# ICIS opamp circuit - current gain derivation

My textbook simply gives the formulas w/o derivations. Using node voltage method, I'm able to successfully prove the formulas for VSVS, ICVS, and VCIS. But I couldn't do the same for the current gain of the ICIS circuit shown :

I tried to apply KCL at the 3 nodes shown (thick black dots) in the figure. But I'm getting complicated expressions and they are not simplifying to the given formula. My work :
1) At input node (call it $V_-$ ) $$-i_{in} + \dfrac{V_{-} - V_x}{R_2} = 0$$

2) At outut node (call it $V_{out}$ ) $$V_{out} = -A_{VOL}V_{-}$$

3) At the 3rd node (call it $V_x$ ) $$\dfrac{V_{x} - V_-}{R_2} + \dfrac{V_{x} - V_{out}}{R_L} + \dfrac{V_x}{R_1} = 0$$

Solving these 3 equations is giving me a really scary looking expression for $i_{out}$. Are the above 3 equations look okay ? I've ignored the currents going into the inputs of opamp because I thought they're negligible.. I feel I'm doing something terribly wrong. Appreciate any help. Thanks!

• Ahh @Andyaka I think I see.. Opamp works so that $V_{-}\approx 0$ is negligible. Right ? That does give $V_x = -I_{in}R2$. Let me grab my notes and give it a try.. I'll get back shortly Ty :) May 24 '18 at 14:44
• @Andyaka I think setting $V_- = 0$ is not working here as it eliminates $A_{VOL}$ from the equations. The formula has $A_{VOL }$ .. ? May 24 '18 at 14:48
• I mean what expression should we use for the voltage right after the opamp output ? It was $V_{out} = -A_{VOL} V_{-}$. I cannot make use of this if I set $V_- = 0$ in other equations right ? May 24 '18 at 14:52
• Using your hint gives the current through $R_1$ resistor = $i_{in}\dfrac{R_2}{R_1}$. It flows up. May 24 '18 at 14:58
• Then I guess $i_{out}$ = (current through R2) + (current through R1) May 24 '18 at 15:02

You're not doing anything wrong at all, it may just be that you still need to solve for $i_{out}$ from $V_{out}$ and $V_x$, which could be a bit more tedious.

I suggest adding the variable $i_{out}$ and adding an extra equation. The equations are very similar, but it allows you to solve for $i_{out}$ directly.

$$-i_{in} + \frac{V_- - V_x}{R_2} = 0$$

$$V_{out} = -A_{VOL}V_-$$

$$\frac{V_x - V_-}{R_2} - i_{out} + \frac{V_x}{R_1} = 0$$

$$\frac{V_{out} - V_x}{R_L} = i_{out}$$

Solving this for $i_{out}$ (I did this with a CAS):

$$\frac{i_{out}}{i_{int}} = -\frac{A_{VOL}R_2+(1+A_{VOL})R_1}{R_L + (1+A_{VOL})R_1}$$

When assuming that $A_{VOL} \gg 1$, you get the expression from your book.

APPENDIX

It is still possible to get the right answer from your equations as well. Solving your equations will give you:

$$V_x = -i_{in}\frac{R_1(A_{VOL}R_2-R_L)}{R_L+(1+A_{VOL})R_1}$$

$$V_{out} = -i_{in}\frac{A_{VOL}R_2(R_L+R_1)+A_{VOL}R_1R_L}{R_L+(1+A_{VOL})R_1}$$

$$i_{out} = \frac{V_{out}-V_x}{R_L} = -i_{in}\frac{A_{VOL}R_2+(1+A_{VOL})R_1}{R_L+(1+A_{VOL})R_1}$$

This is identical.