Discontinuous current operation (Boost Converter)

Discussion in 'Homework Help' started by jegues, Oct 19, 2012.

  1. jegues

    Thread Starter Well-Known Member

    Sep 13, 2010
    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?
    Last edited: Oct 19, 2012