Blog for students taking ECEN5817 Resonant and Soft Switching Techniques in Power Electronics, ECEE Department, University of Colorado at Boulder, Spring 2012
Thursday, April 15, 2010
Hw10, pb2a
Hello,
Did anyone figure out how to solve part a yet ? I am trying to figure out α - ς. My final equation for JL is:
Vojkan, I agree with your J=F/(4*pi)*[...], above. This contradicts J=F/(2*pi)*[...] in L34 p8, since the 4*pi is not specific to the two transistor version. I suppose we're in agreement that the 2*pi in L34 is an error?
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5 comments:
J=F/(4*π)*[(JL1-JL4)*θ+(JL2-JL3)*δ]
Θ=α+ζ (Q1/D1 conduction angle),
δ=Q2/D2 conduction angle
HW10 pb2 defines D=(α + ζ)/(ωo*Ts) while L35 slide 11 defines D=(α + ζ+ξ)/(ωo*Ts)
Should dead time ξ/(ωo*Ts) be included in D?
Sorry for the confusion. I have just posted on the course website the definitions for D and various angles to apply in this problem
Vojkan, I agree with your
J=F/(4*pi)*[...], above.
This contradicts J=F/(2*pi)*[...] in L34 p8, since the 4*pi is not specific to the two transistor version. I suppose we're in agreement that the 2*pi in L34 is an error?
1/(Ts*2*wo)=fs/(4*pi*fo)=F/(4*pi)
J=F/(4*pi)*[...] is in the notes posted on the schedule page L33: QSWZVSnotes.pdf page 13
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