Electronic – Designing an efficient PSU

Switched mode power supplies do not always use isolated topologies just for safety. Even when isolation is not required a tranformer-based design might be attractive since transformers are very convenient components for power supplies, especially when high power levels are involved.

Your assumption that transformers must be bulky heavy things is not entirely correct. Mains transformers are, because the frequency is so low that large amounts of steel are needed for preventing saturation at reasonable power levels. Transformers operating at 10 kHz or more are not necessarily bigger than inductors required for handling similar power levels.

A 1 kV buck or boost converter is not practical, as the switching transistors would have to conduct at least 42 A when on and block 1 kV when off. Applicable converter topologies would be the forward converter, the push-pull converter, and the full bridge converter, all being transformer based. Designing a 12 V DC to 1 kV DC converter is relatively straightforward, but the opposite is going to be a much bigger headache as you would need transistors able to block 1 kV + any transients, not to mention that you would have to find suitable input filter capacitors, design gate drivers, consider safety etc.

Finally you cannot reliably use small signal cables for carrying 1 kV. The isolation might be fine from milliseconds to days but it will eventually break down and short out.

The PA3856.006NL can be used as a flyback transformer. Note 2 in the data sheet states this: -

2. For flyback topology applications, it is necessary to ensure that the transformer will not saturate in the application. The peak flux density (Bpk) should remain below 2700Gauss. To calculate the peak flux density use the following formula: Bpk (Gauss) = K1_Factor * Ipk(A)

Given that the peak primary current is calculated as 10 amps on the lowest input volts, Bpk (Gauss) = 41.7 x 10 = 417 gauss and this is well short of 2700 gauss.

Now, is my approach correct ?

Your approach does seem correct but you forgot to mention operating frequency however, i reasoned that the on time is 4.72 us if that helps.

where is the turns ratio specified ?

The turns ratio is 0.47:1 as shown by the little numbers on the picture of the transformer in your question.

Also, what is the actual advantage of have a dual output in the secondary ? Does it have to do with the current on the output and its effects on the transformer saturation current.

Dual output means you can wire them in series or in parallel. Saturation of the core is caused by primary current reaching too high a value.