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schematic

simulate this circuit – Schematic created using CircuitLab

I have several about this op-amp curcuit:

  1. i1 flows from Node 1. Does it flow into the op-amp? Or Another current will flow from the output of the op-amp? (I don't think i1 will flow into it, because Node 1 where i1 flows from has 0V and V0 is a positive number.)

    1. I am confused with the use of this ground symbol. It is not a physical ground where charge can flow out of. It is just a reference point, so we have one less node to calculate the voltage. Potential doesn't matter, the difference does. But i2 and i3 both flow out of the ground/reference. And nothing flows into it. Should either i2 or i3 flow into the reference to preserve the conservation of charge?
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    \$\begingroup\$ Notice the opamp's power supply isn't displayed in this diagram. For conservation of charge you need to consider the opamp's power supply. \$\endgroup\$
    – Criticizing Israel not allowed
    Commented Sep 26, 2017 at 22:45

3 Answers 3

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Both Martin and immibis have it correct. Here's a slightly more complete circuit, showing how I1 is absorbed by the op-amp output, then flows to the op-amps' negative supply voltage (the op-amp's positive supply voltage is not shown, but is still required).
You cannot accurately measure I1 at the -ve power supply, because there is additional quiescent current to bias the op-amp's internal transistors. I1 flows from op-amp output to either +ve or -ve supply voltage source

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  • \$\begingroup\$ So current will never flow out of op amp's ouput? \$\endgroup\$
    – most venerable sir
    Commented Sep 28, 2017 at 21:51
  • \$\begingroup\$ @user132522 I've shown it flowing in, for your circuit example. Current can flow out, being sourced from the +ve supply pin. Be aware that most op-amps include a current limiter circuit, often kicking in at 20 - 30 ma - it works for both directions. \$\endgroup\$
    – glen_geek
    Commented Sep 28, 2017 at 23:26
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Yes to your first question, opamp's output is sinking or sourcing current as needed to maintain its output voltage (which is determined by difference on its inputs).

And no to your second question. Circuit is closed by output of opamp. This current comes (indirectly) from/to circuit's ground. It flows from opamp's power rails which are in turn referenced to circuit ground (either it is directly ground or voltage supply referenced to the ground).

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  • \$\begingroup\$ What current is flowing into the op amp? How? Node 1 has a voltage of 0, and the V ouput of the opamp is a positive number. \$\endgroup\$
    – most venerable sir
    Commented Sep 28, 2017 at 21:49
  • \$\begingroup\$ Why do you think opamp output is positive (with respect to ground)? It is exactly opposite, as you have just shown. :) Your circuit is inverting, so output voltage has opposite sign compared to input. And yes, it can not work in reality unless negative rail for opamp is lower than ground potential. \$\endgroup\$
    – Martin
    Commented Sep 28, 2017 at 22:19
  • \$\begingroup\$ Just to add bit more clarification. If there were no current leaving node 1, than voltage at this node would rise to about 13V. Node 1 is negative input terminal of opamp and this voltage is clearly higher than voltage at positive input, thus negative input difference and outupt goes toward negative rail. \$\endgroup\$
    – Martin
    Commented Sep 28, 2017 at 22:31
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enter image description here

Figure 1. Internals of the ancient 741 opamp. Source: Wikipedia.

From the internal schematic of the 741 op-amp it should be clear that the output can source current from the \$ V_{S+} \$ rail via Q14 or sink current to the \$ V_{S-} \$ rail via Q20. Nearly all (there are probably exceptions) opamps will have a similar push-pull arrangement on the output.

These current paths complete the circuit as required.

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