The left side are the test leads to poke around on something on a vehicle. I for the life of me cannot remember what this was used for. I do know 100% that it is used to test something on the electrical system of a motorcycle. Everything is in parallel except for the diode.

  1. 8.2M
  2. 224k 100v
  3. 224k 110v
  4. 2A 224j
  5. 104k 100v
  6. diode N1 40?


  • \$\begingroup\$ Or even what this type of thing might do could help with pointing me in the right direction. \$\endgroup\$
    – Dan
    Commented Aug 3, 2017 at 2:10
  • \$\begingroup\$ It has redundant caps for no particular reason to read a DC or positive pulse and store it for reading Vpk - 0.6V. No idea why that is useful. \$\endgroup\$ Commented Aug 3, 2017 at 2:22
  • 2
    \$\begingroup\$ Now I remember! I was to poor at the time (student) to buy a peak voltage adapter to test the ignition coil on my motorcycle. So I made this based of pictures I found. I can't believe I couldn't remember. I did however only use it once. Thank you for taking the time. \$\endgroup\$
    – Dan
    Commented Aug 3, 2017 at 2:59
  • \$\begingroup\$ You really need to post a picture, now what you believe the components are. \$\endgroup\$
    – pipe
    Commented Aug 3, 2017 at 8:45
  • \$\begingroup\$ In a circuit diagram like this, one typically draws the negative rail at the bottom. \$\endgroup\$
    – Mels
    Commented Aug 3, 2017 at 15:08

1 Answer 1


If you sum the capacitors to just one, then it looks like an envelope detector.

enter image description here

If you put capacitors in parallel then you can sum them together to just one. $$C_{tot}=C_1+C_2+C_3+C_4$$

And then you got the exact schematic as in the picture above.

However, your R is really huge and I'm not exactly sure about the capacitance values you're using in the schematic, but it looks like your schematic is made for finding the \$V_{peak}\$. See image below.

enter image description here

It's still an envelope detector, but now the capacitor bleeds out much slower, so when you read the output you will see the peak voltage of the sine wave. If you add the forward voltage of the diode (assume 0.7V) and then divide by \$\sqrt{2}\$, you'll get a very very good approximation of the RMS value. IF it is a sine wave. RMS voltages is really good to know when working with AC.

So if I was too fast with my words, here's an equation of what I just said: \$RMS_{approximation}=\frac{V_{peak}+0.7}{\sqrt{2}}\$

  • \$\begingroup\$ Thank you! I am not able to up vote but this is it. \$\endgroup\$
    – Dan
    Commented Aug 3, 2017 at 3:09

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