amplifiercurrentcurrent-sourcegainoperational-amplifier
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.
ConfusedCheese, here is a simple method for simulating the loop gain for an operational amplifier with feedback. This simulation scheme provides DC feedback and a stable operational point (for double and symmetrical voltage supply only!).
The dashed box contains the feedback factor, if any!.
Without this box (direct connection to the inv. input) you simulate the open-loop gain Aol of the opamp only.
The loop gain is simply T(s)=V(node2)/V(node1). Of course, the frequency must be swept over the desired range (ac analysis).
simulate this circuit – Schematic created using CircuitLab
The trick you missed (I think) is that, since this opamp is in a negative feedback loop, you can assume that the voltage between the + and - inputs of the opamp is zero.
You can theoretically explain this by assuming that the opamp has an infinite gain so Vout = (Vin+ - Vin-) * infinite. This results in that the output voltage can have any value for input voltage = zero.
With that assumption (voltage at Vin- = zero, because Vin+ is grounded) and realizing that no current flows into or out of the opamp's inputs it is now much easier to calculate the currents and then the transfer function.
Best Answer
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})\$.