Op-Amp circuit. Load current is the one through Rl

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.

  • \$\begingroup\$ 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. \$\endgroup\$ – placeholder Aug 24 '14 at 21:12
  • \$\begingroup\$ Do you know node analysis? \$\endgroup\$ – user34920 Aug 24 '14 at 21:24
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    \$\begingroup\$ @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. \$\endgroup\$ – Cholo Serrano Aug 24 '14 at 21:40
  • \$\begingroup\$ 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. \$\endgroup\$ – Andy aka Aug 24 '14 at 23:22
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    \$\begingroup\$ @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. \$\endgroup\$ – 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})\$.

  • \$\begingroup\$ If I do the simulation with spice using the Universal OpAmp and having the current source as 5mA, Rt = 10k, Rl = 100 and Rc = 100, I get the current through Rl is approx. -25mA, which tells me the gain should be about -5. I know one shouldn't always trust the simulator, but this is an ideal case so I don't see how the simulator should fail. \$\endgroup\$ – Cholo Serrano Aug 24 '14 at 21:37
  • \$\begingroup\$ @CholoSerrano I have just finished simulation in Multisim (it is really horrible program) and it says 505 mA flowing into the op-amp. Perhaps your model has too low output voltage swing, it takes -100.5 V at the op-amp output. \$\endgroup\$ – venny Aug 24 '14 at 22:44
  • \$\begingroup\$ It's weird, even if I put +/-200V as supply to the Universal OpAmp 2, I was getting the same results, but then I switched to the one that requires no supplies and it worked. Thanks! \$\endgroup\$ – Cholo Serrano Aug 24 '14 at 23:32

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