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Timeline for Easiest way to extend a pulse

Current License: CC BY-SA 3.0

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Apr 25, 2012 at 17:32 comment added Jason S I mentioned it won't work for short pulses, and I mentioned putting a diode in the circuit. Your comments are correct, but are getting tiresome.
Apr 25, 2012 at 16:37 comment added Telaclavo But you need a diode there. Without it: a) for short pulses it won't work, and b) for slightly longer than short pulses it won't add the correct amount of time, because the state of charge of the capacitor will depend on the pulse time, and you don't want that. You want the capacitor to start fully charged, at the beginning of the exponential, no matter what the pulse length has been.
Apr 25, 2012 at 16:34 comment added Jason S but he also says "When the igniter box thinks a coil fired it grounds the 5V wire from the compputer. So it dips this 5V signal to about .8V for about 1.5ms." which implies he's talking about an active low pulse that he wants extended. I can't figure out which, so he'll have to choose, and I gave him both choices.
Apr 25, 2012 at 16:31 comment added Telaclavo Sorry, I didn't read all your text. The OP wrote "and extend the falling edge of it about 1.7ms". Anyway, you need a diode there.
Apr 25, 2012 at 16:29 comment added Jason S Read my post again: the circuit I drew extends an active low pulse. To extend an active high pulse, move the 2nd HC132 to the input. If you like I can post that circuit too. I couldn't tell from the OP's post whether he wants an active low pulse or an active high pulse, so I responded with an answer that would let someone do both.
Apr 25, 2012 at 16:24 comment added Telaclavo The OP said to delay the falling edge. This thing delays the rising edge. So, in fact it shortens the pulses.
Apr 25, 2012 at 16:21 history answered Jason S CC BY-SA 3.0