Electronic – How to find the secondary inductance value in flyback transformer

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.

1. With a transformer (which a flyback converter's coils technically are a transformer), you don't look at each set of windings as a separate inductor, but instead as parts of the same inductor. When voltage is applied across the primary, current through the primary increases and a shared magnetic field forms across all windings (primary and secondary). The same with disconnecting power to the primary, it generates a shared magnetic field. As long as the secondary windings are connected to a path electricity can flow through (i.e. are not open-circuit), the current in the primary winding can instantly drop to 0 & the voltage across either set of windings will only spike high enough to force the transformed current through the load attached to the secondary side.

2. The formula Vsw=Vin+Vout*(Np/Ns) is showing you a calculation for the voltage at the switching transistor when it opens the circuit. In any inductor, when a current-source is disconnected, the voltage will rise until a path is found to dissipate current through. Here, your switching transistor will see the "natural" open-circuit voltage of the power source it is switching, along with the additional voltage from the breaking-down field in the transformer. That additional voltage is dependent on the transformer's turns ratio, and the impedance of the load attached to the secondary coils. "How fast the current is decreasing" depends on the impedance of the path the current leaving the coils finds. Current through any inductor that is discharging will always be decreasing as a function of inductance versus back-emf (the voltage across the load), so measuring/calculating the output voltage gives you your breakdown speed.
   To calculate Vo at the moment the switching transistor "opens" (considering the output capacitor's reactance as part of Rload), Vo=Ip*(Np/Ns)*Rload. Meaning that the output voltage will rise until the load resistance can be forced to accept the current that just stopped flowing through the primary windings (current-transformed by the turns ratio of the transformer).

3. The zener diode path (clamping circuit) limits the maximum spike voltage to protect your switching transistor from exceeding its maximum voltage in case your load impedance is too high, or the output is open-circuit. Once Vo*(Np/Ns) passes the avalanche voltage of the zener (plus the voltage-drop of the series diode), current will pass through the "clamping circuit," rather than the voltage being allowed to continue building until another part (usually the switching transistor) failed.
   Under normal operation, the zener diode path should appear as an open-circuit to the transformer, only allowing current to pass in "emergency situations" (when something goes wrong, and power has to be dissipated to avoid catastrophic failure).