# Deriving PU / PD given a sketch of a PMOS

For the PMOS given below I can derive the function f, such that f inverted in its variables corresponds to the expression of PMOS(f) and f inverted equals NMOS(f).

For this specific problem I have however two contradicting solutions and thus would be happy, to have a 2nd opinion.

My solution:

$$\ f(a,b,c) = (\bar{a} + \bar{b})(\bar{a}\bar{b} + \bar{c}) \$$

$$\ PMOS(f(a,b,c)) = (a + b)(a b + c) \$$

$$\ NMOS(f(a,b,c)) = a b + c(a + b) \$$

Suggested solution:

$$\ f(a,b,c) = \dots \$$, not available

$$\ PMOS(f(a,b,c)) = (\bar{a} + \bar{b})(\bar{a}\bar{b} + \bar{c}) \$$

$$\ NMOS(f(a,b,c)) = \overline{a b + c(a + b)} \$$

The differences between the expresseions are not much, nonetheless I would like to know, what is correct, what is wrong.

In the image below, Erdung can be translated to ground or earthing. PMOS is equivalent to fitting a Not Gate on every input.

I reached this by making the truth table and simplifying, the main differences in your equations look to be from missing the simplification with B*C. when they are both true, you do not care about A.

NMOS = AB+BC PMOS = A'*B'+B'*C' • Doubt that: You have $\bar{f} = NMOS(f) \Rightarrow f = \overline{AB + BC} = \bar{A} \bar{B} + \bar{B} \bar{C}+ \bar{A} \bar{C}$ $\Rightarrow PMOS(f) = AB+BC+ AC \neq \bar{A} \bar{B} + \bar{B} \bar{C}$ – Imago Sep 9 '19 at 16:28