PMP8740

Dear Anjana,
Yes, I was considering in my case Vtran = 1.6V. The lower you select it, the bigger the output capacitor will result.
Please check out also the switching frequency. If you meant to have 100KHz on the bridge (switching frequency of each FET of the bridge), then you have to select 200KHz on "Inductor (Lout) Switching Frequency" cell.
Regards,
Roberto

Dear roberto, 

What is use of R98 in PMP8740? how to calculate it?

Anjana

Dear Anjana,
R98 is a snubber resistor. It wastes the energy stored in the leakage inductance.
Actually it burns part of the energy, while a part is recovered into the load.
The procedure to calculate it, refers to Vin = 390V, full load (62.5A) and both maximum Vout (32V) and short circuit condition.
We must burn the energy in the leakage inductance of the transformer: this is Llk = 3.5uH, in my case. It must be reported to output, therefore divided by square of turns ratio (for me N = 9.5).
Now the energy is 1/2 * Llk/N * (Ipeak)^2, where Ipeak is "Ips" in the excel sheet, peak secondary current on the transformer. The power is calculated by multiplying by the switching frequency, which is 200 KHz (on each diode of D13 is 100 KHz). Therefore 1/2 * 3.5uH/9.5 * (60.3A)^2 * 200KHz ~ 15.4W. Now on the anodes of D13 we have a square wave which ranges 0V to 2 * VSpeak, where VSpeak = secondary voltage on the transformer = Vin /N = 390V/9.5 = 41.05V. That means, the square wave (without spike) on anode of D13 = 2 * 41.05 = 82.1V.
Because of the leakage inductance, we will have a spike. The 82.1V + the spike must be well below of the maximum rating voltage of the sync. FET, which is 200V. Usually I take the square wave peak voltage (here 82.1V) and I increase it by 50% in order to discharge the energy into the spike but not "steal" energy during normal conduction. 1.5 * 82.1V ~ 123V.
This voltage, minus Vout, will be seen on R98. The worst case condition, happens in short circuit, where Vout = 0.
That means that R98 can be calculated so that 123V are across it and to dissipate 15.4W. Since P = V^2 / R or R = V^2 / P we have R98 = 123V^2 / 15.4W = 982 Ohm...therefore 1 KOhm!
Regards,
Roberto