comparing two ways to find loop gain in feedback circuits rob fox university of florida

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.