Calculate PSS of CMOS inverter

Jan 02, 2014 1 Replies

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:

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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


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