I found this schematic on the Web and I built the circuit but, due to space constraints, I removed some components and it work well.

I've not mounted the two diodes \$D_3\$ and \$D_4\$ and the capacitors.
For the logic gates I used a 74LS00.

The questions:

  1. What is the purpose of \$D_3\$ and \$D_4\$?
    To protect the supply in case of fault in case I connect the probe at high voltage?
  2. And what is the use of the two capacitors?

enter image description here

Thanks in advance.


1 Answer 1


The point of both diodes and capacitors is stretch the output to make LEDs detect even very short pulses. Even if there is a very short "high" pulse, then C2 is quickly discharged through D3, and once the pulse disappears, Cw charges through L1 causing it to glow. The same happens with "low" pulses and C1/D4.

Since you have removed these components, theoretically, you will no longer be able to detect very short pulses (less than 10 microseconds). I do, however, question the design -- 0.1uF driving 10mA LED gives time constant on the order of tens of microseconds, and since 74LS series do not drive very strongly, a sub-microsecond pulse will not discharge the capacitor much. So I am not sure you are missing much.

  • \$\begingroup\$ Then it is used only to make visible short pulse, i.e only enlarging the time that the current flow through the diode due to the charging time of the capacitor, right? \$\endgroup\$
    – Antonio
    Nov 25, 2016 at 15:44
  • 1
    \$\begingroup\$ Yes, and it does not even do this very well \$\endgroup\$
    – theamk
    Nov 28, 2016 at 17:52

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