I have difficulties calculating the gain of a degenerated common-source stage, with the output resistance of the MOSFET taken into account. I came up with the SS-model below. I calculated the following:

The gain is defined as \$ A = \frac{v_{out}}{v_{in}} \$, whereas \$ v_{in} = v_1 + v_s \$. I started writing an expression for the current.

\$ i = g_m v_1 + \frac{v_{out} - v_s}{r_{ds}}\$

Next I used that expression to solve for \$ v_{out} \$.

\$ v_{out} = g_m v_1 R_d+ \frac{R_d}{r_{ds}} v_{out} - \frac{R_d}{r_{ds}} v_{s} \$ \$ v_{out} = (g_m v_1 R_d - \frac{R_d}{r_{ds}} v_{s})/(1 - \frac{R_d}{r_{ds}}) = (g_m v_1 R_d - \frac{R_d}{r_{ds}} v_{in} + \frac{R_d}{r_{ds}} v_{1})/(1 - \frac{R_d}{r_{ds}})\$

And for the gain: \$ v_{out}/v_{in} = (g_m v_1 R_d - \frac{R_d}{r_{ds}} v_{in} + \frac{R_d}{r_{ds}} v_{1})/((1 - \frac{R_d}{r_{ds}})(v_1 + v_s))\$

But now I have again \$ v_s \$ in the denominator, which I cannot substitute for anything useful. I feel like running in circles. Can someone help me out? What am I doing/approaching wrong?


simulate this circuit – Schematic created using CircuitLab


1 Answer 1


The current equation you show is the source current:

$$i_s=g_mv_1+\dfrac{v_{o}-v_s}{r_{ds}} $$

And \$i_s=\dfrac{v_s}{R_S}\$

So you have:

(1) $$\dfrac{v_s}{R_s}=g_mv_1+\dfrac{v_{o}-v_s}{r_{ds}}$$

It's hard for me to follow what you did but you could use the equation that includes the current through \$R_d\$. That is,


$$ \dfrac{v_o}{R_d}+g_mv_1+\dfrac{v_o-v_s}{r_{ds}}=0$$

You could use either equation to start the process to find \$\dfrac{v_o}{v_i}\$

If you take equation 1, you can solve for \$v_o\$ to find:

$$ v_o=v_s\dfrac{r_{ds}}{R_s}+v_s-g_mv_1r_{ds}$$

As you noted, \$v_i=v_1+v_s\$, so the previous equation becomes (after some algebra):

(3) $$v_o=v_s\bigg(\dfrac{r_{ds}+R_S+g_mr_{ds}R_s}{R_s}\bigg)-g_mr_{ds}v_i $$

Everything looks good except for the fact that I still have a \$v_s\$ term and need to get rid of it so that we can solve for \$\dfrac{v_o}{v_i}\$.

You can use the second equation (2), solve for \$v_s\$ and plug that in equation (3).

Solving for \$v_s\$ in equation (2) results in:


You still have to plug this value of \$v_s\$ in equation (3), but now you will have everything in terms of \$v_o\$ and \$v_i\$ so you should get the result you've been looking for.

This is not the only way to do this but it kind of follows the path you took.


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