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The material hard saturates at about 400 mT at 25 degC and it will get hot if driven at this level (see red line in picture) and then the saturation level drops to about 340 mT and things just carry on getting hotter.

If instead of driving it at 400 mT you drove it at 200 mT (green line). Now it is fairly temperature stable should the core get hotter due to (say) copper loss. You can also say that maybe 300 mT is a good level but I initially tend to stick at 200 mT for this type of material.

So, getting back to the root problem. The primary inductance is driven with a voltage and will produce a ramping up and down of current (and flux) and, the peak of that current is the point at which flux is highest (and hence flux density is highest).

Note that in the BH diagram above the H value is about 23 ampere turns per metre to deliver 200 mT.

With 18 uH and (say) 12 volts applied over half the switching period (dt), current (di) is 3.333 amps. That is the peak current both positively and negatively. So the peak ampere-turns is 3.333 x 3 = 10 At.

I should say that I'm explaining a process rather than giving a solution that is suitable for the OP. Moving on...

The material hard saturates at about 400 mT (at 25 °C) and it will get hot if driven at this level (see red line in picture). As it warms the saturation level drops to about 340 mT (at 100 °C) and things just carry on getting hotter. This must be avoided.

So, instead of driving it at 400 mT consider driving it at 200 mT (green line). Now it is fairly temperature stable should temperature rise due to (say) copper loss. You can also say that maybe 300 mT is a good level but I initially tend to stick at 200 mT for this type of material.

Getting back to the root problem; the primary inductance is driven with a voltage and will produce a triangle wave of current (and flux) and, the peak of that current is the point at which flux is highest (and hence flux density is highest).

Note that in the BH diagram above, the H value is about 23 ampere turns per metre to deliver 200 mT.

With 18 uH and (say) 12 volts applied over half the switching period (dt), current (di) is 3.333 amps. That value is the peak current both positively and negatively. So the peak ampere-turns is 3.333 x 3 = 10 At.

I should say that I'm explaining a process rather than giving a solution that is suitable for the OP....

Choosing a core can be very difficult and a bit of try-this, try-that theoretically is par for the course.

And, of course, with the split primary design you have, you'll need 6 + 6 primary turns because only one half is active at any time. I have done a similar design for about 200 watts and to keep efficiency as high as I could, I hard switched the primary as per your design but the centre tap I fed from a 95% efficient synchronous buck converter to do the regulation.

I'm not saying that this is the best route to go for you but I am saying that for my design this suited what I needed to achieve AND I ran at an operating frequency much higher than 100 kHz i.e. 600 kHz. Running higher than 100 kHz is probably what you'll need to do so maybe consider 250 kHz and choose a slightly more exotic ferrite material such as 3F3.

Choosing a core can be very difficult and a bit of try-this, try-that theoretically is par for the course.

So