Circuit Diagram of the Problem

Hi All,

I imagine this is a pretty basic question to ask but I'm struggling to get my head around this. I'm trying to find \$V_{out}\$ of the circuit above.

Now I understand why \$V_{out}\$ can be \$-g_mV_{GS}R_2\$ (as taking the Norton equivalent of the circuit on the left just gives a current source of \$-g_mV_{GS}\$ in series with 0 resistance). Thus this current causes a voltage drop over \$R_2\$ which gives \$V_{out}\$. (This is of course equivalent to saying the output from a voltage source does not change if it is in parallel with a resistor).

What a fail to understand is how the \$V_{out}\$ could not be \$g_mV_{GS}R_1\$. Assuming the bottom of the circuit is grounded, surely there must be a voltage drop due to the current from the current source over \$R_1\$ ? Due to the fact the circuit is parallel, surely this voltage drop should be necessarily equivalent to \$V_{out}\$ ?

I'd appreciate any advice on how to best understand this problem & any tips and tricks you may have for the future!

Many thanks!

  • \$\begingroup\$ Don't let yourself be confused by the way the circuit is drawn. There's nothing in parallel with anything there - Its a simple series circuit. \$\endgroup\$
    – brhans
    Apr 15 '15 at 17:11
  • \$\begingroup\$ You're right in saying that \$R_1\$ has \$g_mV_{GS}R_1\$ across it. However, all this does is change the voltage across the current source. The current source "takes up the slack," if you will: no matter what the voltage across \$R_1\$ is, the current source will compensate by changing its own voltage by the same amount. \$\endgroup\$
    – Zulu
    Apr 15 '15 at 17:16
  • \$\begingroup\$ Thank you for the speedy responses! I completely forgot to register that a current source could have a potential difference across it! :) \$\endgroup\$
    – Imran
    Apr 15 '15 at 17:50

\$V_{\text{out}}\$ is equal to the sum of the voltage across \$R_1\$ and the voltage \$V_{CS}\$ across the current source (it is, of course, also equal to the voltage across \$R_2\$). In order for \$V_{\text{out}} = g_mv_{gs}R_1\$, you would have to have \$V_{CS} = 0\$. However, an ideal current source will support any voltage across itself so you cannot assume that \$V_{CS} = 0\$. In this case, for KVL to hold true the voltage across the current source is the difference between the voltage across \$R_1\$ and the voltage across \$R_2\$:

$$V_{\text{out}} = -g_mv_{gs}R_2$$


$$V_{\text{out}} = V_{CS} + g_mv_{gs}R_1$$

Setting the equations equal to each other and solving for \$V_{CS}\$:

$$V_{CS} = -g_mv_{gs}R_2 - g_mv_{gs}R_1$$

  • \$\begingroup\$ Many thanks @Null , the idea of a current source having a P.D. across it completely slipped past my mind! I'll be careful not to assume \$V_{CS}\$ = 0 in the future! Thanks! :) \$\endgroup\$
    – Imran
    Apr 15 '15 at 17:50

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.