Why the slope compensation usually use down slope of inductor current
is any problem using turn on slope?
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Hi,
You can figure out downslope compensation directly and quickly from a plot in general. If you are ok to figure out either way, then it is ok for you to use either compensation - the effect is the same. Actually the implementation is upslope in practice.
Hi,
Let's start without slope compensation (remove the -m)
1. let m1 hit the threshold (the dashed line),
2. then use -m2 to reach T. You can see the current cannot come back to start of m1.
3. Then you need shift -m2 end until reach the same start of m1.
4. Then draw a line from start of m1 to start of shifted -m2,
5. extend to T, that line is -m.