# Homework Help: Calculate impedance of a feedback circuit with Blackman's technique

1. Jul 21, 2015

### Waxterzz

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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 in to 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

2. Jul 22, 2015

### 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.

Last edited: Jul 23, 2015
3. Jul 23, 2015

### 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.

4. Jul 23, 2015

### 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.

5. Jul 23, 2015

### donpacino

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.

6. Jul 23, 2015

### rude man

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.

7. Jul 23, 2015

### 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.

8. Jul 23, 2015

### rude man

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.

Last edited: Jul 23, 2015
9. Jul 23, 2015

### 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.

10. Jul 23, 2015

### Staff: Mentor

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...

11. Jul 23, 2015

### rude man

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.

12. Jul 23, 2015

### donpacino

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

13. Jul 23, 2015

### The Electrician

These could be depletion mode FETs.

14. Jul 23, 2015

### donpacino

so i said op amp when i meant amp, it doesnt 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

15. Jul 23, 2015

### donpacino

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

edit: actually let me think on that...

16. Jul 23, 2015

### The Electrician

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

Last edited: Jul 23, 2015
17. Jul 23, 2015

### The Electrician

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?

Last edited: Jul 23, 2015
18. Jul 23, 2015

### rude man

@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

19. Jul 23, 2015

### The Electrician

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.

20. Jul 23, 2015

### rude man

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

21. Jul 23, 2015

### 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.

22. Jul 24, 2015

### 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 cant be zero?

23. Jul 24, 2015

### Waxterzz

Thanks for info.

Can you post your blackmans analysis?

This evening I try Vtest Itest again.

24. Jul 24, 2015

### The Electrician

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.

25. Jul 24, 2015

### 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 id = k(Vgs - VT)2
but Vgs = AV
and (Vdd - V)/R = k(AV - VT)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 Vdd, R, k, A and VT as you can imagine.

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