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PSRR of CMOS inverter

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

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  1. How to calculate the PSRR 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 PSRR. I created the small signal equivalent as shown in the image:

    With that I calculate the PSRR:

    PSRR = 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.

    PSSR = -5.8859

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

    Thanks in advance!
  2. Ralph Barone

    Ralph Barone Guest

    Why can't that be true? It's an inverter with gain, so if dVdd produces
    something which could be considered a positive voltage excursion at the
    input, then you should get a negative excursion at the output.
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