How many flip flops is/are required for 10 state count in ripple counter?

This question was previously asked in
JSSC JE Re-Exam Official Paper-I (Held On: 03 Nov, 2022)
View all JSSC JE Papers >
  1. Four
  2. Two
  3. Three
  4. One

Answer (Detailed Solution Below)

Option 1 : Four
Free
JSSC JE Full Test 1 (Paper 1)
5.7 K Users
120 Questions 360 Marks 120 Mins

Detailed Solution

Download Solution PDF

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.

Note: A MOD-N counter is also called as a divide by N counter as the input frequency is divided by the number of states of the counter.

Calculation:

Number no. of flip – flops are required to construct mod-10 counter, must satisfy:

2≥  10

The minimum value of n satisfying the above is:

n = 4

Latest JSSC JE Updates

Last updated on Sep 23, 2024

-> JSSC JE Additional Result has been released for the Jharkhand Diploma Level Combined Competitive Examination-2023 (Regular and Backlog Recruitment) This is for the Advertisement No. 04/2023 and 05/2023. 

-> The JSSC JE notification was released for 1562 (Regular+ Backlog) Junior Engineer vacancies.

-> Candidates applying for the said post will have to appear for the Jharkhand Diploma Level Combined Competitive Examination (JDLCCE).

-> Candidates with a diploma in the concerned engineering branch are eligible for this post. Prepare for the exam with JSSC JE Previous Year Papers.

More Digital Electronics Questions

Get Free Access Now
Hot Links: teen patti wink teen patti refer earn teen patti real cash 2024