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Calculate PSS of CMOS inverter

Discussion in 'Electronic Design' started by Martin Gruber, Jan 2, 2014.

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  1. How to calculate the PSS (power supply sensitivity) for a CMOS inverter?
    I'm struggling a bit because I do not get meaningful result values.

    Let explain what I have. I have the following transistor parameters from a simulation result.

    High-side PMOS:
    rds2 = 11.67k
    gm2 = 879.4 uS

    Low-side NMOS:
    rds1= 20.35k
    gm1 = 1.659 mS

    With that I want to calculate the PSS. I created the small signal equivalent as shown in the image:
    http://s7.directupload.net/images/140102/pu4dzkw6.jpg

    With that I calculate the PSRR:

    PSS = dVout / dVdd

    For doing that I took Kirchhoff's law.

    dVout = UR1 = IR1 * R1

    IR2 - IR1 - gm2 * Vgs2

    In small signal equivalent Vgs2 is dVDD.

    IR1 = IR2 - gm2 * dVdd

    IR2 = (dVdd - UR1) / R2
    IR2 = (dVdd - dVout) / R2

    This can be inserted in the equation before:

    dVout = ((dVdd - dVout) / R2 - gm2 * dVdd) * R1

    dVout/dVdd = (R1/R2 - gm2 * R1) / (1 + R1/R2)

    Inserting now the number values from above unfortunately yields a negative result which can't be true.

    PSS = -5.8859

    Can somebody please help me to do it the right way?

    Thanks in advance!
    Martin
     
  2. Kids returned to school!

    Jamie
     
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