GATE | GATE-CS-2007 | Question 85

In a look-ahead carry generator, the carry generate function G_i and the carry propagate function P_i for inputs A_i and B_i are given by:

Pi = Ai ⨁ Bi and Gi = AiBi 

The expressions for the sum bit S_i and the carry bit C_i+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 P_i and G_i 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