MadeEasy Test Series: Digital Logic - Carry Generator

Question

The number of AND gates are present inside a 5-bit carry look ahead generator circuit are ______.

Answer 1

For n bit cla ,the no of AND gate is n*(n+1)/2

So,for 5 bit, the no of AND gate is 15

Comments

I saw the carry generator circuit. And I can see that it requires n*(n+1)/2 and gates. But that requires, Gi and Pi to be calculated. So I think we should add one AND gate for each bit for Gi. Total = n*(n+1)/2 + n.

Please help me in this _/\_

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@Rishabh Gupta 2

G_i and P_i are not calculated inside carry generator, these are calculated independently before giving it to carry generator circuit

if total AND gates were asked in whole carry-look ahead adder, than it would be n*(n+1)/2 + n.

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Thank you. That makes sense.

But why we don't consider Gi generation in a carry generator? It is a requirement to generate the carry.

Answer 2

We know the equation carry look ahead generator circuit is

$c_{i}=g_{i-1}+c_{i-1}.p_{i-1}$

$c_{1}=g_{0}+c_{0}.p_{0}$

$c_{2}=g_{1}+c_{1}.p_{1}$

$\Rightarrow c_{2}=g_{1}+\left ( g_{0}+c_{0}.p_{0}\right )p_{1}$

$\Rightarrow c_{2}=g_{1}+g_{0}.p_{1}+c_{0}.p_{0} .p_{1}$

$c_{3}=g_{2}+c_{2}.p_{2}$

$\Rightarrow c_{3}=g_{2}+\left (g_{1}+g_{0}.p_{1}+c_{0}.p_{0} .p_{1}\right )p_{2}$

$\Rightarrow c_{3}=g_{2}+g_{1}.p_{2}+g_{0}.p_{1}.p_{2}+c_{0}.p_{0} .p_{1}.p_{2}$

$c_{4}=g_{3}+c_{3}.p_{3}$

$\Rightarrow c_{4}=g_{3}+\left (g_{2}+g_{1}.p_{2}+g_{0}.p_{1}.p_{2}+c_{0}.p_{0} .p_{1}.p_{2}\right )p_{3}$

$\Rightarrow c_{4}=g_{3}+g_{2}.p_{3}+g_{1}.p_{2}.p_{3}+g_{0}.p_{1}.p_{2}.p_{3}+c_{0}.p_{0} .p_{1}.p_{2}p_{3}$

$c_{5}=g_{4}+c_{4}.p_{4}$

$\Rightarrow c_{5}=g_{4}+\left (g_{3}+g_{2}.p_{3}+g_{1}.p_{2}.p_{3}+g_{0}.p_{1}.p_{2}.p_{3}+c_{0}.p_{0} .p_{1}.p_{2}p_{3}\right )p_{4}$

$\Rightarrow c_{5}=g_{4}+g_{3}.p_{4}+g_{2}.p_{3}.p_{4}+g_{1}.p_{2}.p_{3}.p_{4}+g_{0}.p_{1}.p_{2}.p_{3}.p_{4}+c_{0}.p_{0} .p_{1}.p_{2}p_{3}p_{4}$

Total 15 AND gates.

if total AND gates were asked in whole carry-look ahead adder, than it would be n*(n+1)/2 + n.

But here it is asking, The number of AND gates are present inside a 5-bit carry look ahead generator circuit are $15$