Inductor Network Calculator
This inductor network calculator combines up to five inductors connected in series or parallel into a single equivalent inductance (Leq). If you also provide a current value, it estimates the magnetic energy stored in that equivalent inductance. It’s a practical helper when you are selecting standard parts to hit a target inductance, sanity-checking a prototype, or comparing series vs. parallel arrangements in filters, resonant tanks, and power electronics.
What the calculator outputs
- Equivalent inductance of the network (Leq, in henries).
- Stored energy (optional) if you enter current (E, in joules), computed from Leq and I.
Units and entering values
Inductance is measured in henries (H), but most real inductors are specified in smaller units:
- 1 mH = 10−3 H
- 1 µH = 10−6 H
- 1 nH = 10−9 H
Enter all inductors in henries. Examples:
- 47 µH = 47 × 10−6 H = 0.000047 H
- 2.2 mH = 2.2 × 10−3 H = 0.0022 H
| Given value | Convert to H | Result (H) |
|---|---|---|
| 10 µH | 10 × 10−6 | 0.000010 |
| 330 µH | 330 × 10−6 | 0.000330 |
| 4.7 mH | 4.7 × 10−3 | 0.0047 |
| 1 H | — | 1 |
Formulas used
The calculator treats the set of inductors as an ideal series or ideal parallel network (see limitations below). Use the formulas that match your wiring:
Inductors in series
For series inductors, the same current flows through each inductor and the voltages add. The equivalent inductance is the sum:
In plain terms: add up every non-blank inductor value you entered.
Inductors in parallel
For parallel inductors, each branch sees the same voltage and the current splits among branches. The reciprocals add:
1 / Leq = Σ (1 / Li)
Then:
Leq = 1 / Σ (1 / Li)
Energy stored (optional)
If you enter a current, the tool estimates stored magnetic energy using the equivalent inductance:
E = ½ · Leq · I2
Where E is joules (J), Leq is henries (H), and I is amperes (A).
How to interpret the results
- Series increases inductance: adding inductors in series increases Leq linearly (two identical inductors in series → double the inductance).
- Parallel decreases inductance: adding parallel branches decreases Leq (two identical inductors in parallel → half the inductance).
- Energy depends strongly on current: because energy scales with I2, doubling current quadruples stored energy for the same Leq.
Worked example (parallel + energy)
Three inductors in parallel:
- L1 = 10 mH = 0.010 H
- L2 = 20 mH = 0.020 H
- L3 = 40 mH = 0.040 H
Compute the reciprocal sum:
- 1/L1 = 1/0.010 = 100 H−1
- 1/L2 = 1/0.020 = 50 H−1
- 1/L3 = 1/0.040 = 25 H−1
Sum = 100 + 50 + 25 = 175 H−1
Take the reciprocal:
Leq = 1 / 175 ≈ 0.005714 H = 5.714 mH
If the network current is I = 2 A, the stored energy estimate is:
E = ½ · 0.005714 · (2)2 = 0.5 · 0.005714 · 4 ≈ 0.011428 J
So the network stores about 0.0114 joules at 2 A under the ideal assumptions below.
Series vs. parallel: quick comparison
| Aspect | Series inductors | Parallel inductors |
|---|---|---|
| Equivalent inductance | Adds: Leq = L1 + L2 + … | Reciprocals add: 1/Leq = 1/L1 + 1/L2 + … |
| Identical parts (N of them) | Leq = N·L | Leq = L/N |
| Current sharing | Same current through each inductor | Current splits between branches (may be uneven in real life) |
| Common motivation | Increase inductance without sourcing a single large value | Reduce inductance and/or increase current capability via multiple parts |
Assumptions and limitations (important)
- Ideal inductors: ignores winding resistance (DCR), core losses, parasitic capacitance, and skin/proximity effects.
- No mutual coupling: assumes inductors do not magnetically couple. If coils are near each other or share a core, coupling can increase or decrease the effective inductance and the simple series/parallel rules may not apply.
- No saturation: real inductors lose inductance as current increases toward saturation. The energy result is an estimate and can be optimistic at high current.
- Frequency dependence not modeled: inductance and losses can vary with frequency; for RF or fast-switching converters, parasitics can dominate.
- Tolerance not included: component tolerances (e.g., ±10%, ±20%) and temperature drift can materially change the true equivalent inductance.
- Parallel edge cases: extremely small entered inductance values (approaching 0 H) will dominate the reciprocal sum and drive Leq toward 0. Physically, this corresponds to a near-short inductive reactance at the modeled frequency, which is rarely “ideal.”
- Using fewer than five inductors: leave unused fields blank; only the values you enter should be considered part of the network.
If you need accuracy for tightly-coupled inductors, gapped cores, or high-current designs, treat these results as a first-pass calculation and validate with datasheets, measurement, and (when applicable) a coupled-inductor model.