 100, VBEQ ≈ 0.7V and VA1 =
10kΩ
150kΩ
VA2 = 100V for the BJTs. Note that i
1
this circuit can be simulated using
+ R =
SPICE.
S
C 1=
1kΩ 5µF
Solution
4.7kΩ
47kΩ
v1
1.) Topology identification. We see
R1 =
that the circuit is series-shunt,
100Ω
negative feedback. Also, note that RS is “outside” the feedback circuit.
2.) Closed loop small-signal model is shown below.
R2
+
v1
RS
RB
-
-
v2=vo
R3
Rif i
b1
C3 =
Q1 5µF
C5 = 10µF
R2 =
4.7kΩ
+
R4 =
10kΩ
C4 =
50µF
4.7kΩ
33kΩ
i2
C6 =
5µF
Q2
C2 =
50µF
v2
-
Fig. 280-02
β1ib1
rπ1 +
vfb R1 R3
+
vi
-
i2
rπ2
+
v2=vo
-
ib2
R4
β1ib2
-
-
-
RC2 =
4.7kΩ
R2
+
Q2
R4
47kΩ
i2
i1
Q1
++
ve+
vi
vfb R1
+25V
Fig. 280-03
Rif
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 280 – Series-Shunt and Shunt-Series Feedback (3/15/02)
Page 280-4
Example 1 – Continued
3.) Break open the loop by finding the loading effects of the feedback network on the
amplifier. This involves finding h11f and h22f.
v1f
|
h11f = i1f v2f=vo= 0 = R1||R2
i2f
|
h22f = v2f i1f=ii= 0 = 1/(R1+R2)
β1ib'1
4.) Open-circuit small-signal model: Ri ib1'
i2' Ro
R1
vfb'
+
+ rπ1
ib2'
R2
f = vo' = R1+R2
R 1 R2
rπ2
R3
vi'
+ R4 v2' = vo'
100
1
'
v
fb
β1ib2'
R1
= 4700+100 = 48 = 0.0208
Fig. 280-04
-βR
vo' vo' ib2' ib1'
a = v ' = i ' i ' v ' = (-β2[R4||(R1+R2)] R +r
i
b2 b1
i
3




















3 
π2


1
= (-1.916x105Ω)(-83.735)12.364kΩ = 1298V/V
ECE 6412 - Analog Integrated Circuit Design - II

1

rπ1+(1+β)[R1||R2]
⇒
1298
T = 48 = 27
© P.E. Allen - 2002
Lecture 280 – Series-Shunt and Shunt-Series Feedback (3/15/02)
Page 280-5
Example 1 – Continued
5.) Input resistance, v1/i1.
Ri = rπ1+(1+β)[R1||R2] = 12.364kΩ → Rif = Ri(1+T) = 12.364kΩ·28 = 334.3kΩ
However, v1/i1, can be found as
v1/i1 = Rin = RS + RB||Rif = 1kΩ + (150kΩ||47kΩ||334.3kΩ) = 33.33kΩ
6.) Voltage gain, v2/v1. First find vo/vi.
vo
a
1298
=
=
vi 1+af 1+27 = 46.3V/V
 32.33kΩ
v2 vo vi  vo Rif||RB 




=
=
=
46.3
  33.33kΩ = 44.9V/V
v1  vi v1  vi RS+Rif||RB 


7.) Ouput resistance, v2/i2.
Ro = R4||(R1+R2) = 1.916KΩ
Ro
1.916KΩ
Rof = 1+af =
= 68.3Ω
28
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 280 – Series-Shunt and Shunt-Series Feedback (3/15/02)
Page 280-6
Shunt-Series Feedback including Source and Load Resistances
Configuration:
io
is
RS
+
vi
-
i1
g22a
g11a
g12avo
+
v1
g21avi
Basic Amplifier
-
RL
i2
Shunt-Series
Feedback
Network
+
v2
Fig. 280-06
g22f
i1 = g11v1 + g12i2
g21fvi
g12fio
v2 = g21v1 + g22i2
Feedback Network
where for the new basic amplifier,
i1 |
io
g
=
11 v1 i2=0 = GS + g11a + g11f
+
g22a
vi
RS
is
g21avi RL
i1 |
g22f
g
g
11f
11a
g
=
12 i2 v1=0 = 0
New Basic Amplifier
v2 |
g21 = v i2=0 = g21a
1
g12fio
v2 |
Fig. 280-05
g22 = i2 v1=0 = RL + g22a + g22f
New Feedback Network
io i2
(-g21a/g11g22)
-g21a
a
=
=
A
=
=
⇒
a
=
is i 1
g11g22 and f = g12f
1+af 1+(-g21a/g11g22) g12f
g11f
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 280 – Series-Shunt and Shunt-Series Feedback (3/15/02)
Page 280-7
Example 2 – Shunt-Series Feedback Amplifier
VDD
For the amplifier shown, find v2/v1, v1/i1, and v2/i2.
Assume that all MOSFET transconductances are 1mS.
io
R4 =
R2 =
Solution
i2
500Ω
10kΩ
1.) Topology identification. We see that the circuit is
ie M1
+
shunt-series, negative feedback. Also, note that R1 is i1 = ii
M2
“outside” the feedback circuit.
+ R1 =
v2
2.) Closed loop small-signal model is shown below.
1kΩ
=
R
i
R
=
v1
5
fb
3
ifb
i1 = ii ie
R3
+vgs2 -
+
Rif
vgs1
gm1vgs1
-
gm2vgs2
i2
io
R4
R2
1kΩ
50kΩ
-
-
Fig. 280-07
+
v2
Rof
-
R5
Fig. 280-08
3.) Break open the loop by finding the loading effects of the feedback network on the
amplifier. This involves finding h11f and h22f.
i1f
v2f
|
|
g11f = v1f i2f=io= 0 = 1/(R3 + R5)
and
g22f = i2f v1f=vi= 0 = R3||R5
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 280 – Series-Shunt and Shunt-Series Feedback (3/15/02)
Example 2 - Continued
4.) Open-circuit small-signal model:
ifb'
R5
f = i ' = R +R
o
3 5
1
1
= 1+50 = 51 = 0.0196
Page 280-8
Ri
i 1 ' = i i'
i e'
R3
+
R5
gm2vgs2'
+vgs2' -
vgs1'
gm1vgs1'
-
R2
R3
ifb'
R5
R4
i2'
i o'
+
v2'
Ro
-
Fig. 280-09
io' vgs2' vgs1'
-gm1R2
io'
a = i ' = v ' v ' i ' = (-gm2) 1+ g (R ||R ) (R3+R5)
i
gs2 gs1
i
m2 3 5




















= (-1x10-3S)(-5.0495)(51x103Ω) = 257.5A/A




⇒
257.5
T = 51 = 5.05
5.) Input resistance, v1/i1.
Ri
51kΩ
Ri = R3+R5 = 51kΩ → Rif = 1+T = 6.05 = 8.43kΩ
However, v1/i1, can be found as
v1/i1 = Rin = RS + Rif = 1kΩ + 8.43kΩ = 9.43kΩ
6.) Voltage gain, v2/v1. First find io/ii.
 500Ω 
io
v2 io R4 
a
257.5




=
=
=
42.56A/A
→
=
=
42.56
  9.43kΩ = 2.257V/V
ii 1+af 1+5.05
v1  ii Rin 


ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 280 – Series-Shunt and Shunt-Series Feedback (3/15/02)
Page 280-9
Example 2 - Continued
7.) Ouput resistance, v2/i2.
First find Ro.
Ro = ∞ because it includes the drain of M2 and rds2 = ∞.
v2
∴
i2 = R4||(Rof - R4) = 500Ω||∞ = 500Ω
What about i2/i1? Assume the output is shorted, then
i2/i1 = -io/ii = -42.56A/A
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 280 – Series-Shunt and Shunt-Series Feedback (3/15/02)
Page 280-10
SUMMARY
• Series-Shunt:
- Increases the input resistance with feedback
- Decreases the output resistance with feedback
• Shunt-Series:
- Decreases the input resistance with feedback
- Increases the output resistance with feedback
• Analysis of transistor feedback circuits:
1.) Identify the topology.
2.) Draw the closed-loop small signal model.
3.) Break the loop by calculating the new basic amplifier with the new basic feedback
network consisting of only a controlled source.
4.) Draw the open-loop small-signal model using primed variables.
5.) Find a, f, Ri, and Ro.
6.) Calculate A, Rif, and Rof.
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002