# Resistance of CMOS

1. Apr 16, 2012

### nobrainer612

1. The problem statement, all variables and given/known data

I drew a NAND in the picture.

3. The attempt at a solution

I know when A and B are both high ( value 1), resistance will be Rn + Rn because those 2 NMOS will be turned on and resistance will added up since they are in series.

Also when A and B are both low ( value 0), both PMOS will be turned on. My first question is, 1.) is the resistance become Rp*Rp/(Rp + Rp) because they are in parallel?

My second question is, if A is high (value 1) and B is low ( value 0) , output should be VDD ( value 1). But what will be resistance be? I think they are neither in series nor parallel. So will the resistance only become Rp ?

Hope somebody can share their ideas. Thank you

2. Apr 16, 2012

### Staff: Mentor

Hi nobrainer612! Those would be my answers, too.

Can you simplify this expression: Rp*Rp/(Rp + Rp)

3. Apr 16, 2012

### nobrainer612

sorry but I don't get what you mean.

Can you tell me if those I assumed are correct? because I am interested what the resistance is .

simplify this expression: Rp*Rp/(Rp + Rp) : isn't that equal (Rp^2) / 2*Rp = Rp/2?

So what I assumed:

if A is high (value 1) and B is low ( value 0) -----> resistance = Rp only?
Also when A and B are both low ( value 0) -----> resistance = Rp/2 ? :shy:

Last edited: Apr 16, 2012
4. Apr 16, 2012