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Buck converter : L and C values

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jayce3390

New Member
Hello,

I want ot design a buck converter with the following specifications :

Vsupply :70V
Vload : from 65V to 20V
Iload : from 1.8A to 0.2A
Ripple : lower as possible
switching frequency : 1000 kHz

How to get L & C values from these specifications?

I tried to find in some books, but I don't find clearly L= and C= .
 

Chippie

Member
The L and C values are determined by Ton and Toff, which is a function of the switcher frequency, in this case the frequency is 1MHz..As BRE has suggested have a look at simple switchers, or look at the 78S40 data sheet...there are worked examples...Although I'm not sure if it will switch that high...TL494 is another...
 

jayce3390

New Member
I found this attached file on the 78S40 datasheet.

If I apply to my application I find :

C = (4x1000ns)/(8x0.1)=5uF

L=((70-5-60)/4)x500ns = 0.6uH

Do you think if I could use it to design my personnal buck converter?
 

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smanches

New Member
Duty D = Vo_min/Vin_max = 20V/70V = .286
Von = Vin_max - Vout_min = 70V - 20V = 50V
L = (Von x D) / (r x Iout x f) = (50V x .286) / (.4 x 1.8A x 1Mhz) = 19.8uH
 

jayce3390

New Member
Thank you Smanches, 19.8uH for the inductance, where is the mistake in the formula I used?
And what about C value? Maybe C is not necessary (L and the load make a filter?
 

smanches

New Member
The formulas from that datasheet make some assumptions about the circuit. They are designed for that chip alone, not a generic buck converter.

Plus, you always have to use the worst case values. In the case of a buck, it's max Vin, and min Vout and max Iload.

I don't have the formula for the cap on me. But I've seen it on the boards here and much more simple than the inductor. I'll see if I can find it.
 
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