Why do flyback transformers have such a large Primary Inductance when the design really only needs a fraction of that inductance?

Question

I am working on a flyback converter design. I am new to this type of topology and I am trying to learn the fundamentals. The converter I am designing for is 120 W, with 24 Vdc @ 5 A output. The input is 400-900 Vdc. I am on trying to understand the transformers at this point.

I took apart a few commercial power supplies and measured the primary Inductance with the secondary open circuited. I did this with my Extech (1 kHz) LCR meter, and found that the primary inductance was 2.3 mH with a switching frequency of 30 kHz. The input range of this power supply was 400 Vdc (if I remember correctly). This left me wondering how could the current in the primary ever reach the required current based on the duty cycle at the operating frequency if the inductance is that high? According to the standard equation$$ V_t = L\frac{di}{dt} $$

Putting that aside I also calculated the primary inductance based on the output power in all ideal conditions. I am using 50 kHz in my design at the moment $$ energy /cycle = Watts / fsw $$ $$ 120 W / 50 kHz = 2.4 mJ $$ which means the primary must store this energy and transfer all of it to the secondary during the off time (1-D) based on: $$ Joules = \frac{1}{2}LI^2 $$ I chose a 3 A limit at max duty cycle based on the part that I chose having a 3 A switch limit in current mode control. This works out to needing $$ 2.4 mJ*2/3^2 = 533 \mu H$$Assuming my logic and/or math is correct I next looked at the winding details. A PC47PQ32/20/Z has an AL value of $$ \frac{L}{N^2} = 7310 nH$$Solving for N $$ N = \sqrt{\frac {533 \mu H}{7310 nH } }= 8.53 -> 9 turns$$Duty Cycle was calculated using$$ V_{out} = \frac{N_s}{N_p}\frac{D}{1-D}V_{in} $$I want the turns ratio to be low so that the flyback voltage is low as possible. So I chose a ratio of 8 to start and used the above formula to estimate duty cycle for various input voltages while sweeping the duty cycle. I added a reference line at 24 V output that I want. I can see that I need a duty cycle of 32% at 400 Vdc input However, using specifications of the core, with 50 kHz fsw and the formula $$ Ae = 170 mm^2$$$$B_{sat} = 0.25Tesla$$$$N = \frac{Vdc * DutyCycle}{B_{sat}A_ef}$$$$ N = \frac{400 Vdc * 0.32}{0.25mT*170mm^2*50kHz} = 60 Turns $$ Using the AL value with this amount of turns: $$ \frac{7310nH}{N^2} * 60^2 = 26.3 mH $$ Based on this, my understanding is that this is a multiple iteration process. The minimum number of turns is dictated by the voltage applied such that Bmax isn't violated and the core saturates. I plotted duty cycle vs. turns needed to avoid saturation at various input voltages. I computed the average inductance at the midpoint of the curves.

So with all that, how is the transformer ever able to reach the required current? Do we just gap the core so the AL value decreases so much that at the 60 turns the inductance is lowered to the correct value? Or do we change the frequency to a lower value so that the core has more time to charge up during the switch on time?

Answer 1

Do we just gap the core so the AL value decreases so much

I've never seen a flyback design of the power level that you want that doesn't require a gap. And yes, it is usually an iterative process.

You have also made an incorrect assumption about the duty cycle formula. Your design (as tested on my basic website) shows that you are operating in DCM and not CCM. This makes a difference to the formula you used to estimate duty cycle. I estimate it to be 19.99% (as seen below): -

If you want to operate in CCM the primary inductance will need to be bigger circa 1.5 mH on a 400 volts DC supply and, around 2.1 mH if you want to remain in CCM at 900 volts.

Answer 2

The frequency used in commercial converters is higher than 30kHz.

The gapping is the right answer.

I reused many transformers from old power supplies (flybacks) and all EI cores need a gap, otherwise they go into saturation at very low currents or even worst they lose energy during transition from charging to discharging.

The only cores that don’t need a gap are some toroidal cores (rings) but they are used very very rarely. Even legendary yellow-white toroids (#26 from Micrometals) don’t work correctly as flyback transformers.