Skip to content
Related Articles

Related Articles

Improve Article

GATE | GATE-CS-2007 | Question 85

  • Last Updated : 20 Nov, 2018

In a look-ahead carry generator, the carry generate function Gi and the carry propagate function Pi for inputs Ai and Bi are given by:

Pi = Ai ⨁ Bi and Gi = AiBi 

The expressions for the sum bit Si and the carry bit Ci+1 of the look-ahead carry adder are given by:

Si = Pi ⨁ Ci and Ci+1 = Gi + PiCi , where C0 is the input carry. 

Consider a two-level logic implementation of the look-ahead carry generator. Assume that all Pi and Gi are available for the carry generator circuit and that the AND and OR gates can have any number of inputs. The number of AND gates and OR gates needed to implement the look-ahead carry generator for a 4-bit adder with S3, S2, S1, S0 and C4 as its outputs are respectively:
(A) 6, 3
(B) 10, 4
(C) 6, 4
(D) 10, 5


Answer: (B)

Explanation: let the carry input be c0

Now,

c1 = g0 + p0c0 = 1 AND, 1 OR
c2 = g1 + p1g0 + p1p0c0 
   = 2 AND, 1 OR

c3 = g2 + p2g1 + p2p1go + p2p1p0c0 
   = 3 AND, 1 OR
c4 = g3 + p3g2 + p3p2g1 + p3p2p1g0 + p3p2p1p0c0 
   = 4 AND, 1 OR

So, total AND gates = 1+2+3+4 = 10 , OR gates = 1+1+1+1 = 4



So as a general formula we can observe that we need a total of ” n(n+1)/2 ” AND gates and “n” OR gates for a n-bit carry look ahead circuit used for addition of two binary numbers.


Quiz of this Question

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.

My Personal Notes arrow_drop_up
Recommended Articles
Page :