Lecture 260 – Shunt-Shunt Feedback (3/5/02)
Page 260-1
LECTURE 260 – SHUNT-SHUNT FEEDBACK
(READING: GHLM – 563-569)
Objective
The objective of this presentation is:
1.) Show how to identify the type of feedback topology
2.) Illustrate the analysis of shunt-shunt feedback circuits
Outline
• Feedback identification procedure
• Shunt-shunt feedback with nonideal source and load
• Examples
• Summary
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 260 – Shunt-Shunt Feedback (3/5/02)
Page 260-2
IDENTIFICATION OF THE FOUR, SINGLE-LOOP FEEDBACK TOPOLOGIES
Two-Terminal Representation of a Single-Loop, Negative Feedback System
ii
Signal
Source
+
vi
io
ie
+
ve
Mixing
Network
-
-
Amplifier
of forward
gain = a
+
Sampling
Load
Network vo Network
-
ifb
+
vfb
-
ECE 6412 - Analog Integrated Circuit Design - II
Feedback
Network
of reverse
gain = f
Fig. 260-01
© P.E. Allen - 2002
Lecture 260 – Shunt-Shunt Feedback (3/5/02)
Page 260-3
Feedback Topology Identification Procedure
1.) Identify the feedback loop by tracing around the feedforward and feedback path.
Also check to see if the feedback is positive or negative.
2.) Identify whether or not the mixing network is series or shunt. If the signal source
has one terminal on ac ground then:
a.) If the input active device has one of its input terminals on ac ground, then the
mixing network must be shunt.
b.) If the signal and feedback sources are applied to different input terminals of the
input active device, then the mixing network is series (this includes differential
amplifiers where two devices form the input active device).
c.) If the signal source does not have one of its input terminals on ac ground or to
check the above steps, try to assign the variables xi, xfb, and xe on the schematic
in such a manner as to implement the equation,
xe = xi ± xfb
If this equation can be written using voltages (currents) then the mixing circuit is
series (shunt).
ifb
ii
+ +
ve
+
vi
vfb
-
+ +
ie
- +
ve
vfb
-
vi
-
Series
Shunt
Series
Fig. 260-02
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 260 – Shunt-Shunt Feedback (3/5/02)
Page 260-4
Feedback Topology Identification Procedure - Continued
3.) Next identify the sampling circuit as series or shunt. If the load is grounded then,
a.) If the out active device has one of its two possible output terminals grounded,
then the feedback is shunt.
b.) If the output active device has neither of its output terminals on ground and if the
output signal is taken from one of its output terminals and the fed back signal
from the other output terminal, then the feedback is series.
c.) If the load is not grounded or to check the above test, identify the load resistor,
RL, and apply the following test:
i.) If xfb becomes zero when RL = 0, then the sampling network is shunt.
ii.) If xfb becomes zero when RL = ∞, then the sampling network is series.
io
ifb
ifb
+
+
ifb
+vfb
-
RL
vfb
-
Series
ECE 6412 - Analog Integrated Circuit Design - II
RL
Shunt
+
vfb
vo
-
-
+
+
vo
-
Shunt
Fig. 260-03
© P.E. Allen - 2002
Lecture 260 – Shunt-Shunt Feedback (3/5/02)
Page 260-5
Transistor Examples of Negative Feedback Topology Identification
1K
i2
i2
10K
+
Q1
i1
10K
+
v1
-
Q1 10K
+
v2
100K
v1
1K
+
M2
i1
1K
10K
v2
1K
-
-
Circuit 2
Circuit 1
i2
i2
Q2
i1
M1
+
v1
-
3K
10K
1K
M2
v2
10K
v1
10K
+
10K
Q1
+
5K
2K
i1
+
v2
1K
-
-
-
Circuit 3
Circuit 4
Fig. 260-04
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 260 – Shunt-Shunt Feedback (3/5/02)
Page 260-6
Shunt-Shunt Feedback including Source and Load Resistance
Configuration:
i1
is
RS
+
vi
-
y22a
y11a
y12avo
y21avi
Basic Amplifier
y22f
y11f
RL
+
vo
-
+
v1
-
i2
Shunt-Shunt
Feedback
Network
+
v2
Fig. 260-06
i1 = y11v1 + y12v2
Feedback Network
i2 = y21v1 + y22v2
where for the new basic amplifier,
+
+
i1 |
vi
RS
is
vo
=
y
11 v1 v2=0 = GS + y11a + y11f
y11f y11a y21avi y22a y22f RL
i1 |
New Basic Amplifier
y12 = v v1=0 = 0
2
y12fvo
i2 |
y21 = v v2=0 = y21a
Fig. 260-05
New Feedback Network
1
i2 |
y22 = v v1=0 = GL + y22a + y22f
2
vo v2
(-y21a/y11y22)
-y21a
a
=
=
A
=
=
⇒
a
=
is
i1
y11y22 and f = y12f
1+af 1+(-y21a/y11y22) y12f
y12fvo
y21fvi
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 260 – Shunt-Shunt Feedback (3/5/02)
Page 260-7
Example 1 – Inverting Op Amp
Find the closed-loop transfer function, A, the closed-loop input
resistance, Zif, and the closed-loop output resistance, Zof of the
shunt-shunt configuration shown. The op amp has a differential
input resistance of zi, voltage gain of av, and output resistance of zo.
Solution
Equivalent circuit:
Feedback
R
Network F
Equivalent
Feedback
Network
ifb
Zif
+
ii
v1
-
Op Amp
zi
avv1
Zof
+
zo
RL
vo
-
ii'
Zi
RF
ii
Lecture 260 – Shunt-Shunt Feedback (3/5/02)
+
RL
Fig. 260-07
Basic Amplifier
+
zo
v1' zi
RF
'
- avv1
Zo
+
ifb' =
RL vo'
RF
v'o
RF
(Use primed variables to indicate open loop) Fig. 260-08
i1 ' |
-1
It is easy to show that f = y12f = v2' v1'=0 = RF
The forward gain, a, is
  ziRF 
vo' vo'v1'  -av(RF||RL)  ziRF  
-avRFRL
a = ii' = v1' ii'  = zo+(RF||RL)zi+RF = zoRF+zoRL+RFRLzi+RF









ECE 6412 - Analog Integrated Circuit Design - II
vout
where ifb' = 0
© P.E. Allen - 2002
Page 260-8
Example 1 – Continued
The loop gain is

  zi 
avRFRL
T = af = zoRF+zoRL+RFRLzi+RF



The closed-loop gain is

  ziRF 
-avRFRL



 z R +z R +R R   z +R 
(-avRF2RLzi)
vo
a
o L F L i F
 o F
  zi  = (zoRF+zoRL+RFRL)(zi+RF) + avRFRLzi
ii = 1+af = 
avRFRL
1+z R +z R +R R z +R 
o L F L i F
 o F
The closed-loop input impedance is
ziRF
Zi
zoRF+zoRL+RFRL
zi+RF
≈
Zif = 1+T = 
  zi 
avRFRL
avRL
1+zoRF+zoRL+RFRLzi+RF



The closed-loop output impedance is
Zo
zo||RF||RL
Zof = 1+T = 
  zi 
avRFRL

1+zoRF+zoRL+RFRLzi+RF



If av = 200,000, zi = 2MΩ, and zo = 75Ω, RF = 1MΩ, and RL = 10kΩ, then T = 133,333,
Zi = 2||1 = 0.667MΩ → Zif = 5Ω, Zo ≈ 75Ω → Zof = 0.563mΩ and A = -999,992Ω
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 260 – Shunt-Shunt Feedback (3/5/02)
Page 260-9
Example 2 – Transistor Feedback Amplifier
VCC
For the amplifier shown, find v2/v1, v1/i1, and v2/i2. Assume that Fig. 260-09
R3=
gm = 5mS and rds = ∞ for the MOSFET and rπ1 = rπ3 =1000Ω
20KΩ
M2
and βF1 = βF3 = 100 for the BJTs.
i1
i2
Solution
Q1
+
+
1.) Find the feedback topology and polarity of feedback.
Q3
v2
The loop consists of base-collector of Q1, gate-drain of M2, and v1
R1=
R2=
R4=
base-collector of Q3. A positive change at the base of Q1 gives
1KΩ
100Ω
10KΩ
a “-“ and the gate of M2, which gives a “+” at the base of Q3, which gives a “-“ at the base of Q3. ∴ feedback is negative.
2.) The mixing circuit is shunt because only the base terminal of Q1 is connected to the
input and feedback. Note that iC3 = ifb and iB1 = ie ⇒ ie = ii –ifb.
3.) The feedback circuit is shunt because the output transistor (M2) has one of its
possible output terminals on ac ground. Also, if RL = R4 goes to zero, ifb = 0.
i2
4.) Draw the
βib3
ib3
closed-loop circuit i1=ii ie
+
rπ3
and small-signal
Q1
ifb
M2
+
R1
ifb
βib1
ie=ib1
i2
R3 =
i1=ii
model.
v
v1
-
Q3
R1 =
1KΩ
R2 =
100Ω
20KΩ R =
4
10KΩ
2=vo
+
v1
-
-
+
rπ1
R2
R3
vgs2
-
R4
+
v2=vo
gm2vgs2
-
Fig. 260-10
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 260 – Shunt-Shunt Feedback (3/5/02)
Page 260-10
Example 2 – Continued
5.) AC open-loop model is drawn by,
a.) Looking back into the feedback network (Q3,R1) to the left with v1 = 0.
b.) Looking back into the feedback network (Q3,R1) to the right with v2 = 0.
The result is,
i1'=ii'
ib1'
+
rπ1
βib1'
i2'
+
ib3'
v1
R2
-
R3
vgs2'
gm2vgs2'
+
rπ3
v2'=vo'
ifb'
βib3'
R1
R4
Fig. 260-11
6.) Next, find a and f.
vo'  vo' vgs2'
a = i ' = v ' i '  = -gm2{R4||[rπ3+(1+β3)R1]}(-β3R3) = (-45.54)(-2x106) = 91.07MΩ
1
 gs2   1 
vo'
ifb'
β3
f = vo' , iB1’ = rπ3+(1+β3)R1 → ifb' = β3iB3’ → f = rπ3+(1+β3)R1 = 0.98mS (≈ 1/R1)
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002
Lecture 260 – Shunt-Shunt Feedback (3/5/02)
Page 260-11
Example 2 – Continued
7.) Calculate Ri and Ro.
v1'
Ri = i ' = rπ1+(1+β1)R2 = 11kΩ
1
and
v2'
Ro = i ' = R4||[rπ3+(1+β3)R1] = 10kΩ||102kΩ = 9.107kΩ
2
8.) Find v2/v1, v1/i1, and v2/i2.
vo
91.07MΩ
a
1
=
=
=
1019.2Ω
(Note:
3
i1 1+T 1+89.287x10
f = 1020Ω)
v1
Ri
11kΩ
Rif = i1 = 1+T = 1+89.287x103 = 0.1243Ω
v2
Ro
9.107kΩ
Rof = i = 1+T = 1+89.287x103 = 0.102Ω
2
Now,
v2 v2 i1  v2/i1 1019.2Ω
v1 =  i1 v1 = Rif = 0.1243Ω = 8204.5V/V
ECE 6412 - Analog Integrated Circuit Design - II
Lecture 260 – Shunt-Shunt Feedback (3/5/02)
© P.E. Allen - 2002
Page 260-12
SUMMARY
• We have shown how to identify the type of feedback topology
• Illustrate the analysis of shunt-shunt feedback circuits
• Procedure
1.) Identify the type of feedback topology
2.) Draw the schematic for the closed-loop amplifier
3.) Break the feedback loop by replacing the feedback network with the resistance
seen looking into each end of the feedback network with the other end shorted.
4.) Draw a schematic of the open-loop amplifier with all variables primed.
5.) Solve for a, f, Ri, and Ro.
6.) Calculate A and Rif and Rof by the following formulas:
Ri
Ro
a
A = 1+T ,
Rif = 1+T , and
Rof = 1+T
7.) If necessary, use the results of 6.) to find the desired closed-loop transfer function.
ECE 6412 - Analog Integrated Circuit Design - II
© P.E. Allen - 2002