Inductor Ripple Current Calculator

Calculator inputs

Formula used

Inductor ripple current is estimated from the inductor voltage during the switch on-time. For a switching period T = 1/f_s and duty ratio D, the on-time is T_on = D/f_s.

• Buck: ΔI_L ≈ (V_in − V_out) · D / (L · f_s)
• Boost: ΔI_L ≈ V_in · D / (L · f_s)
• Buck‑Boost (inverting): ΔI_L ≈ V_in · D / (L · f_s)

This calculator uses ideal duty relationships in auto mode and a continuous‑conduction approximation. For discontinuous modes, include detailed inductor current waveforms from your controller and load.

How to use this calculator

1. Select the converter topology matching your power stage.
2. Enter input and output voltages, plus inductance and switching frequency.
3. Choose duty cycle mode. Use auto for quick sizing checks.
4. Provide load current to compute average, peak, and ripple percent.
5. Click Calculate. Results appear above the form instantly.

Example data table

Example values are for illustration. Your design may require different inductance and frequency targets.

Professional article

Ripple current as a design metric

Inductor ripple current, ΔIₗ, is the peak‑to‑peak swing around the average inductor current. In power converters it drives copper loss, core loss, and heat. Designers target 20–40% ripple of average inductor current to balance size and efficiency while keeping current stress reasonable.

Interpreting ΔIₗ and ripple percentage

Ripple percentage compares ΔIₗ to the average inductor current (or to load current in buck stages). A 30% ripple means the inductor current traverses ±15% around its average. Lower ripple reduces voltage ripple and EMI, but it increases inductance, cost, and slows transient response.

Buck converter ripple relationship

For a buck stage in CCM, the inductor sees Vₗ,on = Vᵢₙ − Vₒᵤₜ and Vₗ,off = −Vₒᵤₜ. Using duty D ≈ Vₒᵤₜ/Vᵢₙ, the ripple is ΔIₗ ≈ (Vᵢₙ − Vₒᵤₜ)·D/(L·fₛ). Higher fₛ or L directly reduces ΔIₗ.

Boost and buck‑boost nuances

In a boost converter, D ≈ 1 − Vᵢₙ/Vₒᵤₜ and the inductor current equals the input current, which can exceed output load current. Ripple is often estimated as ΔIₗ ≈ Vᵢₙ·D/(L·fₛ). Buck‑boost stages share the same timing idea, but current ratings must include efficiency and conversion ratio.

Choosing L and switching frequency targets

Inductance and switching frequency trade size against switching loss. Doubling fₛ halves ΔIₗ, but may increase MOSFET switching loss and driver power. Increasing L can be safer for reliability. Use the calculator to sweep L and fₛ and observe peak and RMS current.

CCM versus DCM boundaries

Continuous conduction mode (CCM) assumes inductor current never reaches zero. If the computed Iₗ,min drops to or below zero, the converter enters discontinuous conduction (DCM) and ripple equations change. A quick check is Iₗ,avg > ΔIₗ/2. Staying in CCM improves predictable control and reduces peak currents.

Impact on device and capacitor stress

Peak inductor current sets the minimum ratings for the switch, diode, and sense resistor. Higher ripple increases inductor RMS current, raising I²R loss. Output capacitors see higher ripple current when ΔIₗ is large, so verify ripple ratings and ESR to keep output ripple and heating within limits.

Validation with measurements

After calculating, validate with a current probe or sense resistor waveform. Measure peak‑to‑peak current at steady load and compare to predicted ΔIₗ. Differences often come from inductor tolerance, temperature‑dependent inductance, and switching node ringing. Use measured data to refine L selection and confirm margins.

FAQs

What ripple current percentage is commonly used?

Many designs aim for 20–40% peak‑to‑peak ripple relative to average inductor current. Lower ripple can reduce noise, but increases inductance. Higher ripple may shrink the inductor, but raises peak current and losses.

Does increasing switching frequency always reduce ripple?

Yes, ΔIₗ falls as fₛ rises, but higher frequency increases switching loss, EMI challenges, and driver power. Pick fₛ based on efficiency, thermal limits, and allowable component size.

When should I use manual duty cycle instead of auto?

Use manual duty when you have a known fixed duty from control limits, non‑idealities, or measured waveforms. Auto is best for first‑pass sizing using ideal relationships between voltages.

What if I do not know converter efficiency?

Enter a conservative estimate such as 0.85–0.95. Efficiency mainly affects average inductor current for boost and buck‑boost cases. You can sweep efficiency to see the impact on peak current margins.

How do I check if the converter is in CCM?

If the calculated minimum inductor current is above zero, it is in CCM. A quick rule is Iₗ,avg > ΔIₗ/2. If not, DCM behavior may require a different model.

How can I avoid inductor saturation?

Compare the calculated peak inductor current to the inductor’s saturation rating at your operating temperature. Add margin for load transients and tolerance. If peak current is close, increase L or choose a higher‑rated inductor.

For boost converters, which current should I enter as load current?

Enter the output load current. The calculator converts it to average inductor current using the conversion ratio and efficiency. Remember the inductor typically carries input current, which can be significantly higher than Iout.