Unsure of finding Vout for this Opamp

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Discussion Overview

The discussion revolves around calculating the output voltage (Vout) for two operational amplifier (op-amp) circuit configurations. Participants explore different methods for determining Vout, including voltage division and superposition principles. The context includes theoretical understanding and practical application of op-amp circuits.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant attempts to calculate the equivalent resistance (Req) for resistors in parallel and applies the gain formula but arrives at an incorrect Vout for the first image.
  • Another participant suggests using a voltage divider approach to find the gain and proposes that the overall gain is 2 for the first figure.
  • In the second figure, a participant uses the superposition principle but calculates an incorrect total output voltage, leading to confusion about the correct method.
  • Multiple participants provide KVL (Kirchhoff's Voltage Law) equations for both figures, arriving at Vout values of 2 and 3, respectively, while emphasizing the importance of equating the voltages at the op-amp inputs.
  • One participant expresses appreciation for the clarity of the explanations and indicates intent to apply the discussed methods to other op-amp circuits.

Areas of Agreement / Disagreement

While some participants agree on the final answers for Vout (2 for the first image and 3 for the second), there is no consensus on the methods used to arrive at these values, as different approaches are presented and debated.

Contextual Notes

Participants express uncertainty regarding the application of different methods and the assumptions underlying their calculations. There are unresolved questions about the necessity of Req in the context of the first figure and the validity of the superposition approach in the second figure.

whatphysics
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Poster has been reminded to post all schoolwork-type questions in the HH forums
Hi there, these aren't homework questions. I came across them during revision and couldn't solve them. So any help would be appreciated. The ans for the 1st image is 2 and second image is 3.

For the first image, I attempted to calculate the Req for 10k and 20k resistors connected to V1 since they are parallel. And then to get Vo, I took (1+R2/R1)(V1) but I got (1+2)(1)=3 instead. Where have I gone wrong?

For the second image, i tried using superposition principle by killing 2V to get Voutput due to 1V then killing 1V to get Voutput due to 2V. I got Vo(due to 1V)=(1+20/20)(1)= 2 and Vo(due to 2V)=(1+20/20)(2)= 4 so adding them would be 6V but the answer is 3V.Where have I gone wrong?

Thank you for reading this and helping me!
 

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First Figure: For which purpose do you need Req?
There is a simple voltage divider at the non-inv. input (20/30=2/3) - connected to an amplifier with the gain(1+2)=3.
Hence, the overall gain is (2/3)*(3)=2.

Second figure: With superposition (same principle with voltage division) we have
2V*(1/2)*(1+1)=2V and 1V*(1/2)*(1+1)=1V.
Hence, Vo=3V.
 
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whatphysics said:
The ans for the 1st image is 2 and second image is 3.

That's the answers, so i guess the question was "what is Vo?" as in title of thread ?

The way that always works for opamps is this
Write KVL for voltage at opamp's + input.
Write KVL for voltage at opamp's - input.
Set them equal, solve for Vo.

1st image:
Vat +input = 2/3 , do you see why ?
Vat - input = Vo/3 , do you see why ?
equate them, 2/3 = Vo/3 ,
multiply both sides by common denominator of 3 and you get 2= Vo, swap sides to get Vo=2 .
It's that easy .

2nd image
Vat +input = 1.5 , do you see why ?
Vat - input = Vo/2 , do you see why ?
equate them, 1.5 = Vo/2
multiply both sides by 2 and you get 3= Vo, swap sides to get Vo=3 .

The KVL equations for V at + and - inputs are seldom so convenient as in the examples you gave,
but that method always works and you ought to practice until it becomes intuitive.

Thoreau didn't know he was giving electronics advice when he wrote "Simplify Simplify."

old jim
 
LvW said:
First Figure: For which purpose do you need Req?
There is a simple voltage divider at the non-inv. input (20/30=2/3) - connected to an amplifier with the gain(1+2)=3.
Hence, the overall gain is (2/3)*(3)=2.

Second figure: With superposition (same principle with voltage division) we have
2V*(1/2)*(1+1)=2V and 1V*(1/2)*(1+1)=1V.
Hence, Vo=3V.

You make it look so simple! Thanks for all your help, I'll try the method out with other op amp circuits!
 
jim hardy said:
That's the answers, so i guess the question was "what is Vo?" as in title of thread ?

The way that always works for opamps is this
Write KVL for voltage at opamp's + input.
Write KVL for voltage at opamp's - input.
Set them equal, solve for Vo.

1st image:
Vat +input = 2/3 , do you see why ?
Vat - input = Vo/3 , do you see why ?
equate them, 2/3 = Vo/3 ,
multiply both sides by common denominator of 3 and you get 2= Vo, swap sides to get Vo=2 .
It's that easy .

2nd image
Vat +input = 1.5 , do you see why ?
Vat - input = Vo/2 , do you see why ?
equate them, 1.5 = Vo/2
multiply both sides by 2 and you get 3= Vo, swap sides to get Vo=3 .

The KVL equations for V at + and - inputs are seldom so convenient as in the examples you gave,
but that method always works and you ought to practice until it becomes intuitive.

Thoreau didn't know he was giving electronics advice when he wrote "Simplify Simplify."

old jim

If only my teacher had been as clear as you, thank you for all your help! I'll try to solve other op amp circuits now. Thank you!
 
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