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In the circuit provided by the link, would having an open secondary coil (not having it connected to any ground) damage the it? The coil is connected to pulsed 240v ac at 50hz. If it is not arcing, (meaning it has an open secondary coil) where will the electric field collapse, and will it damage anything? Furthermore, if it can damage anything in the circuit, will it do so if it is under powered with a very low current?

https://www.instructables.com/id/Variable-voltage-ignition-coil-power-supply/

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    \$\begingroup\$ Some ignition coils can be damaged by not having a spark plug connected that will naturally restrict the output voltage. \$\endgroup\$
    – Andy aka
    Commented Jan 21, 2019 at 11:09
  • \$\begingroup\$ Alright thanks. Will this still happen with a very low current? \$\endgroup\$
    – Steve
    Commented Jan 21, 2019 at 11:16
  • \$\begingroup\$ Once secondary winding insulation breaks down, how can you judge what current will flow? \$\endgroup\$
    – Andy aka
    Commented Jan 21, 2019 at 11:20
  • \$\begingroup\$ Sorry, I mean for example if a low value capacitor, around 5nf instead of the recommended minimum of 0.1 micro farad, is used which limits the current which can reach the coil. The article says the circuit will not work with values less than 0.1micro farad, I’m assuming because the current is too low. Will that damage the coil with such a low current? \$\endgroup\$
    – Steve
    Commented Jan 21, 2019 at 11:24
  • \$\begingroup\$ Pulsed could mean flyback, in which case there is a high risk of arcing. On the other hand, pulsed could still mean normal voltage transformer and the secondary is rated for the voltage. I would assume the former rather than the latter. \$\endgroup\$
    – winny
    Commented Jan 21, 2019 at 11:40

2 Answers 2

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Your ignition transformer appears to be a voltage transformer, and if this is the case, an open secondary will do no damage, however, be aware that current transformers exist as well, and must never be allowed to have an open secondary while operating.

A voltage transformer maintains a relationship between input and output voltage, so any current that flows on the secondary will be determined by the impedance of what is connected.

A current transformer maintains a relationship between input and output current, and when presented with an impedance on the output, it will produce whatever voltage is necessary to maintain this relationship. This means if the secondary is open circuit (Resistance tends toward infinity), it will attempt to produce a voltage that tends toward infinity to maintain output current, causing arcing or dielectric breakdown and damaging components.

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  • \$\begingroup\$ Alright cool. So is the coil in the link a voltage transformer or a current transformer? How can you tell? autopartssupply.com.au/… \$\endgroup\$
    – Steve
    Commented Jan 22, 2019 at 0:19
  • \$\begingroup\$ I would guess it is a voltage transformer because of multiple references to specifically 30000 V, but I wanted to include mention of CTs because you might not know either and the difference is important, also because the question is specifically about an open circuit on a secondary, which is a very dangerous state for a CT. \$\endgroup\$
    – K H
    Commented Jan 22, 2019 at 0:34
  • \$\begingroup\$ Ok thanks. Could damage be avoided to a current transformer if a low enough current was used? \$\endgroup\$
    – Steve
    Commented Jan 22, 2019 at 0:42
  • \$\begingroup\$ Usually when one is opened while operating you would have an arc of a voltage that depended on the actual resistance of the "open circuit" and a current that was proportional to that on the primary, so the only power limitation is the capability of the system on the primary side and however much the resistance/impedance of the arc is limited. If the primary is 600V 500A and it's a 10:1 CT, it will produce a 5A output at up to 60000V. Hmm.... Maybe it would actually work if current limited this way, but I think yours is a VT anyway. \$\endgroup\$
    – K H
    Commented Jan 22, 2019 at 0:54
  • \$\begingroup\$ Alright thanks. Is there any way if I can tell if I have damaged the coil? \$\endgroup\$
    – Steve
    Commented Jan 22, 2019 at 2:39
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The larger range of capacitors have lower impedance, and can conduct more primary current but secondary arcing, if sustained can melt the secondary fine wire but this can be protected by using carbon ignition wire used by sparkplugs.

Since Copper has a PTC tempco., the weakest section of the secondary may rise in temp then resistance rises possible leading to thermal runaway then fusing open.

If you want to minimize damage risks, then using a carbon ignition wire for secondary used for spark plugs and always use plastic caps the larger size reduces resonant frequency and increases current.

The sunflower oil improves the insulation breakdown of air by ~ 5x due to impurities in oil and components but in transformer grade oil up to 25x and also distributes cooling to hotspots.

This kit was not "engineered" with coil inductance and resistance but appears to work. As I recall , I could put 1V from a 50 Ohm signal generator and get 1kV out sine up to about 20kHz with no load, meaning a coil that looks like this one can have a turns ratio of 1:1000.

The magnet wire has various insulation breakdown ratings and impossible to tell what your model is rated for but ought to handle 30kV easily. However, I doubt oil immersion affects that rating being sealed.

The oil is safer due to protect from creapage surface breakdown as you are applying about 300Vp 100Hz pulses to the primary coil intended for 12V pulses. So always start from the dimmest setting.

A 1A fuse and foot safety switch might be suggested so if an accident occurs, you fall over and turn off the power.

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  • \$\begingroup\$ 0.5mA is safe in NA, 15mA is considered safe in EU but hurts and >30mA may induce heart arrhythmia. \$\endgroup\$
    – D.A.S.
    Commented Jan 21, 2019 at 19:49
  • \$\begingroup\$ That makes sense. I don’t have any carbon wire, so could I reduce risk using a low value capacitor to limit the current? \$\endgroup\$
    – Steve
    Commented Jan 22, 2019 at 0:21
  • \$\begingroup\$ carbon wire is readily avail, in stores or car junk yards \$\endgroup\$
    – D.A.S.
    Commented Jan 22, 2019 at 1:43
  • \$\begingroup\$ Alright I’ll have to have a look. Is there anyway I can tell if I have damaged the coil? \$\endgroup\$
    – Steve
    Commented Jan 22, 2019 at 2:38
  • \$\begingroup\$ MEasure resistance \$\endgroup\$
    – D.A.S.
    Commented Jan 22, 2019 at 2:39

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