Generation of Clock Signals (Determine the pulse - width of the output waveform)

this is one of my college exercise.

the question is
A circuit designer converts a 50 kHz, 70 % duty cycle input waveform to a 50 kHz, 50 % duty cycle using IC 74LS221. Determine the pulse-width of the output waveform.

and my answer is

T = 1/F = 1/50k = 0.02ms
Percentage Duty cycle = (TH / T) × 50% = 1ms

but the correct answer is 10 micro s.

can someone explain to me ? please....

can someone explain to me ? please

Click to expand...

Yes, careless arithmetic.

T = 1/F = 1/50k = 0.02ms

Click to expand...

Correct.

Percentage Duty cycle = (TH / T) × 50% = 1ms

Click to expand...

Wrong

but the correct answer is 10 micro s

Click to expand...

Quite right.

JimB

then how u calculate ? the 74LS221 is non retriggerable one shot

here's my solution.

period (T) = 1 / F = 1 / 50 kHz = 0.02 ms

since duty cycle is 50 % 0.5 * (PW / T ) = 0.5 * (PW / 0.02 ms)

hence, the pulse width (PW) is 10 micro seconds

That looks better!

JimB