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!