Buck Converter with Variable input Voltage

Discussion in 'Homework Help' started by notoriusjt2, Oct 19, 2010.

  1. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    [​IMG]
    i dont know what values to use
    [​IMG]

    before this was mysteriously deleted, i think TNK stated to use the lower voltage with the higher power. is that correct?
     
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    You misrepresent me.

    I said the lower power (75W) higher input voltage (60V) condition.
     
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    I think this is readily verifiable. My argument would be as follows:

    If the minimum current just reaches zero amp then the minimum value of L to permit this would be

    L=(Vs-Vo)*D*T/Imax

    If one "normalizes" the value of L with respect to the switching period T then ...

    L/T=(Vs-Vo)*D/Imax

    L/T increases as Imax decreases and as (Vs-Vo)*D increases.

    These two conditions give the highest required inductance at the lowest power + highest input voltage case.
     
  4. notoriusjt2

    Thread Starter Member

    Feb 4, 2010
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    so

    L=(Vs-Vo)*D*T/Imax

    L=[(60-20)*(20/60)*(1/100000)]/(75/60)
    L=[(40)(.333)(0.00001)]/1.25
    L=[0.00013]/1.25
    L=106.6uH


    somethings still not right...

    Imax=75/60 right? or would I use the higher wattage for Imax
     
  5. t_n_k

    AAC Fanatic!

    Mar 6, 2009
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    Keep in mind that the mean inductor current is (Imax+Imin)/2

    At 75W and 20V Imean = 3.75A

    If Imin just hits zero amps (the limiting condition) then Imax=2*3.75=7.5A.
     
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