Parallel Inductors Calculator

Calculate total inductance for 2-6 inductors in parallel, with current distribution, reactance, energy storage, and mutual coupling analysis.

2 to 6 inductors in parallel
Hz
A
0 = no coupling, 1 = perfect (only for 2 inductors)
µH
µH
µH
Total (Parallel)
5.9977 µH
1/L = 1/L₁ + 1/L₂ + ...
Total (Series)
79.0000 µH
L = L₁ + L₂ + ... (for comparison)
Inductive Reactance (X_L)
0.038 Ω
At 1000 Hz
Energy Stored
3.00 µJ
E = ½LI² at 1A
Series ÷ Parallel Ratio
13.17×
Series is always larger than parallel

Current Distribution

L1
0.600A (60.0%)
10 µH
L2
0.273A (27.3%)
22 µH
L3
0.128A (12.8%)
47 µH

Most current flows through the smallest inductor (lowest impedance path).

Individual Inductor Details

#Value (µH)X_L (Ω)Current (A)EnergyCurrent %
L110.000.060.6001.8 µJ60.0
L222.000.140.2730.8 µJ27.3
L347.000.300.1280.4 µJ12.8

Standard Inductor Values Reference

ValueTypical TypeCurrent RatingApplication
1-10 nHAir core / PCB trace0.5-5 AGHz RF matching
10-100 nHChip inductor0.1-3 ARF / high-speed digital
0.1-10 µHSMD power1-30 ADC-DC converter
10-100 µHDrum core0.5-10 AEMI filter / buck converter
100 µH - 1 mHToroidal0.1-5 APower line filter
1-10 mHBobbin wound0.01-1 AAudio crossover
10-100 mHLaminated core0.01-0.5 AChoke / relay drive
0.1-10 HIron core0.001-0.1 AAudio output transformer
Planning notes, formulas, and examples

About the Parallel Inductors Calculator

The **Parallel Inductors Calculator** computes the total equivalent inductance when 2 to 6 inductors are connected in parallel. Like resistors in parallel, the reciprocals add: 1/L_total = 1/L₁ + 1/L₂ + ... — so the parallel combination is always less than the smallest individual inductor.

This calculator goes beyond the basic formula to calculate **current distribution** through each inductor (using the current divider principle), **inductive reactance** at a given frequency, **energy stored** at the total current, and optionally the effect of **mutual coupling** between two inductors. The series configuration total is also shown for comparison.

Power electronics engineers use parallel inductors to share current in multi-phase converters. RF designers combine inductors to achieve precise values from standard parts. This calculator helps visualize how current divides inversely with inductance — with more current flowing through smaller inductors. The current split view is useful for spotting saturation risk and for checking whether nearby windings are changing the parallel total through mutual coupling.

When This Page Helps

Parallel inductor configurations appear in multi-phase power converters, EMI filters, RF matching networks, and current-sharing circuits. This calculator determines the total inductance, current split, and impedance — information that requires careful manual calculation with reciprocals.

The current distribution visualization is particularly valuable because the inverse relationship (more current through smaller inductors) is counterintuitive for many designers. The mutual coupling option for two inductors addresses a common practical concern in compact PCB layouts.

How to Use the Inputs

  1. Select the inductance unit (nH, µH, mH, or H).
  2. Choose how many inductors (2 to 6) to combine in parallel.
  3. Enter the signal frequency and total current for reactance and energy calculations.
  4. Optionally set the coupling coefficient k (for 2 inductors with mutual inductance).
  5. Enter each inductor's value in the chosen unit.
  6. Read the total parallel inductance, reactance, and stored energy.
  7. Check the current distribution chart to see how current splits between inductors.
Formula used
Parallel Inductance (no coupling): 1/L_total = 1/L₁ + 1/L₂ + ... + 1/L_n With mutual coupling (2 inductors, aiding): L_parallel = (L₁L₂ − M²) / (L₁ + L₂ − 2M) where M = k√(L₁L₂) Current divider: I_k = I_total × (L_total / L_k) Reactance: X_L = 2πfL Energy: E = ½LI²

Example Calculation

Result: Total = 6.00 µH, X_L = 0.038 Ω

1/L = 1/10 + 1/22 + 1/47 = 0.100 + 0.0455 + 0.0213 = 0.1667. L_total = 1/0.1667 = 5.998 µH. Current splits: I₁ = 0.60A (60%), I₂ = 0.27A (27%), I₃ = 0.13A (13%) — most current through the smallest inductor.

Tips & Best Practices

  • For two equal inductors in parallel: L_total = L/2. For N equal inductors: L_total = L/N.
  • When paralleling inductors in power circuits, match DCR (DC resistance) as well as inductance to ensure equal current sharing.
  • Most current flows through the smallest inductor — ensure it can handle the higher current without saturating.
  • To minimize mutual coupling, orient parallel inductors at 90° angles on the PCB.
  • Shielded inductors are preferred for parallel configurations to prevent unwanted coupling.
  • For the series comparison: series inductance is L₁ + L₂ + ... — the sum, not the reciprocal sum.

Current Sharing in Power Applications

In multi-phase DC-DC converters, each phase has its own inductor but they share the output capacitor in parallel. Current sharing depends on both inductance matching and DCR (DC resistance) matching. A 10% inductance mismatch causes roughly 10% current imbalance at the switching frequency, while DCR mismatch affects the DC current distribution. High-performance designs use coupled inductors on a single core to improve both current sharing and transient response.

Magnetic Coupling Effects

When parallel inductors share magnetic coupling, the effective inductance changes. For aiding connections (fields in same direction), the parallel inductance increases above the uncoupled value. For opposing connections (fields opposite), it decreases below it. Transformer-coupled inductors intentionally maximize coupling (k → 1) while filter inductors minimize it (k → 0) to maintain independent operation.

Saturation Considerations

Inductors saturate when the core magnetic material reaches its flux density limit, causing inductance to drop sharply and current to spike. In parallel configurations, the inductor carrying the most current saturates first. Once it saturates, its inductance drops, causing even more current to flow through it — a positive feedback that can be destructive. Always verify that each individual inductor operates within its saturation limits at peak current conditions.

Sources & Methodology

Last updated:

Frequently Asked Questions

  • Parallel inductors reduce total inductance, share current loading, and can improve saturation behavior. In multi-phase converters, parallel inductors on separate phases handle higher total current. They're also used to achieve non-standard inductance values from standard parts.