# RMS Current through output diode in a flyback transformer with multiple outputs

I clearly do not understand where does this formula comes from ... (I did not find it on internet) $$\mathrm{ I_{diode_{rms}} = \frac{\sqrt{1+\frac{A²}{3}}}{\sqrt{1-D}}*I_{out_{avg}} }$$

where Iout is the output current drawn by the load, D is the duty cycle and A is defined as follow (it seems to be the transfer function of a flyback working in CCM but it is not exactly right) :

$$\mathrm{ A = \frac{(DVin)^2}{2L_{primary}P_{in}F_{sw}} }$$

where Fsw is the switching frequency of the flyback, Lprimary is the primary inductance of the flyback, Pin is the input power.

If anybody is able to demonstrate this formula even for a simple transformer with only one output, it would be nice !

Thank you very much and have a nice day !

• Where did you find the formulas if not on the internet? Commented Aug 11, 2020 at 13:30
• I cannot answer that - you have formulas made by some guy who might be in-contactable and therefore there is little go on. BTW, it's "current through" and "voltage across". Not helpful to you I know LOL. Commented Aug 11, 2020 at 14:02
• Thank you ! I am completely bad in english ... I think that It will be hard to answer to this question ... I will probably close it
– Jess
Commented Aug 11, 2020 at 14:51
• Maybe wait for Mr @VerbalKint - he might know or recognize something that might help? Commented Aug 11, 2020 at 14:59
• Yes, I am not yet roasted by the hot summer in France and I gave an answer below : ) Commented Aug 11, 2020 at 20:39

I don't know where this formula comes from and I have never seen it. I derived the formula for the secondary-side diode rms current in my book published in 2014, page 755. I have built a quick Mathcad sheet to test it against your formula. This is a 72-W ac-dc adapter, no leakage inductance (otherwise the peak sec. current would be reduced) and low conduction losses in the MOSFET:

So for a 120-V input voltage, the sheet computes a peak current of 2.15 A and a valley current of 0.95 A in the primary. The rms current in the diode is computed to slightly less than 5 A.

To test these results, I have captured a quick flyback circuit with SIMPLIS and ran a simulation:

Once the simulation is done, pressing F3 in the curve windows gives you the diode rms current:

The simulated current is 4.98 A, very close to what the calculation gives. If I now use the formula you mentioned:

As you can see, there is no significant error with the simulation and my Mathcad sheet. So your formula is correct and elegantly arranged : )

One way to get the diode rms current is to determine the ripple in the magnetizing inductance of the flyback converter. You can follow the calculation details in a series of articles I published in How2Power in 2010. For a multi-output converter, precisely calculating all rms current can be somewhat adventurous considering all the leakage inductances at play which heavily distort waveforms (see here for the cantilever transformer model and Fig. 9 for the sec. waveforms).

• Thank you very much for your answer and your simulation ! Your formula is only valid for a flyback operating in CCM, isn't it ? Apparently the formula that I gave you works also in DCM. I will take a look on it and see how to demonstrate it ! And I will do some simulations with LTspice (or SIMPLIS ??)
– Jess
Commented Aug 12, 2020 at 8:39
• Yes, the (1-D) factor only works in CCM as, in DCM, you have the dead-time $d_3T_{SW}$ which must be evaluated to get $d_2T_{SW}$, the time duration during which the current circulates in the output diode. Commented Aug 12, 2020 at 9:10
• Thank you very much ! :D
– Jess
Commented Aug 12, 2020 at 11:19