Half bridge smps transformer turns calculation.

[codemaster11]

dear, i calculated the numbers of turns for a half bridge smps using EI33a core

https://tahmidmc.blogspot.com/2013/02/ferrite-transformer-turns-calculation_22.html

according to me minimum input Vin(min) = 100*1.414 = 141Vdc

N(prim) = (141*0.5*100000000)/(4*50000*1.3*1500) = 18 turns


at minimum voltage duty cycle will be max, duty cycle = 0.98, at max duty cycle average voltage to transformer

vin(min) = 141Vdc

duty cycle = 98%

desired output voltage Vsec = 22 volt

average voltage to xformer (Vprim) = 0.98*0.5*141 = 69.29 volts

the ratio (Vpri/Vsec) = 69.29/22 = 3.15, that's equal to turns ratio (Nprim/Nsec) = N = 3.15.

here Nprim = 18 turns, Nsec = Nprim/N = 18/3.15 = 5.7 turns

for auxiliary Vaux = 18 volts, auxiliary voltage to (secondary voltage + diode drop) N = Vaux/(Vsec + vd) =

18/(22+0.5) = 0.8


Naux = Nsec/N = 5.7/0.8 = 4.6 turns.

here another video https://www.youtube.com/watch?v=TLD6hUV64YU&t=10s

who calculate turns for half bridge xformer, i 'm confused here who one is correct?

because he take the half of full 311Vdc while calculating Nprim also used another method for finding Nsec & Naux?

can someone help me regarding this calculation whether it is correct or not ?