Min. Common Mode Input MOS Diff Pair with Current Mirror

jegues · Oct 14, 2011

See first figure attached for problem statement.

I've solved everything in this circuit, I just have some confusion about solving $V_{CMmin}$ when I have a current mirror acting as my current source connected to the sources of Q1 and Q2.

Usually when I have this setup with an ideal current source instead of a current mirror I can solve for $V_{CMmin}$ by writing a KVL from the common mode input down to the -Vss.

Like so,

$V_{CMmin} = V_{GS} + V_{CS} - V_{SS}$

Where $V_{CS}$ is the minimum voltage required across the current source.

(See 2nd figure attached for example)

How do I do this now with my current mirror in place?

It looks as though my $"V_{CS}"$ is going to be replaced by the voltage $V_{DS3}.$

Computing $V_{DS3}$ ,

First note that,

$V_{GS} = -V_{S}$

Where $V_{S}$ is the voltage at the source of Q1 and Q2.

$-V_{S} - V_{t} = V_{ov}$

$V_{S} = -V_{ov} - V_{t} = -0.7V$

Thus,

$V_{DS3} = V_{S} + V_{SS} = 0.5V$

Now writing my KVL,

$V_{CMmin} = V_{GS} + V_{DS3} - V_{SS} = 0V$

I'm not entirely sure if my reasoning is correct so if somebody could point out any mistakes or confusions I'm having it would be greatly appreciated.

Thanks again!