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Min. Common Mode Input MOS Diff Pair with Current Mirror

  1. Oct 16, 2011 #1
    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]


    [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!

    Attached Files:

  2. jcsd
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