# Calculate PSS of CMOS inverter

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

1. ### Martin GruberGuest

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:

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