Yes, it didn't occur to me, but it will also work that way (both modes aren't mutually exclusive).Strikes me that it may be a form proportional base drive.
This article doesn't cover the calculation of transformersStrikes me that it may be a form proportional base drive. Consider Q3. When it turns on top 1 turn dot goes positive enhancing the base drive to Q3... proportional to current? Same for Q4 the other/same way up. Not sure about CR3/C1 other than it might result in a clamped -Vbe drive for turn off of the non-active device.
PROPORTIONAL DRIVE CIRCUITS FOR BIPOLAR TRANSISTORS
Gosh I'm old.
Current transformers were used in self-oscillating converter.T1 works as a base drive transformer with the PWM source, and as a current transformer with the feedback from the output, probably to make sure that each transistor has completely turned off before the other one is allowed to turn on, even if the PWM signal is already active.
This mixed mode of operation makes the calculation somewhat more complicated than for either of the modes.
If you are unable to carry out the calculations by yourself, drop it because nobody is going to embark on such a chore for free. DIY help has some limits
This article doesn't cover the calculation of transformers
I think you got wrong calculation,Ae=5mm^2Moan, moan, moan. Sigh.
Assume it is proportional base drive.
Refer to the On characteristics.
Hfe is about 10 at IC=5A. Your secondary turns ratio becomes about 10, ratio of Ic/Ib, which you already have, 10Tbase/1Tcoll. If Hfe were 5 then 5Tbase/1Tcoll or 10Tbase/2Tcoll.
Your primary PWM has to overcome this current to switch the secondary device Off.
Assume the peak collector current is 5A. Base current is 500mA. With 50Tpri/10Tbase the required primary current is 100mA. If your bias voltage is 15V then, ignoring voltage drops, R3 needs to be about 150R to provide this current, pick 120R.
Vbeon is about 1.2V. Referred to the primary this will be about 6V. Say your switching frequency, Fs, is 50KHz. The period is 1/2Fs or 10uS. Your transformer has to support this volt-second product, 6V 10uS, without saturating.
Npmin = Vin.Ton/Bsat.Aemin
Aemin = Vin.Ton/Bsat.Npmin
Aemin = 6V x 10uS/0.3T x 50
Aemin = 40mm^2
That's the basics. You also have to worry about wire sizes and whether they will fit. All things are variable so you have to pick your own numbers and then iterate to find a core that works.
Oh.. If you want to be hard you might use the fact that push-pull operates in two quadrants so you can double Bsat.