A __________ counter can be implemented using three flipflops.a)mod-6b...
Explanation:
In digital electronics, a counter is a sequential circuit that counts the number of occurrences of an event. A counter can be implemented using flip-flops, which are basic building blocks of sequential circuits.
Modulus Counter:
A modulus counter is a type of counter that counts up to a specific modulus value before resetting back to zero. In this case, we are looking for a counter that can count up to six before resetting.
Implementing a Mod-6 Counter:
To implement a mod-6 counter using flip-flops, we need to use three flip-flops. Each flip-flop will represent a binary digit, and the combination of these flip-flops will represent the count.
Binary Representation:
The modulus value can be represented in binary as 110, which means we need three bits to represent the count from 0 to 5.
Using D Flip-Flops:
D flip-flops are commonly used to implement counters. Each D flip-flop has two inputs - D (data) and CLK (clock). The output of one flip-flop is connected to the CLK input of the next flip-flop, creating a ripple effect.
Implementing a Mod-6 Counter using D Flip-Flops:
Here is how we can implement a mod-6 counter using three D flip-flops:
- Connect the Q output of the first flip-flop to the D input of the second flip-flop.
- Connect the Q output of the second flip-flop to the D input of the third flip-flop.
- Connect the Q output of the third flip-flop to the D input of the first flip-flop.
Working:
The counter starts at 000. On the rising edge of the clock signal, the first flip-flop increments its count by 1. When the count reaches 2 (010 in binary), the second flip-flop increments its count by 1. When the count reaches 4 (100 in binary), the third flip-flop increments its count by 1.
Reset:
When the count reaches 6 (110 in binary), the counter resets back to 000 and starts counting again. This creates a mod-6 counter.
Conclusion:
Therefore, a mod-6 counter can be implemented using three flip-flops.
A __________ counter can be implemented using three flipflops.a)mod-6b...
Concept:
For a counter with ‘n’ flip flops:
- The total number of states = 2n (0 to 2n – 1)
- The largest number that can be stored in the counter = 2n – 1
To construct a counter with any MOD number, the minimum number flip flops required must satisfy:
Modulus ≤ 2n
Where n is the number of flip-flops and is the minimum value satisfying the above condition.
Calculation:
Given:
n = 3
Modulus ≤ 2n
Modulus ≤ 23
Modulus ≤ 8
The most appropriate answer is option 1 i.e. Mod 6.