Normally, I learned choosing right Buck converter's inductor value is based on the below equation.
This is also in the LM61460's datasheet page 31.
But page 32, there are two more equations about choosing inductor value.
So, I am confused.
In my system
Vin : 24V
Vout : 19V
fsw : 1.18Mhz(from the webench tool)
k = 0.3
Iout(max) = 6A
If I calculate with first equation then L value is about 1.86uH.
If I calculate with equation 10. L > 0.32*Vout/fsw, then L value is about 5.25uH
If I calculate with equation 11. L > 0.2*Vout/fsw, then L value is about 3.22uH.
Refer to the datasheet,
" Equation 10 assumes that this design must operate with input voltage near or in dropout. If minimum operating voltage for this design is high enough to limit duty factor to below 50%, Equation 11 can be used in place of Equation 10."
But, I cannot understand what this mean. My system's Vin is 24V, Vout is 19V. So is this input voltage near the operation?
Please, guide me for this.
If I choose inductor value for 5uH, then this becomes same results as I assume that K(inductor ripple) for the 10%. I think this is very tight.
What is appropriate L value for this system?
