Electronic – Understanding the slayer exciter circuit base voltage


I am quite new to electronics and I have a bit of trouble understanding part of the slayer exciter circuit.

If we take this schematic from electroboom here:

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I don't understand how the secondary would send a negative voltage to Q1. From the flow of current through the primary, it would seem to me like it would actually send positive current; I don't see at what point Q1 is interrupted.

How does the secondary create a negative voltage?

Best Answer

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Figure 1.

  • The main trick to understanding this is to think of capacitors as voltage maintainers - a bit like a short-term battery.
  • The parasitic capacitance of the coil to ground at (1) tends to hold (1) at a constant voltage.
  • When voltage is induced in the secondary the voltage across the coil rises.
  • If the top of the coil - the positive end - is held at constant voltage then the voltage at the bottom of the coil will be driven in a negative direction. This will continue until (2) reaches about -0.7 V at which point D turns on.
  • D has a much lower impedance than C does so now (2) is held at -0.7 V.
  • The still-increasing coil voltage therefore drives (1) more positive.

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Figure 2. Slayer Exciter simulation done by LTSPICE. Source: ElectroBoom which explains the operation in further detail. (Click image for full resolution view.)