Q.Explain the working of a four bit synchronous up counter using relevant Diagrams.

In a four-bit synchronous up counter all flip-flops are clocked simultaneously, and each higher flip-flop toggles only when all the lower ones are 1, so the count advances by one on every clock. …

[!TLDR]

A 4-bit synchronous up counter clocks all four J-K flip-flops together; each stage toggles when all lower stages are 1, counting 0000→1111.

Logic diagram: four J-K flip-flops FF1–FF4 with the clock applied to all four simultaneously — this is what makes the counter synchronous (there is no ripple, so the propagation delays do not add up and the counter is fast). The J-K inputs are steered by AND gates:

• FF1: $J_1=K_1=1$ (logic HIGH) → FF1 is always in toggle mode, so $Q_1$ changes on every clock pulse.
• FF2: $J_2=K_2=Q_1$ → FF2 toggles whenever $Q_1=1$.
• FF3: $J_3=K_3=Q_1Q_2$ (an AND gate) → FF3 toggles whenever $Q_1=Q_2=1$.
• FF4: $J_4=K_4=Q_1Q_2Q_3$ (an AND gate) → FF4 toggles whenever $Q_1=Q_2=Q_3=1$.

Working: on each clock pulse FF1 toggles; FF2 toggles only when $Q_1$ is already 1 (i.e. every second pulse); FF3 toggles only when $Q_1Q_2=1$ (every fourth pulse); FF4 toggles only when $Q_1Q_2Q_3=1$ (every eighth pulse). This is exactly binary up-counting.

Counting sequence (status $Q_4Q_3Q_2Q_1$):