I was looking at this StackExchange question:

Why does this rising edge detector using a capacitor and a resistor work?

I ran the simulation and I'm confused by the behavior. On the rising edge of the clock the left side of the capacitor charges to a positive voltage. Shouldn't the right side be negative, since current is flowing out of that side?

And then a side question: I set up the circuit on a breadboard and ran it, with a clamping diode, and did not get a suppression of the negative spike. In fact, I get the same behavior with or without the diode.

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    \$\begingroup\$ for very fast edges, the capacitor acts like a short(the voltage takes time to change) and both plates of the capacitor will rise together (and fall together). \$\endgroup\$ – analogsystemsrf Oct 6 '19 at 23:30
  • \$\begingroup\$ Oh, that makes sense. How am I not suppressing the negative spikes with the diode? I'm using 1N4148 across the resistor with anode to ground. \$\endgroup\$ – Mike Oct 6 '19 at 23:44
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    \$\begingroup\$ R1N4148 ~= 10 Ohms @ -0.7V , so with cathode on output and Anode to Gnd, it will charge up cap with negative spike and restore DC level after AC cap. -0.6V negative spikes may still be visible with 100 Ohm load and shifts according to R ratio. Long probe grounds may induce noise (false glitches) so must be short (pref 1cm) using tip& barrel of 10:1 probe. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Oct 7 '19 at 6:22

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