I need to collect at least 60uA via capacitive coupling between a high tension Cu cable with PVC insulation and a connector that is acting as the opposite face of the capacitor (can be a copper plate wrapped around the wire insulation or some turns of Cu wire around the insulation, or any other way).

Here the details:

  • The cable insulation is made of PVC (dielectric constant 3.19) and is 3mm thick.
  • The inner cable is made of copper 1mm thick.
  • The voltage provided to the cable is 14 000 V.
  • the voltage is provided by an ignition coil, with about 1000 spark each minute

How can I act? is it possible to collect the 60uA in any other way?

  • 2
    \$\begingroup\$ What is the frequency of the voltage? \$\endgroup\$ Commented Sep 25, 2018 at 13:23
  • \$\begingroup\$ I've updated the question! \$\endgroup\$
    – Luigi
    Commented Sep 25, 2018 at 13:26
  • 2
    \$\begingroup\$ What is the risetime of the spark? What is the load through which this 60 uA must flow? Do you require the spark waveform to be preserved? You need to tell us a lot more about what you're trying to accomplish here! \$\endgroup\$
    – Dave Tweed
    Commented Sep 25, 2018 at 13:34
  • \$\begingroup\$ I think this is an interesting question and I'm not sure why it's downvoted. Maybe it's because I simply don't know anything about the subject so I don't know what's missing. @DaveTweed Do you think you could enumerate what's missing, or is it too much to list? \$\endgroup\$
    – pipe
    Commented Sep 25, 2018 at 13:52
  • \$\begingroup\$ @pipe: Too much to list. We need to have some idea of where the OP is headed with this to even ask reasonable clarifying questions. \$\endgroup\$
    – Dave Tweed
    Commented Sep 25, 2018 at 13:55

1 Answer 1


Back of envelope calculation coming, very low precision but the right ballpark.

1000 sparks/minute = 16 per second.

A current of 60uA = 60uC/second = about 4uC per spark.

To get 4uC from a 14kV pulse, we need a capacitance of about 4u/14k = 300pF.

That's a lot of capacitance to achieve by wrapping a bit of wire round your cable. In fact, it's equivalent to the capacitance of about 3m of 50ohm coaxial cable, which has broadly the dimensions you've specified in your OP.

So cover the 3m of the ignition lead with alli foil. Or replace the lead with 3m of 50 ohm coax, and take your coupled output from its screen.

Of course, an ignition waveform is not a nice sine wave, but has very steep edges. The current during the edge will be much, much higher than your target 60uA. But that's the capacitance you need if you want 60uA average. If you actually want 60uA peak, then a) you don't need as much capacitance to couple and b) you need to know a lot more about your ignition waveform, specifically the risetime into your load.

You might find that 300pF loads your ignition coil down somewhat. It might be better to extract your power from the coil's primary circuit, rather than the secondary.

  • \$\begingroup\$ Many thanks for the precise answer! I need to activate a TIC106D SCR [link] (alldatasheet.com/…) with the current collected by this trigger, so probably, but correct me if I'm wrong, I need only a 60uA peak current. I don't have 3 meters of cable, but less then one! \$\endgroup\$
    – Luigi
    Commented Sep 25, 2018 at 13:43
  • 1
    \$\begingroup\$ If you only need 60uA peak, then If you actually want 60uA peak, then a) you don't need as much capacitance to couple and b) you need to know a lot more about your ignition waveform. I suggest you wrap a few inches and try it. Then a few more, until it works. Then add 50% extra for luck. \$\endgroup\$
    – Neil_UK
    Commented Sep 25, 2018 at 13:45
  • \$\begingroup\$ Thanks, I'm not such an expert to have more info on the ignition waveform.. Just to have a more complete view: I need this for a engine timing light for my '70s car. here the complete circuit (not working now) electronics.stackexchange.com/questions/396216/… \$\endgroup\$
    – Luigi
    Commented Sep 25, 2018 at 13:48
  • \$\begingroup\$ @Luigi See your earlier question, I've commented there. \$\endgroup\$
    – Neil_UK
    Commented Sep 25, 2018 at 13:51

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