I've started studying power electronics and am currently designing a 45W flyback converter with a 15V, 3A output. The input voltage can vary widely, from 20V to 120V. I'm following classical flyback design guides, but I'm struggling to reduce the secondary peak current — it's excessively high. How can I lower the secondary peak current? Should I consider using a different flyback topology, such as a two-switch flyback, or are there other approaches within the flyback category? I'm required to design it as a flyback converter, so alternatives like forward or push-pull are not an option. Any help would be greatly appreciated.
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2\$\begingroup\$ It's a numbers game so please provide the numbers. Operating frequency, primary inductance, turns ratio, calculated duty cycle, load current minimum etc.. \$\endgroup\$Andy aka– Andy aka2025-04-10 12:58:55 +00:00Commented Apr 10 at 12:58
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\$\begingroup\$ Do you have to be in DCM? How much can you increase the inductance before you enter BCM? The 20-120 V input would be the culprit of a flyback converter at this power level. Not that it can't be done, but you need to pay for such a wide range on other parameter(s). \$\endgroup\$winny– winny2025-04-10 13:04:12 +00:00Commented Apr 10 at 13:04
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\$\begingroup\$ Vin may vary between 20v and 120v, Vout=15v Ioutmax=3A, Ioutmin=2.4A, switching frequency is 100kHz, output voltage ripple is %1, I assumed an efficiency of %88, turns ratio = 6:1 , max duty cycle is 0.823, Pinmax=51.136w, Iprimarypeak=6.22A , Lpri=26.46uH, transformer core was chosen to be EE30/15/7 , Bmax=0.3T , Np=12, Ns=2, Iprimaryrms=3.258A, Isecondarypeak=37.32A, Isecondaryrms=9.065A I have tried decreasing turns ratio but secondary peak is still too high I dont have to be in DCM but it is 45w,low power so i tought DCM would be better. @Andyaka \$\endgroup\$Ali Rasim Kuş– Ali Rasim Kuş2025-04-10 13:31:42 +00:00Commented Apr 10 at 13:31
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\$\begingroup\$ I would also like to ask, when you say "it's excessively high", what is making you say that other than just looking at the value? DCM vs CCM? \$\endgroup\$RogerDodger– RogerDodger2025-04-10 15:43:49 +00:00Commented Apr 10 at 15:43
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That sounds about right for a DCM design with such a low primary inductance. In my own flyback modeller I get around the same: -
I see a secondary peak of 35 amps at a duty cycle of 77% approximately so, it's not far off (37.32 amps and 82.3% are your figures).
It's the average secondary current that feeds the load and, the average of the high peak value (35 amps) multiplied by 1.7179 μs and divided by 10 μs is 3 amps. Yes these are my figures from my "ideal" calculator but they are reliable. Your peak current is a tad higher and your duty is a tad longer but, it's well-within the same ballpark.
If you want a lower peak output current you need to redesign it. More to come....
- Begin with the turns ratio and adjust it to 3:1. This drops the peak secondary current to 50% (17.5 amps) and puts you in CCM with a duty of ~70%.
- Also consider increasing the primary inductance to 100 μH. This reduces peak secondary current to ~12 amps also at a duty of ~70%.
I dont have to be in DCM but it is 45w,low power so i tought DCM would be better. @Andyaka
CCM will have a lower peak secondary current. The above design will drop into DCM at the higher input voltages. You could also reduce the turns ratio to 2:1 and get further improvements in peak secondary current (~8 amps at a duty of 60%). It remains in CCM at the highest input voltage too.
However, it's fine line between transformer design and output peak current. You need more inductance to make inroads and that can cause the core to saturate but, the big take-away should be operating with a 3:1 and not 6:1 turns ratio.
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\$\begingroup\$ Firstly Thank you Andy. Secondly How should I choose the rectifier diode? Should I consider the peak secondary current and choose a diode rated 37A or Secondary rms current and choose a diode rated 9A. And I will redesign it with reduced turns ratio and higher inductance. @Andyaka \$\endgroup\$Ali Rasim Kuş– Ali Rasim Kuş2025-04-11 03:58:52 +00:00Commented Apr 11 at 3:58
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\$\begingroup\$ @AliRasimKuş diodes have a peak rating an an average/RMS rating so choose the diode that meets both criteria. Good luck with your design and, if we are done here, please take note of this: What should I do when someone answers my question. If you are still confused about something then leave a comment to request further clarification. \$\endgroup\$Andy aka– Andy aka2025-04-11 07:54:20 +00:00Commented Apr 11 at 7:54