 # How to Select Inductor for Boost Converter

## 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)

Example:

Dutycycle = 1 – (Vin / Vout) = 1 – (12V / 24V) = 50%

#### e. Know the Input Current (Iin)

Input current = Vout X Iload_max / Vin

Example:

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 ]

Example:

L = [ (0.5 / 300kHz) X 12V ] / [ 2A ] = 10uH

Where;

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) ]

Example:

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

Exmple:

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.

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