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....
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....
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