How Do You Calculate the Voltage at Node C in an Op Amp Circuit?

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Discussion Overview

The discussion revolves around calculating the voltage at node C in an operational amplifier (op amp) circuit. Participants explore various methods for analysis, including Kirchhoff's laws, Thevenin's theorem, and nodal analysis, while addressing the implications of negative feedback in the circuit.

Discussion Character

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

Main Points Raised

  • One participant attempted to use Kirchhoff's laws and Thevenin's theorem to find the voltage at node C, arriving at a value of 6V but expressing uncertainty about the correctness of their method.
  • Another participant asserts that the correct voltage at node C is 6V and suggests using nodal analysis, noting that due to negative feedback, the voltages at points A and B must be equal, with both being 0V.
  • A later reply emphasizes the importance of using Kirchhoff's Current Law (KCL) for node C and provides equations for the currents at that node, indicating a need to solve simultaneous equations involving node B as well.
  • Participants discuss the implications of the ideal op amp's infinite input impedance, which leads to the conclusion that certain currents in the circuit must be equal, although there is uncertainty about the setup of the equations.

Areas of Agreement / Disagreement

There is a general agreement on the voltage at node C being 6V, but participants express differing opinions on the methods used to arrive at this conclusion and the correctness of their approaches. The discussion remains unresolved regarding the best method to analyze the circuit.

Contextual Notes

Participants mention various methods and equations without fully resolving the mathematical steps or assumptions involved in their analyses. There are indications of missing information or potential errors in the setup of equations, particularly concerning the currents at node B.

lam58
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Hello, I'm stuck on this question, it asks to find the voltage at node c (on the circuit below) then use this value to work out other values of the op amp circuit.

I tried using kirchhoffs laws but that didn't work, then I used thevenins. Basically I took out the 6k and 8k resistors and calculated the voltage difference between the two 3k resistors and got 10.5V. Then I took away the two 3k resistors and put back the 8k and 6k then using the 10.5v found the voltage drop across the 6k resistor i.e. (6k x 10.5v)/(6k + 9k) and got 4.5v. I then took 4.5 from 10.5 and got 6 volts. However I don't have answers to this question so I'm unsure if I've got the right answer and/or my method is correct, which I doubt.

Any help on how to approach this question would be much appreciated.
 

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The correct answer is 6V. Why you don't use nodal analysis? Also are you aware that thanks to negative feedback action the voltage at point B must be equal the voltage at point A? If so VA = 0V so the VB also must be equal to 0V.
 
Last edited:
Jony130 said:
The correct answer is 6V. Why you don't use nodal analysis? Also are you aware that thanks to negative feedback action the voltage at point B must be equal the voltage at point A? If so VA = 0V so the VB also must be equal to 0V.

Mainly because I'm an idiot.
 
lam58 said:
Mainly because I'm an idiot.
Don't be so harsh on yourself, new year is coming.

I think that you should try again but this time use KCL.

For this diagram

attachment.php?attachmentid=65257&stc=1&d=1388503829.jpg


We can write for node C the KCL

I1 = I2 + I3 + I4 and additional

I1 = (21V - Vc)/3K

I2 = Vc/6K

I3 = (Vc - Vd)/8K

I4 = ??

Next do the same thing for B node and next instead VB put VB = 0V and solve this simultaneous equations.
 

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Jony130 said:
Don't be so harsh on yourself, new year is coming.

I think that you should try again but this time use KCL.

For this diagram

attachment.php?attachmentid=65257&stc=1&d=1388503829.jpg


We can write for node C the KCL

I1 = I2 + I3 + I4 and additional

I1 = (21V - Vc)/3K

I2 = Vc/6K

I3 = (Vc - Vd)/8K

I4 = ??

Next do the same thing for B node and next instead VB put VB = 0V and solve this simultaneous equations.


I'm missing something here I'm putting I4 as (Vc-Vb)/3k

and for node B

(Vc-Vb)/3k = (Vb-Vd)/5k.

But this doesn't seem to be right.
 
lam58 said:
I'm missing something here I'm putting I4 as (Vc-Vb)/3k
Very good

and for node B

(Vc-Vb)/3k = (Vb-Vd)/5k.

But this doesn't seem to be right.
Why? I don't see any mistake. The ideal opamp has infinite input impedance. That means its inputs don't draw any current at all. And this is why I4 = I5
So you have all information needed to solve this problem.
 
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