Digital Electronics Test 4

Concept:

The MOD number of a counter is equal to the number of complete states that a counter goes through before it recycles back to its starting state. To construct a counter with MOD X, it requires a minimum number of N flip-flops.

 Such that, 2^N ≥ MOD.

Calculation:

The time period of the clock pulse \(= \frac{1}{{10\;MHz}} = {10^{ - 7}}sec\) 

Propagation delay of each FF = 15 × 10^-9 sec

Number of F.F required \(= \frac{{{t_{clk}}}}{{{t_{pd}}}} = \frac{{{{10}^{ - 7}}}}{{15 \times {{10}^{ - 9}}}} = 6.67\) 

∴ Number of FF required = 6

2^N ≥ MOD

Modulus of counter = 2⁶ = 64.

Common mistake:

We should not take N = 7

If we take N = 7, Using 7 Flip Flops with delay 15ns. The total delay becomes 105ns, which is greater than the time period of the CLOCK (100 ns),is not a desirable operation.

 Using 6 Flip flops allows us to keep a delay under 100ns (90ns).