M
Martin Gruber
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
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
