# 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:

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?

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.

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

• So basically, the coil voltage goes down because the capacitor doesn't want to go up... I am no expert about transformers, but it means that in this case, current will be induced in the secondary in a way opposite to the one in the primary? – Sixela963 Sep 28 '17 at 18:29
• I listened to the video and also thought a little about the circuit. $C_1$ as shown is not the parasitic capacitance of the $L_2$ coil. If it were, it would be effectively across the two leads of $L_2$ (parallel.) However, as wired, it's between one end and a different node (ground.) So this must be the trace capacitance between $L_2$ and that ground part of the circuit, which is a different kind of trace capacitance. There will also be a trace capacitance across $L_2$, but that's not shown. Makes me think that protoboard might be "required." I'd vary the distance to test. – jonk Sep 28 '17 at 19:00
• It's capacitance to the power supply, if you use a wall wart it's capacitance to ground (return path via the Y capacitor and supply wires) if you use a battery it's capacitance to the battery – Jasen Apr 28 '19 at 1:59

The secondary doesn't send a negative voltage to Q1, but a negative voltage can build up through the current demanded by the secondary winding.

Initially there is no current anywhere. On power-up current flows from the battery through R into the Base-Emitter of Q. This means that Q can conduct from Collector to Emitter. Current and voltage result in a magnetic field building up in the transformer. The secondary is subjected to this field and will try to counteract it ("inductors do not like changes to their magnetic field").

As indicated by the polarity dots, the secondary will need current from the base node of Q and it will try to charge the parasitic capacitor. As Q can not provide this current, it must come through R or through D (and it is not likely to be provided by a parasitic because R and D are easier to get the current from). As the base voltage is higher than ground, the current will come through R, and it will therefore start to lower the base voltage. At some point the current in the primary of T can no longer increase because BE will decrease. If the secondary demands current beyond $V_{batt}/R$, the diode D will start to conduct and provide most of the extra current.

This is where the voltage on the base of Q can be negative.

As Q is now "closed", the current in the primary must either be 0 or a voltage might be building up in the primary to keep it running (there is no "flyback" on the primary of T). I'll suppose the current has ceased.

Now the voltage that has been built up over the secondary will (have) start(ed) reversing the current in the secondary. Therefore the current in the primary will also be reversed - it can flow through the BC diode of the transistor into the resistor or through the battery and the diode if the voltage is high enough and/or the secondary winding.

The voltage BE will be positive again, and it all starts over.

In the schematic current into BE is limited by the resistor at first, so that current in the primary will be limited also at first, but current coming from the secondary coil is not limited the same way - it is limited by the energy that was "charged" in the secondary winding (and "parasitic").

If the secondary winding can not conduct, then the primary winding should see its current rise up to $\beta$ of Q times the base current limited by R. Voltage over the primary will drop to 0 as there is no longer a current change needed to sustain a voltage.

Anyway, I hope you can see how the voltage on the base can be negative.

• Not only did you make a lot of things clearer, you also forced me to search about transformer polarity. I didn't even notice the dots before... now I understand the circuit. Thanks! – Sixela963 Sep 29 '17 at 16:23

If you really want to know more details about it, please refer to this patent. There are details of the principle about Slayer Exciter.

Replace the transformers in this figure with Tesla coils. The GND side terminal of the secondary winding of the transformer is the detection circuit of the current phase. It can be understood that if the current phase is directly used as the switching signal of the switching means (transistor of Slayer Exciter), the principle is exactly the same as this figure. The circuit of Slayer Exciter is very simple, but it is necessary to frequency analyze of circuit simulation to understand its function. Please look carefully at Figures 11 through 16 and their associated descriptions in this patent. If you understand them, you can understand the principle of Slayer Exciter. This patent has been declared free use for amateurs and enthusiasts who study Tesla Coil.

The voltage induced at the transistor base when the transistor conducts is -positive-, as the dots clearly indicate. Most of the descriptions of this circuit on the internet are incorrect.