In a flyback converter why do we assume that $\Delta I_{on}=\Delta I_{off}$?

 

In a flyback converter why do we assume that $\Delta I_{on}=\Delta I_{off}$?

So that the average input current remains constant.
I assume you are referring to a boost converter with a single-winding inductor, not an isolated flyback converter with a multi-winding inductor.

 

What would hapoen if it was not the case?

Suppose dI/dt x t(on) was greater than dI/dt x t(off). What would the current look like over time?

 

Yes something else as well:

I am trying to understand how

So that the average input current remains constant.
I assume you are referring to a boost converter with a single-winding inductor, not an isolated flyback converter with a multi-winding inductor.

No im refering to a flyback converter with a isolated output.

 

I have one more question:I have been studying for a exam for quite some time and here are my notes


The result in the last equation according to the videos I have been watching is wrong it should be ND/1-D.Where am I wrong?I cant find my error.

 

I have one more question:I have been studying for a exam for quite some time and here are my notes

View attachment 315259
The result in the last equation according to the videos I have been watching is wrong it should be ND/1-D.Where am I wrong?I cant find my error.

First thoughts say that it is because there is no current in the primary when current flows in the secondary, and vice versa, but I‘ll have to think about it!