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 '18 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 '18 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 '18 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 '18 at 14:58
  • \$\begingroup\$ Then I guess \$i_{out}\$ = (current through R2) + (current through R1) \$\endgroup\$
    – AgentS
    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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.