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I am generating 500V from a 24V source. Attaching schematics. This circuit is generating stable 500V.

I am attaching a load of 7 kohm. In the schematics the buck regulator is supplying 19V to the transformer. What happens is at no load, the current sunk from buck converter is 0.6Amp, but at loaded condition the current becomes 3.2Amp and voltage of buck convertor drops to 2V and still the 500V output remains stable.

I need to limit the current because I am not allowed to sink 3.2Amp.

Please have a look at the schematics and suggest if there is any technique to reduce losses of this circuit?

enter image description here

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  • \$\begingroup\$ Peak current limitation should work as long as input and output voltage remains the same. Good enough or do you need to limit the inrush behavior too? Do you softstart? \$\endgroup\$ – winny Apr 3 at 16:34
  • \$\begingroup\$ Is this a thermal issue from high RdsOn or a supply issue limiting current? Or ? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Apr 3 at 16:45
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    \$\begingroup\$ What exactly is the buck converter doing for you? Why can't you run the transformer directly from 24V? \$\endgroup\$ – Dave Tweed Apr 3 at 17:17
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    \$\begingroup\$ Hang on- You have a load of 7K on the 500V, which is about 36W. At the input of the flyback (say you have 85% efficiency, could be less) you will need 42W. If the buck is 90% efficient your 24V input will have to supply 46.7W. At 24V that's <2A. If you're drawing 3.2A then your efficiency is way off. Are you sure you're not saturating your transformer? (Coupled inductor actually). And as @DaveTweed said, why do you need the buck converter? \$\endgroup\$ – John D Apr 3 at 17:19
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    \$\begingroup\$ Why use an MCU instead of an SMPS chip? \$\endgroup\$ – AltAir Apr 3 at 18:10
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You need to take a look at what you are expecting.

You want about 35 W out at 500V. That's 500V into a load of 7000 ohms.

Given 3.2 A at 24V, you are putting 77W in to get 35 W out. Your efficiency is about 58%.

Your smaller power supply can 2.1A * 24V, so that's 50W going in.

But, wait. You are reducing that to 19V so you've got closer to 40W going through the transformer.

Assume 45W to the transformer. You would need an efficiency of at least 80% in the boost section to get your desired output.

That 80% needed efficiency is a long ways from the 58% you are getting.

Why are you introducing losses into the system by reducing the voltage going into the transformer?

Limiting the current is the wrong solution.

You need to be more efficient.

  1. Lose the buck converter, drive the transformer directly from the 24V source.

  2. Check the signal driving the FET. If it has a slow rise time, you will lose power.

  3. Ditch the 7805, and use a buck converter to power the processor. The 7805 is throwing away power that could be going to the booster. It may only be a couple of watts, but you seem to be pretty close to the edge with your power requirements.

  4. Use a better FET. Check the Rdson for 5V, and see if there are FETs available with a lower "on" resistance.

  5. Your "schematic" shows no resistor between the processor and the FET. That might cause ringing on the gate, which will also reduce your power output.


I'm no expert on switching power supplies - I have never built one.

The pointers above are based on the other questions about switching power supplies that I have read on this site.

If I have gotten something wrong, I hope someone will point out my mistakes.

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  • \$\begingroup\$ Thanks for detailed suggestions, let me look into these points except 1. I'll update results here. \$\endgroup\$ – Sidk Apr 3 at 17:50
  • \$\begingroup\$ Power the processor temporarily from another source until you get your booster working. That means: lose that buck converter. Improve your booster until it no longer clobbers your 24V supply. At that point you can go back to powering the processor from the 24V. \$\endgroup\$ – JRE Apr 3 at 18:03
  • \$\begingroup\$ Also, using a 5V output buck converter to power to processor gives you more "working space." The 7805 requires around 7.5V in to provide 5V out. A buck converter can deliver 5V out from much closer to 5V in. \$\endgroup\$ – JRE Apr 3 at 18:05
  • \$\begingroup\$ @AltAir I am switching at 60KHz varying duty cycle from 5% to 50%. \$\endgroup\$ – Sidk Apr 4 at 13:06
  • \$\begingroup\$ Sorry I missed t mention in schematics I use TC4427V Gate Driver IC to drive mosfet with resistor on gate. \$\endgroup\$ – Sidk Apr 4 at 13:06

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