How does Vout (DC) track VX (DC)?

Question

I examined the both Vx and Vout with the Aspect Ratio in all transistors. I found that both are equal and I don't understand how this happens.
Could someone please explain with details, because I want to understand.

Answer 1

\$V_{sig}\$ complicates the issue, but if you ignore Vsig, there is a simple explanation for why \$V_{out}\$ tracks \$V_x\$.

Assuming all of the P-channel MOSFETs are identical, and all of the N-channel MOSFETs are identical, the current through each MOSFET will be 80 uA. This is because MB1, MB2 and M1 form a 2-output current mirror, and MB3 and MB4 form a current mirror.

Since all the drain currents are the same, and all the MOSFETs identical, all the N-channel MOSFETs will have the same \$V_{DS}\$ and the P-channel MOSFETs will have the same \$V_{DS}\$ as well.

However, this relies upon the identical behavior of each MOSFET in a group. This is achieved easily in a simulation, but you may not see this behavior, or at least not as perfectly, in an actual circuit.