Coil Inductance Calculator
Calculate the inductance of a single-layer air-core coil from its dimensions.
Uses Wheeler formula for solenoid inductance.
Coil Inductance determines how much energy a coil stores in its magnetic field. This calculator uses the Wheeler formula for single-layer solenoid (air-core) coils, which is accurate to within 1% for coils where the length is at least 0.4 times the diameter.
Wheeler’s Formula:
L = (d² × n²) / (18d + 40l) (in microhenries, with d and l in inches)
Where:
- L = Inductance (µH)
- d = Coil diameter (inches)
- n = Number of turns
- l = Coil length (inches) — the winding length, not wire length
Metric Version:
L = (d² × n²) / (45.72d + 101.6l) (in microhenries, with d and l in centimeters)
Factors Affecting Inductance:
| Factor | Effect |
|---|---|
| More turns | Inductance increases as n² |
| Larger diameter | More inductance (more flux area) |
| Longer coil (same turns) | Less inductance (weaker field) |
| Core material | Ferrite/iron cores multiply inductance |
| Spacing between turns | More spacing = less inductance |
Reactance at a Given Frequency:
X_L = 2π × f × L
Where X_L is inductive reactance in ohms, f is frequency in Hz, and L is inductance in henries.
Common Inductance Values:
| Application | Typical Range |
|---|---|
| RF coils | 0.01 – 100 µH |
| Power supply chokes | 1 – 100 mH |
| Audio crossovers | 0.1 – 10 mH |
| Antenna loading coils | 10 – 500 µH |
Practical Example: A coil with 20 turns, 1 inch (2.54 cm) diameter, and 2 inches (5.08 cm) length: L = (1² × 20²) / (18×1 + 40×2) = 400 / 98 = 4.08 µH.
At 7 MHz: X_L = 2π × 7,000,000 × 4.08×10⁻⁶ = 179 ohms.
Tips:
- Close-wound coils have maximum inductance for a given number of turns.
- Stretching the coil (increasing length) decreases inductance.
- For higher inductance in a small space, use a ferrite core.
- Wire gauge affects resistance but not inductance significantly.
Do not use this for cored coils
The Wheeler formula describes an air-core solenoid and nothing else. You will see advice to take the air-core answer and multiply it by the core material’s relative permeability, and that advice is wrong by orders of magnitude. Drop a ferrite rod of µr = 2,000 into a coil and you do not get 2,000 times the inductance. You might get 30 or 50 times, because a rod is an open magnetic path and the field has to close through air at the ends. The effective permeability depends on the rod’s length-to-diameter ratio, not on the material figure printed in the catalogue.
For a closed core, a toroid or a gapped pot core, skip the geometry entirely and use the A_L value from the core datasheet:
L = A_L × n²
with A_L in nanohenries per turn squared. A core with A_L = 250 nH/n² wound with 40 turns gives 250 × 1,600 = 400,000 nH = 400 µH. This is what core manufacturers publish precisely because calculating it from dimensions is unreliable.
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