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I'm designing a circuit that requires a high voltage pulse in order to fire a spark plug. I'm thinking of using an inductor to provide this pulse (via inductive kickback), but I'm struggling to figure out how to actually make it work without destroying my power supply.

I understand that spark plugs in a car are fired in a similar way using an ignition coil (which is actually like a transformer, so the high voltage side is isolated from the low voltage side). However, I haven't been able to figure out how the battery is protected from the massive voltage spike. From what I can tell, a flyback diode on the primary also prevents spikes on the secondary, so it seems near impossible to actually get a spike.

My design doesn't use a transformer as it seemed like there was no difference in using an inductor vs a transformer. So, I am currently designing the circuit with an inductor although I could change it to a transformer if that's the only way to do it. The circuit looks like this:

enter image description here

I'm simulating the spark plug as a high resistance (100GOhm) element. The 1.4 kOhm resistor is being used to tune the output voltage, as my understanding is that the output voltage is given by V=I*R where I is the current through the inductor before switching, and R is the resistance of the RL circuit. The 1 Ohm resistor is being used to limit the current through the resistor.

This circuit does technically do what I want it to do. It produces a very high voltage (I think about 7 kV) spike when the switch is opened. However, the switch is the real problem. If I use a solid state switch (MOSFET, SSR, etc.), there would be major issues with the voltage rating of the switches as it is going to be very hard to find a switch that is ok with 7 kV being dropped over it. The easier alternative is using a relay, but I need to prevent the relay contacts from sparking. The only way I know to prevent this issue is to use a flyback diode, but that will also reduce the voltage produced by the inductor so the spark plug is unlikely to fire.

The only solution I've come up with is to use a PPTC resettable fuse to block high currents (as anything greater than (5 V/1 Ohm =) 5 A would be unusual in the above circuit), but that would also rely on the fuse being rated for very high voltages. Are there any better ways of doing this? How do cars do this without having a huge potential difference between the positive terminal of the battery and the positive terminal of the inductor in the ignition coil?

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    \$\begingroup\$ I'm not sure why you want to avoid using a transformer. Using a transformer would allow you to use lower voltage (but higher current) components for switching. \$\endgroup\$ Nov 13, 2020 at 1:05
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    \$\begingroup\$ falstad also has a spark gap element under "passive components" \$\endgroup\$
    – BeB00
    Nov 13, 2020 at 1:13
  • \$\begingroup\$ @MathKeepsMeBusy There's no particular reason to not use a transformer (except maybe space). The biggest issue I'm facing is that the primary coil on the transformer reacts in the exact same way as the inductor does so it doesn't seem to make a difference. Or are you saying that a transformer could allow the low-V side to have a low-V spike (that the power supply/switch can hopefully handle), and isolate the high-V spike on the high-V side? \$\endgroup\$
    – JolonB
    Nov 13, 2020 at 1:16
  • \$\begingroup\$ @JolonB. Yes, exactly. You switch the current on the low-V side of the transformer, and use the high-V side, for whatever you need high voltage spikes for. For example, if you had a 1:20 transformer, instead of your switch needing to withstand 7KV, it would only need to withstand 7000/20 = 350V. If you can get an even higher transformer ratio, your switch's max V requirements would be reduced accordingly. \$\endgroup\$ Nov 13, 2020 at 1:22
  • \$\begingroup\$ A flyback "transformer" is more of a coupled inductor (usually with a gapped core) than a transformer. But the output voltage will be reflected back to the input winding (or tap) by the turns ratio as in normal transformer action. That allows a few hundred volt switch to control thousands of volts output voltage. \$\endgroup\$ Nov 13, 2020 at 18:40

2 Answers 2

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How can I use the voltage generated from inductive kickback?

The current flowing through an inductor, upon closure of switch contacts, would result in energy being stored in its magnetic field. Disruption of the current, caused by the switch contacts opening, would lead to a collapse of the magnetic field and consequent induction of a high voltage that would spark across the open contacts to transfer energy back to the source.

Should that inductor form the primary of an ignition coil with a turns ratio of 1:100, the voltage of around 200 V induced by it would be stepped up by the secondary to around 20 kV and fed to the spark plug via a high tension cable.

enter image description here

A capacitor across the contacts would provide a path for the induced current to to weaken the spark and minimizing pitting.

In a capacitor discharge ignition system, a capacitor, charged to a high voltage would be discharged through the primary of the ignition coil to generate extra high voltage at the secondary.

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  • \$\begingroup\$ It seems that most vehicle ignition systems (that use an ignition coil) also use a ballast resistor. Is there any reason that's not included in your diagram? What's the purpose of the switch connected to the battery? Surely there will be sparks over that one too \$\endgroup\$
    – JolonB
    Nov 16, 2020 at 2:09
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    \$\begingroup\$ Hi JolonB, A ballast resistor would be used with an ignition coil, designed to work at the lowered battery voltage available during starting. It would be bypassed during starting and connected in series during normal running. I have left it out for simplicity. The ignition switch, shown in series with the battery, would be turned on during engine start and remain on during running. It would spark, while being opened to turn off the engine. \$\endgroup\$
    – vu2nan
    Nov 16, 2020 at 3:51
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Are there any better ways of doing this?

You use a step up transformer.

If the step-up ratio is (say) 50:1 and the output voltage is (say) 10 kV before it sparked, then the primary voltage back emf is: -

$$\dfrac{10,000}{50} = 200\text{ volts}$$

Note that when the secondary output produces a 10 kV spark it causes a primary back-emf of 10 kV divided by the reverse turns ratio i.e. it then operates as a step-down transformer.

That is how flyback transformers can generate big voltages without an excessive back-emf on the primary.

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  • \$\begingroup\$ This is also how millions of small engines drive their spark plugs : the kick on the "low voltage" primary is limited to a few hundred V by the secondary circuit's weak point (hopefully the plug!) breaking down at 10s of kV. The only difference is that the primary is driven by a couple of magnets flying past the gap, right when the contacts open, rather than a battery \$\endgroup\$ Nov 13, 2020 at 13:45

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