I am learning op amps and am confused about one concept. The output current of the op amp. Assuming ideal op amps the input side would have infinite resistance and therefor 0 amps. However the output side would have 0 resistance.

Using V=IR V/R = I

if R = 0 then its undefined...

Does it have something to do with R2 in the drawing? The resistor values are just arbitrary values.


simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ The feedback resistor is the load - see my answer. \$\endgroup\$ – Andy aka Nov 19 '15 at 14:55

The output current from the op-amp (as depicted in the picture in the question) is that current needed to keep the inverting input at ground potential.

So, with 1V at R1 (left hand side), there has to be -1V at the output to make the inverting input zero volts.

This means the current is -1V/100R = -10 mA.

If R2 were (say) double the value (200 ohms), the voltage at the output is -2V but the current remains the same.

In fact the current flowing in R2 is that current flowing thru R1 due to the input voltage. Remember the op-amp forces the inverting node always equal the non-inverting node and, if the non-inverting node is 0V then R1's current is simply input volts/R1. This is what is meant by a virtual earth amplifier.

  • \$\begingroup\$ Yes, your right I forgot that the input terminals are ideally the same potential... \$\endgroup\$ – MadHatter Nov 19 '15 at 14:56
  • \$\begingroup\$ Okay that all makes sense. I need to look at more problems now of a more complicated nature. \$\endgroup\$ – user125621 Nov 19 '15 at 15:49

Andy aka is correct, the current at the point in your diagram is -10mA. Keep in mind however that the current through R2 is +10mA and flows from left to right since the left side of the R2 is positive.

Also keep in mind that once you add a load to the circuit, the current at the denoted point will change. Say you add a 100\$\Omega\$ load from the output to ground, this load impedance is now in parallel with R2 due to the virtual ground at the inverting input of OA1. Thus, the new current at the indicated point in the circuit becomes \$i_O=v_O(\frac{R_2+R_L}{R_2R_L})=-1\text{V}(\frac{100\Omega +100\Omega}{100\Omega*100\Omega})=-20mA\$

The current through R1 and R2 will remain unchanged.


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