# output resistance and transconductance in the calculation of NMOS common source voltage gain

For a NMOS, the transconductance gm is defined as id/vgs at a fixed VDS. However when we calculate the small signal gain of a common source amplifier, we use vds = -id x RD and then vds = -gm x vgs x RD.

Why we can substitute id as gm x vgs? according to the definition of gm, it is defined as id/vgs at a fixed VDS, but here the VDS has the change in vds in the equation.

A similar issue is that the output resistance ro, defined as id/vds at a fixed VGS. but why we can still use ro for calculation when the equation contains vgs?

Do i miss anything here?

• I think this is because, these are small-signal quantitites, i.e. they apply for a given operation point and for small deviations thereabout. Oct 25, 2021 at 7:08
• Also questions like this would be more readable, if you used MathJax notation, i.e. instead of gm x vgs, write $g_m V_{gs}$ to obtain: $g_m V_{gs}$ Oct 25, 2021 at 7:23

Why we can substitute id as gm x vgs?*

Because, if you draw the small signal equivalent model of the common source:

Image source.

you should "see" that $$\i_D = gm * v_{GS}\$$.

Note how I write $$\i_D\$$ and not $$\I_D\$$.

$$\i_D\$$ is the small signal current so actually the derivative of $$\I_D\$$.

$$\I_D\$$ is the actual drain current that you would calculate with for example: $$\I_D = K_p \frac W L (V_{GS}-V_t)^2\$$)

gm, it is defined as id/vgs at a fixed VDS, but here the VDS has the change in vds in the equation

Yes correct BUT you should separate:

small signal parameters ($$\gm, i_D, v_{GS}\$$) and

large signal parameters ($$\I_D, V_{GS}\$$)

The small signal ones are the ones where we:

• assume the transistor is biased at a certain operating point (a certain $$\I_D\$$ and $$\V_{GS}\$$)

• assume that we apply small signal ($$\v_{GS}\$$) so that the circuit operates in a linear fashion (no distortion)

Then we can use the small signal model and parameters.