To determine the minimum number of flip-flops required to design a mod-10 counter, we must consider the properties of binary counters: A mod-N counter requires enough flip-flops to represent at least N different states.
The number of flip-flops (n) required is determined by the formula:
2n ≥ N
For a mod-10 counter, N = 10. We solve for n:
Thus, the minimum number of flip-flops required is 4.
Flip-flops (n) | 2n Values |
---|---|
3 | 8 |
4 | 16 |
Match the LIST-I with LIST-II
LIST-I (Logic Gates) | LIST-II (Expressions) | ||
---|---|---|---|
A. | EX-OR | I. | \( A\bar{B} + \bar{A}B \) |
B. | NAND | II. | \( A + B \) |
C. | OR | III. | \( AB \) |
D. | EX-NOR | IV. | \( \bar{A}\bar{B} + AB \) |
Choose the correct answer from the options given below:
Match List-I with List-II:
List-I (Counters) | List-II (Delay/Number of States) |
---|---|
(A) n-bit ring counter | (I) Number of states is \( 2^n \) |
(B) MOD-\(2^n\) asynchronous counter | (II) Fastest counter |
(C) n-bit Johnson counter | (III) Number of used states is \( n \) |
(D) Synchronous counter | (IV) Number of used states is \( 2n \) |
Choose the correct answer from the options given below:
A MOD 2 and a MOD 5 up-counter when cascaded together results in a MOD ______ counter.