Calculate total inductance for 2-6 inductors in parallel, with current distribution, reactance, energy storage, and mutual coupling analysis.
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 tool 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. Check the example with realistic values before reporting. Use the steps shown to verify rounding and units. Cross-check this output using a known reference case.
Parallel inductor configurations appear in multi-phase power converters, EMI filters, RF matching networks, and current-sharing circuits. This calculator instantly 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.
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
Result: Total = 5.92 µH, X_L = 0.037 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.
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.
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.
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.
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.
At AC or transient conditions, current divides inversely with inductance — the smallest inductor carries the most current. This is opposite to resistors where the largest resistance carries the least current. For DC steady-state, current divides based on DC resistance (wire gauge), not inductance.
When two inductors are close together, their magnetic fields interact. The coupling coefficient k (0 to 1) quantifies how much flux from one passes through the other. k = 0 means no coupling (well-separated), k = 1 means perfect coupling (like a transformer). Mutual coupling changes the effective parallel inductance.
If inductors are physically close and especially if wound on the same core, coupling matters significantly. For PCB-mount components more than a few centimeters apart, coupling is typically negligible. Shielded inductors also minimize coupling. When in doubt, orient inductors at right angles to minimize mutual inductance.
Without mutual coupling, yes — the parallel combination is always less than the smallest individual inductor, just like parallel resistors. With aiding mutual coupling (k > 0), the effective inductance can be slightly higher than the uncoupled parallel value.
Yes, but be careful. In AC conditions, more current flows through the smaller inductor. Each inductor must handle its share of the current plus any DC bias. Check that no individual inductor exceeds its saturation current, especially in power applications.