is the duty cycle the same as the duty ratio?
if so then D=Vo/Vs
D=25/50
D=0.5
7.6μS / 10μS = 76% duty cycle, yes.right so if you add up all the time that current flows and then divide it by the period you would have a duty ratio
based on the graph
DT + D1 T = total on time
T = period
5uS + 2.6uS = 7.6uS
7.6uS/10uS = .76
total on time/period =
Kindly permit me an observation or two ...recalculating with D=0.316 gives me...
but that answer is incorrect. am I not using the right formula here?
thats a very good idea, i will take that into considerationKindly permit me an observation or two ...
You've been tinkering with these problems for some time now and still seem to be struggling.
As Mik3 & others point out, the particular converter mode of operation and topology dictate the approach required. I'm sure you've grasped that fact.
You asked me some posts back whether I remember the formulas for any given situation. In brief - no I don't. Well that's not exactly correct - I recall the relationship between switching duty and output voltage in a continuous mode buck converter. Rather than remembering a formula which may not be relevant to a particular case, I try to understand what's happening physically and then analyse the circuit.
The present example is a case in point. I came momentarily came 'unstuck' because I didn't initially think it through carefully.
You may derive some benefit from putting the text book aside and systematically analyzing the various cases for yourself - or make sure you understand why a certain formula is relevant for a particular case. Then you may be able to apply the formulas with confidence.
by Duane Benson
by Jake Hertz
by Duane Benson