# Purpose of reverse biased diode in pulse generation circuit

(Caveat: "young player" at large here.)

The circuit below (not my creation) debounces a push button and generates a 1us negative pulse when it's pressed. I breadboarded it and it works as expected, but I am trying to understand the purpose of the reverse biased diode.

My understanding is that as the first inverter goes high, the 100pF capacitor charges through the 10k resistor, which normally pulls the node to logic level 0, giving a short pulse of logic level 1 to the last inverter, creating the desired negative pulse.

But what is the purpose of the diode shown in the schematic? I don't see any way it would ever be anything but reverse biased, nor that it could ever be subjected to its reverse breakdown voltage.

I tried to remove it (it's unspecified in the schematic, I used an 1N4148) and compare single shot references on the scope (measured at the 10k node), but could see absolutely no difference between the signals with or without the diode.

(Actually not really sure about the 1k resistor's role either as I'd expect the HC input to draw so little current that there'd be no voltage drop over it.)

Update: Annotated the schematic below based on the answers, showing how it makes sense that the diode is in fact forward biased in this state.

When the first inverter output goes HIGH, the 100 pF capacitor will slowly charge up to 5 V through the 10K resistor. It will now have nearly 5 V across its plates.

When the inverter output goes LOW, the capacitor's (let's call it) top plate will be connected to near-0 V by the inverter output. This plate is charged to 5 V wrt to the bottom plate, so the bottom plate voltage will now be -5 V.

The diode is now forward biased as its anode is at 0 V and its cathode at -5 V. It conducts, discharging the capacitor quickly.

The effect of this RC and diode circuit is to charge the capacitor slowly and discharge it very quickly. So when the first inverter goes HIGH, there will be a pulse generated at the second inverter input. And when the first inverter goes LOW, there will be no pulse generated. So that RC and diode is an edge filter.

If you work either side of the inverters, you can then see how the circuit produces the final pulse it does and for which switch action (press or release) it does it.

Note that the second inverter has an internal diode connected the same way as this diode, in parallel with it. Stressing the internal diodes is typically avoided to not reduce the reliability of the inverter. The external diode is a significantly higher current part.

• Thanks, somehow I had a hard time picturing current flowing out of ground towards the cap, but after annotating the schematic with the voltages [question updated with this] in this state, it does make sense. Marked this as accepted answer because it also provided an explanation for why I could observe no difference removing the diode (the 74HC14 having a clamping diode itself). Nov 15, 2021 at 22:29

But what is the purpose of the diode shown in the schematic?

When the output of the first Schmitt trigger goes high, the cap develops a voltage across it. When the output of the Schmitt subsequently goes low, the cap will initially still have the voltage across it. This would bring the diode side of the cap well below ground (if there were no diode). The 10 K resistor will discharge the capacitor, but slowly. However, the diode will conduct as soon as that side of the cap goes to -0.6V. Thus the cap will both discharge more quickly because of the diode, and also the diode will clamp the voltage to always be above -0.6V. Which one of these factors is more important in the circuit at hand? My guess is the clamping effect, to protect the 74HC14 from too negative input voltages. The rapid cap discharge being a side effect.