Voltage Output of a Current Source in Parallel and in Series with a resistor 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!

• 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. Apr 15 '15 at 17:11
• 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.
– Zulu
Apr 15 '15 at 17:16
• Thank you for the speedy responses! I completely forgot to register that a current source could have a potential difference across it! :) 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 $R_2$:

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

and

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

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

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

• 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! :) Apr 15 '15 at 17:50