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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.

given_pmos

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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

enter image description here

PMOS = A'*B'+B'*C'

enter image description here

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  • \$\begingroup\$ 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} \$ \$\endgroup\$ – Imago Sep 9 '19 at 16:28

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