Series/Parallel Inductance
Finds the single inductance that can replace up to four inductors wired all in series or all in parallel. Use it to reach an inductance that isn’t stocked, for example in a filter or a loudspeaker crossover, or to reduce a group of chokes before further circuit analysis.
Equations
Series: Leq = L1 + L2 + L3 + L4
Parallel: 1/Leq = 1/L1 + 1/L2 + 1/L3 + 1/L4
L1 to L4 are the individual inductances and Leq the equivalent inductance [1]. Inductors combine like resistors. In series they carry the same current, and their voltages, L·dI/dt, add. In parallel they share one voltage, and their currents add. Two in parallel give L1L2/(L1 + L2). The result is shown in mH.
Inputs and limits
- Each field has a unit picker (µH, mH, H and so on). Leave a field at 0 to leave that inductor out. Inductors used counts the ones that went into the result.
- Negative values are rejected.
- All inductors must be in the same arrangement. For a mixed network, work out each group first and then combine the group results.
- The equations assume no mutual inductance: the coils must not share magnetic flux. Two coils that couple add or subtract 2M in series, where M is their mutual inductance, and the page can’t model that.
Common mistakes
- Mounting inductors close together and side by side, so their fields couple. Space them apart or set their axes at right angles.
- A series total is always larger than the largest inductor. A parallel result is always smaller than the smallest one.
- Ignoring winding resistance and saturation. Real inductors also have DC resistance, which combines by the resistor rules, and a current rating. In parallel the current splits in inverse proportion to inductance, so the smaller inductor carries more of it.
- Mixing up µH and mH: 470 µH is 0.47 mH.
Worked example
A loudspeaker crossover calls for a 1.5 mH air-core inductor, but only 2.2 mH and 4.7 mH coils are on hand. Do the two in parallel come close enough?
- Choose Configuration Parallel.
- Enter L1 2.2 mH and L2 4.7 mH, and leave L3 and L4 at 0.
- Press Solve. The Solution shows:
Equivalent inductance (Leq): 1.49855 mH
Inductors used: 2
Check with product over sum: 2.2 × 4.7 / 6.9 = 10.34 / 6.9 = 1.499 mH, within 0.1 % of the target. Mount the two coils well apart, or at right angles, so that they don’t couple and shift the value.
