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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!

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  • \$\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
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\$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}\$:

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

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  • \$\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

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