Flyback Secondary RMS current

Hi everyone

I confused about flyback secondary side wire calculation. Before wire diameter, I need to find rms of secondary current. The output wil be 120W(24V_5A). The rms of primary current is about 3A. If I calculate it from turns ratio, it will be right or not? turns ratio is 32T/9T. Dmax is 0.45us.

Hi,

Can you draw a chart how the secondary current looks like? = I-t diagram...

Klaus

The rms depends on shape, if you definitely know the pri is 3A rms then the sec current will be 32/9 times this = 10.67 A rms

Note for a flyback in DCM with 50% on/off time and nearly zero dead time, the peak current is 4 x the average and the rms is Ipk * SQRT(D/3) = Ipk * 0.408 for 50% DCM ( CrCM )

So for 5A ave the pk would be 20A and the rms = 8.16 ( 50% pwm DCM / CrCM )

if you definitely know the pri is 3A rms then the sec current will be 32/9 times this = 10.67 A rms

Click to expand...

That's true for a transformer with negligible magnetizing current but not for a flyback, where the output current is sourced by transformer inductance and primary and secondary current waveform are independent of each other.

You need to analyze the secondary waveform to find the RMS current.

Nope - the Ixturns is a match pri and sec - you can test it in theory as well as in practice - at turn off virtually all the Ipk x turns on the pri translates into Ipk x turns on the sec, this then translates into ave and rms as they are linearly related ... in steady state operation ...

virtually all the Ipk x turns on the pri translates into Ipk x turns on the sec, this then translates into ave and rms as they are linearly related ...

Click to expand...

No, both average and rms are modified by the primary to secondary duty cycle ratio.

perhaps you can show an example where they are not related by the turns ratio ... ?

Sure. Any operation point with asymmetric duty cycle. Peak current translates with winding ratio, average and rms does not.

Ahh, I was thinking of constant power ( which a flyback is ) as in Vin x Iin = Vout x Iout, some how I linked this with turns ratio in my head ....

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,

Vin x Iin = Vout x Iout

Click to expand...

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