Determining the value of beta to make the amplifier stable

1. Apr 14, 2014

DODGEVIPER13

1. The problem statement, all variables and given/known data
The open loop response A(jw) of particular op-amp is measured and is shown. The top
figure shows magnitude plot of A(jw) in dB and the bottom figure shows the phase plot
of A(jw). Determine the value of β in order to have a stable feedback amplifier?

2. Relevant equations

3. The attempt at a solution
Ok so I know for it to be stable the phase must be less than 180 degrees and the gain must be less than 1 so I set up that or at least I thought I was told my answer that I have uploaded was incorrect where did I go wrong?

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2. Apr 14, 2014

DODGEVIPER13

I know 20log gives me gain in dB so I set that equal to what I thought was the given stable gain of 90 dB and I found Av which I then used in an equation I found in the book the other way I though of was to set T(jw)=beta(Av) and then beta(Av) < 1 so beta < 1/31622.7766

3. Apr 15, 2014

rude man

Can't make out your writing, & if I guess at it I don't like what I see

What does 'stability' mean? It means absence of oscillation. Even a long sine response is OK long as it tapers off eventually, like exp(-at)sin(wt), a > 0.

So, you got the CL transfer function right: H(jw) = A/(1+βA). So what does it take for the output to blow up with any kind of input, even a tiny bit of noise? Hint: CL gain must be infinite.

A is complex: A = A(jw) = |A(jw)|exp(jø).
β is real if we assume resistor-only feedback (no C or L allowed).

For example, A = k/(jwT+1) = [k/√(1+ w2T2)]exp[j tan-1(wT)].
So |A(jw)| = k/√(1+ w2T2)
and ø = tan-1(wT).

So set H(jw) = infinity, what expression including A(jw) and β can you get?

4. Apr 16, 2014

Staff: Mentor

Why did you choose 90 dB as your stable gain?

5. Apr 16, 2014

DODGEVIPER13

If I set H(jw)=infinity wouldn't I get A=A(jw)=infinity and that would mean beta=infinity I feel as though I missed something? Also what is k in your equation you gave for A? I choose 90 by mistake I should have picked the 60 dB.

6. Apr 16, 2014

rude man

No. Set H = A/(1+Aβ) to infinity. If β = infinity, H would be zero, not infinity.

The k was just part of an example helping you understand complex gain. It has nothing to do with the present problem.

Never mind 90 dB or 60 dB. You're not ready for that. You need to understand the basic situation better first.

7. Apr 17, 2014

Staff: Mentor

So you realized your mistake, and are now using the 60dB gain. Good.

For the stabilty criterion Aβ < 1, what value of β does this give?

BTW, I don't recognize "the equation from the book" you used. It seems to give a value of β well into the unstable region. Can you scan that paragraph from the book and attach it?

8. Apr 17, 2014

DODGEVIPER13

Wait H can only equal infinity when beta=1 or -1 and A=1 or -1, so that I egt a division by zero which would give infinity. Ok now I have Av=1000 and then beta < .001

9. Apr 17, 2014

DODGEVIPER13

As far as the page in the book im not sure its even correct heh I just found it and it looked like it would give me what I needed

10. Apr 17, 2014

rude man

You now have it right. You picked A where the phase shift was 180 deg.and solved for beta.

But you should realize this is a simple case where beta is real (just resistors). Suppose your amplifier circuit had input resistor R1, and feedback resistor R2 paralleled with feedback capacitor C? Beta is now complex, A(jw) of course does not change. Want to try that?

11. Apr 17, 2014

rude man

Not just then. Whenever A < -1/beta. So A has to be real and 180 phase-shifted (that makes it real and negative) so that the equation A = -1/beta can hold.

12. Apr 17, 2014

DODGEVIPER13

Ok so I have the question correct now sweet. I actually had that answer but I read the question as value and not values so I changed it. However, I think he meant to say values and not value. Now I guess I can give that version you said a whirl. So A is -1,1 and when A is < -1/beta and beta is now complex, hmmm what equation would I use when betas complex??

13. Apr 17, 2014

rude man

Don't know what you mean by "So A is < -1,1".

The equation for instability (oscillation) is the same: βA = -1 + j0.

One can therefore also say |βA| < 1 when the angle of βA = π,
or |β| < |1/A|
or |A| < |1/β| when that angle = π. All these are tests for stability.

(It may not be obvious that|β||A| = |βA| but it's true).

If you want to pursue a case when β is complex as well as A, answer this for starters: what is β for the R-C circuit I mentioned in my post #10?

Last edited: Apr 17, 2014
14. Apr 17, 2014

DODGEVIPER13

I can take at whack at it but I will probably miss it. I know this is probably obvious but do you have any more hints? Anyways my guess would be beta = +j and -j.

15. Apr 17, 2014

rude man

No, no. Beta is simply the fraction of the output fed back to the input, expressed in complex notation. Another way to put it is it's the "gain" from the output back to the input.

In your present problem, beta = 0.001 would mean Rf = 1000Ri, or a gain of A = 1000. Actually, 1/beta is the "noise gain" which is the gain from the op amp's + input to the output. So the noise gain A would be 1001.

Now how about you express the "gain" from the output back to the input for the R-C network?

16. Apr 18, 2014

DODGEVIPER13

Wait were you asking me to solve the unstable case this whole time? If thats all then wouldnt the answer be Beta*Av>1 and then Beta > 0.001 if not what is this complex case I cant find anything in the book? I went searching for an op-amp that resembled what you described and I think I found one. The impdance was Z = Rf/(sCRf+1) so if I set that equal to A*R1 where A=1/beta I could find beta but I dont have resistance or capacitance values.

17. Apr 18, 2014

rude man

You're solving for the case where the system just goes into instability. In your case the system is stable if beta < 0.001 and unstable if beta > 0.001.

I need more time to formulate a good example for when beta is complex. But really, you have all the info you need now.

You need to understand that for an op amp circuit with feedback impedance Zf and input impedance Zi , beta = Zi/(Zi + Zf). Just a simple voltage divider from the output back to the input.

18. Apr 18, 2014

DODGEVIPER13

Thanks man no worries yah I think we were too far off from each other for me to understand the complex just yet its probably obvious but oh well I learn it later on but thanks for the help

19. Apr 19, 2014

rude man

You'll be OK. Just work hard! And it's not that obvious. I've worked with professional EE's who didn't understand it all either. And it took me some time too!