Min. Common Mode Input MOS Diff Pair with Current Mirror

Thread Starter


Joined Sep 13, 2010
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.

\(-V_{S} - V_{t} = V_{ov}\)

\(V_{S} = -V_{ov} - V_{t} = -0.7V\)


\(V_{DS3} = V_{S} + V_{SS} = 0.5V\)

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!