Blog for students taking ECEN5817 Resonant and Soft Switching Techniques in Power Electronics, ECEE Department, University of Colorado at Boulder, Spring 2012
Sunday, January 17, 2010
Hw1 Pb3
Hello All,
On Pb 3c (the boost conv with aux circuit), can I get a hint how the aux circuit should work ?
I wonder if the secondary diode will share load current with first diode when M is off ?
The idea is to decrease the diode current down to zero amps with slope di(D)/dt=<10A/usec (no rev recovery losses) and than turn M on (zero current switching). - During the period when M=off and D=on, Ms turns on(Ds=off)per timing diagram in fig3(pb3): current i(Ls) ramps up and i(D) ramps down with d(i(Ls))/dt=--d(i(D))/dt (since I_L=I=constant).
Aux circuit=DCM Boost conv. (with input voltage=Vds(M)) running out of phase with the original CCM boost conv. except when the g and gs intervals overlap.
Thank you again Vojkan. I still have question in regard to when Ds is supposed to be on. During period M and Ms off, can I assume both D and Ds are supposed to be turned on and split the current load I ?
All, Please comment on what is happening during the overlap between g and gs pulses; how it affects rev recovery losses and how long does the overlap period last? Thank you
Possible options: Having a shorted inductor Ls and i(Ls)=I=const, it=0. Upon turn on, mosfet M current is prevented from increasing at a rate of dit/dt=500A/usec…
This is what I think: during overlap, M turns on at a rate of dt/dt=90A/usec. The overlapping time needs to be long enough to let M totally turned on (it=its=I/2). If there is no overlap, when M and Ds are on, M will force ids to increase at 90A/us, which causes reverse recovery of Ds.
Vojkan, You say that when M and Ms are both off Ds is OFF. If that is so that Ls will have one end floating and its current will have no where to go. I think when M and Ms are off Ds will stay ON and since voltage across Ls will be nearly zero the current through Ls will be whatever it was at that start of that portion of the cycle.
That period was used as an example to show that both diodes can not be on at the same time. During the period when both mosfets are off, D=on and Ds=off: Ls inductor current is zero; i(Ls)=0 amps. Inductor Ls works in discontinuous conduction mode (DCM). During DCM it is perfectly normal to have one end of the inductor floating.
You may use this blog as a tool to post questions and comments related to Spring 2012 ECEN5817 homework problems. However, simply posting your solutions or copying someone else's work is not allowed - all work you turn in must be your own. Absolutely no collaboration in any form is allowed on exams.
8 comments:
The idea is to decrease the diode current down
to zero amps with slope di(D)/dt=<10A/usec
(no rev recovery losses) and than turn M on
(zero current switching).
- During the period when M=off and D=on,
Ms turns on(Ds=off)per timing diagram in fig3(pb3): current i(Ls) ramps up and i(D) ramps down with
d(i(Ls))/dt=--d(i(D))/dt
(since I_L=I=constant).
Aux circuit=DCM Boost conv. (with input
voltage=Vds(M)) running out of phase with the
original CCM boost conv. except when the g and gs intervals overlap.
Thank you again Vojkan.
I still have question in regard to when Ds is supposed to be on.
During period M and Ms off, can I assume both D and Ds are supposed to be turned on and split the current load I ?
Ds is on during the period when M=on and Ms=off.
When M and Ms are off, D=on and Ds=off.
Both diodes are not on at the same time.
Thanks. One last question. What is the purpose of M and Ms overlap ?
All,
Please comment on what is happening during the overlap between g and gs pulses; how it affects rev recovery losses and how long does the overlap period last?
Thank you
Possible options:
Having a shorted inductor Ls and i(Ls)=I=const, it=0.
Upon turn on, mosfet M current is prevented from increasing at a rate of dit/dt=500A/usec…
This is what I think: during overlap, M turns on at a rate of dt/dt=90A/usec. The overlapping time needs to be long enough to let M totally turned on (it=its=I/2). If there is no overlap, when M and Ds are on, M will force ids to increase at 90A/us, which causes reverse recovery of Ds.
Vojkan,
You say that when M and Ms are both off Ds is OFF. If that is so that Ls will have one end floating and its current will have no where to go. I think when M and Ms are off Ds will stay ON and since voltage across Ls will be nearly zero the current through Ls will be whatever it was at that start of that portion of the cycle.
That period was used as an example to show that both diodes can not be on at the same time.
During the period when both mosfets are off, D=on and Ds=off: Ls inductor current is zero;
i(Ls)=0 amps. Inductor Ls works in discontinuous conduction mode (DCM).
During DCM it is perfectly normal to have one end of the inductor floating.
Post a Comment