Electronic – Current Mirror Sense Amplifier

The only real difference is in the mobility of the carriers, where electrons are faster than holes (around 2x) thus giving nMOS transistor a 2 times better performance for everything else equal.

About notation, it's mostly a practical thing, and the small signal model is an abstraction that you make to analyze a circuit. So you can use it the way you understand it better.

For instance, I usually use magnitudes (flipping notations drives me crazy) and then determine the direction of the current and voltage from the circuit. So the nMOS will generally have a current from drain to source, and the pMOS from source to drain, both with positive sign.

pMOS's \$I_{SD} \$ will be proportional to \$V_{SG} - |V_T| \$, where nMOS's \$I_{DS} \$ will be proportional to \$V_{GS} - |V_T| \$. Note that for the pMOS, you can flip SG and SD and still obtain the right values, as long as you use the absolute value of Vt.

The gm of input transistor M2 is loaded by the rds of mirror transistor M4. Remembering that rds = 1/gds, we see that the input stage gain is gm_M2 / gds_M4. I will let you figure out how to include or exclude the factor of 2 that may result from the current mirror. The second stage has amplifying transistor M7 loaded by current source transistor M8. So its gain is gm_M7 / gds_M8. Now all you need to do is calculate the gm's and gds's, and perform the multiplications and divisions to get final gain from input to output. As you know, gm depends on drain current. In this circuit the drain current is "IREF" times the transistor size ratios (Wm6/Wm5) and (Wm8/Wm5). IREF is equal the voltage across bias resistor "R" divided by R. Namely, (10 volts - Vth_M5)/R. So choose an IREF that gives the gm's and gds's you need. Then calculate R = (10V - Vth_M5)/IREF.