Suppose the following inverting op-amp.

How does one calculate the voltage output when we connect a DC voltage source of for example 750mV to it.

With an ideal amplifier we can just multiply the gain with the input voltage. But that is not the case with inverting op-amps. But I don't know how to then calculate the output voltage with a inverting op-amp when we give a dc voltage input. I know that in this case the output voltage should be -2.05V. But I do not know how this is calculated.


simulate this circuit – Schematic created using CircuitLab

Since there is a negative feedback there is no current going into the input terminals of the op-amp. Thus the currents of R1 and R2 are the same. U = IR, so I = U/R.

For R1, \$I_1 = 750{\rm\,mV}/100\,\Omega\$, and for R2, \$I_2 = V_{out}/300\,\Omega\$. These are equal to each other.

So we get \$ 750{\rm\,mV}/100\,\Omega = V_{out}/300\,\Omega\$, which leads to \$ V_{out} = 2250{\rm\,mV}\$, which is still not the correct answer.

  • \$\begingroup\$ There are two principles that are used in analyzing such a circuit: 1) When there is a negative feedback then the voltage on the two input terminals is equal. 2) There is no current flow into/from the input terminals. Now, using these principles write down the voltages and currents on this schematic. \$\endgroup\$
    – Eugene Sh.
    Commented Oct 30, 2019 at 19:39
  • \$\begingroup\$ I have edited the post, but with your information i get to 2250mV instead of 2050mV \$\endgroup\$
    – elcie
    Commented Oct 30, 2019 at 19:50
  • \$\begingroup\$ You need to take account which direction current flows, or which way you look voltage levels. 2.25V is from negative input to output pin - so -2.25V to ground level. \$\endgroup\$
    – jalaffo
    Commented Oct 30, 2019 at 19:54
  • 1
    \$\begingroup\$ Apart from the sign, you calculated it correctly -2.25V is the correct answer. Note that \$I_{R1} = -I_{R2}\$. I think your reference "I know that in this case the output voltage should be 2.05V" is incorrect. That answer is lacking a sign too, BTW \$\endgroup\$
    – Huisman
    Commented Oct 30, 2019 at 19:58
  • \$\begingroup\$ You don't really show it in the schematic, but is the op-amp + terminal connected to ground? \$\endgroup\$
    – user4574
    Commented Oct 29, 2020 at 0:11

3 Answers 3


When you say "ideal amplifier", I think you mean "non-inverting amplifier". The term "ideal" refers to the op-amp itself, a hypothetical, perfect-in-every-respect version of something that doesn't exist in reality, in the same way the ideal car, with infinite acceleration, and zero fuel consumption does not exist.

Anyway, here's your circuit, an inverting amplifier, employing an ideal op-amp, with annotations to help:


simulate this circuit – Schematic created using CircuitLab

You said

Since there is a negative feedback there is no current going into the input terminals of the op-amp

which is wrong. The fact that there is negative feedback has nothing to do with the input drawing no current. Neither the inverting input nor the non-inverting input draw current, because they have infinite input resistance.

However, negative feedback does have an important effect; it cause the op-amp's output to adopt exactly the right voltage necessary to make those two inputs have the same potential. Since the non-inverting input is held at 0V, that means the potential at node Q, the non-inverting input, must also be 0V:

$$ V_Q = 0V $$

This is why, for this configuration of inverting amplifier, we call Q a "virtual earth". It's not connected to ground, but it it held at ground potential artificially by op-amp itself, due to negative feedback.

Then you say:

For R1, I = 750mV/100 and for R2 I = Vout/300. These are equal to each other.

This is partially true, current in R1 must equal current in R2, but you have completely neglected current direction, and the polarity of voltages across the resistors.

The thing to remember is that current in a resistor always enters the resistor at the end with the highest potential, and exits the resistor from the end with the lower potential.

Right away, applying that principle to the schematic above, surely you can see that \$V_{OUT}\$ must be lower in potential than \$V_Q\$, since that's the side of R2 where current is emerging?

The formula relating current in R2 to the voltage across it (Ohm's law) cannot just ignore current direction or voltage polarity, it must conform to that principle.

The potential difference across R2 is clearly either \$V_Q - V_{OUT}\$, or \$V_{OUT} - V_Q\$. If you define current to be from left to right through that resistor, then your application of Ohm's law must have the form:

$$ I = \frac{V_Q - V_{OUT}}{R_2} $$

If, however, you define current to be from right to left, then your equation will be:

$$ I = \frac{V_{OUT} - V_Q}{R_2} $$

This same diligence is required when calculating current in R1. Using Ohm's law, you either have:

$$ I = \frac{V_{IN} - V_Q}{R_1} $$


$$ I = \frac{V_Q - V_{IN}}{R_1} $$

You can use either, but whichever you choose, you have to use the same convention for both resistors.

I am going to assume (and I may be wrong, that's OK) that current goes from left to right, in both resistors, and this means I will be specifying the voltage across them as \$V_{LEFT} - V_{RIGHT}\$, so that if \$V_{LEFT} > V_{RIGHT}\$ then I have a positive current, to the right, in accordance with the principle of resistors I described above.

Current in R1 is:

$$\begin{aligned} I &= \frac{V_{IN} - V_Q}{R_1} \\ \\ &= \frac{V_{IN} - 0}{200} \\ \\ &= \frac{V_{IN}}{200} \end{aligned}$$

Now for current in R2:

$$\begin{aligned} I &= \frac{V_Q - V_{OUT}}{R_2} \\ \\ &= \frac{0 - V_{OUT}}{300} \\ \\ &= \frac{-V_{OUT}}{300} \end{aligned}$$

Equate those two, since \$I\$ is the same in both resistors, in both equations

$$\begin{aligned} \frac{V_{IN}}{200} &= \frac{-V_{OUT}}{300} \\ \\ V_{OUT} &= -V_{IN}\frac{300}{200} \end{aligned}$$

The negative sign here emerges because we were very careful to respect the calculation of potential differences, consistent with our assumed current direction. If we guessed "wrong", that current flows leftwards, but remained consistent to that assumption, by calculating potential differences as \$V_{RIGHT}-V_{LEFT}\$ instead, we would obtain the same result, including the negative sign.

The lesson is that it doesn't matter if you guess current direction wrong, or voltage polarity wrong, as long as your application of Ohm's law is consistent:



You basically got it almost right - except the signs of the current.

First, a convention: let the current flowing into a node be positive.

Then, the sum of currents at the (-) node is: $$I_- = (V_1-V_-)/R_1 + (V_{out}-V_-)/R_2.$$ The op-amp adjusts \$V_{out}\$ so that \$V_-=V_+=0{\rm\,V}\$, in this case. So $$I_- = V_1/R_1 + V_{out}/R_2.$$ We also know that the sum of currents at the \$I_-\$ node should be zero for the feedback condition to be satisfied. Thus $$\begin{aligned} 0{\rm\,A} &= V_1/R_1 + V_{out}/R_2 \\ V_{out}/R_2 &= -V_1/R_1\\ V_{out} &= -V_1\frac{R_2}{R_1} \\ V_{out} &= -750{\rm\,mV} \frac{300{\,\Omega}}{100{\,\Omega}} = -\frac{3}{4}\cdot\frac{3}{1}{\rm\,V} = -9/4{\rm\,V} = -2.250{\rm\,V}. \\ \end{aligned}$$

Also, the voltage gain of this circuit \$A_V=3\$.


In this case Opamp's high gain will drive both positive and negative inputs to same potential. it means that you have 0.75V over R1 or 7.5 mA. Same current will also flow thru R2 so 0.0075A*300R = 2.25V (ohms law)..

Output voltage is -Vin * R2/R1.

  • \$\begingroup\$ But the voltage should be 2.05V, not 2.25 \$\endgroup\$
    – elcie
    Commented Oct 30, 2019 at 19:52
  • 1
    \$\begingroup\$ The inverting and non-inverting inputs will only be at the same voltage if the circuit has negative feedback and has not saturated...high gain alone does not guarantee this. Also, the output voltage is not merely the ratio of resistor values. \$\endgroup\$ Commented Oct 30, 2019 at 20:06
  • 1
    \$\begingroup\$ -3 * 0.75 V = -2.25 V \$\endgroup\$
    – jalaffo
    Commented Oct 30, 2019 at 20:10
  • 1
    \$\begingroup\$ -R2/R1 does not equal -2.25 \$\endgroup\$ Commented Oct 30, 2019 at 20:11
  • \$\begingroup\$ @Elliot I corrected the formula (oops), and the statement to mean this case, which I meant. \$\endgroup\$
    – jalaffo
    Commented Oct 30, 2019 at 20:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.