DC-DC Buck converter calculation

Thread Starter

anishkgt

Joined Mar 21, 2017
523
Hey All,
I am trying to design a high current DC-DC buck converter using LT8390. The voltage source is a 12v 82A power supply salvaged from a Server.

In the datasheet from page 18 onwards, to calculate Inductance \[ L = 10*{vout}*R{sense} \]. Then to calculate Rsense
\[ frac2*50{m}V/2*I{out}{(max)+\Delta I{L}}{(buck)} \] So to find delta IL(buck) = Vout x (Vin(max)-Vout)/f x L x Vin(max)

Every formula requires one variable from each other. Can somebody help me find what i've been missing ?
 

Thread Starter

anishkgt

Joined Mar 21, 2017
523
So based on this formula on Page2 Section 3

L=Vout(Vin(max)-Vout)/Delta IL x fs x Vin(max), substituting
= 8.1(24-8.1)/(80 x 150000 x 24)
= 0.00000045. I presume the unit is Henry (H). Searching digikey i can't find any suitable inductor with that spec. Would anyone know how to ?
 

Papabravo

Joined Feb 24, 2006
13,967
You would have to make one from magnet wire and a toroid. 450 nH if I read your string of zeros correctly. Nobody writes strings of zeros like that. Use scientific notation like 4.5e-7 or engineering units like 450e-9. It greatly reduces the probability of error.

I didn't watch the whole thing, but this will set you on the right track.

 

eetech00

Joined Jun 8, 2013
1,947
So based on this formula on Page2 Section 3

L=Vout(Vin(max)-Vout)/Delta IL x fs x Vin(max), substituting
= 8.1(24-8.1)/(80 x 150000 x 24)
= 0.00000045. I presume the unit is Henry (H). Searching digikey i can't find any suitable inductor with that spec. Would anyone know how to ?
Use this formula:

L=(Vout*(VinMax-Vout))/Kind*Fsw*VinMax*Iout

Where Kind= ripple current coefficient and is relative to max output current. A good value is 0.2-0.4

It is from a TI document and is basically the same as LT equation.

https://www.ti.com/lit/an/slva535b/slva535b.pdf
 
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