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I'm studying an interesting circuitry that generate a high voltage pulses to prime a DC/DC isolated regulator: https://www.vicorpower.com/documents/application_notes/an_HV_ChiP_BCM_Reverse_Start_Up_Circuit.pdf

I'm stock with the calculation of the Flyback output voltage because I can't make it match the AN.

enter image description here

The application note states :

  • The switching frequency is 100 kHz (p3)
  • Flyback operate in discontinuous conduction mode (p3)
  • Turn ratio of the transformer is Np:26 and Ns:151 (Wurth 760871135)
  • Comparator switch MOSFET Q1 when VFlybk_sec - (Vr1 + Vd1) is 393V (p7)

I used the calculation depicted here: Transformer turn ratio in DC-DC flyback converter for high voltage output

enter image description here

  • Rl = 400 kΩ
  • D = 0.5 (50%)
  • Vin = 8 VDC
  • Lp= 1250 µH
  • Fsw= 100 kHz

And I get Vout = 180 160 V which is much less than the >393 V expected.

Where is my mistake?

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    \$\begingroup\$ I calculate it as 160 volts so, there's a mistake in your math. \$\endgroup\$
    – Andy aka
    Commented Sep 19 at 11:14
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    \$\begingroup\$ @Andyaka The voltage divider on the secondary side is two series resistance in series, which are approximately 400k (392k + 1.21k) Even with 160volts, it still far from the >393V expected. \$\endgroup\$
    – St7ven
    Commented Sep 19 at 12:16
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    \$\begingroup\$ @andyaka It's not my schematics. It's an application note from VICOR Power. I'm trying to understand why our calculation don't match the application note. They gave 393V for the output. I did not understand this part of your answer : "If you downvoted the answer you linked, I would consider revising that action please." What answer ? did not downvoted anything. \$\endgroup\$
    – St7ven
    Commented Sep 19 at 14:03
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    \$\begingroup\$ The resistance can be found on the bottom right : R4 and R5. I think the schematic is not that bad. \$\endgroup\$
    – St7ven
    Commented Sep 19 at 14:09
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    \$\begingroup\$ OK, the link in your question was to a previous answer I gave and, it received a downvote since your wrote your post. If it wasn't you then you have nothing to answer. I'm glad it wasn't you. \$\endgroup\$
    – Andy aka
    Commented Sep 19 at 14:43

2 Answers 2

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@Andyaka The voltage divider on the secondary side is two series resistance in series, which are approximately 400k

A load of 400 kΩ on the output of your flyback converter (producing 160 volts), has to dissipate 0.064 watts. That is inevitable. Then, if you calculate how much energy the primary inductor has to store to service that load, you'll find that it's 640 nJ per switching cycle. That's 0.064 watts divided by 100 kHz.

Given that the primary inductance is a whopping 1250 μH, that's a peak current of 32 mA.

How long does it take for the inductor (1250 μH) current to rise to 32 mA on an 8 volt supply? Use \$V = L\frac{di}{dt}\$ rearranged to get 5 μs.

This is exactly the duty cycle of 50% that you state. In other words there's nothing wrong here except your knowledge about flyback converters operating in DCM.

If you want to get over 300 volts you need to seriously consider dropping the primary inductance to around 330 μH or running at a duty cycle of over 97%.

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    \$\begingroup\$ Thank you for the very informative answer. Yes, I took this application note from VICOR to better understand flyback and my lack of knowledge on the topic. I though if I could match my calculation with the Application Note, then I could say that I did understand, but unfortunately, it don't match. In conclusion, would you say their application note is wrong ? Or I miss understood it ? \$\endgroup\$
    – St7ven
    Commented Sep 19 at 14:07
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    \$\begingroup\$ But, I would certainly consider revising the transformer you used. If it was recommended in the app note then they are certainly pushing things too far. \$\endgroup\$
    – Andy aka
    Commented Sep 19 at 14:45
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    \$\begingroup\$ Aha @St7ven you are operating the transformer in reverse hence, the "new" primary inductance is only \$(\frac{28}{136})^2\$ of 1250 uH i.e. 53 uH. I can see that in your now deleted answer. You would be able to get possibly over 700 volts out now at D=0.5. \$\endgroup\$
    – Andy aka
    Commented Sep 19 at 15:07
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    \$\begingroup\$ I read 15 and 13 turns in series @St7ven \$\endgroup\$
    – Andy aka
    Commented Sep 19 at 19:45
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    \$\begingroup\$ @St7ven I would expect that the ratio of primary leakage to total primary inductance will be the same as the ratio of secondary leakage to total secondary inductance. That means the leakages are proportional to the turns ratio squared. Yes, using the term Ls is confusing. \$\endgroup\$
    – Andy aka
    Commented Sep 26 at 11:30
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I have also run a simulation on LTSPICE

enter image description here

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