This is a question I found in one of the textbooks I have. The textbook didnt explain thoroughly about the concept or theory of the op-amp, and that left me confused. So please correct me if I'm wrong. I was just wondering if the voltage of the noninverting pin of the op-amp is zero, as shown in the picture. Does that mean the gain is infinity? (Assuming that the op-amp is ideal.) Or how can we find its gain? Any help is greatly appreciated.

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There are Three gains involved in your diagram (I'm assuming you mean the DC gain).

The gain of the circuit with respect to the V1 input, the gain of the circuit with respect to the V2 input, and the gain of the op-amp itself.

From that diagram, if we assume the gain of the op-amp to be infinite, then we assume the + and - opamp inputs will be the same voltage, and we can calcuate the circuit gains as R/R1 and R/R2.

We cannot calculate the opamp gain from that circuit, there is not enough information, which is why we assume it to be infinite.

If we have more information, like the part number, say TL071, then we can look up in the datasheet, where it says gain is typically 200V/mV, or a minimum of 25V/mV, depending on loading and type.

Obviously having a finite gain means that the inputs will not be at the same voltage, if there's an output voltage. 1v output would typically need a few uV across the input to support it. Given that there could be several mV across the inputs due to input offset voltage, the contribution from finite gain is irrelevant.

Having a finite gain also means the calculated gains will be lower. If the ideal (infinite opamp gain) circuit gain is 10.00000, then the gain with a real opamp would be 9.999....whatever, you figure the exact number of 9s! With 1% resistors, the gain could be anywhere from 9.8 to 10.2, so the error from finite opamp gain is irrelevant.

That's why for most purposes, we assume infinite opamp gain.

  • \$\begingroup\$ And there is the opamp noise gain also. \$\endgroup\$ – Andy aka Oct 6 '16 at 12:09
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    \$\begingroup\$ I look forward to reading an answer that brings the noise gain into the picture without confusing the bejaebers out of an OP who doesn't even realise that one assumes the open loop gain of the opamp when doing HelloWorld level exercises on simple amplifiers. I chose not to include frequency effects, or temperature effects, for the sake of a short answer that addressed what I intuited to be the specific confusion of the OP. He didn't ask that actual question because, I suspect, he lacked the answer, experience or vocabulary. Please go ahead and tell him about noise gain. \$\endgroup\$ – Neil_UK Oct 6 '16 at 13:13

An ideal op amp is assumed to have infinite gain. Basically, it will drive the output voltage to whatever is necessary to make its inputs equal.

The way to solve op amp circuits like this one is to assume the op amp inputs are at the same voltage and the current flowing in to them is zero, then figure out what the op amp output voltage has to be to satisfy those conditions.

  • \$\begingroup\$ So I carry out the work and I get \$V_0 = R(\frac{V_1}{R_1} + \frac{V_2}{R_2}) \$ But I was not sure how to calculate the gain. Is the definition of the gain \$ \frac{V_{out}}{V_{in}} \$? If so, does that mean the gain is infinity since the voltage \$V_{in}\$ across the noninverting and inverting pin of the op-amp is \$0\$ and thus makes \$\frac{V_{out}}{V_{in}}\$ infinity? \$\endgroup\$ – IgNite Oct 6 '16 at 4:40
  • \$\begingroup\$ You actually have 2 gains here since this circuit is an inverting summer. One would be Vo/V1, the other would be Vo/V2. In this case, you use superposition and assume the other inputs are set to 0. \$\endgroup\$ – alex.forencich Oct 6 '16 at 4:42
  • \$\begingroup\$ It seems I have a lot more to read. Could I ask you more about the definition of the "gain"? Isn't the gain defined as the \$ \frac{V_{out}}{V_{in}} \$, where \$ V_{in} \$ is the voltage across the inverting and noninverting pin of the op-amp ? \$\endgroup\$ – IgNite Oct 6 '16 at 4:47
  • \$\begingroup\$ Well, the gain of the op amp is assumed to be infinite, so calculating that then that's your starting assumption is pointless. In this case, you have two gains: one with V1 as Vin, one with V2 as Vin. Calculate these one at a time, assuming the other input is tied to 0 volts. \$\endgroup\$ – alex.forencich Oct 6 '16 at 4:51

You have I1, and you have I2. None of either is going through the op amp input, so they must be going through R.

Both op amp inputs are supposed to be held at the same voltage when the op amp is operating correctly.

Since you have the voltage on one side of the resistor, and you have the current flowing through it, you can calculate the voltage on the other side of the resistor. Now just put it in terms of the input voltages and you get the gain of the circuit.

  • \$\begingroup\$ I was wondering that, since the lower pin (+) of the op amp is connected to the ground, doesn't that force the upper pin (-) to have the same voltage, which is \$ 0 \$, too? \$\endgroup\$ – IgNite Oct 6 '16 at 5:06
  • \$\begingroup\$ It doesn't "force" it per se, but with proper feedback the op amp will try to make it so that they have the same voltage. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 6 '16 at 5:11

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