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Question

a).A modulo16 ripple counter uses JK flipflops. If the propagation delay of each flipflop is p ns and the maximum clock frequency that can be used is 5 MHz, then which of the following represents value of p?

a. 30
b. 40
c. 50
d. 55

b).Two binary counters are cascaded, where first counter has n flipflop and second counter has mflipflop.If the input signal has frequency of 30 MHz and n = 6, m =4, then the output frequency of output signal will be _______ kHz.

c).An eight stage ripple counter uses a flipflop with propagation delay of 75 nanoseconds. The pulse width of the stroke is 50 ns. The frequency of the input signal which can be used for proper operation of the counter is approximately.

I am not able to solve the frequency related problem with flip-flops please guide what approach should i used to solve these kind of problem .?

Comments

Helpful Video Solution of GATE ECE 1987 Ripple Counter Question: https://youtu.be/S7iUh6vbtis 

Answer 1

a) In MOD-M ripple counter no. of JK flip-folp n is given by:
$M \leq 2^n$
$16 \leq 2^n$
$\therefore n = 4$ flip-flops
frequency f of a ripple counter with propagation delay p,
$f = \frac{1}{np} = \frac{1}{4 \times p}$
$p = \frac{1}{4 \times f} = \frac{1}{4 \times 5 \times 10^6} $
$p = 50$ ns

b) Output frequency in cascaded MOD-M and MOD-N counter,
$f_{output}= \frac{f_{input}}{2^m \times 2^n}$ (here m and n are no. of flip flops not MOD)
$f_{output}= \frac{30 \times 10^6}{2^6 \times 2^4}$
$f_{output}= 29.29$ KHz

c) eight stage means 8 - bit ripple counter and when strobe signal is given you have to consider
$T_{clk} = n \times T_p + T_s$ (here $T_p$ is propagation delay of each flip flop and $T_s$ is strobe delay)
$\therefore f = \frac{1}{T_{clk}} =\frac{1}{n \times T_p + T_s} = \frac{1}{8 \times 75 + 50} = \frac{1}{650} = 1.538$ MHz

Answer 2

1. For ripple counter...we have Freq=1/nTp

Here 'n' is no of counters. Tp is propagation delay. For Mod-16...'n' will be 4. Apply the formula. You will get Tp= 50 n's

2. When a Mod-n counter is Cascaded with Mod-m counter, we get a Mod-m*n counter.

So required Freq will be 30 MHZ / (2⁶*2⁴)