Opamp Doesn't Satisfy Ohm's law

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Discussion Overview

The discussion revolves around the behavior of operational amplifiers (op-amps) in circuit analysis, particularly in relation to Ohm's law and current flow at the terminals of op-amps. Participants explore the implications of ideal op-amp assumptions and their application in solving circuit problems.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant asserts that no current enters or leaves the terminals of an op-amp, referencing the virtual ground concept and attempting to calculate current through feedback paths.
  • Another participant clarifies that while no current flows into the inputs of an ideal op-amp, the output can supply current, suggesting that the assumption of "infinite" current is misleading.
  • A participant questions the conditions under which the output voltage (Vo) can be treated as having no current, noting a previous successful calculation under that assumption.
  • Another response indicates that zero output current is a special case and not a general rule.
  • One participant challenges the accuracy of the initial assumptions about current flow, proposing that the relevant equations should reflect the infinite input impedance and zero output impedance of an ideal op-amp.
  • A later reply suggests modeling the op-amp as a dependent voltage source, emphasizing that the output will adjust to maintain its voltage, thus discouraging the use of output current in circuit analysis.

Areas of Agreement / Disagreement

Participants express differing views on the treatment of current flow in op-amps, particularly regarding when it is appropriate to assume no current at the output. There is no consensus on the assumptions that should be applied in these scenarios.

Contextual Notes

Limitations include the reliance on ideal op-amp assumptions, which may not hold in all practical situations. The discussion reflects varying interpretations of circuit analysis principles and the implications of feedback in op-amp circuits.

jaus tail
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Homework Statement


upload_2018-1-17_18-20-50.png


Homework Equations


No current enters of leaves Opamp Terminals

The Attempt at a Solution


Red line is current path.
Inverting terminal of Opamp -1 is at ground due to virtual ground concept, so Current = 2/1 = 2mA.
This current goes through feedback path and then right toward inverting terminal of opamp-2
And then it goes up as shown in arrow.
But at output of Opamp -1, let the V be Vo.
So Vo / 5 (feedback resistor of opamp 1) is not same as Vo / 1 (inverting terminal resistor of opamp 2).
Why is this so?
I got answer by: Vo = -Rf/ R1 for opamo 1. And then I found current as Vo1/1k. And then using KCL at inverting terminal of Opamo-2
But why is the underlined part wrong?

Using KCL at inverting terminal of Opamp-2
2mA = (0-2)/1k + (0 -Vo)/1k
Which gives Vo = 4V.
 

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jaus tail said:
No current enters of leaves Opamp Terminals
Not at the inputs. The output can have a current. "Infinite" current for an ideal opamp.
 
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upload_2018-1-17_18-43-7.png

But I solved this other example using the assumption no current flows out of opamo 1 output and i got correct answer of 6V
When to take Vo has no current and Vo has current output?
 

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It can happen that you have zero output current, but that is a special case.
 
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jaus tail said:
When to take Vo has no current and Vo has current output?
Never make that assumption, it is totally false in general.
jaus tail said:

Homework Equations


No current enters of leaves Opamp Terminals
No, that is not accurate for the Relevant Equations for opamp circuit problems.

Instead, the Relevant Equations / Assumptions should be more like these:
  • The input impedance of the + and - inputs to an opamp is infinite (no current flows in or out of the inputs)
  • The output impedance of the opamp is zero (you can draw lots of current out of the output)
  • The negative feedback of an opamp's external circuit keeps the - input at the same voltage as whatever the + input's voltage is
Now, using those rules, please write the KCL equations for each of the nodes in your opamp circuit, and solve for the unknown voltages. Please show your work. Thanks! :smile:
 
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When you are working with op amps it can also help to think of it as the below model. The output is simply a voltage source dependent on the input. In an ideal op amp, Rin is inf and Rout is zero.

So this voltage source will act like any other voltage source, it will produce whatever current it has to in order to maintain it's voltage. Therefore you do not want to use the op amp current to solve the circuit, rather you want to use the voltage.

250px-Op-Amp_Internal.svg.png


If you have used nodal analysis, you are told to never use the opposite side of a voltage supply as a node, the same principal applies here, for the same reason.
 

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