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While designing converter I got to the point of calculating secondary coil. Most guides have these equations for it:
enter image description here
Wait a second, didn't I calculate Nps already at the very beginning? Yes, I did! And I've used it extensively throughout the process.

Nevertheless, I use these formulas and get a different Nps. I thought I am going crazy, until I noticed tiny detail: I did not use (1-Dmax) in the beginning. Instead, I used (0.8-Dmax) to leave 20% dead time before CCM boundary.

Question: Is the above equation for reflected voltage applicable in DCM? Or should I just use the Nps that I already have?

This tiny change unfortunately resolves into 30% in coil turns, the difference that does not fit on the core.

UPDATE:

In the equations and text:
Vr - voltage reflected to primary when the switch is off
Vin(min) - minimum input voltage
Vo - output voltage
Vf - forward voltage drop on rectifier
Dmax - maximum duty cycle
Nps - primary-to-secondary turns ratio

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  • \$\begingroup\$ Perhaps you'd define what Vr and Nps, and the other terms, mean. Some people unfamiliar with the terminology as used by your particular source might want to think about this question. \$\endgroup\$ – Neil_UK Jun 11 '18 at 6:31
  • \$\begingroup\$ @Neil_UK I am so used to see exactly the same terms in two dozen of sources I found that I assumed anyone who can help is already familiar with them. Added them anyway, just in case. \$\endgroup\$ – Maple Jun 11 '18 at 7:11
  • \$\begingroup\$ If you are leaving 20% spare time to avoid entering CCM shouldn't your "off time" be proportionately bigger i.e. 1-D should be more like 1.4 - D? \$\endgroup\$ – Andy aka Jun 11 '18 at 9:10
  • \$\begingroup\$ I've used equation from the book: Ton(max) = [(Vo+Vf) * Nps * 0.8 * T] / [Vin + (Vo+Vf) * Nps] and resolved it for Nps (since Ton comes from Dmax). As I understand the logic behind it, Ton(max) + Tr + Tdt = T, and the reflected voltage is proportional to the ratio of "on" time (Dmax) and "reset" time (T - Tdt - Ton i.e. 0.8-Dmax), since there is no reflected voltage during dead time. \$\endgroup\$ – Maple Jun 11 '18 at 9:25

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