# 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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i dont know what values to use

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
209
0
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
5,448
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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.