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I am considering a simple circuit like the one below. Let us say that the voltage source suddenly turns on (from 0V to 1V), then current will through through the resistor R1, correct? But assuming an ideal op-amp (drawing no current) and an ideal voltmeter (drawing no current), where does the current flow to (in order to satisfy Kirchhoff's current law)?

Simply put, what is the behavior of this circuit after the voltage source is activated?

Thanks all.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ It is impossible to answer this question properly until the power supply connections are added to the schematic. \$\endgroup\$ – Brian Drummond Apr 1 '13 at 15:38
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    \$\begingroup\$ @BrianDrummond not true, the opamp can be considered ideal and treated as such \$\endgroup\$ – clabacchio Apr 1 '13 at 16:07
  • \$\begingroup\$ Even with no negative supply, and bearing mind the TL082 is far from being a rail to rail opamp? \$\endgroup\$ – Brian Drummond Apr 1 '13 at 16:08
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    \$\begingroup\$ If you read "correctly powered" into "assuming an ideal opamp", I have to withdraw my critical comment; I just thought it's worth pointing out the issue in case the questioner later goes on to build real hardware. \$\endgroup\$ – Brian Drummond Apr 1 '13 at 16:56
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But assuming an ideal op-amp (drawing no current) and an ideal voltmeter (drawing no current), where does the current flow to (in order to satisfy Kirchhoff's current law)?

The inputs of an ideal op-amp draw no current.

But the output behaves like an ideal voltage source --- it can source or sink as much current as needed to respond to signals at the inputs. For purposes of Kirchoff's current law you can imagine the other side of this voltage source being connected to ground inside the op-amp symbol.

Kirchoff's current law is satisfied by the output pin sinking current so that the R2 current equals the R1 current.

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for a ideal op-amp, the current pass through R1 and R2, and I(R1) = I(R2).

I like use Nullor to analyze op-amp circuit. nullator, and norator.

you can drawing a equivalent circuit yourself, no matter how it is complex. you will feel it is very very easy to analyze op-amp circuit.

The Nullator, is the output of a op-amp, it's a current source or current sink, this depend on the direction of current.

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    \$\begingroup\$ Could you try to expand over that statement, which is by the way correct? \$\endgroup\$ – clabacchio Apr 1 '13 at 16:15
  • \$\begingroup\$ I'm assuming you mean I(R1) = I(R2) and not I*R2 etc.. \$\endgroup\$ – Andy aka Apr 1 '13 at 16:36
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Assume the input changes from 0V to +1V instantly. The op-amp will instantly have an imbalance on it's inputs and this "forces" the op-amp's output in a negative direction. Some short time later, the op-amp output will attain a level of -1V and this will totally counteract the initial imbalance caused by the input rising to +1V.

Here it will sit in a kind of equilibrium.

What should be noted is that the op-amp (in this type of circuit) will always try and maintain the voltage difference between it's two input pins as 0. This is the effect of negative feedback and the fact that in this simple analysis, the op-amp gain can be assumed to be infinite.

I also said "some short time later" in the first paragraph and this was to try and illustrate the reactionary mechanism of negative feedback. It does take a finite length of time to deal with instant changes and although in all other considerations I've assumed a perfect op-amp, in this area I have not.

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The op amp will do whatever it needs to do to keep the inverting input of the op-amp at zero volts (or, less specifically, the same as the potential at the noninverting op amp, which is grounded here).

The voltage will cause a current to go through R1, \$I = \frac{(V1-0)}{R_1}\$. As you've already noted, this current can't go into the amplifier, so it MUST go through R2. Thus $$ \frac{0-V_{M1}}{R_2} = \frac{V_1-0}{R_1} $$ or $$\frac{V_{M1}}{V_1} =\frac{-R_2}{R_1} $$

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You will have to make up your mind: are you talking about an ideal circuit, or are you considering things like stray capacitance, finite-speed opamps, etc?

If the circuit is ideal and the voltage of V1 suddenly changes, the voltage at the negative input will stay 0 (first and only law of ideal OpAmps), hence the current that flows into R1 will escape through R2, hence the OpAmp must output exactly opposite voltage compared to V1.

When you want to consider the dynamic behavior things get much much more complicated. Don't go there until you fully understand the static cases (and even then maybe not. I certainly don't.)

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