I am looking to assemble a power supply with an output of roughly 600 watts. One of the components required for such an apparatus is a 40 kVDC, 20 mA flyback transformer that I managed to find and hack in order to work for this specific purpose. I noticed that in this process, many people were calling for a transistor based circuit that seemed to be for transformer feedback. This was the circuit that I was considering:

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Thus, I was first of all curious if such a circuit would be required for the flyback. If so, I was also curious if the shown parts and their descriptions would be capable of withstanding an input of 60 VDC and 10 A. Thank you.

  • \$\begingroup\$ How about 65kV boginjr.com/electronics/hv/flyback-driver-1 \$\endgroup\$ – Sunnyskyguy EE75 Aug 9 '18 at 1:42
  • \$\begingroup\$ I'm not sure about the power input/output of this device, although it looks pretty low. Also, I am personally aiming for a transformer and driver set with an output of 20-40 kV, as I am sending it through a C-W multiplier. \$\endgroup\$ – Super Nerds Team Aug 9 '18 at 1:49
  • \$\begingroup\$ @TonyEErocketscientist: perhaps I am wrong, but it seems to me that this “flyback” circuit is a BJT blocking oscillator in disguise. If it is not so, what are the differences? \$\endgroup\$ – Daniele Tampieri Aug 9 '18 at 7:09
  • \$\begingroup\$ The flyback driver circuit and your described oscillator circuit work in the same way and perform similar tasks. for the flyback the circuit basically works by providing a given frequency of DC to the flyback and uses the feedback loop to determine what the optimal current and frequency is for the flyback. \$\endgroup\$ – Super Nerds Team Aug 9 '18 at 12:42

Look at the data sheet for the 2N3055 and note that the maximum power dissipation is ~115 W with an excellent heat sink, and that the maximum current rating is 15 A. 12 VDC at 15 A could theoretically produce 180 W out, but that would be for a "perfect" switching device with no dissipation. This circuit does not use the most efficient switching mode, so considerable power is wasted in heating the transistor.

That said, the circuit you show, apparently from Instructables, should work, actually producing a few thousand volts at a milliamp or two... far less than your suggested 600 W. This is a good thing, as one could get serious RF burns from a 600 W RF supply.

To get higher voltage, bypass the resistors with capacitors, as shown at PocketMagic. To get yet more voltage, the primary of the flyback transformer should resonate with the secondary, using an experimentally-determined capacitor.

If you make it, use a larger heat sink than that shown -- perhaps a heavy aluminum fry pan, with bottom filed flat. Use thermal compound, as well. In any case, the transistor will likely fail, particularly if you use more than 12 VDC. If it doesn't fry from overheating, high-voltage transients on the primary will likely destroy it. Still, this is a simple circuit, provides a learning opportunity, and is fun as long as it lasts.

More power can be output by this two-transistor push-pull circuit from RimStar, or the more complex one at Silicon Junction. The LM555 circuit could easily be made with frequency variable to determine if tuning affects spark length and current. BTW, a rough estimate of output voltage: between rounded conductors, ~1 kV/mm for the intial spark length.

  • \$\begingroup\$ Greetings, would there by any chance be a driver that would be able to last longer without putting components such as the transistors at risk? \$\endgroup\$ – Super Nerds Team Aug 9 '18 at 1:51
  • \$\begingroup\$ Sorry if I seemed too negative (no pun intended). The circuit is worth trying but may not last long. Alternatives have been added. In your circuit, using Zeners across B-C and B-E will help prevent high-voltage damage. \$\endgroup\$ – DrMoishe Pippik Aug 9 '18 at 2:19
  • \$\begingroup\$ Your local computer store probably often has 50-200W heat sinks on sale. Ones for older sockets falling into disuse can be quite cheap if you decide to go this way with things. \$\endgroup\$ – K H Aug 9 '18 at 4:04

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