0
\$\begingroup\$

I have a question about one of the current mirrors introduced in the "Bandgap References" chapter of Razavi's Design of Analog ICs.

He has this:

enter image description here

My confusion is about Equation 12.1.

I know that for a basic current mirror, $$I_{out} = (W/L)(M_2)/(W/L)(M_1) * I_{ref}$$ is valid.

From the looks of it, he seems to have ignored the voltage drop across R1 and just taken it to be the supply voltage Vdd. Shouldn't there be another term?

Here is my answer:

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ He's not calculating Iout, he's calculating the change in Iout wrt Vdd or dIout/dVdd. \$\endgroup\$ – John D Dec 28 '19 at 18:13
  • \$\begingroup\$ @JohnD Ah yes. Thanks very much. I differentiated by Iout function and got the same answer. (had to add the 1/gm resistance). \$\endgroup\$ – AlfroJang80 Dec 28 '19 at 19:02

Your Answer

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

Browse other questions tagged or ask your own question.