I´m trying to figure out the transfer function of the attached op amp circuit. I came to the conclusion that v_out = - v_in as as the two resistors in the upper path are equally sized, and the non inverting input is on ground potential. However this is a task from one of our exams which I try to solve and it seems that my solution isn`t right according to the following tasks, however I got no solutions to this exam. Do I miss something?

Kind Regards


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  • \$\begingroup\$ Marc - it is correct; the gain is "-1". \$\endgroup\$
    – LvW
    Jun 17, 2018 at 8:05

1 Answer 1


Your conclusion that Vout = - Vin is correct.

Capacitor C and resistor R2 are just there to confuse you. As there is no current flowing into (or out of) the + input of the opamp, Vin+ = 0

The opamp will try to make Vin- = 0 as well if possible. Let's assume the opamp can manage that so we assume Vin- = 0

That means the current Iin = Vin / Rin with In flowing in the direction as shown. This Iin cannot flow into the opamp's - input, all of it has to go through the other resistor R1. The left side of R1 will be at 0 V, the current will flow from left to right and that means that Vout has to be negative with a value of Vout = Iin * R1.

So Vout = - Vin.

  • \$\begingroup\$ Thanks, I also figured out that my solution was right I just didn't get the next subtask right. \$\endgroup\$
    – Lukas
    Jun 17, 2018 at 10:05
  • \$\begingroup\$ Bimpelrekkie`s assumptions (in particular Vin-=0) are correct, of course, as long as we are allowed to treat the opamp as IDEAL. However, the circuit desription does not say anything about ideal/real. Therefore, for real amplifiers the parts R2 as well as C are not superfluous. R2 can provide a kind of input bias compensation and C (if it is large enough) can make the (small) ac drop across R2 to zero (above the corresponding corner frequency). \$\endgroup\$
    – LvW
    Jun 17, 2018 at 13:19

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