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6 votes
6 votes

In a $8$-bit ripple carry adder using identical full adders, each full adder takes $34$ ns for computing sum. If the time taken for $8$-bit addition is $90$ ns, find time taken by each full adder to find carry.

  1. $6$ ns
  2. $7$ ns
  3. $10$ ns
  4. $8$ ns
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2 Answers

10 votes
10 votes
Consider n-bit Ripple Carry Adder.
Total Delay = Delay_Sum + (n-1) Delay_Carry
Here, n=8.
90ns = 34 ns + 7 * Delay_Carry
56ns = 7 * Delay_Carry
Delay_Carry= 8ns
8 votes
8 votes
Sum in ripple carry adder is --> $max( Sum, Carry)$ $for$ $MSB+$ $delay$ $to$ $produce$ $carry$ $of$ $N-1bits$

now given sum takes $34ns$ it can be clearly seen that for the last bit $Max(sum,carry)$ is $sum$ only

So the carry generation for $N-1=8-1=7bits$ has taken $90-34=56ns$

so time to produce carry $\dfrac{56}{7}=8ns$

$D$ $is$ $the$ $answer$
Answer:

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