# PSRR of CMOS inverter

Started by January 2, 2014
```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

Martin

```
```Martin Gruber <schilling.ro@gmail.com> wrote:
> 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
>
>
> Martin

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.
```
```On Thu, 2 Jan 2014 11:13:32 -0800 (PST), Martin Gruber
<schilling.ro@gmail.com> wrote:

>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
>
>
>Martin
>

I don't know quite _why_ one would want to calculate PSRR for an
inverter, but perhaps look at it this way... with input voltage fixed
(relative to ground), raising VDD applies more gate drive to the PMOS
while the NMOS device would stay with constant drive... so, for an
UNBUFFERED inverter one would expect _positive_ gain from VDD to the
output.

For a _buffered_ inverter it could be anyone's guess... lots of gain
and who knows what size-staggering was used.... so I could expect
positive and negative values dependent of the designer's whims and the
input voltage.

...Jim Thompson
--
| James E.Thompson                                 |    mens     |