in Digital Logic recategorized by
1,751 views
1 vote
1 vote

To build a mod-$19$ counter the number of flip-flop required is

  1. $3$
  2. $5$
  3. $7$
  4. $8$
in Digital Logic recategorized by
by
1.8k views

1 Answer

1 vote
1 vote

For a mod N counter the number of flip flops required is less than or equal to 2 raised to power n where n is a positive integer.

N<= 2^n

Hence,

For a mod 10 counter, 10< 2^4. So 4 flip flops are required.

For a mod 16 counter, 16=2^4. So again 4 flip flops are required.

For a mod 19 counter, 19<2^5. So again 5 flip flops are required.

For a mod 32 counter, 32=2^5. So 5 flip flops are required.

Thus 5 flipflops are required.

Option B. 5

Answer: