Comparing two ways to find loop gain in feedback circuits
Rob FoxUniversity of Florida
R3R1
vo
vs
R2
Analyzing FB circuits.
The old way.
R3R1
vo
vs
R2
Analyzing FB circuits.
The old way. (I think it’s the wrong way.)
R3R1
vo
vs
R2
Analyzing FB circuits.
The old way.
First, recognize this connection is series shunt.
This determines which of four methodswe’ll use.
R1
vsRi RoGmvi
vi
voR2
R3
Replace the amp with itssmall-signal equivalent circuit
R1
R2
vsRi RoGmvi
vi
voR2
vf
R3 R1
Because it’s shunt at the output, we short the right side of R2
to find the circuit for loading at the input side.Series feedback at the input means we cut Ri out but leaveR1 at the right (output) side.
R1
R2
vsRi RoGmvi
vi
voR2
vf
R3 R1
Because it’s shunt at the output, we short the right side of R2
to find the circuit for loading at the input side.Series feedback at the input means we cut Ri out but leaveR1 at the right (output) side.
Analyze to find A = vo/vs and β = vf/vo.
The loop gain T should be Aβ = vf/vs.
R1
iovs
R2
R3
Now it’s series series.
Note also that there are other places wecould define series outputs, giving differentresults.
R1
vsRi RoGmvi
vi
io
R2
R3
Replace with equivalent circuit.
Series series.
R1
R2
vsRi RoGmvi
vi
R2
vf
R3
io
R3 R1
Changing output to series meanswe now leave R3 in series with R2
at the input side.
This gives a new value for the loop gain.
Series series.
R1
io
isR2
R3
Now it’s shunt at the input, series at output.
R1is
Ri RoGmvi
vi
io
R2
R3
Now it’s shunt at the input, series at output.
R1
R2
Ri RoGmvi
vi
R2
R3
io
is
if
R3
Because it’s shunt at the input, weshort the right side of R2 at the output.
Yet another value for Aβ.
Now it’s shunt at the input, series at output.
R1is
voR2
R3
Last permutation is shunt shunt.
R1is
Ri RoGmvi
vi
voR2
R3
Last permutation is shunt shunt.
R1
Ri RoGmvi
vi
R2
isR2
R3
This gives a fourth circuit and a fourth equation for Aβ.
How does the circuit know it’s supposed to havedifferent loop gains depending on where we putthe inputs and define what we want to use asoutputs??
voR2
if
Last permutation is shunt shunt.
R1
Ri RoGmvi
vi
R2
isR2
R3
This gives a fourth circuit and a fourth equation for Aβ.
Are the results the same for the four methods?
Let’s try it with numbers for the four cases.
voR2
if
Last permutation is shunt shunt.
R1
R2
Ri RoGmvi
vi
R2
R3
io
is
if
R3
Shunt series.
Aβ = R1 PRi P R2 +R3( )⎡⎣ ⎤⎦⋅Gm ⋅ Ro PR3 PR2
⎡⎣ ⎤⎦⋅1
R2
=15.466
Let R1=10kR2=9kR3=5kRi=20kRo=2kGm=25 mA/V
R1
Ri RoGmvi
vi
R2
isR2
R3
voR2
if
Shunt shunt.
Aβ = R1 PRi PR3
⎡⎣ ⎤⎦⋅Gm ⋅ Ro PR3 PR2⎡⎣ ⎤⎦⋅
1R2
=13.116
Let R1=10kR2=9kR3=5kRi=20kRo=2kGm=25 mA/V
R1
R2
vsRi RoGmvi
vi
voR2
vf
R3 R1
Series shunt.
Aβ =
Ri
Ri + R1 PR2( )⋅Gm ⋅ Ro PR3 P R2 +R1( )⎡⎣ ⎤⎦⋅
R1
R1 +R2
=14.135
Let R1=10kR2=9kR3=5kRi=20kRo=2kGm=25 mA/V
R1
R2
vsRi RoGmvi
vi
R2
vf
R3
io
R3 R1
Series series.
Aβ =Ri
Ri + R1 P R2 +R3( )⎡⎣ ⎤⎦⋅Gm ⋅ Ro PR3 P R2 +R1( )⎡⎣ ⎤⎦⋅
R1
R1 +R2
=13.535
Let R1=10kR2=9kR3=5kRi=20kRo=2kGm=25 mA/V
R1
R2
vsRi RoGmvi
vi
R2
vf
R3
io
R3 R1
Four analyses; four different results for loop gain.Shunt series: T = 15.466Shunt shunt: T = 13.116Series shunt: T = 14.135Series series: T = 13.535
Let R1=10kR2=9kR3=5kRi=20kRo=2kGm=25 mA/V
R1
R2
vsRi RoGmvi
vi
R2
vf
R3
io
R3 R1
Four analyses; four different results for loop gain.
Are any of these actually results correct?
No. The actual loop gain is
T =
R1 PRi
R2 + R1 PRi( )⋅Gm ⋅ Ro PR3 P R2 + R1 PRi( )( )⎡
⎣⎤⎦
R1
R2
vsRi RoGmvi
vi
R2
vf
R3
io
R3 R1
Four analyses; four different results for loop gain.Shunt series: T = 15.466Shunt shunt: T = 13.116Series shunt: T = 14.135Series series: T = 13.535
Actual value: T = 13.928
This is the value for T that correctly gets the closed-loop gain and all series and shunt resistances.
Let R1=10kR2=9kR3=5kRi=20kRo=2kGm=25 mA/V
R3R1
vo
vs
R2
Here’s the better way to find loop gain.
R3R1
R2
First kill the input sourceand remove the definitionof the output variable.
Note that you can no longer tell whichof the four topologies this is.
R3R1
R2
The circuit has a closed signal path so there is feedback.The feedback is negative. (Odd # of inversions.)As the schematic is drawn, the feedback signal flows clockwise.The feedback signal flow is unidirectional.
R3R1
R2
Break the signal path in any branch, so long as it breaksall signal flow.
R3R1
R2
Break the signal path in any branch, so long as it breaksall signal flow.
R3R1
R2
Break the signal path in any branch, so long as it breaksall signal flow.
R3R1
R2
Break the signal path in any branch, so long as it breaksall signal flow.
R3R1
R2
Let’s use this breakpoint.
R1
Ri RoGmvi
vi
R2
R3
Substitute the small-signal equivalent circuit.
R1
Ri RoGmvi
vi
R2
R3
x y
Break the branch.
Call the input side the x portand the output side the y port.
R1
Ri RoGmvi
vi
R2
R3
vx
yZin
Temporarily short the y portand find the input impedanceZin looking into the x port.
R1
Ri RoGmvi
vi
R2
R3vx
yZin
Zin = R1||Ri
Temporarily short the y portand find the input impedanceZin looking into the x port.
Here, Zin = R1||Ri.
R1
Ri RoGmvi
vi
R2
R3vx
vy
Zin = R1||Ri
Zin
Now place a copy of Zin
across the y port, andfind the loop gain as
|T| = |vy/vx|.
R1
Ri RoGmvi
vi
R2
R3vx
vy
Zin = R1||Ri
Zin
Now place a copy of Zin
across the y port, andfind the loop gain as
|T| = |vy/vx|.
Don’t worry about the sign – we already know the feedback is negative.
R1
Ri RoGmvi
vi
R2
R3
x y
What about the special case thatRi = 0? (Maybe the amp is current-controlled.)Alternate algorithm using currentgain:
R1
Ri RoGmvi
vi
R2
R3
x y
Zout
First short the x port and find Zout
looking into the y port.
Alternate algorithm using currentgain:
R1
Ri RoGmvi
vi
R2
R3
x y
Zout
Zout = R2 + R3||Ro
Here, Zout = R2 + R3||Ro.
R1
Ri RoGmvi
vi
R2
R3
iy
Zout = R2 + R3||Ro
Zout ix
Put a copy of Zout at the x port.
Now short the y port and find |T |= |iy/ix|.
Same result we got before using vy and vx.
We started with a voltage-involtage-out amplifier, but the firstthing we did was ignore that information.Same T if we’d started with current-incurrent-out or any other combination.
R1
io
isR2
R3
R3R1
R2
The Aβ approach requires a new analysis if you use adifferent input type or redefine the output variable.The Aβ approach is ambiguous unless the circuit is anamplifier. (What if it’s just a bias circuit?)The Aβ approach gives different results for differentassumptions about the feedback topology.Those results can differ significantly from the actualvalue of the loop gain.
R3R1
R2
The value of T found as shown here is the one that gives correct results for the impedances looking into various ports using RCL = ROL*(1 + T) for seriesand RCL = ROL/(1 + T) for shunt.The value of T found as shown here does not change depending on where the inputs and ouputs are. You get the same value no matter where you break the loop or whether you use current or voltage test sources.
R3R1
R2
This loop-gain-based approach is much easier to teach and learn.It doesn’t require a table listing the tricks to use for each of four amplifier configurations. One rule for all.Compatible with the A∞ approach for closed-loop gain – no need for β’s with different dimensionality for each configuration.
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