# Designing flyback converter - Current ripple and turns ratio

This would be a long question, I apologize in advance.

In the process of learning about flyback I've calculated same LT8302-based converter using five methods. Four from generic (as in: not related to any specific chip) application notes and guides, one straight from LT8302 datasheet primer.

All five produced different results. :( Generic methods were at least close (Lp 2.7~4.3uH) but the fifth was way out of line (Lp 23.4uH). Here is heavily simplified description of generic methods:

1. Calculate input power Pin from output requirements.
2. Either choose turn ratio proportional to voltages and calculate D from there or set D to 50% and calculate turn ratio from it. In either case D <= 0.5 and Nps is about 1:5.
3. Arbitrary choose current ripple factor Kr within 0.25~0.5.
4. Either directly calculate Lp from D, Kr, Pin and Vin(min) or first calculate average current from D and swing from Kr and then get Lp from those. In both cases Fsw assumed to be max possible and the results are essentially the same.

Every time I end up with Isw going over max rated current at Vin = min, and with Fsw going above maximum possible at Vin = max. So, I have to loop back and increase Lp until Isw gets into acceptable range (at about 5uH). As I understand, this decreases Kr well under 0.25.

Now, the method in LT8302 datasheet goes in opposite direction:

1. Ignore actual power requirements and go straight with Vsw(max) - safety margin.
2. Calculate Nps from that, which gives something like 1:1.3
3. Calculate D from above, which produces 0.8~0.89 (?!)
4. Calculate Lp from above using Toff(min) and Isw(min); result is 24uH (?!)

Surprisingly, the end result is that Isw stays under max in all Vin range, Fsw is around half of max FQ, and Pout is comfortably above required at any Vin. Although Kr is higher this time at 0.5~0.9.

I've been reading the book by Pressman and Billings (which was an eye opener BTW, "flyback transformer is not a transformer, it is inductor with multiple coils" 8O ). From there and other sources Dmax must stay under 0.5 to avoid oscillations. Looks like LT8302 makers do not worry about this at all.

Question: Which way to follow? (Obviously) Also, is Kr < 0.25 a bad thing?

List of documents used for calculations:
Analysis and Design of Multioutput Flyback Converter
Fairchild AN-4137
National Semiconductor AN-1055
Texas Instruments U-165
LT8302 datasheet

Update:

Here are couple calculations using "generic" methods. Both had Ipeak over max, so I had to reduce Kf to 0.2 to adjust.

And here are calculations based on LT8302 datasheet:

• Consider where the energy starts, & where the energy ends up. The inductor energy (0.5 * L * I * I) is input energy. The output energy is 0.5 * C * Co * Vo. – analogsystemsrf Jun 3 '18 at 3:40
• I have no idea what your second formula means and how is that supposed to answer the question, sorry – Maple Jun 3 '18 at 4:26
• This might help a bit. – Andy aka Jun 3 '18 at 10:42
• @Andyaka I've read that. In fact, before posting this question I've read about three dozen documents like it. Anyway, that document is about DCM converter. If I am not mistaken all the calculations I described are for CCM. – Maple Jun 3 '18 at 15:55
• @Maple I have a suspicion that only the LT8302 data sheet is about CCM and, inevitably, it will use a bigger value inductor. But this gets to the root of your problem namely that there are no links to the other formula examples and no mention of DCM or CCM in your question. Most people assume that a flyback operates in DCM because there is basically no benefit to operating in CCM either in power throughput, core saturation or EMI generated. – Andy aka Jun 3 '18 at 15:59