0
\$\begingroup\$

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?

\$\endgroup\$
2
  • 2
    \$\begingroup\$ I think this is because, these are small-signal quantitites, i.e. they apply for a given operation point and for small deviations thereabout. \$\endgroup\$
    – tobalt
    Oct 25, 2021 at 7:08
  • \$\begingroup\$ 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}\$ \$\endgroup\$
    – tobalt
    Oct 25, 2021 at 7:23

1 Answer 1

0
\$\begingroup\$

Why we can substitute id as gm x vgs?*

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

enter image description here

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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.