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I keep coming back here searching for an RC-based negative edge detector for TTL levels, that generates a negative pulse on a falling edge. The only place I find this solution is a certain spot in a certain video by Ben Eater. But I keep forgetting it and there is no clear entry on this site about it, and I'm tired to go back to a video to jog my memory.

Note: I have already answered the question with the only way I know. But it has problems, and therefore I am not accepting my own answer, but the answer that has fewer issues, and is simpler than mine.

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  • \$\begingroup\$ Be very careful with this kind of thing: it often leads to unreliable circuits - you typically have to consider "and then what..." type questions, such as "and what if the signal bounces right back?" Or "what if the transition is slower?" Really you have to consider the purpose and not just build something because someone talked it up on youtube. \$\endgroup\$ Sep 11, 2020 at 16:38
  • \$\begingroup\$ The 74121 comes immediately to mind. It works on either rising or falling edges, memory serving. (You pick.) They are probably not as readily available as they once were. I've used them. The 74123 (which I've not used) also springs to mind. Then there is the 555 timer, which I believe can be configured this way. And I'm sure a variety of cheaper if a little more finicky approaches. If you would carefully descreibe your experiences and problems, as best you see them anyway, it might help a lot. I should not need to go back to other questions to learn -- just re-write here. \$\endgroup\$
    – jonk
    Sep 11, 2020 at 17:12

3 Answers 3

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How about this simple circuit?

schematic

simulate this circuit – Schematic created using CircuitLab

Where \$RC\$ time constant should be much lower than the input impulse duration.

Or use a NAND monostable gen

Monostable out of NAND gates

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    \$\begingroup\$ Add a diode across the resistor, so that you don't overdrive the CMOS input on the low-to-high transition. \$\endgroup\$
    – Dave Tweed
    Sep 11, 2020 at 16:24
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    \$\begingroup\$ @DaveTweed CMOS gate does not have this diode already built-in? \$\endgroup\$
    – G36
    Sep 11, 2020 at 16:29
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    \$\begingroup\$ It does have a diode (actually a pair of them), for ESD protection. However, they are not designed to handle significant amounts of current injected deliberately on a regular basis. \$\endgroup\$
    – Dave Tweed
    Sep 11, 2020 at 16:32
  • \$\begingroup\$ I'd use a Schmitt trigger buffer here since I want that negative pulse. But my concern is the signal A cannot sink current fast enough to drain C1? And I think you are right that I don't need my R5 in my answer. Except in the final circuit I need R5 anyway because I use that n-MOSFET to work around the problem of A not being able to sink enough current. \$\endgroup\$ Sep 11, 2020 at 16:39
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Riffing off the clock doubler (with an XOR gate) in, I think it was Peter Alfke's classic "Six Easy Pieces", you can generate short pulses off a single edge.

schematic

simulate this circuit – Schematic created using CircuitLab

Shortly after the negative edge on the input, both inputs to the OR gate are low, until the inverter output goes high, and with a time constant = RC, so does the delayed input to the OR gate. If the logic level differentiating '0' and '1' is Vcc/2 the delay will be the half-life ofthe R-C circuit or about 0.7*RC, or 70ns with the values given, but in practice it will depend on the logic family.

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The only place where I find this answer documented is in Ben Eater's video "Using an EEPROM to replace combinational logic" at this particular time linked here.

This needs to be shown in schematics. I happen to want his ~600ns pulse width in my application too, so I use his exact values for R6 = 680 ohm and C3 = 1 nF for a 680 ns pulse. And the R5 resistor is there to re-charge the capacitor. Please excuse that I am showing the schematics in a certain application of a ramp generator that I'm building. Falling edge of a counter is supposed to reset the ramp. But the circuit in question here is all aroun R5, R6, C3, and D1, along with the rectangle wave from V2. The gate of M1 is simply the output pulse I want.

enter image description here

I added D1 to Ben Eater's circuit to chop off the +10V pulse on the rising edge. It may not matter with my MOSFET application, but it probably does matter on a TTL input pin to run it out of spec.

Here is the detail view of the negative pulse, and don't worry about the red trace, which is the charging of C1.

enter image description here

Here is the trace showing the positive edge pulse too, chopped off by D1. The problem of this is that it works with the switch application that Ben Eater uses, but it might not work driven directly by a TTL device since the momentary current it needs to sink to discharge C3 (a whopping full ampere) is not possible with a TTL device.

enter image description here

So I wonder if I need to deploy another MOSFET, an n-channel to do the discharging.

enter image description here

and this seems to work. Not sure if it is the simplest solution, but it is doable.

enter image description here

without that n-MOSFET M2 the resistance of the pulse generator would not be enough to generate that negative pulse.

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