Design complementary static CMOS circuits with minimized number of transistors to realize the following Boolean functions (hint: you may want to try logic minimization before implementing it in circuits): F = (ABC + D(A+B)) F = AC + BD F = ABCD + ABC + ABCD + ABC + ABCD+ABCD.

as pull up network. the metteing point of pull down and pull up is the point where we take the output

note 1: if two n-mos are connected in series it gives logical AND and p-mos paralle gives logical-AND

note 2: if two n-mos are connected in parallel it gives logical OR and p-mos series gives logical-OR

note 3: output is always complement of what we implement

example Y= (AB)'

image attached

A) F = (ABC + D(A+B) )'

pulldown:

this can be realize by takeing three n-mos in series which gives ABC ,two n-mos are parallel which in series with another n-mos whic gives D(A+B), now connect ABC and D(A+B) in parallel

pull up

this can be realize by takeing three p-mos in parallel which gives ABC ,two p-mos are series which is in serires with

another p-mos whic gives D(A+B), now connect ABC and D(A+B) in series

the out put will be (ABC + D(A+B) )'

so we require total 6-mos and 6-pmos total 12mos transistors

B) F = AC + BD

pull down

this can be realize by takeing two n-mos in series which gives AB ,two n-mos are in series

which whic gives BD, now connect AC and BD in parallel

pull up

this can be realize by takeing two p-mos in parallel which gives Ac ,two p-mos are in parallel

which whic gives BD, now connect AC and BD in series

the output is (AC+BD)'

to avoid the complement we have to connect the output to c-mos inverter then we get AC+BD

so we require 5-nmos, 5-pmos total 10 mos transistors

answer:

(b)

explanation: