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Official PYQ · 2015Set 2 · Q48Computer ArithmeticNAT1 mark

GATE CS 2015 Set 2 Q48 — Computer Arithmetic

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A half adder is implemented with XOR and AND gates. A full adder is implemented with two half adders and one OR gate. The propagation delay of an XOR gate is twice that of an AND/OR gate. The propagation delay of an AND/OR gate is 1.21.2 microseconds. A 44-bit-ripple-carry binary adder is implemented by using four full adders. The total propagation time of this 44-bit binary adder in microseconds is ______.
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Answer: 19.2

To find the total propagation delay of the 44-bit ripple-carry binary adder, we first determine the delays of the individual gates:

  • Propagation delay of AND/OR gate (tAND=tORt_{\text{AND}} = t_{\text{OR}}) = 1.2 μs1.2\ \mu\text{s}
  • Propagation delay of XOR gate (tXORt_{\text{XOR}}) = 2×1.2 μs=2.4 μs2 \times 1.2\ \mu\text{s} = 2.4\ \mu\text{s}

A Full Adder (FA) is implemented using two Half Adders (HA) and one OR gate. The carry propagation path for a single Full Adder stage is:

tcarry=tXOR+tAND+tORt_{\text{carry}} = t_{\text{XOR}} + t_{\text{AND}} + t_{\text{OR}}

Substituting the values:

tcarry=2.4 μs+1.2 μs+1.2 μs=4.8 μst_{\text{carry}} = 2.4\ \mu\text{s} + 1.2\ \mu\text{s} + 1.2\ \mu\text{s} = 4.8\ \mu\text{s}

For a 44-bit ripple-carry adder, the carry must propagate through all 44 full adders sequentially. Therefore, the total propagation delay is:

Total Delay=4×tcarry=4×4.8 μs=19.2 μs\text{Total Delay} = 4 \times t_{\text{carry}} = 4 \times 4.8\ \mu\text{s} = 19.2\ \mu\text{s}

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