# Calculate gain of degenerated common-source stage using small-signal model

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

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,

(2)

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

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:

$$v_s=\dfrac{(R_d+r_{ds})v_o+g_mv_ir_{ds}R_d}{(g_mr_{ds}+1)R_d}$$

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.