So in this video from Ben Eater he makes a rising edge detector using a capacitor and a resistor like the one below.


simulate this circuit – Schematic created using CircuitLab

In the video Ben said that the circuit would produce a quick pulse whenever the clock signal switched to high. I simulated the circuit and found out that the circuit produces a positive voltage when the clock signal switches to high and a negative voltage when it switches to low. Here's an image of the simulation.

Voltage goes negative when clock goes low.

In the video the circuit worked as if the negative pulses were nonexistent. Why did it work? In the video the circuit is connected to the pins 4B and 3B of the chip SN74LS08. Could be something with the chip?


1 Answer 1


In the video the circuit worked as if the negative pulses were nonexistent.

Yes, that effect is due to the chip. Let's have a look at the SN74LS08's datasheet and look at the circuits on the chip:

enter image description here

Notice the two Schottky diodes between the inputs A and B and GND.

These diodes are needed to protect the sensitive transistors in the chip. Nearly all chips have such ESD protection diodes.

These diodes will conduct when the voltage on the inputs becomes less than about -0.2 V

That "clamps" the voltage to -0.2 V, the voltage will not get much lower than -0.2 V.

In your simulation, add a diode and observe the same effect!

  • 2
    \$\begingroup\$ Is it a good idea to rely on ESD protection diodes to obtain the desired effect? \$\endgroup\$ May 1, 2019 at 16:49
  • 3
    \$\begingroup\$ @RobertHarvey I would say no, it's not a good idea. In fact, I would not rely on internal ESD protection diodes for anything at all, including ESD protection. The reason is that they're not actually diodes. They're parts of transistors. Almost everything on an IC is part of a transistor because you practically get the rest of that structure for free anyway. And those transistors can cause problems if you activate them. Like shorting the power supply, for example, until it stops delivering current. \$\endgroup\$
    – AaronD
    May 1, 2019 at 17:28
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    \$\begingroup\$ @RobertHarvey I would take a milder opinion than AaronD. If the current through the ESD diodes is limited then there is no issue. In your circuit, there is no current limit except the series resistance of the source making the square wave. That resistance could be 50 ohms. If the ESD diodes are proper ESD diodes then they can handle A LOT of current but only for a very short moment. During an ESD pulse as much 4 Amps can flow through an ESD diode. They are designed to handle that but only when this does not happen often (like an ESD event would not happen often). \$\endgroup\$ May 1, 2019 at 17:41
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    \$\begingroup\$ If I would want to use this in a product that's supposed to be reliable then for sure I would not rely on the ESD diodes. I would simply add an external Schottky diode. \$\endgroup\$ May 1, 2019 at 17:42
  • \$\begingroup\$ Thank you! This explains it really well. I thought about diodes having to do something with it but for some reason I came to the conclusion that it would not work. I made a working circuit using a diode now and it is very clear now. \$\endgroup\$
    – intoo
    May 1, 2019 at 19:07

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