The purpose of the circuit below is to produce an extremely high voltage impulse over the inductor so that I can place a spark gap in parallel to it and create a very large spark. According to simulation this should work as designed, but I am concerned about protecting the capacitor. The highest voltage rated cap I could find was 30k, but the simulation predicts spikes of -220 kV. I tried placing a diode in parallel to the inductor, but that drastically reduces the voltage spikes. The source controlling the switch is just a simple pulse function switching every 1.2 seconds.


Impulse fn

  • \$\begingroup\$ You don't need a capacitor, just use the back emf. You will need to take care that the relay (or whatever else you have) can take the voltage. The contacts don't just jump to the open position, they take the whole way from contact to open, which means there will be an air gap that will have a finite amount of resistance, and that resistance is variable. \$\endgroup\$ Commented May 14, 2021 at 15:52
  • \$\begingroup\$ You could put capacitors in series, with balancing resistors to share the voltage across them. \$\endgroup\$ Commented May 14, 2021 at 15:52
  • \$\begingroup\$ You won't get 220kV. To accurately simulate this you need to ensure the parasitic capacitance of the components, especially the inductor are included. The voltage in any practical circuit will be much, much less. It will also be limited by breakdown in the switch. \$\endgroup\$ Commented May 14, 2021 at 16:43
  • \$\begingroup\$ If using the cap to charge the inductor (with voltage) to make a spark, instead think in terms of charging the inductor (with current), then interrupting that current. For that is what creates the spark (the near-instantaneous collapse of the inductor's magnetic field.) \$\endgroup\$
    – rdtsc
    Commented May 14, 2021 at 16:49
  • \$\begingroup\$ The simplest solution is a dry contact relay switch to an energized L with low Parasitic C and rapid velocity to delay ionization breakdown at 3kV/mm. ESR and DCR with switch and wiring R’s will limit current. But a spark plug between two step up transformers will create the Tesla corona \$\endgroup\$ Commented May 14, 2021 at 17:39

1 Answer 1


To answer your question: This is not possible, the circuit won't work in reality. As the commenters have already pointed out, there is no switching device (electronic or mechanical) that can handle such high voltages and dV/dt.

To solve your problem: You should instead try to create a transformer-based circuit. A flyback transformer is very well suited for the task of generating long sparks (I've done this a lot as a child).

Your flyback transformer should have a large turns ratio (i.e. 1:50), a ferrite or iron core and an air gap to increase the maximum energy that can be stored in the core. Then you can build a circuit that charges the transformer by applying a voltage to its primary winding. You can do this with an N-channel MOSFET switching to ground. While the transformer is charging, you have to measure the current through it with a shunt resistor. Once the current reaches a certain value (i.e. the saturation current of the transformer's primary), you switch off the transistor. Now the transformer momentarily generates a very high voltage at its secondary winding (which will create a spark). There will also be a voltage spike at the input (but a much smaller one), which is why you should choose a MOSFET that can handle high voltages (i.e. a 600V MOSFET).

Measuring the primary current and switching off the MOSFET has to be done electronically since this has to happen within a few microseconds. You might be able to use a simple NE555 timer for this task. The MOSFET also has to be switched off as fast as possible in order to create the largest possible output voltage.

You also have to be very careful when winding your transformer and use LOTS of insulation, otherwise the spark will form within the transformer and destroy it. (I do have a lot of experience with destroying transformers this way, unfortunately...)


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