Understanding Op-Amp Circuits: Simple Circuit Help & Calculation Tips

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Discussion Overview

The discussion revolves around understanding the calculations and behavior of a simple operational amplifier (op-amp) circuit. Participants seek clarification on output voltage calculations, the role of feedback, and the implications of saturation in op-amp outputs.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant initially proposes that the output voltage (Vout) can be calculated using the formula (V2-V1)(Aol), expressing uncertainty about its correctness.
  • Another participant suggests considering output saturation as a factor in the calculations.
  • A different participant points out that op-amps are not typically used in open-loop configurations and emphasizes the need for feedback to create an effective op-amp circuit.
  • One participant corrects the initial formula, proposing (2*V2 - V1) * Aol as a more accurate representation, while noting that this formula applies even in feedback scenarios.
  • Several participants engage in calculations using both the original and corrected formulas, expressing confusion over discrepancies in their results and questioning the influence of source voltages.
  • Another participant highlights that the op-amp's output may not reach the supply voltage limits, suggesting that this could explain unexpected output values.

Areas of Agreement / Disagreement

Participants express differing views on the correct formula for calculating Vout and the implications of using op-amps in open-loop versus closed-loop configurations. The discussion remains unresolved regarding the exact calculations and the reasons behind the output discrepancies.

Contextual Notes

Participants mention potential limitations related to the assumptions made in their calculations, such as the type of op-amp used (e.g., rail-to-rail output) and the effects of feedback on the circuit's behavior.

Petrucciowns
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I have a simple op-amp circuit that I need help understanding:
http://img200.imageshack.us/img200/3939/opamp.jpg

I know that the positive input is noninverting and the negative input is inverting.

I thought Vout was calculated by (V2-V1)(Aol), but I guess I am wrong. Can anyone help me figure out how to calculate Vout on at least one of the lines so I can figure out the rest.

Thanks.
 
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Have you considered that the output might saturate?
 
Petrucciowns said:
I have a simple op-amp circuit that I need help understanding:
http://img200.imageshack.us/img200/3939/opamp.jpg

I know that the positive input is noninverting and the negative input is inverting.

I thought Vout was calculated by (V2-V1)(Aol), but I guess I am wrong. Can anyone help me figure out how to calculate Vout on at least one of the lines so I can figure out the rest.

Thanks.

Opamps are not used open-loop. You need to add feedback to have an opamp "circuit".
 
Last edited by a moderator:
Petrucciowns said:
I have a simple op-amp circuit that I need help understanding:
http://img200.imageshack.us/img200/3939/opamp.jpg

I know that the positive input is noninverting and the negative input is inverting.

I thought Vout was calculated by (V2-V1)(Aol), but I guess I am wrong. Can anyone help me figure out how to calculate Vout on at least one of the lines so I can figure out the rest.

Thanks.

You made a slight mistake in your formula. It's a common mistake and I've noticed experienced engineers in this forum making the same mistake. The formula should be (2*V2 - V1) * Aol though opamps are rarely used open loop. This formula holds even for a standard inverting amplifier with feedback. For instance in your circuit, if you put a 1k resistor between V1 and the inverting input and another 1 k resistor between the output and the inverting input the amplifier becomes an amplifier with a gain of -1. If we put 5V at V2 and 4 V at V1, what is the output voltage? Is it 5V - 4V = 1V or is it 2*5V - 4V = 6V? Try it.

In your example, because the gains are so high your outputs were always at the + or - rail except once, when you got -200mV. Let's plug your input voltages into the formula and calculate the result. (2*1uV - 3uV) * 200K = (2uV - 3uV) * 200k = -200mV. Okay?
 
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I am still having problems with both formulas. For example let's do the first two. First with my formula and then with yours.

V1=4mv V2=1mv

my way- (1mv-4mv)* 200k= -600 volts
your way - (2x1mv-4mv)* 200k = -400 volts

So why is the answer 9 volts? Does it have something to do with the sources?



V1=2 micro v V2=1 micro volt

my way- (1 micro volt- 2 micro volts)*200k = -200mv
This actually is the answer that is written down why?

Your way (2X 1 micro volt- 2 micro volts)*200k = -400mv
Which is not the answer written down.


Can someone try to explain to me what I am not getting?
 
berkeman said:
Opamps are not used open-loop. You need to add feedback to have an opamp "circuit".

Yes, this is just academic to see how one works.
 
Petrucciowns said:
my way- (1mv-4mv)* 200k= -600 volts
your way - (2x1mv-4mv)* 200k = -400 volts

So why is the answer 9 volts? Does it have something to do with the sources?

Yep. How would the output of the opamp swing to hundreds of volts, using only +/-10V power supplies?
 
Petrucciowns said:
So why is the answer 9 volts? Does it have something to do with the sources?

It's probably because the opamp isn't a so-called rail-to-rail output opamp. This means that when the opamp output is saturated, the outputs can't swing all the way to the rail (supply) voltages. They can't make it all the way to 10 volts; only 9 volts.
 
Oh, I get it guys. Thanks a lot.
 

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