The Gateway to Computer Science Excellence
+2 votes


I have read somewhere that J-K flip-flop used as divide by 2 frequency counter is it true ??

if not how to solve given problem??

in Digital Logic by Boss (25.7k points) | 196 views
can we do like this :

30Khz / mod of counter  ??


soved :: JK flip flop worked as divide by 2 frequecy counter when j=k=1  that is nothing but mod only...


so just consider mod of counter

1 Answer

+5 votes
Best answer

A counter acts as a frequency divider, So, A mod-n counter divides the incoming frequency by factor of $n$.

$\large \Rightarrow f_{out} = \frac{f_{in}}{n}$

For a JK flip flop $\color{green}{Q_{n+1} = J\overline{Q_n} + Q_n\overline{K}}$

In above combination $Q_{0n} = \overline{Q_1}.\overline{Q_0}$ and $Q_{1n} = Q_0\overline{Q_1}$

So, It acts as a Mod-$3$ Counter $(Q_0,Q_1) : (0,0) \rightarrow (1,0) \rightarrow (0,1) \rightarrow (0,0)$

Thus $f_{out} = \frac{30}{3} = 10KHz$

by Boss (28.8k points)
selected by
@mcjoshi can you tell me what is logic over here or how it works !

what is reason behind dividing mod-n

A mod-n counter has $n$ counting states, and after reaching initial state, we again start counting from $0$(cross n states) and Tclk of repetition is increased by a factor of n and freq = 1/time_period, thus fnew = fold/n

In general any mod-n counter acts as frequency divider by factor of $n$. So, in any of these type of questions the only task is to find the modulus of circuit represented. and then divide original frequency by $n$

but i think it act as frequency divider only in case of asynchronous clock while in the diagram it is synchronous.

Am i right or wrong?
It acts as a frequency divider in both cases.

No related questions found

Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,737 questions
57,365 answers
105,260 users