Parallel Inductance Calculator

Parallel Inductance Formula

Inductors in parallel share the same two nodes, so the same voltage appears across every one of them. Each branch then develops its own rate of change of current, inversely proportional to its inductance.

The branch currents add to the total current, so their rates of change add too:

Each rate of change is the shared voltage divided by that branch inductance, and the common v divides out:

Two inductors and equal inductors

For exactly two coils the reciprocals clear to product over sum, and for n equal coils the formula collapses to a division:

As always, product over sum is for two only. Both forms assume no coupling between the coils — with coupling, the parallel case gains cross terms and is considerably messier than the series one.

Worked example

Why Put Inductors in Parallel?

Less inductance for more parts is rarely the goal, so parallel inductors are almost always about current rather than inductance.

Sharing current and raising the saturation limit

Two identical coils in parallel each carry half the current, so the pair handles twice the current before either saturates — at half the inductance. Where a single part with the required current rating does not exist or costs too much, this is the standard answer.

Halving the DC resistance

Parallel windings put their DC resistances in parallel too, which halves the conduction loss for identical parts. In a switching converter output that loss is often what sets the efficiency.

Spreading heat

Two parts dissipating a quarter of the power each — half the current squared — run far cooler than one part dissipating all of it, and they present more surface area to the air.

Current Sharing Is Not Automatic

The formula assumes ideal inductors, and ideal inductors would share current in inverse proportion to their inductance. Real ones share it in inverse proportion to their DC resistance at DC, and to their impedance at frequency — which is not the same split.

So two nominally identical coils with a 20% tolerance spread will not carry equal currents. The one with the lower resistance takes more, heats more, and — since copper resistance rises with temperature — partly self-corrects. Partly. For parallel inductors carrying significant current, use identical parts from the same batch and derate for the imbalance rather than assuming a clean half each.

Coupling in Parallel Inductors

Two coupled coils in parallel are a much worse problem than two in series. In series, coupling shifts the total predictably by ±2M. In parallel, coupling creates a path for circulating current between the two branches, and if the winding sense is wrong that circulating current can be large while contributing nothing to the load.

The practical rules are the same as for series and matter more: keep the parts apart, use shielded or toroidal types, or mount them at right angles. The calculator above assumes no coupling.

Common mistakes

• Adding the inductances. Parallel inductors use reciprocals; series inductors add.
• Forgetting the final reciprocal, so the answer is inverted.
• Using product-over-sum for three or more coils. It is exact for two only.
• Accepting a total larger than the smallest inductor, which is impossible in parallel.
• Assuming parallel coils share current equally. The split follows resistance and impedance, not the nominal inductance.
• Paralleling coupled windings without checking the sense, which can drive a large circulating current between the branches.

Frequently asked questions

How do you calculate inductors in parallel?

Add the reciprocals and invert: 1/Leq = 1/L1 + 1/L2 + … + 1/Ln. For two coils use Leq = L1×L2/(L1+L2). The result is always smaller than the smallest inductor, assuming the coils do not couple.

Does parallel inductance decrease?

Yes. Parallel inductors behave like parallel resistors, so the total is below the smallest branch. Two equal coils in parallel give exactly half the inductance of one.

Why put inductors in parallel if it reduces inductance?

For current. Two coils in parallel each carry half the current, so the pair handles twice as much before saturating, with half the DC resistance and much less heating in each part. The reduced inductance is the price, not the purpose.

Do parallel inductors share current equally?

Not exactly. Sharing follows DC resistance at DC and impedance at frequency, not the nominal inductance, so tolerance spread makes one branch carry more than the other. Use identical parts and derate rather than assuming a clean split.

What is the formula for two inductors in parallel?

Leq = (L1 × L2) / (L1 + L2), the product over the sum. It is exact for two uncoupled coils and wrong for three or more, where the reciprocal formula is needed.