# Non-ideal Op Amp voltage

1. Nov 21, 2012

### Pepjag

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

I have to use the basic equivalent model of an op-amp to develop an expression for Vo from the given op-amp. I also have to simplify the expression by treating the given op-amp as an ideal op-amp.

2. Relevant equations

Vo=A(Vp-Vn)
Vo=G*Vs

3. The attempt at a solution

Vp=6v
Vn=6-Vs

Vo = A(6-6-Vs) = -A*Vs. Am I on the right track? I don't even know how to start on the other half of the problem. I've only had half a lecture on op-amps.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### IMG_20121121_004041.jpg
File size:
26.9 KB
Views:
88
2. Nov 21, 2012

### CWatters

Looks right so far.

For an ideal op-amp you can essentially ignore the two 6v sources. For a real world op-amp you might have to check out the common mode rejection ratio.

3. Nov 21, 2012

### Pepjag

So for an ideal op-amp, would Vo just end up being Vs?

4. Nov 21, 2012

### Staff: Mentor

I think you have a couple small sign errors in your equations, but you are on the right track.

The 6V source and Vs look to add going into the - opamp input, but you have written that equation as a subtraction.

And then When you calculate Vo, you do not account for the fact that Vs is going into the - opamp input. Your answer is almost correct, you just need to fix those two sign errors.

5. Nov 21, 2012

### Staff: Mentor

Unless, by this equation:

Vo = A(6-6-Vs) = -A*Vs

you mean Vo = A[6 - (6+Vs)] = -A*Vs

If so, then you don't have a sign error.

6. Nov 21, 2012

### rude man

The attached circuit shows an op amp in open-loop operation.

That means that, ignoring offsets, output = AVs.

It also means the output is infinty times Vs if the op amp is ideal.

Very unsuitable circuit!

7. Nov 22, 2012

### CWatters

For an ideal opamp the output would be -A*Vs where A => ∞ as others have said.

Is there more text to go with the diagram? The title of the thread says "Non-ideal Op Amp voltage". To answer your question fully we would need to know what was Non-ideal about it.

8. Nov 23, 2012

### aralbrec

well you might start to call it a comparator instead of an amplifier.

9. Nov 23, 2012

### CWatters

as long as you don't mind some spikes on the output as the input transitions through Vs=0V.