Op Amp Voltage Expression & Simplification Homework Solution

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SUMMARY

The discussion focuses on deriving and simplifying the output voltage expression (Vo) for an operational amplifier (op-amp) circuit using the ideal op-amp model. The key equations referenced are Vo = A(Vp - Vn) and Vo = G * Vs, with Vp set at 6V and Vn expressed as 6 - Vs. Participants confirm that for an ideal op-amp, Vo approaches -A * Vs, where A approaches infinity, indicating that the output is significantly influenced by the input voltage Vs. Additionally, the conversation highlights the importance of correctly accounting for signs in the equations and suggests that the circuit may function more like a comparator than a traditional amplifier.

PREREQUISITES
  • Understanding of operational amplifier fundamentals
  • Familiarity with ideal vs. non-ideal op-amp characteristics
  • Knowledge of voltage source configurations in circuits
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the concept of common mode rejection ratio in op-amps
  • Learn about the differences between ideal and non-ideal op-amps
  • Explore the application of op-amps as comparators
  • Investigate the effects of feedback in op-amp circuits
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Students of electrical engineering, circuit designers, and anyone seeking to understand operational amplifier behavior and applications in electronic circuits.

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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