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Explain the working of 3 bit asynchronous counter with proper timing diagram
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In asynchronous counter, a clock pulse drives FF0. Output of FF0 drives FF1 which then drives the FF2 flip flop. All J and K inputs are connected to Logic 1. Therefore, each flip flop will toggle with negative transition at its clock input.

The 3 bit MOD-8 asynchronous counter consists of 3 JK flipl flops. Overall propagation delay time is the sum of individual delays. Initially all flip flops are reset to produce 0. The output conditions is Q2Q1Q0 = 000.

When the first clock pulse is applied, the FF0 changes state on its negative edge. Therefore, Q2Q1Q0 = 001. On the negative egde of second clock pulse flip flop FF0 toggles. Its output changes from 1 to 0. This being negative change, FF1 changes state. Therefore, Q2Q1Q0 = 010. Similarly, the output of flipflop FF2 changes only when there is negative transition at its input when fourth clock pulse is applied.

The output of the flip flops is a binary number equivalent to the number of clock pulses received. The output conditions are as shown in the truth table.

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On the negative edge of eighth pulse, counter is reset.

the counter acts as a frequency divider. FF0 divides clock frequency by 2, FF1 divides clock frequency by 4, FF2 divides clock frequency by 8. If n flip flops are cascaded, we get $2^n$ output conditions.

the largest binary number counted by n cascaded flip flops has a decimal equivalent of $2^n-1$. MOD-8 counter has count of the largest binary number 111 which has decimal equivalent of $2^3-1=7$

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