Blog for students taking ECEN5817 Resonant and Soft Switching Techniques in Power Electronics, ECEE Department, University of Colorado at Boulder, Spring 2012
Wednesday, January 20, 2010
Switching loss
I am getting this in order of hundreds of Watts. This doesn't sound right.
With the new 90A/us is lower than before but still int he high tens which gives efficiency <90%....I guess this is not a synchronous converter so lower efiicinecy is expected. Hope I am correct.
The first interval you can treat it like the textbook example like you told me before (because Vg is constant across)
The second interval you treat with two linear functions. Find the equation of the line between these two pts: (0, 400V), (30ns, 0) for v(t). For the current, I assumed a straight line with points (0, -2.7A) and (30ns, 0)
The 2.7A is 90*30ns. Multiply those two functions to get a polynomial and integrate from 0 to 30ns.
Guys, I am getting much higher losses. ~69W for M and ~20W for D and eff ~85%. I think my method is correct but I'm sure I'm making a math error somewhere.
In the tb interval, I got linear equations for v(t) and i(t) for both D and M and and integrating the product of v(t) amd i(t) gave me the energy which I then multiplied by fs to get P. I mde one approximation, the product gives a quadratic inegral of which has cube and square terms which I ignored as they would be very small.
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6 comments:
With the new 90A/us is lower than before but still int he high tens which gives efficiency <90%....I guess this is not a synchronous converter so lower efiicinecy is expected. Hope I am correct.
Hey Yusef, what approach did you use for the time interval tb? Did you integrate the product of i(t)*v(t)?
There are two intervals in the transistor:
The first interval you can treat it like the textbook example like you told me before (because Vg is constant across)
The second interval you treat with two linear functions. Find the equation of the line between these two pts: (0, 400V), (30ns, 0) for v(t). For the current, I assumed a straight line with points (0, -2.7A) and (30ns, 0)
The 2.7A is 90*30ns. Multiply those two functions to get a polynomial and integrate from 0 to 30ns.
Let me know if you agree.
Hello All,
I just finished calculating the switching loss using new info and want to check if it's sound right.
Ploss_switch = 34.69 W,
Ploss_diode 1.08 W (during period tb only, assume linear i(t) and v(t) as yusuf suggested)
Conv Eff = 92.85%
Diode losses compare with mine.....the transistor losses are higher than mine. I got 23.4W.
Guys, I am getting much higher losses. ~69W for M and ~20W for D and eff ~85%. I think my method is correct but I'm sure I'm making a math error somewhere.
In the tb interval, I got linear equations for v(t) and i(t) for both D and M and and integrating the product of v(t) amd i(t) gave me the energy which I then multiplied by fs to get P. I mde one approximation, the product gives a quadratic inegral of which has cube and square terms which I ignored as they would be very small.
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