# Peculiar Opamp circuit behavior

1. May 2, 2013

### The Electrician

An opamp circuit was given in the homework forum:

There the circuit was analyzed with the usual assumptions of opamp ideality.

But, if the circuit is analyzed with one change to ideality, namely the opamp voltage gain is set to 7 rather than ∞ (this brings with it the change that v+ ≠ v- any more), the calculated overall circuit gain, Vout/Vin, is greater than 7.

How can this be?

I think I know the answer, but I pose this for the readers to think about and discuss.

Edit: I forgot to mention this: Exchange the + and - opamp inputs (or set the opamp gain to -7).

#### Attached Files:

• ###### OpAmpCircuit.jpg
File size:
27.7 KB
Views:
161
Last edited: May 2, 2013
2. May 6, 2013

### milesyoung

I'm not sure I follow. I think you're saying there's something odd about the gain from Vin to Vout being greater than the open-loop gain of the opamp. If that's the case, why do you think that?

Also, for an opamp open-loop gain of 7 (with its terminals swapped), I get a gain of -27.2 from Vin to Vout.

3. May 6, 2013

### jim hardy

That's kinda how self saturating magnetic amplifiers work and it's a nice basic concept to have in your bag of tricks..

Positive feedback will increase the gain of an inherently low gain device.
If you take that positive-feedback-enhanced device and surround it with some negative feedback, you have treated it as an op-amp.
The old timers called them "Servomechannisms".

4. May 6, 2013

### Staff: Mentor

When the OP-AMP here is ideal and has a gain of precisely 7 3/7 then
I think we find that Vout/Vin computes to infinity.

Looking good!

5. May 7, 2013

### The Electrician

Why don't you think that?

6. May 7, 2013

### milesyoung

Because that's not all that unusual for systems with positive feedback. For an ideal opamp with its output fed back into its noninverting input and its inverting input used as the system input, we have:

$$V_\mathrm{out} = A(V_+ - V_-) = A(X V_\mathrm{out} - V_\mathrm{in}) \Leftrightarrow \frac{V_\mathrm{out}}{V_\mathrm{in}} = \frac{A}{A X - 1}$$
where X, 0 < X ≤ 1, is the positive feedback gain and A > 0.

For Vout/Vin > A, you'd just need 0 < AX - 1 < 1 ⇔ 1/A < X < 2/A.

For the circuit you posted, we have:
$$V_\mathrm{out} = A(V_+ - V_-) = A(X V_\mathrm{out} - Y V_\mathrm{out} - Z V_\mathrm{in}) \Leftrightarrow \frac{V_\mathrm{out}}{V_\mathrm{in}} = \frac{A Z}{A(X - Y) - 1}$$
and for Vout/Vin > A, you'd need 0 < A(X - Y) - 1 < Z ⇔ 1/A < X - Y < (Z + 1)/A.

You must then have X > Y, i.e. net positive feedback.