LTspice - Flyback simulation misunderstanding

Question

I am trying to understand how to compute the rms current through the output diode of a flyback with multiple outputs. To make it simpler, I did not add the leakage inductance of each windings. Here is my schematic :

Here is the result of the simulation :

What I do not understand is why I(D3), the output diode of Vout1 take all the current from the primary when the voltage at the secondaries (across L4 and L2) is superior to Vout1 and Vout2. If it is the case the two output diodes should conduct as the voltage across the diode are forward bias. The simulation said that the two output diodes conduct when vout1= vout2 ??? Is someone is able to understand what is happening ?

I(D5) -> Vout2

I(D3) -> Vout1

Thank you very much and have a nice day :)

Answer 1

What I do not understand is why I(D3), the output diode of Vout1 take all the current from the primary when the voltage at the secondaries (across L4 and L2) is superior to Vout1 and Vout2. If it is the case the two output diodes should conduct as the voltage across the diode are forward bias.

The main reason is that the load current on Vout1 is twice that on Vout2 but you also need to consider that the diodes are not ideal in the real world.

So, you are not thinking about diodes in enough detail. If you had a variable power supply, a voltmeter and an ammeter and you plotted the forward voltage of a diode against the current it takes you would find this approximate rule: -

$$\boxed{\text{A 100 mV reduction in the applied forward voltage reduces the current ten times}}$$

In a comment you say this: -

The both should conduct if the voltage across the diode is foward biased whatever if one output voltage is lower than the other ? The difference between the two outputs is about 50 mV

A 50 mV difference in the forward bias voltage of two identical diodes will result in about a 3:1 difference in forward current. Look at the data sheet for ON semiconductor's 1N4148: -

Picture from this answer.

Concentrate on the left diagram - at 0.5 volts applied, the current is 100 uA. Now drop down 100 mV to an applied forward voltage of 0.45 volts and can you see that the current has dropped to about 33 uA?

That's a 50 mV drop causing a 3:1 reduction in current.