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

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