Op Amp Voltage Expression & Simplification Homework Solution

AI Thread Summary
The discussion revolves around developing and simplifying an expression for the output voltage (Vo) of an op-amp using its basic equivalent model. The user attempts to derive Vo based on given voltages, initially expressing it as Vo = -A*Vs, but is uncertain about the correct approach. Participants point out potential sign errors in the user's equations and clarify that for an ideal op-amp, Vo should be expressed as -A*Vs, with A approaching infinity. They also note that the circuit may function more like a comparator than a traditional amplifier, especially if it is non-ideal. The conversation emphasizes the importance of accurately accounting for input voltages and the implications of ideal versus non-ideal op-amp behavior.
Pepjag
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Homework Statement



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.

Homework Equations



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

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.
 

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

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.
 
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.
 
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!
 
Pepjag said:
So for an ideal op-amp, would Vo just end up being Vs?

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.
 
rude man said:
Very unsuitable circuit!

well you might start to call it a comparator instead of an amplifier.
 
as long as you don't mind some spikes on the output as the input transitions through Vs=0V.
 

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