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Maybe my first question Dc-Dc converter from 12v Input 400v Output was too broad for what I was actually trying to solve, which is why I am reformating my question.

I want to charge a capacitor (C=0.68μF) at 400v for a CDI (Capacitor Discharge Ignition) System. I have thought to use a dc-dc flyback converter going from 12v, can also be 24v, to 400v. The discharge will be happening when I press a trigger button, which is why I am thinking of charge current, output current for the dc-dc converter, of 10mA.

I have not been able to come with the correct design for this project. Could anyone give me some tips or hints on how to solve this?

Edit: The idea, for now, was using this transformer: Myrra 94094. But thanks to some of your answers, it doesn't look to be the best idea.

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  • \$\begingroup\$ Link to datasheet please. What makes you think the transformer is limited to 265 V output? \$\endgroup\$ – winny May 28 '19 at 8:40
  • \$\begingroup\$ @winny I am thinking of using the transformer with the 12v as Primary side and the 265v as Secondary side, kind of as a Step-up transformer. \$\endgroup\$ – Sebe May 28 '19 at 8:48
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    \$\begingroup\$ However, it just elevates the voltage to 265V maximum No it does not.The 265V is listed in the datasheet as the maximum input voltage in the example usage case. It is not a rating or guarantee. You intend to use this transformer "in reverse". Nothing is said about using it like that in the datasheet. If you magnetically charge this transformer (using the 12 V side if you like) and then abruptly cut the charging current, you will get a voltage spike at the high voltage winding of the transformer. That is a property of all transformers. You can get more than 265 V. \$\endgroup\$ – Bimpelrekkie May 28 '19 at 8:55
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    \$\begingroup\$ I could obtain the 400v out of it? I think so, I see no reason why that would not be possible. Would it not damage the transformer in any way? No, it should not. However, there is no guarantee as your usage case is not what this transformer is designed for. So you will just have to try and see. But usually the highest voltage across a winding is limited by the isolation and the minimum distance between the contacts. These should be more than capable to handle 400 V on this transformer. \$\endgroup\$ – Bimpelrekkie May 28 '19 at 9:21
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    \$\begingroup\$ My recommendation would be to use a COTS for this. Next step above that is a flyback with custom transformer. \$\endgroup\$ – winny May 28 '19 at 10:10
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Flyback transformers are best viewed as two inductors, wound on the same core. The primary winding charges up the core, and the secondary dumps the charge into the load. Using the data sheet on your link, we can go through some rough calculations for this transformer.

enter image description here

The winding ratio is 135:9, and inductance is proportional to the square of the turns for a given core. Since these are wound on the same core, the ratio of inductance is 18225:81. Since the 135-turn coil has an inductance of 2100 uH, the 9-turn coil on the same core can be approximated at 9.3 uH.

Your load is 450 volts at 0.1 amps, or 4.5 watts. One watt is one joule/second, and this transformer appears to be happy at 132Khz, so neglecting losses, each time the transformer switches, it must pass 4.5 watts/132,000 or 34 microjoules to the load.

Since the energy stored in the 9.3-uH coil is (1/2)Li^2, this means that the 9.3-uH coil must reach a peak current of 2.7 amps each time it switches in order to have the 34 microjoules to dump into the load.

So, the transformer coil, rated at 1.5A, will likely either heat up or saturate. You can apply this approach to other transformers you find. You can see that you will have to add switching losses and the diode drop effect, but this will get you in the ball park.

Good Luck!

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  • \$\begingroup\$ Really good explanation! Thanks a lot for the time to help :) If you could maybe check the question again, I had to edit it. \$\endgroup\$ – Sebe May 29 '19 at 10:35

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