Calculate impedance of a feedback circuit with Blackman's technique

[Waxterzz]

Homework Statement

[/B]
Calculate impedance of the node shown in this feedback circuit with Blackman and the Vtest/ Itest method.

Homework Equations

[/B]
Zin (with feedback) = Zin(with feedback circuit disabled) * (1-RRshort)/(1-RRopen)

RR = return ratio

In this form blackman theorem to us in the course I've been taken

3. The Attempt at a Solution

a) I put Vbias to zero, to the ground. I cut feedback loop and the + terminal connected to ground

I find Z into be equal to R

b) I connect Zin to the ground (short circuit), I apply test voltage to + terminal

RRshort = 0

c)

Same circuit, but now open circuit

vtest * A * R / 10 R is the voltage on the second lower gate
vg2 = vtest * A * R / 10 R

The current throught gate 1 and gate 2 has to be the same so

vgs1 = vgs2

the gate on first mos is connected to ground so vg1 = 0
the source on second mos is connected to ground so vs2 = 0

the source on first mos is connected to drain of the second

vs1 = vd2

vout equals vd1

If I combine all this I get vout/vtest = -A /10 for RRopen

d) Now Blackman

Z in = R * (1-0) / (1- (-A/10))

Z in = 10 R / (10+A)

This is what I get with Blackman

Is this correct?

How can I proceed with Vtest/ I test because I get nonsense If I do that, but first, Blackman.

Thanks

 

[rude man]

This circuit is impossible to analyze; you have not even stated the polarities of the two MOSFETs (n vs. p channel).
If both are n channel they are both active or saturated, as is the op amp output so the input Z would be low but impossible to compute without given the devices' parameters and the value of g.

 

[Waxterzz]

The first drawing was the only thing I got, it is not mine, but from a student who did an earlier exam and posted the questions on a board..

I suppose it are two NMOS's.

So my blackman computings are wrong? :)

Blackman is poorly explained in my course.

 

[rude man]

Well, confession time: I never heard of Blackman.
But I do know how to analyze circuits, most of them, and so I have to leave it where I posted last time - insufficient data to proceed.

 

[donpacino]

Homework Statement

[/B]
Calculate impedance of the node shown in this feedback circuit with Blackman and the Vtest/ Itest method.

Homework Equations

[/B]
Zin (with feedback) = Zin(with feedback circuit disabled) * (1-RRshort)/(1-RRopen)

RR = return ratio

In this form blackman theorem to us in the course I've been taken

3. The Attempt at a Solution

a) I put Vbias to zero, to the ground. I cut feedback loop and the + terminal connected to ground

I find Z into be equal to R

b) I connect Zin to the ground (short circuit), I apply test voltage to + terminal

RRshort = 0

c)

Same circuit, but now open circuit

vtest * A * R / 10 R is the voltage on the second lower gate
vg2 = vtest * A * R / 10 R

The current throught gate 1 and gate 2 has to be the same so

vgs1 = vgs2

the gate on first mos is connected to ground so vg1 = 0
the source on second mos is connected to ground so vs2 = 0

the source on first mos is connected to drain of the second

vs1 = vd2

vout equals vd1

If I combine all this I get vout/vtest = -A /10 for RRopen

d) Now Blackman

Z in = R * (1-0) / (1- (-A/10))

Z in = 10 R / (10+A)

This is what I get with Blackman

Is this correct?

How can I proceed with Vtest/ I test because I get nonsense If I do that, but first, Blackman.

Thanks

You need to do V short and V open with Vbias active. this will give you the correct answer. note that at vshort your answer will remain the same.

EDIT: actually you are right
rudeman. With feedback the impedance of the fets do not matter. due to the feedback the circuit will correct itself to the proper value regardless.

 

[rude man]

You need to do V short and V open with Vbias active. this will give you the correct answer. note that at vshort your answer will remain the same.

EDIT: actually you are right
rudeman. With feedback the impedance of the fets do not matter. due to the feedback the circuit will correct itself to the proper value regardless.

That would be true if the circuit worked with feedback to begin with. It can't. The drain voltage on the upper MOSFET would have to be zero to bring the op amp out of saturation, which of course would shut down the entire FET chain due to lack of bias voltages. So the drain voltage on the upper FET would go to some finite positive value, with the op amp output saturated positive, unless "g" was too low to turn on the bottom FET in which case the op amp output is still saturated positive.

 

[donpacino]

thats a good point. However I think that situation works itself out. The entire principle of the two terminals of an op amp being equal is dependent upon A being inf. The gain of the amplifier is taken into account with the circuit anlysis Op did above. With this circuit topology, and a typical operational amplifier, the voltage will be REALLY close to zero, but not exactly. But we don't know the gain of this amplifier, which is why it is a variable in the equation for Zin.

 

[rude man]

thats a good point. However I think that situation works itself out. The entire principle of the two terminals of an op amp being equal is dependent upon A being inf. The gain of the amplifier is taken into account with the circuit anlysis Op did above. With this circuit topology, and a typical operational amplifier, the voltage will be REALLY close to zero, but not exactly. But we don't know the gain of this amplifier, which is why it is a variable in the equation for Zin.

There's no way the two FETs could saturate to within the offset voltage of any op amp I've ever encountered. Furthermore, a saturated MOSFET has zero transconductance gain. Furthermore still, the - input to the op amp has a 50-50 chance of having to be negative to operate any loop. None of those things are possible.

If the amplifier had a very low gain (on the order of 1-10, i.e. in no way an op amp) then I acknowledge the circuit could possibly work. That would allow several volts of drain voltage of the upper FET. But without compensation, oscillations are likely.

 

[donpacino]

nowhere does it state that the fets have to be in saturation...
no, the op amp input does not have to be negative for this to work. furthermore no-one said it was an op-amp. for all we know it could be an ideal amplifier with a gain of 2.

you are missing the point of the OPs question. he wanted to know how to use blackmans theorm for analyzing impedance in a feedback circuit, not debating the finer points of circuit design. It is an analytical exercise.

 

[berkeman]

I agree with rude man on this -- that negative rail has to be some -Vdd instead of ground for the circuit to bias up correctly...

 

[rude man]

furthermore no-one said it was an op-amp. .

Oh? maybe look again at your post 7? We both assumed it was an op amp.

As I said, I never heard of Blackman's theorem, and I still claim that there is nowhere near enough info given to solve for input impedance.
Even knowing the MOSFET characteristics & the amp gain, this would be a challenging analysis.

 

[donpacino]

I agree with rude man on this -- that negative rail has to be some -Vdd instead of ground for the circuit to bias up correctly...

This is not a real amplifier system where you have to deal with noise, non-ideal supplies, etc.
lets say the gain of the amplifer is 1 to make it simple...
the positive input of the amp would be x volts in steady state.. x/10 volts would be drive the gate of the lower fet, which would in turn effect the curret through the fet, and then x (through feedback).

with x/2 voltage drop across each fet, i don't see how there is an issue

 

[The Electrician]

I agree with rude man on this -- that negative rail has to be some -Vdd instead of ground for the circuit to bias up correctly...

These could be depletion mode FETs.

 

[donpacino]

Oh? maybe look again at your post 7? We both assumed it was an op amp.

so i said op amp when i meant amp, it doesn't change my point later.
the point of this type of feedback is the actual properties of the fets can vary (within reason) and the circuit can work. as a disclaimer, if it was a real circuit you would need to take into account the maximum allowable gate voltage, etc

 

[donpacino]

These could be depletion mode FETs.

they can be, but due to the nature of the feedback it wouldn't effect Zin

edit: actually let me think on that...

 

[The Electrician]

d) Now Blackman

Z in = R * (1-0) / (1- (-A/10)) Should be: Z in = R * (1+0) / (1+ (-A/10))

Z in = 10 R / (10+A)

This is what I get with Blackman

Is this correct?

How can I proceed with Vtest/ I test because I get nonsense If I do that, but first, Blackman.

Thanks

There's a short writeup on Blackman's theorem at: https://en.wikipedia.org/wiki/Blackman's_theorem

It would appear that you have made a sign error in his formula. I added to the quote above to show the error.

Edit: There isn't exactly a sign error in the above. See post #19

 

[The Electrician]

How can I proceed with Vtest/ I test because I get nonsense If I do that, but first, Blackman.
Thanks

I think that for both Blackman and Vtest/Itest, if you want a solution that allows the opamp gain A to be finite, you need to deal with the gain of the bottom FET. Assuming that biasing is not a problem, maybe with a negative rail, or who knows what, and assuming that both FETs have infinite Ro and some transconductance gm, you can show that the voltage gain of the FET stage is -gm*R. You can include that in the loop gain and get a somewhat reasonable result.

I notice that there are absolute value bars around two of the subexpressions in the Blackman's formula at Wikipedia. I solved this problem with the Vtest/Itest method and I don't get the same result as Blackman's formula unless I ignore the absolute value bars. Edit: I made a mistake in my Vtest/Itest analysis. See post #19.

Here's another very detailed discussion of all this: http://users.ece.gatech.edu/~pallen/Academic/ECE_6412/Spring_2004/L290-ReturnRatio-2UP.pdf

On page 10, this author gives Blackman's formula without absolute values. Could Wikipedia be wrong?

 

[rude man]

I agree with rude man on this -- that negative rail has to be some -Vdd instead of ground for the circuit to bias up correctly...

@berkeman, If the op amp were a low-gain amp I think the circuit could theoretically work.

I would compute Zin by (1) establishing the equilibrium amp input v, (2) forcing a voltage v + Δv at that input, then computing Δv/i where i is the current thru an auxiliary wire forcing the input voltage to v + Δv. May try that soon. The upper fet looks like a cascode connection so the analysis could ignore it & let the lower fet determine the total current thru R and the auxiliary wire.

rm

 

[The Electrician]

d) Now Blackman

Z in = R * (1-0) / (1- (-A/10)) Should be Z in = R * (1+|0|) / (1+|(-A/10)|)

Z in = 10 R / (10+A)

This is what I get with Blackman

Is this correct?

How can I proceed with Vtest/ I test because I get nonsense If I do that, but first, Blackman.

Thanks

I have to take it back. After redoing my Vtest/Itest calculation, I think the absolute values are needed as shown above. The OP's result is the same as what one gets taking into account the absolute values.

 

[rude man]

I'm sending berkeman my computation for Zin by PM.
If anyone else wants to compare results, send me a PM also.

 

[The Electrician]

Without giving away anything else, I will say that I get the result the OP does for the Blackman method if the assumption is made (as he essentially does) that the voltage gain of the FET stage is unity.

As I mentioned earlier, it's not too unreasonable to assume a transconductance for the FETs of gm, for a better result.

Waxterzz, you say "How can I proceed with Vtest/ I test because I get nonsense If I do that, but first, Blackman."

At this point, you should post your attempt so we can see why you're getting nonsense.

 

[Waxterzz]

Hi all.

Need to read all these posts first.

Rudeman You can send your calculations also to me if you want.

Donpacino. In the short introduction of Blackman in my course it was stated I had to put the input voltage to zero. You say Vbias can't be zero?

 

[Waxterzz]

I think that for both Blackman and Vtest/Itest, if you want a solution that allows the opamp gain A to be finite, you need to deal with the gain of the bottom FET. Assuming that biasing is not a problem, maybe with a negative rail, or who knows what, and assuming that both FETs have infinite Ro and some transconductance gm, you can show that the voltage gain of the FET stage is -gm*R. You can include that in the loop gain and get a somewhat reasonable result.

I notice that there are absolute value bars around two of the subexpressions in the Blackman's formula at Wikipedia. I solved this problem with the Vtest/Itest method and I don't get the same result as Blackman's formula unless I ignore the absolute value bars. Edit: I made a mistake in my Vtest/Itest analysis. See post #19.

Here's another very detailed discussion of all this: http://users.ece.gatech.edu/~pallen/Academic/ECE_6412/Spring_2004/L290-ReturnRatio-2UP.pdf

On page 10, this author gives Blackman's formula without absolute values. Could Wikipedia be wrong?

Thanks for info.

Can you post your blackmans analysis?

This evening I try Vtest Itest again.

 

[The Electrician]

Thanks for info.

Can you post your blackmans analysis?

My analysis is the same as yours, except where you get -A/10 for RRopen, I get -A*gm*R/10. the term -gm*R is the voltage gain of the bottom FET which you took to be -1.

If it should happen that the value of gm were 1/R, the gain of the FET would be -1, and my result would be identical to yours.

 

[rude man]

By permission of our staff mentor I am posting the following:

1. compute bias operating point:
Let V = voltage at amp input; amp has gain A.
Also ignore the R-gR network, it's part of A.
MOSFET equation i_d = k(V_gs - VT)²
but V_gs = AV
and (V_dd - V)/R = k(AV - V_T)².
Solve for V (this is very messy; Wolfram Alpha comes in handy, I won't repeat what it gave but it's a straight-forward function of V_dd, R, k, A and V_T as you can imagine.

2. Then force V → V + dV, which requires injection current di_i. Amp output voltage changes by A*dV.
This changes fet current by di_d and current thru R by di_R :
di_d = di_i + di_R .
Then 1/Z_in = di_i/dV = di_d/dV - di_R/dV but di_R < 0
So 1/Z_in = 2k(AV - V_T) + 1/R.
Sorry I never heard of the "Blackman method" but I wonder if it's all that important to know ...

 

[rude man]

Rudeman You can send your calculations also to me if you want.

See my post #25.

 

[LvW]

Gentlemen, I didn`t follow the discussion in detail - but I have realized that the main subject is Blackmans method.
Perhaps it helps to know the correct solution (although the circuit is described only very roughly).
I did not use Blackmans method (by hand calculation) - instead, I have used a symbolic analysis program with the following result:

Zin=R/(0.1*gm*A*R + 1)=10R/(10+gm*R*A) or
1/Zin=1/R + 0.1*gm*A
with A=opamp gain and gm=transconductance of the FET chain.
This calculation assumes that the FET chain is an ideal current source (infinite r,out).

 

[rude man]

Gentlemen, I didn`t follow the discussion in detail - but I have realized that the main subject is Blackmans method.
Perhaps it helps to know the correct solution (although the circuit is described only very roughly).
I did not use Blackmans method (by hand calculation) - instead, I have used a symbolic analysis program with the following result:

Zin=R/(0.1*gm*A*R + 1)=10R/(10+gm*R*A) or
1/Zin=1/R + 0.1*gm*A
with A=opamp gain and gm=transconductance of the FET chain.
This calculation assumes that the FET chain is an ideal current source (infinite r,out).

@LvW, this looks persuasive, since g_m for a MOSFET in active mode is 2k(V_gs - V_T). Where did the 0.1 come from? What did you assume for the MOSFET gain (k)?

 

[The Electrician]

@LvW, this looks persuasive, since g_m for a MOSFET in active mode is 2k(V_gs - V_T). What did you assume for the MOSFET gain (k)?

Whatever the value of k, it's subsumed in gm.

Where did the 0.1 come from?

There's a resistive voltage divider at the output of the opamp. The top resistor looks like it's labeled "gR", but it's actually "9R". The .1 comes from that resistive voltage divider.

 

[LvW]

There's a resistive voltage divider at the output of the opamp. The top resistor looks like it's labeled "gR", but it's actually "9R". The .1 comes from that resistive voltage divider.

Yes.

 

[LvW]

I must admit that - up to now - I didn`t need Blackman`s theorem. So, I took this opportunity to become familiar with it.
If somebody is still interested to see how it woks for the present task, here is the solution:

The impedance Z between two points of a feedback system is:
Z=ZD(1+Tsc)/(1+Toc)
with ZD=Impedance between both points without feedback , and
Tsc=Loop gain magnitude with a short across the selcted points; Toc=Loop gain magnitude with an open circuit across both points.

For the present case: We need the resistance between the pos. opamp input (point 1) and ground (point 2).

Therefore: ZD=R and
Tsc=0 because A=0 (both opamp input terminals at ground potential) and |Toc|=A*0.1*gm*R.

This gives

Z=R*(1+0)/(1+A*0.1*gm*R)=R/(1+A*0.1*gm*R)
(confirmed by symbolic analyzer, see my previous post)

 

[rude man]

I must admit that - up to now - I didn`t need Blackman`s theorem. So, I took this opportunity to become familiar with it.
If somebody is still interested to see how it woks for the present task, here is the solution:

The impedance Z between two points of a feedback system is:
Z=ZD(1+Tsc)/(1+Toc)
with ZD=Impedance between both points without feedback , and
Tsc=Loop gain magnitude with a short across the selcted points; Toc=Loop gain magnitude with an open circuit across both points.

For the present case: We need the resistance between the pos. opamp input (point 1) and ground (point 2).

Therefore: ZD=R and
Tsc=0 because A=0 (both opamp input terminals at ground potential) and |Toc|=A*0.1*gm*R.

This gives

Z=R*(1+0)/(1+A*0.1*gm*R)=R/(1+A*0.1*gm*R)
(confirmed by symbolic analyzer, see my previous post)

Thanks for your research! Only thing I would point out is that g_m is a strong function of V_gs = 0.1AV where V is the quiescent + amp input voltage ("point 1"). So the operating point of the circuit still has to be solved for, then g_m = 2k(0.1AV - V_T) can be determined. Of course, you'd need k also.

 

[LvW]

Yes, of course. My calculation (blindly) assumes an operating point in the linear FET region.

 

[rude man]

Yes, of course. My calculation (blindly) assumes an operating point in the linear FET region.

OK. Now you'd be subject to V_ds modulation but I'd say enough of this, thanks for enlightening us all on the (in?)famous Blackman method!
r m

 

[The Electrician]

Waxterzz, are you going to try get a result with the Vtest/Itest method?