enter image description here

According to my textbook $$\frac{v_o}{i_s} = -R_1\left(1+\frac{R_3}{R_1}+\frac{R_3}{R_2}\right)$$

The current on the terminal branches are both 0 for an ideal op-amp. Using KCL we see R1 and R2 have a current of \$i\$ going to the left of them and R3 has no current on it.


simulate this circuit – Schematic created using CircuitLab

Since this is an ideal op-amp, the voltages at the input terminals are both 0 with respect to the ground. And Vo is the voltage between R1 and R2 because the current through R3 is 0. This means Vo is the output voltage of a voltage divider where Vin is 0. Thus, $$\frac{v_o}{i_s} = 0$$

I have been getting wrong answers when I'm using KCL so I assume I'm somehow not able to apply it correctly but I don't exactly know from where does the error come. Isn't this a valid use of KCL?


2 Answers 2


Given this schematic:\$\quad\$enter image description here


$$\begin{align*} \frac{V_1}{R_1} &=\frac{V_2}{R_1}+i_s \\\\ \frac{V_2}{R_1}+ \frac{V_2}{R_2} +\frac{V_2}{R_3} &=\frac{V_1}{R_1}+ \frac{V_o}{R_3} \\\\ \frac{V_o}{R_3}&=\frac{V_2}{R_3}+i_o \\\\ V_1&=0\:\text{V} \end{align*}$$

Those four equations solve out for \$\{V_1, V_2, i_o, V_o\}\$. And here you should get: \$\frac{V_o}{i_s}=-R_1-\frac{R_1}{R_2}R_3-R_3\$. Which is about the same thing you say the answer is supposed to be.

So go back up there and see what I did and compare it to what you did. Feel free to move things around, of course. (I drew them up by putting outflowing currents on the left and inflowing currents on the right. But you can do whatever you want, just so long as we each get to the same place in the end.)


The output of the Op-Amp can sink current, so the assumption that the current through R3 must be zero is incorrect.

Let's call the node at the top of R3 Vx. Then:

$$V_x = -i_s R_1 $$

because of the virtual ground that you identified. By Ohm's law, the current through R2 is:

$$V_x/R_2 = -i_s \frac{R_1}{R_2}$$

So the current through R3, by KCL, must be:

$$i_s + i_s \frac{R_1}{R_2}$$

Thus the voltage at v0 is given by:

$$v_0 = V_x - i_3 R_3 = -i_s R_1 - (i_s + i_s \frac{R_1}{R_2}) R_3 $$

And this simplifies to the expression given by your textbook.


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