Following an existing circuit I have been analyzing.

Here, there is no confusion with the working of buck regulator output 3.3 V. But with the blue-rectangle section, input to the LDO (Vout2) is taken. When it measures, Vout3 = 5 V. So obviously Vout2 would be greater than 5 V. I am trying to figure out how this circuit will work and how to calculate the voltage Vout2 theoretically. Coupled inductor (N1=N2) inductance is 10µH.
Can anyone suggest how to calculate the input voltage (Vout2) to the LDO and the working of this section.
Thanks for the support.
MOD:tidied your images.E

 

I doubt you can with the partial circuit shown.
There is a lot missing. Post the full circuit please.
And where do you get a 10uF transformer?

 

Sorry, that was typo
Could you please mention what are the missing information. Basically this is the shareable version of the schematic I have.

 

The circuit as you show it is just a part of the full job. Do you have a real circuit?
Then there is the question of why you are trying to go from 12V to 3.3V then 5V.
It would make a lot more sense to me to have a 12V to 5V switcher, then the linear reg from 5V to 3.3V.
The output of the transformer secondary, Vout2, will depend a fair bit on the loading of the 3.3V supply I think.
Still, the collapsing magnetic field that is used to generate the 3.3V also induces a voltage in the secondary that is added to the 3.3V and rectified by the Shottky diode to drive the LDO reg. So if the transformer is 1:1, the Vout2 is probably around 6.6V
At least I think that is how your setup will work.

 

Well this is all got as circuit. This is from a reference design and I am not sure how this LDO come form 12 V to 3.3V then 5V. But your answer clarifies my answer. Input to the LDO would be 6.6 V (= 3.3 V +3.3 V). Thanks a lot.