Min. Common Mode Input MOS Diff Pair with Current Mirror

Discussion in 'Homework Help' started by jegues, Oct 14, 2011.

  1. jegues

    Thread Starter Well-Known Member

    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!
  2. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    Still looking for help on this one.