1. Select Inductance Value
a. Define the switching frequency for the boost converter
Example: Fsw = 300kHz
b. Define the input and output voltage
Example: Vin = 12V, Vout = 24V
c. Determine the duty cycle
Dutycycle = 1 – (Vin / Vout)
Dutycycle = 1 – (Vin / Vout) = 1 – (12V / 24V) = 50%
d. Know the Maximum Load Current, (Iload)
Example: Iload_max = 10A
e. Know the Input Current (Iin)
Input current = Vout X Iload_max / Vin
Input current = Vout X Iload_max / Vin = 24V X 10A / 12V = 20A
Above equation is just a quick method; not that accurate but more than enough to do the job.
f. Assume Ripple Current, (ripple)
The rule of thumb is 10%-30% of the input current.
Example: ripple = 10% X Iin = 0.1 X 20A = 2A
g. Compute the Inductance Value using Below Equation
L = [ (Dutycycle / Fsw) X Vin ] / [ ripple ]
L = [ (0.5 / 300kHz) X 12V ] / [ 2A ] = 10uH
Dutycycle – boost converter duty cycle
Fsw – Boost converter switching frequency
Vin – Boost converter input voltage
Vsw_drop – voltage drop of the switch (can be assumed 0 for ideal)
Ripple – ripple current of the inductor
Vout – Output voltage of the boost converter
2. Selecting the Inductance Value Based on Catalogue Part and Recompute the Ripple Current
In case the resulting inductance value is not a standard value, you need to select a standard one. Then, rewrite the inductance equation to get the corresponding ripple current based on the selected inductance.
The computed inductance above is already a standard value.
3. Compute the Peak Current (Ipeak)
The peak current will be the basis of the inductor saturation current rating. The actual inductor must have a saturation current rating higher than the computed peak current.
Ipeak = [ Dutycycle X ripple – 2 X Iload – ripple ] / [ 2 X (Dutycycle – 1) ]
Ipeak = [ 0.5 X 2A – 2 X 10A – 2A ] / [ 2 X (0.5 – 1) ] = 21A
4. Compute the RMS Current
The RMS current is needed to size accordingly the continuous current rating of the inductor.
Irms = [ ripple / sqrt(3) ] + Ipeak – ripple
Irms = [ ripple / sqrt(3) ] + Ipeak – ripple = [ 2A / sqrt(3) ] + 21A – 2A = 20.15A
5. Final Part Selection
Select an inductor with inductance of 10uH. Select the one that has the smallest tolerance. The inductor rms current must be higher than 20.15A. Consider a maximum stress of 75%. The inductor saturation current rating must be higher than 21A. The 75% maximum stress will do.