Coil & Inductor Design
Estimates the inductance of three simple air-core shapes: a straight solenoid, a toroid, and a single straight round wire. For the solenoid and toroid it also gives the flux density B inside the winding at a given current. Electronics designers and hobbyists use it to size a winding for a filter, a sensor or a test coil, and to estimate the stray inductance of a lead.
Equations
Solenoid: L = μ0N²A/l B = μ0NI/l
Toroid: L = μ0N²A/(2πr) B = μ0NI/(2πr)
Straight wire: L = (μ0l/2π)·[ln(4l/d) − 1]
L is the inductance and B the flux density inside the winding. N is the number of turns and I the current. A is the cross-sectional area of one turn (of the former, or of the toroid’s core), l the length of the winding or of the wire, r the toroid’s mean radius (to the center of its cross-section) and d the wire diameter. μ0 = 4π × 10−7 H/m is the permeability of free space [1]. The wire gauge list fills d from the AWG diameters [2].
Limits and common mistakes
- The solenoid equation is exact only for a coil much longer than it is wide. A shorter coil has less inductance: multiply by Nagaoka’s factor K from Grover’s tables [1], which is 0.92 at a length of five diameters and 0.69 at one diameter.
- All three are for air or another nonmagnetic core. An iron or ferrite core multiplies L by its effective relative permeability, which depends on any gap and on the current.
- The wire equation leaves out the field inside the wire, so it suits high frequencies. At DC, use −0.75 in place of −1. It also needs a wire much longer than it is thick, and the page refuses one too short for it.
- A is in cm², not the diameter. A 20 mm former has A = π × 1² = 3.14 cm².
Worked example
A test coil of 200 turns is wound over 10 cm of a 20 mm plastic tube and will carry 2 A. Estimate its inductance and the field inside it.
- Choose Geometry Air-core solenoid.
- Enter Turns (N) 200 and Current (I) 2 A.
- Enter Cross-sectional area (A) 3.14 cm² and Coil length (l) 10 cm, and press Solve. The Solution shows:
Inductance (L): 157.834 µH
Magnetic flux density (B): 5.02655 mT
Check: N·I/l = 400 / 0.1 = 4000 A/m, and 4π × 10−7 × 4000 = 5.03 mT. The coil is five diameters long, so its real inductance is about 0.92 × 157.8 = 145 µH.
