Table of Contents
- 1 Is it possible to use D flip-flop to implement the counter?
- 2 How many FF are required for mod counter?
- 3 How many natural states will there be in a 5 bit ripple counter?
- 4 How many FF are required for 4 bit up down counter?
- 5 What is the difference between a Mod-2 and Mod-4 counter?
- 6 What is the modulus of a 3 Flip-Flop counter?
Is it possible to use D flip-flop to implement the counter?
A D-Type Flip-Flop Circuit can be used to store 1 bit of information. It has two input pins (Called D (Data) and E (Enabler) and two output pins (Q and Q = NOT Q). When the enabler input E is set to 0, the output Q cannot be changed.
How many FF are required for mod counter?
3 D- Flip flops are needed for implementation of Mod-6 counter because you need 3 bits in order to represent 6. In the same way, we require 4 flip flops for implementation of Mod-11 counter i.e., For 11 we require 4 bits.
What is the modulus of 5 bit ripple counter?
Explanation: The minimum number of flip-flops used in a counter is given by: 2(n-1)<=N<=2n. Thus, for modulus-5 counter: 22 <= N <= 23, where N = 5 and n = 3. Explanation: There are 10 states, out of which MSB is high only for (1000, 1001) 2 times.
What is the maximum modulus of a 4 bit counter?
Exercise :: Counters – General Questions
| 54. | A 4-bit counter has a maximum modulus of ________. |
|---|---|
| A. 3 B. 6 C. 8 D. 16 Answer: Option D Explanation: No answer description available for this question. Let us discuss. Workspace Report errors Name : Email: View Answer Discuss |
How many natural states will there be in a 5 bit ripple counter?
Each bit has propagation delay = 12ns. So, 5 bits = 12ns * 5 = 60ns.
How many FF are required for 4 bit up down counter?
In a 4-bit up-down counter, there are 4 J-K flip-flops required. For modulus-10 counter, N = 10. Therefore, 23 < = 10 < = 24. Thus, n = 4, and therefore, we require 4 FFs.
How many flip-flops are required to design a mod 5 counter?
The number of Flip-flops required can be determined by using the following equation: where, M is the MOD number and N is the number of required flip-flops. Therefore, to design a MOD 5 Counter, 3 flip-flops would be required. In this case inputs of the flip-flops are: D A, D B & D C Therefore, D A = Q B . Q C Therefore, D C = Q A/ .
How many D-FFS are needed to design a 4 bit up counter?
The characteristic equation for the D-FF is: Q+ = D We need to design a 4 bit up counter. So, we need 4 D-FFs to achieve the same. Let’s draw the state diagram of the 4-bit up counter
What is the difference between a Mod-2 and Mod-4 counter?
Then we have seen that a MOD-2 counter consists of a single flip-flop and a MOD-4 counter requires two flip-flops,allowing it to count in four discrete steps. We could easily add another flip-flop onto the end of a MOD-4 counter to produce a MOD-8 counter giving us a 2 3 binary sequence of counting from 000 up to 111, before resetting back to 000.
What is the modulus of a 3 Flip-Flop counter?
The modulus of a counter is given as: 2 n where n = number of flip-flops. So a 3 flip-flop counter will have a maximum count of 2 3 = 8 counting states and would be called a MOD-8 counter. The maximum binary number that can be counted by the counter is 2 n –1 giving a maximum count of (111) 2 = 2 3 –1 = 7 10.