Equation for Q factor in AN-57

Hi treez,

Thank you for reaching out to Power Integrations and for your trust in our devices.

The 0.15 that you saw on the AN57 app note is an assumption, based on our practical applications we suggest a q-factor of 0.1 to 0.3.

Regarding DER243, upon computation, the q-factor here is only 0.3 (not including the ESR and DCR). 

The computation for approximating the Q-factor is finding the effective inductance of the transformer using the formula in equation 13 of Appendix A page 11. After finding the effective inductance, compute for the total output capacitance which in DER243 is around 940uF. The effective inductance as per computation is 83uH.

Using the formula sqrt(L/C), the Q-factor is only 0.3 not including the ESR of Cout and DCR of Leffective. If you happen to find the resistance as seen by the output, you can use the formula (1/R)(sqrt(L/C)). 

Remember that the q-factor mentioned in the app note only pertains to the tank circuit formed by the Cout and Leffective.

Thank you.

Kind Regards

Thanks, but SQRT(L/C) is the characteristic impedance formula with units of Ohms...the Q factor is unitless.

Also,  for DER243 the L(effective) at 230VAC is 24uH and at 115VAC its 40uH.

The Q factor is needed for the equation 15, page 11 of AN57.

As such, please could you advise what formula is used for Q factor, and how to calculate Q factor for DER243?

Hi Treez,

Thank you for your patience. 

After reading through the materials and comparing with your answer, the initial response I provided is for a series tank circuit. The correct approach is looking into the tank circuit as seen by the load which makes it a parallel tank circuit.

The initial effective inductance I provided is using a 0.61 of duty cycle which is the max expected duty cycle and I noticed that you considered the actual duty cycle per line using the formula ((Np/Ns)*Vo)/(Vin+(Np/Ns)(Vo)) = D, using this duty cycle and substitute it to get the secondary effective inductance using the formula (L*(Ns/Np)^2)/(D')^2 = Le, I arrived also with the effective inductance as you mentioned in your reply.

Using the formula for the parallel tank circuit with consideration for the energy transfer which happens during the off-time, hence, D' (1-D) should be added to compensate for the off-time which makes it D'*Ro*sqrt(C/L) which I also arrive for a 26 Q-Factor for the DER243 at 230VAC. 

In response to your question of how does the 0.15 is computed on the app note, using the same approach above doesn't arrive for a 0.15 Q-Factor. Kindly allow me to reach out internally to see on how they computed for the 0.15 Q Factor. 

Thank you for your understanding.

Regards