In a look-ahead carry generator, the carry generate function Gi and the carry…

2007

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:

  1. A.

    6, 3

  2. B.

    10, 4

  3. C.

    6, 4

  4. D.

    10, 5

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Correct answer: B

Key idea: express carries in terms of the available generate (G) and propagate (P) signals, count the product terms (each implemented by an AND gate) and count one OR gate per carry output.

Carry expressions (using C0 as input carry):

  • C1 = G0 + P0*C0 → product terms requiring AND gates: P0*C0 (1 AND). OR gates for C1: 1

  • C2 = G1 + P1*G0 + P1*P0*C0 → product terms requiring AND gates: P1*G0 (1), P1*P0*C0 (1) → 2 ANDs. OR gates for C2: 1

  • C3 = G2 + P2*G1 + P2*P1*G0 + P2*P1*P0*C0 → product terms requiring AND gates: P2*G1 (1), P2*P1*G0 (1), P2*P1*P0*C0 (1) → 3 ANDs. OR gates for C3: 1

  • C4 = G3 + P3*G2 + P3*P2*G1 + P3*P2*P1*G0 + P3*P2*P1*P0*C0 → product terms requiring AND gates: four multi-input products → 4 ANDs. OR gates for C4: 1

Total AND gates = 1 + 2 + 3 + 4 = 10.

Total OR gates = 1 (for C1) + 1 (for C2) + 1 (for C3) + 1 (for C4) = 4.

Note: The sum bits S_i = P_i ⊕ C_i can be implemented using XOR gates (or expanded into AND/OR/NOT), but the look-ahead carry generator gate counts asked here refer to the carry logic, giving the final answer:

Answer: 10 AND gates and 4 OR gates.

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