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?
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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