CMOS Resistance: Drawing NAND & Solutions

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Homework Statement



I drew a NAND in the picture.

302tlld.jpg




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
 
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nobrainer612 said:
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
Hi nobrainer612! Those would be my answers, too.

Can you simplify this expression: Rp*Rp/(Rp + Rp) :wink:
 
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? :bugeye:
Also when A and B are both low ( value 0) -----> resistance = Rp/2 ? :shy:
 
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