MadeEasy Test Series: Digital Logic - Adder

Question

A 1-bit full adder circuit takes 5 ns to generate the carry-out bit and 10 ns for the sum-bit. When 4, 1-bit full adders are cascaded, the maximum rate of additions per second will be _______ × 107.

Comments

This might help :https://gateoverflow.in/83500/digitals

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see you need to wait for each adder for 1 ns i.e for only carry 

and at last adder you need to take sum time all other will overlap .

so (n-1)* carry time + sum time = 3*5 + 10 = 15+ 10 = 25ns 

so this will  give = 1/25ns = 4*10⁷

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Same Question

https://gateoverflow.in/83500/digitals

Answer 1

5 is right .

Comments

In question it is given 4 bit adder not 3 bit so answer should be 4*10^7

Answer 2

T =  ( n-1 ) * carry propagation delay + 1 * sum propagation delay  (if sum pd > carry pd)

   =     3 *  5ns + 1 * 10ns = 25ns

That means 20ns per addition.

So additions per second =  1 / 25ns = 4 * 10^7.

so ans is  4 (given 10^7).

Answer 3

Let me answer without formula first

4 Full adders are there

F$_1$-->F$_2$-->F$_3$-->F$_4$

to calculate sum are of Full adders need two bit to sum and a carry bit from previous Full adder

    so F$_1$ takes 5ns to calculate carry and then  passes it on to F$_2$ during same time calculates sum also 

• at 6^th ns F$_2$ starts to calculate carry and sum and in the same time F$_1$ is working on its sum 

after 10ns F$_1$ produces its sum and F$_2$ produces its carry

• at 11^th ns F$_3$ starts to calculate carry and sum and in the same time F$_2$ is working on its sum 

after 15ns F$_2$ produces its sum and F$_3$ produces its carry

• at 16^th ns F$_4$ starts to calculate carry and sum and in the same time F$_3$ is working on its sum 

after 20ns F$_3$ produces its sum and F$_4$ produces its carry

• at 25^th ns F$_4$ finishes its sums also

so it took 25ns to complete 1 sum

1 sum time = 25 * 10^-9 seconds

=>in 1 sec 1/25 * 10^-9  sums will get executed 

=> 10⁹ /25 = 4 * 10⁷ 

Method 2

let T(n) is the time taken by n^th full adder as it has to wait for 5ns for previous adder to calculate its carry 

so we can form a recurrence

T(n)  =  T(n-1) + PD$_C$  (Propagation delay of carry)

T(1) = (PD$_s$) Propagation delay of sum (as for only 1 full adder time taken = 10ns(=(PD$_s$)

solving by back substitution 

T(n) = T(n-2) + (PD$_c$)+ (PD$_c $)

 T(n) = T(n-2) + 2*(PD$_c$)

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 T(n) = T(n-k) + k*(PD$_c$)

if n-k=1 ,k=n-1

 T(n) = T(n-(n-1)) + (n-1)*(PD$_c$)

 T(n) = T(1) + (n-1)*(PD$_c$)

 T(n) = (PD$_c$) + (n-1)*(PD$_c$)  we can derive this formula by ton a ways 

here n=4 (PD$_c$) = 5ns (PD$_s$) =10ns

 T(4) = 10 + (4-1)*(5)

T(4) =25ns

1 sum time = 25 * 10^-9  seconds

=>in 1 sec 1/25 * 10^-9   sums will get executed 

=> 10⁹  /25 = 4 * 10⁷