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 :

enter image description here

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!

  • \$\begingroup\$ 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 :) \$\endgroup\$
    – AgentS
    May 24, 2018 at 14:44
  • \$\begingroup\$ @Andyaka I think setting \$V_- = 0\$ is not working here as it eliminates \$A_{VOL}\$ from the equations. The formula has \$A_{VOL }\$ .. ? \$\endgroup\$
    – AgentS
    May 24, 2018 at 14:48
  • \$\begingroup\$ 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 ? \$\endgroup\$
    – AgentS
    May 24, 2018 at 14:52
  • \$\begingroup\$ Using your hint gives the current through \$R_1\$ resistor = \$i_{in}\dfrac{R_2}{R_1}\$. It flows up. \$\endgroup\$
    – AgentS
    May 24, 2018 at 14:58
  • \$\begingroup\$ Then I guess \$i_{out}\$ = (current through R2) + (current through R1) \$\endgroup\$
    – AgentS
    May 24, 2018 at 15:02

1 Answer 1


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.


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.


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