1
\$\begingroup\$

All MOSFETs (not JFETS)

Have I calculated the small-signal gain correctly? enter image description here

IGNORE body effect etc. but not channel length modulation

I was basically able to get rid of M2 apart from its resistance.

\$\endgroup\$
  • \$\begingroup\$ Quick analysis in my head: if \$r_{o1}, r_{o2}, r_{o3}\$ are infinitely high impedances I expect the gain to be \$-g_{m1} / g_{m3}\$. Your formula has an extra "1 +" in the denominator which I do not expect.Since the \$r_o\$ only make the gain lower I do not expect their values to be present in the numerator. Also: draw the small signal equivalent circuit. \$\endgroup\$ – Bimpelrekkie Oct 2 '18 at 11:36
1
\$\begingroup\$

This is common source amplifier for which the gain is given by \$-g_{1} * Resistance_{seen by the small signal VCCS }\$
The small signal resistance seen by the small signal VCVS of \$M_{1}\$ is \$r_{o1}||r_{o2}||r_{o3}||\frac{1}{g_{3}}\$
So the small signal gain of the system is \$-g_{1}(r_{o1}||r_{o2}||r_{o3}||\frac{1}{g_{3}})\$

And for sanity check \$r_{o1}=r_{o2}=r_{o3}=infinity\$ the gain is equal to \$\frac{-g_{1}}{g_{3}}\$

\$\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.