I want to find the gain in this MOSFET small-signal model.

KVL in loop 1:

\$+V_{in}-V_{gs}-g_mV_{gs}R_s=0 \iff V_{in}=V_{gs}(1+g_mR_s)\$

KVL in loop 2:

\$+V_{out}-g_mV_{gs}R_s=0 \iff V_{out}=g_mV_{gs}R_s\$


In the answer \$V_{out}\$ should have a negative sign, \$V_{out}=-g_mR_s \cdot V_{in}\$. What have I missed?

The final answer is \$V_{out} / V_{in}=-g_mR_d/(1+g_mR_s)\$. (*)

How can the current through \$R_d\$ be \$g_mV_{gs}\$? Shouln't there instead be a KCL in \$V_a\$, so the current through \$R_d\$ is \$V_{out}/R_d\$?


simulate this circuit – Schematic created using CircuitLab

Update with KCL in \$V_a\$:


where \$i_d=g_mV_{gs}\$ and \$I_r=V_{out}/R_d\$, so


I'm stuck here, how can I find equation (*) above?

  • \$\begingroup\$ Your KVL loop2 is incorrect, Vout is not gmVgsRs, you made Vs = Vout. You should not use a voltage loop as there's a current source (gmVgs) in the loop. Use the current law instead and make sure Rd is in the expression because it is crucial. The overall transfer (Vout/Vin) will have a negative sign and both Rs and Rd must be in the expression. \$\endgroup\$ Apr 13, 2017 at 14:30
  • \$\begingroup\$ Where is Iout going? (Is there a loop there)? \$\endgroup\$
    – Tyler
    Apr 13, 2017 at 15:11

2 Answers 2


Donsert you can find the solution in the following figure

enter image description here


KVL in loop 2 is missing the voltage across the current source. Because that's an unknown, instead combine your KCL expression with your KVL loop 1 expression:

$$ V_{in} = V_{gs}(1+g_mR_s) \\ I_{out} = g_mV_{gs}+V_{out}/R_d \implies V_{out} = (I_{out} - g_mV_{gs})R_d \\ \therefore \frac{V_{out}}{V_{in}}=\frac{I_{out}R_d - g_mV_{gs}R_d}{V_{gs}(1+g_mR_s)}=\frac{I_{out}R_d/V_{gs} - g_mR_d}{1+g_mR_s} $$

Now if and only if \$I_{out} = 0\$, the expression reduces to:

$$ \frac{V_{out}}{V_{in}}=\frac{-g_mR_d}{1+g_mR_s} $$

which is the result you're after.


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