2022-10-27 17:53 Status: ## Using a Transistor Replace Rc with an nmos transistor. In the operating region, there is zero current flowing in the transistor as it is coupled to the compensation capacitor. Thus it operates in the triode region.  Thus the resistance seen is:  We know that for a zero at infinity, 1/gm7 = Rc, and since $g_{m7} = \mu C_{OX}(V_{GS}-V_{th})$, we have that $g_{ds10}=g_{m7}$. Thus W10 = W7 if the effective voltages are equal for transistor M10. This bias condition can use the same bias current! Just need two series nFETs with the same bias current as the main bias branch. Make these a fifth of the width of M3 since they have a fifth of the current. This ensures they have the same effective voltages. If all transistors have the same effective voltage, the voltage at M13's drain is the same as the drain voltage of M3. Thus the source voltage of transistor M10 is the same as that of the source voltage of M12 because the gate and drain voltages are the same in the current mirror. Thus we have exactly achieved that the effective voltage for M10 is the same for M7 and we have satisfied the condition for perfect lead compensation!  [^1] --- # References [^1]: [vr-4602-wk05-sc04-opampsizebias](../../Spaces/University/ELEC4602%20–%20Microelectronics%20Design%20and%20Technology/Lectures/W5/vr-4602-wk05-sc04-opampsizebias.mp4)