#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# The minimum number of NAND gates required to implement A+AB'+AB'C ?Plz explain aslo...

Saurabh Koranglekar
one year ago
Dear student

Minimal expression is A
No. Of nand gates required is 0

Regards
Vikas TU
14149 Points
one year ago
Dear student
Above ans is incorrect

There are two addition Terms and Three inputs

Inputs are a,b,c

for b to b’ conversion one NAND gate is required

For adding two numbers 3 NAND Gates are required

Therefore split the term as Y = a + ab’ + ab’c as

Z = b’ —- One NAND Gate

Z1 = ab’ is an and operation . this requires 3 NAND gates of 2 inputs

Z2 = ab’ + ab’c = Z1(1+c) = Z1 since 1 + c in digital is 1

theretofore Y becomes a + Z1 Totally 4 NAND gates of two input is the Minimal requirement

There are other different methods too for implementation of this function