Hi,
I agree with FvM.
Let´s imagine the primary duty cycle is the same as in the picture above. DCM.
Energy stored in the inductor: E = 0.5 x U x I_Peak x t
Now let´s assume the secondary voltage is doubled.
The result will be:
* Peak current will still be te same.
* but the conductive time will be 1/2 than before.
--> you get half of the secondary average current than before. --> Area under the secondary current curve is half.
Edit: not true: and the RMS current will be half, too, because only the conductive time changes and not the amplitude.
I didn´t consider the square root after the mean calculation.....
--> Thus RMS will not be half the value than before.
Calculation according energy:
Primary energy will be secondary energy in either case: (ignoring losses)
E = 0.5 x U_s x I_Peak x t
this also tells, that if you double the secondary voltage --> time will be 50%
Klaus
Added after reading EasyPeasy´s post:
Hi,
for a flyback system one can assume V_in and V_out to be constant (not changing much with time).
Then I_out and I_in are
average values.
so the power is: Vin x
Iin_average
and it is not: Vin x Iin_RMS
(V_in is constant, thus V_avg = V_RMS)
*****
Calculate with I_avg when you want to calculate input or ouput power
Calculate with I_RMS if you want to calculate winding loss.
Klaus