How to Select Inductor for Buck Converter: Key Steps Explained

To select an inductor for a buck converter, first calculate the required inductance using input voltage, output voltage, switching frequency, and desired ripple current. Then choose an inductor with a current rating above the maximum load current and low resistance to improve efficiency.

Syntax

The main formula to calculate the inductor value L for a buck converter is:

• L = (V_out * (V_in - V_out)) / (V_in * f_s * ΔI_L)

Where:

• V_in = Input voltage
• V_out = Output voltage
• f_s = Switching frequency
• ΔI_L = Desired peak-to-peak inductor ripple current

This formula helps determine the inductance needed to keep current ripple within limits.

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L = (Vout * (Vin - Vout)) / (Vin * fs * delta_IL)

Example

This example calculates the inductor value for a buck converter with 12 V input, 5 V output, 100 kHz switching frequency, and 30% ripple current of 2 A load current.

python

Vin = 12.0  # Input voltage in volts
Vout = 5.0   # Output voltage in volts
fs = 100000  # Switching frequency in Hz
I_load = 2.0 # Load current in amperes
ripple_percent = 0.3

delta_IL = ripple_percent * I_load  # Ripple current

L = (Vout * (Vin - Vout)) / (Vin * fs * delta_IL)  # Inductance in Henry

print(f"Calculated Inductance: {L*1e6:.2f} uH")

Output

Calculated Inductance: 41.67 uH

Common Pitfalls

Choosing too small inductance causes high ripple current, increasing losses and noise. Too large inductance slows response and increases size and cost.

Another mistake is selecting an inductor with a current rating below the peak current, which can cause saturation and failure.

Also, ignoring the inductor's DC resistance (DCR) leads to efficiency loss.

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## Wrong: Inductor current rating less than peak current
inductor_current_rating = 1.5  # A, less than load peak current 2.6 A

## Right: Choose current rating above peak current
inductor_current_rating = 3.0  # A, safe margin above peak current

Quick Reference

Parameter	Description	Typical Value/Tip
Inductance (L)	Calculated from formula	20-100 µH common range
Ripple Current (ΔI_L)	Usually 20-40% of load current	Lower ripple improves efficiency
Current Rating	Must exceed max load current + ripple	Add 20-30% margin
DC Resistance (DCR)	Lower is better for efficiency	Check datasheet
Saturation Current	Inductor must not saturate at peak current	Higher than max current

Key Takeaways

Calculate inductance using input/output voltage, switching frequency, and ripple current.

Select an inductor with current rating above peak load current to avoid saturation.

Keep ripple current between 20-40% of load current for balance of size and efficiency.

Choose inductors with low DC resistance to improve converter efficiency.

Avoid too large inductance to maintain fast transient response and reasonable size.