# 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
735
43
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?

File size:
259.4 KB
Views:
37
• ###### Q-_0132_NEW.jpg
File size:
203.7 KB
Views:
36
Last edited: Oct 19, 2012