Parallel Inductance Calculator

Inductors connected in parallel — with both terminals joined to the same two circuit nodes — follow the same reciprocal-sum rule as parallel resistors: 1/L_total = 1/L1 + 1/L2 + 1/L3 + ... The total inductance is always less than the smallest individual inductor. This assumes no magnetic coupling between the inductors; mutual inductance between physically close inductors modifies the formula significantly.

The physical reason: inductance is a measure of a coil's ability to oppose changes in current by storing energy in its magnetic field. When inductors are in parallel, the applied voltage is the same across all of them, but each inductor independently limits the rate of current rise (dI/dt = V/L). With multiple parallel paths, more total current can flow for the same dI/dt, which is equivalent to a single inductor with lower inductance.

Parallel inductors find application in power electronics filter design. DC-DC converter output filters often use parallel inductor stages (interleaved converters) where multiple small inductors — each handling a fraction of the total current — reduce ripple current more effectively than a single large inductor. An interleaved buck converter with two parallel inductors sees ripple cancellation at the output due to phase shift between the two switching phases.

For two inductors in parallel, the product-over-sum shortcut applies: L_total = (L1 × L2) / (L1 + L2). This is the same mathematical form as parallel resistors, reflecting the circuit theory duality between inductors and resistors in parallel configurations.

In RF circuits, parallel inductor combinations form the tank circuit of LC oscillators and bandpass filters. The inductive reactance XL = 2πfL sets the impedance at the operating frequency. Parallel inductors reduce XL for a given frequency — or equivalently, shift the resonant frequency of an LC tank upward by reducing inductance.

When placing real inductors in parallel, designers must ensure the inductors have similar DC resistance (DCR) so current shares equally. If one inductor has much lower DCR, it will carry disproportionately more current, potentially saturating its core. Inductors with ferrite cores are especially susceptible to saturation at high current, making current sharing critical for reliable parallel operation.

Note: this calculator assumes no mutual inductance (M = 0) between inductors. For magnetically coupled inductors, the parallel formula becomes more complex: 1/L_total = 1/(L1 ± M) + 1/(L2 ± M), where the sign depends on whether the mutual coupling is aiding or opposing.