I have been searching for how to calculate Vout in the following operational amplifier circuit, but so far I have come up empty.


I feel it is either an inverting amplifier or a differential amplifier. I have yet to find an equation for an inverting amplifier that incorporates an external voltage source between the non-inverting input and ground. Also, if this is a differential amplifier, then the just-mentioned external voltage source could be the result of a voltage division between R2 and Rg according the following circuit.

differential amplifier

I should then be able to use the following formula to calculate an output voltage.


This gives a Vout of 100 mV. Does this seem right? It is interesting how Vout = V1 = VRg.

Can anyone shed some light of understanding upon this conundrum of mine?

  • \$\begingroup\$ In the first circuit, are the two 100mV sources constant? If not, which one(s) are variable inputs? Based on the drawing I would say the one connected to the non-inverting input is a battery and is thus constant, but it's not clear. \$\endgroup\$
    – Null
    Sep 23, 2014 at 3:16
  • \$\begingroup\$ Both inputs are constant. \$\endgroup\$
    – yohosuff
    Sep 23, 2014 at 3:51
  • \$\begingroup\$ Try this for a nice simulation. \$\endgroup\$
    – yohosuff
    Sep 23, 2014 at 23:56

2 Answers 2


You're correct; this will output 100 mV as drawn. The reason for this is the amplifier outputs whatever voltage is required to set its inputs equal. Since one input is fixed at 100 mV, the other one must be driven to 100 mV. This happens when the output is at 100 mV. Since every node in the circuit is at 100 mV, no current will flow through the resistors (well, ignoring any current flowing in to the op amp input pins).

  • \$\begingroup\$ Why is it that the amplifier outputs whatever voltage is required to set its inputs equal? Is this behaviour true for all operational amplifiers? \$\endgroup\$
    – yohosuff
    Sep 23, 2014 at 3:53
  • \$\begingroup\$ Op amps have very high gains (on the order of 100,000 or more). An 'ideal' op amp has infinite gain. So there will maybe be a few nV of difference between the input pins to generate the required output voltage, but it can be neglected for most normal applications. \$\endgroup\$ Sep 23, 2014 at 4:01
  • \$\begingroup\$ So, given that we know the 'other' input (I assume this is the inverting-input) must be driven to 100 mV, how do we reach the conclusion that the output voltage is 100 mV? (Sorry for my obvious lack of understanding on this.) \$\endgroup\$
    – yohosuff
    Sep 23, 2014 at 4:20
  • \$\begingroup\$ It's alright. It can take a while to get a handle on op-amp circuits. Basically an op amp drives the output to set the inputs equal, and no current flows into the inputs. Under this assumption, both sides of the 10k resistor are at 100 mV. This indicates that the current through the 10k resistor is zero. Since no current flows into the op-amp input pin, the current through the 100k resistor must also be zero. So the voltage across the 100k resistor must also be zero. Therefore, the output voltage is also 100 mV. \$\endgroup\$ Sep 23, 2014 at 4:49
  • \$\begingroup\$ Thanks for the explanation. It makes sense now after walking through it step-by-step. \$\endgroup\$
    – yohosuff
    Sep 23, 2014 at 5:13

Yes, your calculation is correct.

You can think of this as an inverting amplifier in the input is to the 10K, with a gain of -10 and an input-referred offset of +110mV. So an input of 100mV becomes 10*(110mV-100mV)= 100mV.

Of you could think of it as a differential amplifier with a gain of -10 from the 10K and a gain of +11 from the non-inverting input. So 100mV on each gives -1V+1.1V = +100mV.

Note that a real LM324 typically has an offset voltage of 3mV, so the output might be +70mV to +130mV (and, since the offset voltage could be as much as +/-9mV is only guaranteed to be within +10mV to +190mV!).


  • \$\begingroup\$ How do you come by an input-referred offset of +110 mV? \$\endgroup\$
    – yohosuff
    Sep 23, 2014 at 3:56
  • \$\begingroup\$ An input of +110mV (to the 10K) is required to get 0V out. \$\endgroup\$ Sep 23, 2014 at 3:59
  • \$\begingroup\$ Forgive my ignorance, but what does getting 0 V out do for us? \$\endgroup\$
    – yohosuff
    Sep 23, 2014 at 4:13
  • \$\begingroup\$ The definition of input offset is the voltage you would need to apply to get 0V out. See Wiki here. Given that, you can immediately write Vout = (Vin - Voffset) * Gain = -10* (Vin - 110mV) \$\endgroup\$ Sep 23, 2014 at 4:19

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