There are a number of things that I'm confused about, but one of the biggest sources of confusion is how the equation \(\frac{V_o}{V_i}=\frac{1}{1-D}\) is derived. I've read Wikipedia's article on boost converters and understand most of that equation's derivation, but I get confused towards the end.

I know the idea behind a boost converter is that the voltage across an inductor is proportional to the rate of change of current. So, to amplify a voltage, the current level flowing through an inductor needs to be quickly changed.

To show that I really have been working hard to understand this, and so you can see where I'm at, here's my work. Hopefully I did the math correctly, but my calculus is a little rusty so let me know if I didn't To summarize, I get one equation for when the switch is closed and another for when the switch is open:

\(\Delta I_Lon=\frac{V_i}{L}DT\)

\(\Delta I_Loff=\frac{(V_i-V_0)(1-D)T}{L}\)

So, assuming those equations are correct, here's where I'm confused. Wikipedia says:

Then theySo, the inductor current has to be the same at the start and end of the commutation cycle. This means the overall change in the current (the sum of the changes) is zero:

\(E=\frac{1}{2}LI_L^2\)

**equate**the two equations. But the thing is, the voltage is being amplified, right? So, since the voltage has been

*increased*, due to conservation of energy mustn't the current be

*decreased*?

Hopefully someone can clear this up for me, as I there are a few more things I'm confused about and I have to get this done be next Monday

Thanks for your help.