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Anonymous Poster

Isolated inverse sepic converter

05/03/2009 12:36 AM

I was wondering if anyone remembers their school days when they did small ripple equations for an inverse sepic converter. I am using the erickson power electronics book and am having trouble doing problem 6.3 that is an isolated inverse sepic converter. I have the question answered, just wanted to check it with someone.

If interested I can email you info.

Thanks for your time
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Anonymous Poster #1
#1

Re: Isolated inverse sepic converter

03/13/2011 4:29 PM

Were you able to find answer to question 6.3?
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Roger Brewer
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Join Date: Oct 2011
Posts: 1
#2

Re: Isolated inverse sepic converter

10/13/2011 9:04 PM

Hi. I too am now working through that problem and having troubles. In writing volt-sec balance across Lm I come up with D*Vg + D'Vc1/n or Vc1=-D/D'*(nVg). Then writing volt-sec balance across L2 I come up with D*(Vgn-VC1-V)-D'*V=0. If I plug VC1 from above I get VgDn+Vg*(D^2/D')*n=V=Vc2. It looks to me like we have Vg*n acoss the secondary with a KVL reaching us to Vgn=VC1+Vl2+V when the transistor is on. When transistor is OFF, Vl2=-V and capacitor C1 is shorted across secondary since D1 is ON ? So the ripple for i is the slope of Vg/Lm - slope (Vgn-VC1-V)/L2 ? Very confused here ?
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eebrenner
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Join Date: Mar 2012
Posts: 1
#3

Re: Isolated inverse sepic converter

03/05/2012 2:28 PM

Did you ever complete 6.3? I am trying to work through it in preparation for a test and getting stumped.
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