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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??

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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

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)
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+1
@mcjoshi can you tell me what is logic over here or how it works !

what is reason behind dividing mod-n
+1

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

https://www.quora.com/How-do-counters-help-in-frequency-dividers

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 $n$

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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?
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It acts as a frequency divider in both cases.