VoltSecond equations for equivalent buck-boost of flyback

Question

Here is the equivalent buck-boost of flyback converter:

I proved the Volt-Second in a switch-on mode that agrees with the book result (n=22.86.)

In switch-off mode (regarding the above table):

$$t_{off}=(1-D)/f=2.94us$$ $$V_{off}=V_{diode}+V_{OR}=n*V_{D}+n*V_{o}$$

Where \$V_{o}=5V\$

Thus:

1. If I assume (as we assumed the switch has no drop voltage) \$V_{D}=0V)=>Voff=n*V_{o}=114.3\$ and \$Et=Voff*t_{off}=114.3*2.94us=336us\$ which is not equal to 473us.
2. If I assume \$V_{D}=1V=>Voff=n*V_{D}+n*V_{o}=22.86+114.3=137.16\$ and \$Et=Voff*t_{off}=137.16*2.94us=403us\$ which is not equal to 473us.

Where is my mistake?

Answer 1

At equilibrium or steady-state and in CCM operation, the dc transfer characteristic of the isolated buck-boost converter is \$V_{out}=V_{in}\frac{ND}{1-D}\$. From there, you can extract the operating duty ratio and determine the volt-seconds linked to the on- and off-time duration. At steady-state, they should be equal:

In the equation, \$N\$ characterizes the transformer turns ratio \$1:N\$ and, if made 1, you can determine the values for buck-boost converter since the flyback derives from this structure.