Consequences of higher magnetising inductance in flyback converter

I have a design where an ETD44 ferrite flyback transformer is used with a switching frequency of 50 kHz to step up voltage. The rated power is 250 W. The magnetising inductance of the transformer is specified as 14 uH. I found a commercially available transformer (Coilcraft JA4823-CL) which has 28 uH at 150 kHz but has a (smaller) RM14 core. I will be operating in DCM/BCM and only need 200 W output power.

As a flyback works by storing energy and then releasing it again, is my assumption correct that this transformer could store more energy due to the higher inductance and theoretically allow a higher output power? As I don't need such a high output power, I assume the smaller core would work and would cause less core losses?

As a flyback works by storing energy and then releasing it again, is my assumption correct that this transformer could store more energy due to the higher inductance and theoretically allow a higher output power?

• The primary inductor stores only so-much energy in a given time.
• That time is determined by the duty cycle and the operating frequency.

So, when you connect the primary to the DC supply with a MOSFET, the primary current ramps up at V/L amps per second. If L increases then, the current ramps up at a slower rate and, by the time the MOSFET has deactivated, the peak current will be smaller.

Given that energy stored is proportional to current squared you can see that there is a net loss with this idea despite the inductance being greater. Basically, doubling the inductance halves the energy stored per cycle. You would have to operate at a lower frequency to make things equal.

As I don't need such a high output power...

250 watts is a lot for a flyback converter.

I assume the smaller core would work and would cause less core losses?

No, not normally. The core you chose is this one (the JA4635) and, it has a maximum primary current of 10.5 amps so, with your peak current of 17 amps (from comments), it would heavily saturate.

Heavy saturation means that instead of the primary current rising linearly, it rapidly shoots off to a very much higher value and you end up in deeper saturation with the possibility of destroying the MOSFET that tries to control current.

Or, if the control circuit does its job well then you can regard excessive saturation as producing a much lower value of inductance, so you don't get the energy storage you want and you don't meet your specification.

• I went through your answer from electronics.stackexchange.com/questions/305275/… and I calculated that I could transfer 202.3 W with it. I calculated with 28uH, 17A peak current, and 50 kHz frequency. I then calculated what voltage I would need to reach those 17A peak current by calculating $0.000028*\frac{17}{\frac{1}{50000}}$ which gives me 23.8V, which is above my minimum operating voltage. So all good from there I assume and I can continue checking other parameters? Aug 8 at 16:18
• I am not sure that is correct though, I have a series inductor so my converter is supposed to be running as a quasi-resonant converter. That inductor needs to be included in my calculations aswell I assume? That inductor is 2.2uH, so it's $0.0000302*\frac{17}{\frac{1}{50000}} = 25.67 V$ minimum voltage, correct? Or did I miss that my duty cycle is 0.5 maximum? So it's not $\frac{1}{50000}$ but $\frac{0.5}{50000}$, which would mean 51 V, which is above my minimum input voltage, so that would not be okay. Aug 8 at 16:20
• Sorry, 23.8V is not above but below my minimum operation voltage. Aug 8 at 16:31
• I believe I have answered your question but, it seems you want something else. This isn't how this site works. If you want help stick with the original question. This site isn't a forum and once a question has received an answer, that is basically it. So, has your question been answered? Aug 8 at 17:55
• No, the question has not fully been answered. The first part makes sense and is clear, you elaborated and explained why that is. Hoewever, for the second part you simply stated "No, not normally" as answer to my question if the smaller core would work and would cause less core losses. I didn't ask if it "normally" works, but if it works in this specific case (I wrote the smaller core, not a smaller core). That requires most likely calculations, rather than just saying what generally applies. As an answer I would expect pointers in the right direction how to verify if it works for this core Aug 8 at 18:06

That is a lot of power to push through a flyback! You typically will see them 100W, and more typically 40W or less, but it really depends on field you are operating within..

There is a ton of information to unpack because it doesn't seem you have a background in magnetics, so I think the first place and the best place for you to start is with one of the magnetic guru's (Andyaka) post on this platform.

Go down to his answer, and go through each hyperlink he posts, the problem, and his answer to it. I think you will learn a ton, as I did when I started looking into this stuff (still am :) ).

Inductor design tips

Once you go through those hyperlinks, edit your question or post the answer to your own question, and let's go through that. Cheers!

• The problem I have is that it is kinda a given design but it is lacking information on the transformer/how it's manufactured. I went through the posts but they all assume that I know parameters like peak current, RMS current and so on. While I do have an oscilloscope plot of the current through the flyback primary of the original design and I can see, that it has a peak current of about 38A for a few microseconds, the wave form is not a sine, making it super hard to calculate the RMS value from it. Also I am not sure if the 38A now need to be considered when looking at the saturation. Aug 8 at 15:50