Blog for students taking ECEN5817 Resonant and Soft Switching Techniques in Power Electronics, ECEE Department, University of Colorado at Boulder, Spring 2012
Saturday, April 24, 2010
HW11 P2 resonant intervals
In problem 2 it seems like the resonant intervals should be intervals 3 and 5 only, not intervals 2, 3, 5, and 6 as stated in the problem. Anyone agree or disagree?
Intervals 2,3,5 and 6 are all resonant. In interval 2, since all semiconductors are off, a resonant tank is formed with Ll and Cds, with the voltage across these being Vg. In interval 3, when D3 starts conducting, the resonant tank still remains the same, but the current reflected from the secondary to the primary is in series with Ll and tries to discharge Cds. This is equivalent to L38, slide 13. The same is the case with intervals 5 and 6. Hope this helps!
Ishminder is correct for this class of circuits but I also see where Andy is coming from. Note that some of the circuits we are analyzing are exactly like what we did on the midterm and in class weeks back. But in interval 2 (weeks ago), we didn't treat this as a resonant tank but a linear increase in current as Cds was charged. That's the puzzling part......its like we know how the circuit is suppose to behave so we are just trying to get equations to match.
I am still confused as to where the resonance in interval 6 comes from. Doesn't D1 turning on makes v1=0 by definition? Doesn't this make vl as DC value of Vg+V/n for this interval?
This problem was copied from an earlier, and I missed to correct the statement: since filter-component ripples can be neglected, only intervals 3 and 5 should be considered resonant in this problem. Sorry for the confusion
Does that mean that intervals 1 and 2 are both linear with the same slope of Vg/L? Interval 3 is when we have resonance and this is what contributes to the curve now going in the negative direction?
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6 comments:
Intervals 2,3,5 and 6 are all resonant. In interval 2, since all semiconductors are off, a resonant tank is formed with Ll and Cds, with the voltage across these being Vg. In interval 3, when D3 starts conducting, the resonant tank still remains the same, but the current reflected from the secondary to the primary is in series with Ll and tries to discharge Cds. This is equivalent to L38, slide 13. The same is the case with intervals 5 and 6. Hope this helps!
If I interpret my circuit correctly, I agree with Andy.
Ishminder is correct for this class of circuits but I also see where Andy is coming from. Note that some of the circuits we are analyzing are exactly like what we did on the midterm and in class weeks back. But in interval 2 (weeks ago), we didn't treat this as a resonant tank but a linear increase in current as Cds was charged. That's the puzzling part......its like we know how the circuit is suppose to behave so we are just trying to get equations to match.
I am still confused as to where the resonance in interval 6 comes from. Doesn't D1 turning on makes v1=0 by definition? Doesn't this make vl as DC value of Vg+V/n for this interval?
This problem was copied from an earlier, and I missed to correct the statement: since filter-component ripples can be neglected, only intervals 3 and 5 should be considered resonant in this problem. Sorry for the confusion
Does that mean that intervals 1 and 2 are both linear with the same slope of Vg/L? Interval 3 is when we have resonance and this is what contributes to the curve now going in the negative direction?
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