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The fall time can't be < 2us since R3 = 2M.Taking comments above on board, here's a suggested improved debounced circuit. Switching pulse rise and fall times are both < 2uS. In simulation a 10mS train of 0.5 mS pulses at 1kHz rate represents switch bounce and gives rise to a single switching pulse.
The bjt is replaced by a FET.
The pulse is differentiated by the 10nF*150Ω. When the switch is released, the output voltage will go below ground by less than a millivolt (12V*150Ω/2Meg=900uV).The fall time can't be < 2us since R3 = 2M.
The pulse is differentiated by the 10nF*150Ω. When the switch is released, the output voltage will go below ground by less than a millivolt (12V*150Ω/2Meg=900uV).
The slow fall time is on the left side of the capacitor. The 150Ω resistor differentiates this slow exponential.Ron,
When the button is released, the time constant is 10 nF * 1M = 10 ms.
So the FF needs a Schmitt Trigger input to cope wih such a long fall time.
Yes, I was concentrating on the time constant & forgertting about the voltage division due to the 2 resistors.But the long TC is on the switch side of the dif cap. the other side is quite fast. 10nf x 150Ω.
Slow fingers.
When the switch closes, the time constant is 10nF*150Ω.But the long TC is on the switch side of the dif cap. the other side is quite fast. 10nf x 150Ω.
Slow fingers.
I agree.When the switch closes, the time constant is 10nF*150Ω.
When the switch opens, the time constant is ≈20mS. The 150Ω resistor is part of that time constant. The resistive divider, 150/(2Meg+150), means that less than 1mV appears across the 150Ω resistor.
Yeah, I said essentially that in post #24:I agree.
However it is -1mV since the discharge current goes through the 2 M resistor then back to the capacitor via the 150Ω.
The pulse is differentiated by the 10nF*150Ω. When the switch is released, the output voltage will go below ground by less than a millivolt (12V*150Ω/2Meg=900uV).
There were valid criticisms of my first circuit; hence the second one. But I'm a great believer in "If it ain't broke don't fix it".Even though the one I have now works
I would use the MOSFET. Most BJTs don't have enough current gain to predictably sink 240mA when being driven from CD4000 logic, unless you use a Darlington, and then you suffer at least a volt Vce(sat).Thanks everyone for your posts
The circuit that alec_t initially posted works well with the decoupling cap added to it. It turns off and on with every push of the switch. I saw the second diagram is supposedly a better debounce circuit and uses a MOSFET transistor ( i think). Even though the one I have now works, would you guys suggest that one? I couldn't quite follow all of the posts to tell if it is good or not..