Since 4-bit counter is required we will use 4 J-K flip-flops.

**Operation.**

Initially, a short negative going pulse is applied to the clear input of all flip-flops. This will reset all the flip-flops. Hence, initially the o/ps are $Q_3 Q_2Q_1Q_0$ =0000.

But $Q_3$'=1 and since it is copied to $J_0$ it is also equal to 1.

$J_0$ =1 and K=0.....initially.

On the first negative edge of clock arrives at first f/f. o/p of $Q_0$ =1.

after 1st –ve edge clock the o/ps of f/fs will be,

$Q_3Q_2Q_1Q_0$ =0001

On second –ve clock o/p of 2nd f/f will be 1 i.e $Q_1$ =1.

$Q_3Q_2Q_1Q_0$ =0011

Similarly for 3rd –ve edge clock,

$Q_3Q_2Q_1Q_0$=0111

For 4th –ve edge clock,

$Q_3$$Q_2$$Q_1$$Q_0 =1111$ - Now as soon as 5th –ve edge is arrived o/p of 1st f/f becomes 0 i.e $Q_0$=0 i.e $Q_3Q_2Q_1Q-0$ =1110 - This operation continues till the o/p is reached to zero o/p state. i.e $Q_3Q_2Q_1Q_0$ =0000

**Logic diagram:-**

**Waveforms for Johnson’s Counter**