Does this diode help discharge the capacitor once the flip-flop is reset?
Thanks in advanced.
(Image was taken from http://educypedia.karadimov.info/library/oneshots.pdf)
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\$\begingroup\$ Yes, it does. The capacitor will discharge near-instantly through the diode and the output. \$\endgroup\$– Criticizing Israel not allowedCommented Jul 8, 2016 at 18:50
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\$\begingroup\$ More importantly, it prevents the capacitor from charging instantly when the output goes high. \$\endgroup\$– Dave TweedCommented Jul 8, 2016 at 20:10
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1 Answer
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Figure 1. D flip-flop table with the only two conditions of interest highlighted.
Since the D input is permanently high there are only two conditions of interest.
- CL going high will set Q.
- R going high will reset Q.
R and C in your sample circuit will have a time constant given by $$ \tau = RC = 1M \times 0.1 \mu = 0.1~s $$
- On each rising edge of CL, Q will turn on for about 0.1 s, R will go high and Q will turn off again.
- The diode provides a rapid discharge path for C so that the circuit won't be held in RESET mode for the next 0.1 s and is ready to be re-triggered.
- Maximum pulse detection rate without skipping pulses will be in the order of \$ \frac {1}{\tau} = \frac {1}{0.1} = 10~Hz \$.
answered Jul 8, 2016 at 21:19