For this question I have to give the relationship between Vout and Vin. I already have the answer, but I just do not know how I should get there exactly.

The circuit looks like the following:

circuit diff op-amp

To me this looks like a difference amplifier. I know the formula for this kind of amplifier is:

$$V_\mathrm{out} = \frac{R_2}{R_1} \times (V_{2}-V_{1})$$

This would give me the following:

$$V_\mathrm{out} = \frac{10\ \mathrm{ k\Omega}}{2\ \mathrm{ k\Omega}} \times (V_{2}-V_{1}) = 5V_{2}-5V_{1}$$

However, the answer should be: $$V_\mathrm{out}=-5V_{2}-10V_{1}$$

I don't know what assumptions I should make to get to this value, does it have something to do with the load resistance? Or is my reasoning itself flawed?

  • 1
    \$\begingroup\$ What you have is an inverting summing amplifier, not a difference amplifier. Look it up and you'll find lots of resources online. \$\endgroup\$
    – Big6
    Sep 4 at 15:45
  • \$\begingroup\$ The negative input to the opamp is zero volts (assuming a properly constructed negative feedback circuit). Then use KCL at this node. V1/1k + V2/2k + Vout/10k = 0 \$\endgroup\$
    – Mattman944
    Sep 4 at 15:56
  • \$\begingroup\$ Thank you Big6 and @Mattman994, I did the KCL at the node wrong before, thanks for the help. \$\endgroup\$
    – Sydon
    Sep 4 at 19:44

1 Answer 1


Don't blindly memorize formulas, work it out from first principles. There is negative feedback and the non-inverting input is grounded, so we can assume the voltage at the inverting input is very close to zero at balance.

Note that the op-amp power supply connections are shown, and we can certainly assume that the output can't be lower than 0 or higher than 5.0 V, so we'll check that later.

Using KCL we know that

$$\frac{V_1}{R_1} + \frac{V_2}{R_2+R_3} + \frac{V_\mathrm{out}}{R_\mathrm{F}} = 0$$

(\$R_\mathrm{L}\$ doesn't matter ideally; in reality it might exceed what the op-amp can drive).

From the above equation, it's easy to calculate Vout (V1, V2), and the answer agrees with your answer above.

In reality the output must be >0 V and < 5 V, and \$R_\mathrm{L}\$ draws a lot of current for \$V_\mathrm{out} > 0\$ so it might not be able to deliver more than a volt or two at the output, not less than 0 V.

When the op-amp can't balance, then other things will happen since the inverting input will no longer be at 0 V.

  • \$\begingroup\$ Thanks for the comment, I tried recreating the formula for this use case the same way I learned before however, I made the mistake where I applied KCL wrong. I did it twice however both times I either left out the V1/R1 or Vout/Rf, for some reason thinking I couldnt take more, which is quite wrong in hindsight... Thanks for the clear explanation! \$\endgroup\$
    – Sydon
    Sep 4 at 19:43

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