🔧 Inductor Values
📐 Circuit Diagram
⚙️ Formula
📋 Branch Impedance
| Inductor | Inductance | XL (Impedance) |
|---|
📊 Results
Frequency
1000 Hz
Leq
—
XL Total
—
Equivalent Inductance
Total Inductive Reactance (XL = 2πfLeq)
Stored Energy (E = ½LeqI²) at 1A
Inductors in Parallel Calculator — Complete Guide to Parallel Inductance
This inductors in parallel calculator instantly finds the equivalent inductance when multiple inductors share the same two terminals using the reciprocal formula 1/Leq = 1/L1 + 1/L2 + … + 1/Ln. It also shows the current in each branch, total inductive reactance XL = 2πfLeq at any operating frequency, and total stored energy — with a live circuit diagram. Use it for multi-phase DC-DC converters, current-sharing designs, fine-tuning LC resonators, EMI filter design, and any circuit where splitting inductance across parallel paths is needed.
Quick Reference: Parallel Inductor Formulas
| Quantity | Formula | Notes |
|---|---|---|
| Equivalent Inductance | 1/Leq = 1/L1 + 1/L2 + … + 1/Ln | Always < smallest L |
| Two Inductors (shortcut) | Leq = (L1 × L2) / (L1 + L2) | Product over sum |
| n Equal Inductors | Leq = L / n | e.g., 3× 33 µH → 11 µH |
| Voltage (same on all) | V1 = V2 = … = V | Defining property of parallel |
| Current in branch Li | Ii = V / XLi = V / (2πf × Li) | Smaller L → more current |
| Inductive Reactance | XL(eq) = 2π × f × Leq | Total parallel reactance |
| Stored Energy | E = ½ × Leq × Itotal² | Also Ei = ½ × Li × Ii² |
The Parallel Inductor Formula Explained
When inductors are wired in parallel, the same voltage appears across every inductor's terminals. Each branch provides an independent path for current with its own magnetic field. The total opposition to current change decreases because there are now multiple parallel paths sharing the load — exactly analogous to parallel resistors reducing effective resistance:
The result is always less than the smallest individual inductor. This assumes no mutual coupling between inductors (M = 0).
Product-Over-Sum Shortcut (Exactly Two Inductors)
Example: 10 mH and 22 mH in parallel → (10 × 22) / (10 + 22) = 220 / 32 = 6.875 mH
Equal-Inductor Shortcut
Example: Four 4.7 µH inductors in parallel → 4.7 / 4 = 1.175 µH
Current Distribution in Parallel Inductors
Because all parallel inductors share the same voltage, the current in each branch is inversely proportional to its inductive reactance — and therefore inversely proportional to its inductance. Smaller inductors carry more current:
Itotal = V / XL(eq) = I1 + I2 + … (phasor sum at same phase)
Mutual Inductance in Parallel Circuits
When two parallel inductors are magnetically coupled (mutual inductance M), the equivalent inductance changes significantly:
Leq = (L1×L2 − M²) / (L1 + L2 − 2M)
Opposing (fields in opposite direction):
Leq = (L1×L2 − M²) / (L1 + L2 + 2M)
This calculator assumes M = 0 (no mutual coupling).
Worked Examples
Example 1: Two Inductors in Parallel — Product-Over-Sum
Problem: 10 mH and 22 mH in parallel at 1 kHz. Find Leq and XL.
Leq = (10 × 22) / (10 + 22) = 6.875 mH
XL = 2π × 1000 × 0.006875 = 43.2 Ω
Example 2: Three Equal Inductors in Parallel
Problem: Three 33 µH inductors in parallel. Find Leq and compare current sharing.
Leq = 33 / 3 = 11 µH — Each branch carries 1/3 of the total current. ✓
Example 3: Four-Phase Buck Converter
Problem: Four 4.7 µH inductors in a 4-phase buck converter at 500 kHz. Find effective output Leq and XL.
Leq = 4.7 / 4 = 1.175 µH
XL = 2π × 500 000 × 1.175×10⁻⁶ = 3.69 Ω
Each phase carries 1/4 of the load current, keeping individual core temperatures low.
Parallel vs Series Inductors — Complete Comparison
| Inductors in Parallel | Inductors in Series | |
|---|---|---|
| Formula | 1/Leq = Σ(1/Li) | Leq = ΣLi |
| Result vs individual | Always < smallest L | Always > largest L |
| Voltage across each | Same on all | Divides (proportional to L) |
| Current through each | Divides (inversely with L) | Same through all |
| Analogous to resistors | Parallel resistors | Series resistors |
| Analogous to capacitors | Capacitors in series | Capacitors in parallel |
| Used for | Current sharing, multi-phase, fine-tuning L | Increasing L, filter design, EMI chokes |
Inductance Units Reference
| Unit | Symbol | Value in Henries | Typical Use |
|---|---|---|---|
| Henry | H | 1 H | Large power-line chokes, transformers |
| Millihenry | mH | 10⁻³ H | Audio filters, DC-DC converter chokes |
| Microhenry | µH | 10⁻⁶ H | Switching supplies, RF inductors |
| Nanohenry | nH | 10⁻⁹ H | RF circuits, PCB trace inductance |
Practical Applications of Parallel Inductors
- Multi-phase DC-DC converters: Each phase uses its own inductor; all phases effectively appear in parallel at the output, reducing current ripple and spreading heat.
- Current sharing: Distribute high current across multiple lower-rated inductors when no single part meets the full current specification.
- Saturation prevention: Splitting current across parallel inductors keeps each one well below its saturation current threshold, maintaining inductance stability.
- Fine-tuning LC resonators: Combine standard-catalogue inductors in parallel to hit a precise non-standard target value for filter or oscillator design.
- EMI filter redundancy: Parallel common-mode chokes on redundant signal paths provide symmetric filtering while sharing impedance.
- Thermal management: Parallel inductors share resistive losses (I²R) across multiple packages, reducing hot-spot temperatures.
Common Mistakes to Avoid
- Result larger than smallest inductor: If your Leq is bigger than any individual inductor, you've used the series formula — recheck.
- Ignoring mutual coupling: Two parallel inductors on a PCB without shielding will not simply follow the reciprocal formula — their fields interact. Measure Leq or use the M-corrected formula.
- Mismatched saturation currents: If parallel inductors have different saturation thresholds, the one that saturates first forces all the current onto the remaining inductors, potentially causing a cascade failure.
- Mixing units without converting: Convert everything to one unit before calculating. 47 nH and 10 µH in parallel = 0.047 µH and 10 µH, not 47 and 10 of the same unit.
Frequently Asked Questions
Why does parallel inductance decrease?
Parallel inductors provide multiple current paths. More paths means less total opposition to current change — lower effective inductance. This is directly analogous to parallel resistors lowering effective resistance.
Why put inductors in parallel instead of one large inductor?
Three reasons: (1) current sharing — no single component handles the full current; (2) thermal distribution — heat spreads across multiple parts; (3) component availability — two standard values in parallel can hit a non-catalogue target.
Does mutual inductance affect parallel inductors?
Yes. For two magnetically coupled parallel inductors: Leq = (L₁L₂ − M²) / (L₁ + L₂ − 2M) for aiding, or (L₁L₂ − M²) / (L₁ + L₂ + 2M) for opposing. This calculator assumes M = 0.
Can I mix different inductor types in parallel?
Yes, as long as each part can handle its share of the total current without saturating. Air-core, ferrite, and iron-powder inductors can all be combined in parallel (assuming no mutual coupling).
Related Calculators
- Inductors in Series Calculator — Leq = L1 + L2 + L3
- Capacitors in Parallel Calculator — same simple addition as series inductors
- RLC Series Resonance Calculator
- RLC Parallel Resonance Calculator
- Parallel Resistor Calculator — same reciprocal formula
- AC Power Calculator — inductive reactive power