Bettecher,
I have been studying how to make switching power supplies and i always try to check my calculations with others when enough information is given. The only information that you left out was the ferrite material used in your EE42/21/20 core. This makes it hard to confirm the best flux level, but since you gave the flux level you used i could still calculate the primary turns.
Please take a look at my work and compare it to yours. I belive you should have used about 33 primary turns and you used 16 turns.
Np = (Vmax * 10 ^ 8) / (X * F * B * Ae)
The constant X used in the primary turns formula can be any of the following.
1.11 - compensates for the sine wave ratio of RMS to average voltage. Used for standard sine wave transformer design.
1 - Used for half wave square wave operation of a transformer. A flyback converter uses this constant.
2 - This is the square wave constant above multiplied by 2. there are two semi cycles in each full cycle (Hertz refers to full cycles). A half bridge uses this constant.
4 - This is the constant 2 above multiplied by two for topographies that have a flux swing in all 4 quadrants, like a full bridge circuit.
These are the only constants there are for this formula.
UNITS IN FORMULA
Ae = Area of center leg of transformer in CM squared
B = Flux in Gauss, common values of 1400 to 1600
F = Frequency the transformer is switched at in Hz
Vmax = Voltage in volts the transformer sees
GENERAL RULES
For a half bridge driven off of a split capacitor supply the transformer only sees half of the bus voltage.
Vpri minV gives you the minimum turns to use on the primary, this sets the flux number you used in the original calculations as a minimum. As the voltage rises to Vpri maxV the flux rises, possibly to the point of saturation, if you started to high in flux density to begin with.
Vpri maxV gives you the maximum turns to use on the primary, this limits the flux to a maximum of your original number used in the formula. The flux will go down when the voltage goes to Vpri minV. This will make a safe design.
Increasing the Gauss decreases the turns.
As the turns go down the Gauss goes up
NUMBERS TO USE FOR YOUR DESIGN
Ae = 2.4 CM squared
B = 1500 Gauss
The constant 2 as defined above
Frequency 80,000 Hz
Vmax, from schematic, 380Vdc bus / 2 for driving transformer with split capacitor power supply = 190Vdc
Np = (Vmax * 10 ^ 8) / (X * F * B * Ae)
Np = 190Vdc * 100,000,000 / 2 * 80,000Hz * 1500G * 2.4CM squared
Np = 19,000,000,000 / 576,000,000
Np = 32.986 = 33 turns.
Your calculations showed 16 turns. So if i am right you are actually running 3000 Gauss on your transformer.
Let me know what you think.
Wally