Buck/Boost DC-DC Converter Discontinuous Conduction Mode

1. Oct 19, 2012

jegues

EDIT: I have updated the original post to make my confusion more clear!

Hello all,

Attached below are the pages from my textbook for which I am concerned.

On page 245 we can calculate the average inductor current as follows,

$$I_{L} = \frac{1}{2}I_{max}(D+D_{1})$$

Now what I thought an equivalent expression for the diode current would be,

$$I_{D} = \frac{I_{L}D_{1}T}{T} = I_{L}D_{1}$$

Is this incorrect? If so, why?

It is clear from the graph of the diode current on page 245 that indeed,

$$I_{D} = \frac{\frac{1}{2}I_{max}D_{1}T}{T} = \frac{1}{2}I_{max}D_{1}$$

but instead of calculating the area of a triangle, I've always "stretched" that triangle into a rectangle since I know that the current flowing through the diode for that time frame (i.e. D1T) is the average inductor current. With this in mind, the area of my rectangle would be,

$$I_{D} = \frac{I_{L}D_{1}T}{T}$$

Is this intuition incorrect?

This is thought of stretching the area of said triangle into a equivalent rectangle has previously worked for me when doing the analysis for the Buck converter under discontinuous conduction mode. In particular, I wrote

$$I_{s} = I_{L}D$$

where Is is the source current.

Does the source of my confusion make sense? Can you see where I'm going wrong?

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Last edited: Oct 19, 2012