(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I have a book which says that the gate delay for generating C_{i}is 2 log_{r}(n) + 1, where r is the fan-in for each gate and n is the number of bits.

This implies that with a fan-in of 2 and 4 bits, the delay for a generating C_{5}as shown below should be 5 gate delays. How is this possible?

2. Relevant equations

For ann-bit carry lookahead adder, it is well known that the carry out can be determined by examining the 'carry propagate' and 'carry generate' for each of the inputs. This allows the carry out to be expressed solely in terms of the input bits and carry-in.

As an example, the carry-out for a four bit adder is given by:

C_{5}= G_{4}+ P_{4}G_{3}+ P_{4}P_{3}G_{2}+ P_{4}P_{3}P_{2}G_{1}+ P_{4}P_{3}P_{2}P_{1}C

Where

C is the carry in,

Carry propagate P_{i}= A_{i}+ B_{i},

Carry generate G_{i}= A_{i}B_{i}

3. The attempt at a solution

Initially:

We have the inputs

C,

A_{1},

B_{1},

A_{2},

B_{2},

A_{3},

B_{3},

A_{4},

B_{4}

After one gate delay:

The carry-propagate and carry generate for each bit can be determined. So we have

C,

G_{1},

P_{1},

G_{2},

P_{2},

G_{3},

P_{3}, G_{4}, P_{4}

After two gate delays:

We can use 'and' to start the carry propagates and carry generates together. So we have

P_{4}G_{3},

P_{4}P_{3},

P_{2}G_{1},

P_{2}P_{1}

After three gate delays:

Now we can use 'or', and also continue 'anding' together the propogates

So

G_{4}+ P_{4}G_{3},

P_{4}P_{3}G_{2},

P_{4}P_{3}P_{2}G_{1},

P_{4}P_{3}P_{2}P_{1}

After four gate delays:

G_{4}+ P_{4}G_{3}+ P_{4}P_{3}G_{2},

P_{4}P_{3}P_{2}G_{1},

P_{4}P_{3}P_{2}P_{1}C

After five gate delays:

G_{4}+ P_{4}G_{3}+ P_{4}P_{3}G_{2}+ P_{4}P_{3}P_{2}G_{1},

P_{4}P_{3}P_{2}P_{1}C

This is too slow. There is still more work to be done as we need another 'OR' to put the two remaining terms together.

How is it possible to determine the carry out in only five gate delays?

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# Homework Help: Carry lookahead adder - How is this possible?

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