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First of all I'm fairly new to making a more complicated circuit.

I'm trying to make a PCB with an ATmega328PB chip on it. With that chip I want to control a high voltage pulse generator (arc lighter.)

Now I found the 15kV high voltage pulse generator that I want to use because of its size.

15kV arc lighter

I also found a schematic that can control this high voltage pulse generator.

Schematic of high pulse generator

enter image description here

I tried to make the same schematic for this forum but I'm not sure it is correct so please spare me on that.

This schematic is triggered by completing the power circuit.

Now I need some way to get this triggered by a high/low signal from the ATmega328PB (a digital signal if I'm correct.)

schematic

simulate this circuit – Schematic created using CircuitLab

Things I've thought of:

Adding a second MOSFET to the ground line.

schematic

simulate this circuit

Both schematics work that is the first thing. But with the second schematic, once it is turned on it stays on. That is one thing I don't need. I've tried to do it with and without the pull down resistor but nothing works.

I'm sorry if some of the components in the schematic aren't right because the voltage or current they can handle but that is not what I'm trying to ask.

The only thing I want to know is how to control the on and off function of the arc lighter.

Thanks in advance if you need more info or I did something totally wrong please tell me because I need to learn a lot.

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3 Answers 3

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I assume you have drawn the schematic incorrectly since you said that this works, but what you have drawn can never work.

In your first schematic the polarity for the N-Chan IRF530 FET shown is definitely incorrect.

Lets start by correcting the schematic, I think it should look like this when you use a FET:

schematic

simulate this circuit – Schematic created using CircuitLab

There are a couple of problems with the circuit above:

  1. The IRF530 is only rated to 100 V and there is a likelihood of the back EMF generated to hit 5-600 V. If the output breaks over then it would reduce the back EMF generated, but if you move the output terminals far enough apart to prevent breakover, the voltage developed across the FET could rise considerably above 500 V. If you say you have this working, I must assume that the IRF530 is entering avalanche but it seems likely that it would fail eventually.
  2. The VGS(th) for the IRF530 is 4 V maximum, that means there is a likelihood that some devices would not start oscillating at 4.2 V ….and even more in the range of devices may not work down to 3. 6 V (If you are using a LiIon battery).
    If you are using an MCU that works at 5 V you are in a better position and the IRF530 would reliably start with a 5 V supply.

A more suitable device for this application may be the Infineon IPA50R380CE. This device supports a minimum of 500 V across the drain/source and has reasonable avalanche capability. This device would start oscillation down to about 3.5 V so would work for a LiIon battery implementation as well as a 5V implementation.

Providing digital control

To enable an MCU to turn the high voltage generator on/off you could take a couple of approaches:

  1. Switch the power supply on/off (which is what you tried to show in your second schematic, but what you showed would not work). This would require a P-Chan FET to turn the supply off/on.
  2. Clamp the feedback so that the generator cannot start.

I've chosen to use #2 and updated the schematic to show this approach:

schematic

simulate this circuit

Note: This circuit appears to have been originally designed for use with a BJT, and the changes to use a FET complicate the circuit somewhat (since there is no equivalent to the base current). I give no gaurantee that the schematic shown above will work for you, and you'd need lots of details on the transfomer to check exactly what is needed.

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  • \$\begingroup\$ D5 looks to be redundant \$\endgroup\$ Nov 27, 2022 at 21:28
  • \$\begingroup\$ @eugene-bright D5 is NOT redundant. Since the M1 gate is driven by a transformer it is possible to overdrive the gate. D5 prevents that. You could make D6 a Zener and eliminate D5, but the cost would be higher IMO. \$\endgroup\$ Dec 21, 2022 at 19:16
  • \$\begingroup\$ If D6 is not zenner, then one have to apply series of diodes in place of D5 to get the appropriate voltage drop and in effect the supply voltage will be fixed. D6 zenner is more idiomatic \$\endgroup\$ Dec 22, 2022 at 20:58
  • \$\begingroup\$ @eugene-bright The circuit as shown is correct. D6 clips negative excursions and D5 clips positive excursions current limited by R1. The 5V supply will typically have some capacitance though it could be a simple rechargeable battery and still work. Tell me what you think does not work? \$\endgroup\$ Dec 24, 2022 at 19:20
  • \$\begingroup\$ Everything works. Unless you use 20V power source. In this case the gate will be broken down \$\endgroup\$ Dec 26, 2022 at 11:02
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The schematic shows a BJT transistor that has much lower output capacitance compared to FETs, which speeds up the L/R rise time of the positive feedback Oscillator resonant frequency and thus rise time and thus voltage across V=LdI/dt.

Stray capacitance also affects the BDV as does air quality but ignition should be reliable.

(Adding capacitance reduces this by omega=(LC)^-1=2Pi*f. J which you want to avoid using FETs. I would choose a superBeta power transistor, hFE>=500 from Diodes INC.

The inversion of coil primary with inversion of transistor current makes this a simple reliable Oscillator with a step up output voltage ratio. Use about 1kV/mm for wire tip gap voltage measurements for start breakdown so 15kV=15mm roughly for a sharp wire tips.

So stick to the schematic and choice of power transistor they use.

FETs turn on very fast (ns) but the HV action here needs a very fast turn OFF to get high LdI/dt=V and your Coss spec on FET reduces this. ( to micro-s)

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    \$\begingroup\$ ”The schematic shows a transistor because they have low output capacitance compared to FETs” FETs are transistors. \$\endgroup\$
    – winny
    Jul 14, 2019 at 16:32
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    \$\begingroup\$ @winny ok ok a “bipolar junction” transistor. TY for the obvious reminder \$\endgroup\$ Jul 14, 2019 at 16:33
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!High Voltage Peak to Peak750]1

enter image description here

You can also use Mosfet IRF 520N. Pulse amplitude has periodic fluctuation. The switch 2 in the gate circuit can be a push button or one can use a Reed relay contact or any digital control via optocoupler.

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    \$\begingroup\$ Is there meant to be one image or two? Edit your answer and provide an explanation. \$\endgroup\$ Aug 17, 2020 at 11:37

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