# What is the current gain of this op-amp circuit? I just happened to see this circuit in some lecture slides and I thought I could solve it. Unfortunately, I've been trying to check my result with an ideal simulation in LTspice and all my attempts have been wrong.

What is the current gain of this op amp circuit?

$$I_G = \frac{I_l}{I_1}$$

where $I_l$ is the current through Rl.

• What have you worked out so far? We typically don't do homework for people but will help with particular problems. As an aside, to make an opamp work you probably should supply power, and your placement of the ground on the + input is problematic, some drawing programs don't like that, and it is confusing to your readers. – placeholder Aug 24 '14 at 21:12
• Do you know node analysis? – user34920 Aug 24 '14 at 21:24
• @placeholder I did 2 KVL equations: $$0 = I_1R_t + (I_1 + I_l)R_c\\ 0 = I_LR_l + (I_1 + I_l)R_c\\$$ put them together and got something like $\frac{R_t }{R_l}$, but it's clearly wrong. – Cholo Serrano Aug 24 '14 at 21:40
• That circuit is a mess - no supplies, non-inverting input floating - do you really expect folk to help you on this? Listen to what placeholder said. – Andy aka Aug 24 '14 at 23:22
• @Andyaka First, calm down. Of course I used supplies!!! I just withdraw them as I didn't want to clutter everything. The non-inverting input is grounded, I tried putting a cable and ground since the above comment said ltspice might not like it, but I got the same results. – Cholo Serrano Aug 24 '14 at 23:26

At $R_c$, there is voltage of $-R_t\cdot I_1$. Which means that through $R_c$ goes current of $-(\frac{R_t\cdot I_1}{R_c})$.
So current through $R_l$ is sum of currents going through $R_t$ and $R_c$ and that is $-I_1-\frac{R_t\cdot I_1}{R_c}$.
Gain is $-(1+\frac{R_t}{R_c})$.