Flyback primary inductance changes

inductance is probably chosen to get core loss and wiring loss to equal values. possibly the switch I^2R loss, aswell. The problem with those tools is that you never know what they calculate. Of course it might be complete garbage. In the end it is just one of those AI conspiracies to lure you onto the website to look at the ads.

That sounds too high. Did you just take its default value? Did you click the Calc button and notice the waveforms? It recommends CCM. Default is based on 20% current ripple, I think.

Nowhere near what a TinySwitch wants: DCM. Fortunately, you don't need to stick with the values given. You can plug in whatever value you like, and see the waveforms you'll get. :-)

Roughly speaking, to get to 100% ripple (BCM), use 5x less inductance. Less even than that may be desired, because of the wide input voltage range. (Don't forget to check at the voltage extremes and see what happens!)

Tim

Doing the math by hand isn't too hard, here is an instruction guide:

I can't speak for the PI tool, but the Schmidt tool has all the equations listed in the help box. It's just a calculator, nothing more: garbage in, garbage out. If your circuit does not conform to the assumptions made, if you don't get the waveforms seen; that's your problem. :) If you don't fill in the free variables yourself, it will recommend them, based on whatever assumption the writer chose to go with -- as it happens, CCM is assumed for all topologies I think.

Tim

Since it's a flyback it will have to be an air-gapped core of some type, whether distributed or "real", the saturation curve is more gentle compared to a ferrite brickwall. AFAIK there's no law that says you can't run a core a bit into saturation if it helps make the solution size smaller, for some e.g. powdered irons the curve looks like a complementary error function u can run four or five times the "cutoff" of 100% nominal relative mu vs H and only lose about 20% of the mu

D_limit = 0.5 (current mode controller I believe)

turns ratio = n_2/n_1 = (Vin_min * D_limit)/((V_out + V_diode) * (1 - D_limit)) = (120*0.5)/(15.2 * 0.5) ~= 8

D_ccm = ((V_out + V_diode) * 8) / ((Vin_min + (V_out + V_diode)) * 8) = ~ 10%

Magentizing current referred to the primary = I_m = n_1/n_2 * 1/(1 - D_ccm) * I_out = 1/8 * 1/0.1 * 0.5 = 0.07A

delta_Im = 0.2 * I_m = 0.014A

L_m = (Vin_min * D_ccm * T_s)/(2 * delta_Im) = (120 * 0.1 * 7.57e-6)/(2

• 0.014) = 0.00324

Comes out to about 3.2 mH as an estimate for the required primary side inductance for CCM, how'd I do

Should be 1/8 * 1/0.9 * 0.5 = 0.07A