**Other Parts Discussed in Thread:**LM25018

Hello,

attached the TINA circuit I am trying to simulate.

The circuit includes the IC LM25018, a coupled inductors and the design is finalized to realize an isolated DC DC converter with two outputs, one is 12V (not isolated), the second one 24V (isolated), the clock frequency is 250kHz, but planned is to use an higher frequency, e.g. 900kHz in order to use available commercial coupled inductors with Ns/Np= 2. Other details of this design are reported directly on the schematic.

All of the values of the passive components are computed based on the LM25018 data sheet and TI's ANs.

The circuit works, but the coupled inductors seem to be critical. I am using Lp = 150uH (calculated) and Ls = 600uH. The ratio of 2 is obtained as Ns/Np = 2 = (Ls/Lp) EXP (0,5) (i.e. sqrt (Ls/Lp)). M is estimated by the following formula: M = k * sqrt [(Lp x Ls)]. With K = 0.93 we obtain M = 279uH

The problem of this circuit is exactly the value of k.

If k is less than 0.95 the circuit more or less produces Vout2 more than 20V (23,74V with k = 0,93), but if k > 0.95 the circuit does not work!

But if k > 0.95 means also similar to ideal transformer, it should be better, not worse!

I know, it is not easy to understand this circuit and the related values, but I would like to have a circuit less sensitive to the k values. I cannot know, what is the real value of k for commercial coupled inductors.

Any idea if I can improve this circuit? Are the simulated waveforms really correct?

Of course I only simulated the start, with defined values of load current (both 50mA). But planned is also transient of load, different balancing of loads and so on. But first at all I want to understand the dependency from k.

For any assistance or help thank you very much in advance

Federico