# How is the positive input of this op-amp calculated

So suppose the following circuit. $$\ V_{out} \$$ needs to be 0V. The negative voltage input is 100mV.

I now want to calculate the positive voltage input. The gain of the amplifier is -(R3/R1) = -9. $$\ V_{out} \$$ is calculated by

$$\ V_{out} = A(v^+ - v^-) \$$

$$\ V_{out} \$$ is 0, A is -9, $$\v^-\$$ is 100mV.

So we get

$$\0 = -9(v^+ - 100mV)\$$

which eventually leads to

$$\v^+\$$ = 100mV.

But it should be 90mV. I cannot really find my mistake. I am assuming i have a wrong formula for $$\ V_{out} \$$ but this should be the formula for how $$\ V_{out} \$$ is calculated.

simulate this circuit – Schematic created using CircuitLab

• You're kind of jumping around grabbing formulas intended for other circumstances which is why your analysis doesn't work. Do you understand WHY the gain is -R3/R1? As in, do you understand that if the noninverting input is tied to GND, then the opamp tries to drive the inverting input to be zero. So 100mV appears across R1 which means there is a current in it, but that current must also be in R3 and it is this that produces the output voltage of a larger magnitude. Oct 29, 2019 at 23:35
• To elcie. You have 100 mV from the input to the output. There's 10 kOhm from the input to the output. It's a simple voltage divider pair of resistors. What's the voltage at the (-) input? You should be easily able to work that out. Whatever that is? It's the same as what you put onto the (+) input.
– jonk
Oct 30, 2019 at 0:05

## 1 Answer

I now want to calculate the positive voltage input. The gain of the amplifier is -(R3/R1) = -9. $$\ V_{out} \$$ is calculated by

$$\ V_{out} = A(v^+ - v^-) \$$

In this formula, the $$\A\$$ is the open-loop gain of the op-amp, not the closed loop gain of the op-amp circuit with feedback.

$$\A\$$ should be 100,000 or so, not 9.

However, we wouldn't normally use this formula to solve the circuit.

We'd simply assume $$\A\$$ is "very large". From this (and the presence of negative feedback) we can get the rule that

$$v^+ = v^-$$

Then, the rule that (very nearly) no current flows into the op-amp's input pins, tells us that the current through R3 is the same as the current through R1.

Then we have

$$\frac{v_{in} -v^-}{1000} = \frac{v^- - v_{out}}{9000}$$

and we can solve to get the final answer.