(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

See first figure attached for problem statement.

2. Relevant equations

3. The attempt at a solution

I've solved everything in this circuit, I just have some confusion about solving [tex]V_{CMmin}[/tex] 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 [tex]V_{CMmin}[/tex] by writing a KVL from the common mode input down to the -Vss.

Like so,

[tex]V_{CMmin} = V_{GS} + V_{CS} - V_{SS}[/tex]

Where [tex]V_{CS}[/tex] 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 [tex]"V_{CS}"[/tex] is going to be replaced by the voltage [tex]V_{DS3}.[/tex]

Computing [tex]V_{DS3}[/tex],

First note that,

[tex]V_{GS} = -V_{S}[/tex]

Where [tex]V_{S}[/tex] is the voltage at the source of Q1 and Q2.

[tex]-V_{S} - V_{t} = V_{ov}[/tex]

[tex]V_{S} = -V_{ov} - V_{t} = -0.7V[/tex]

Thus,

[tex]V_{DS3} = V_{S} + V_{SS} = 0.5V[/tex]

Now writing my KVL,

[tex]V_{CMmin} = V_{GS} + V_{DS3} - V_{SS} = 0V[/tex]

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!

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# Homework Help: Min. Common Mode Input MOS Diff Pair with Current Mirror

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