# understanding canonical model of a buck converter

Discussion in 'General Electronics Chat' started by kokkie_d, Feb 27, 2013.

1. ### kokkie_d Thread Starter Active Member

Jan 12, 2009
72
0
Hi,

In the attached form I have my full work out of a canonical form of the buck converter. On the final page I have my final version and the version that the book I am working from is showing (the book does not show how to get to this form it just gives the variables).

As can be seen the resulting form is similar but the variables are different.
Now, from the steady state I have the following 3 equations (page 2):
$V=D*V_{g}
I = \frac{V}{R}
I_{g}=D*I
$

By Replacing the following:
$V_{g} = \frac{V}{D} \rightarrow \frac{V_{g}}{D}=\frac{V}{D^{2}}$

I can match the voltage source to the result from the book.

But applying he same thinking to the current source will result in:

$I = \frac{V}{R} \rightarrow D*I = D*\frac{V}{R}
$

I am thinking that the ideal transformer has something to do with it but I can not figure out how.

Is there someone who knows the answer to this?

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2. ### Wendy Moderator

Mar 24, 2008
20,764
2,534
This is not a classic math issue, so much as an electronics question. I have moved it where it will be a better fit.