Inductance Calculator

Inductor Diagram
Air Core Inductor Length (l) Inductance: 0.00 µH Enter values to calculate

Inductance Calculator - Educational Guide

How to Calculate Inductance

Inductance calculation is essential in electronics and electrical engineering for designing circuits with coils, transformers, and inductors. This calculator uses Wheeler's formula to determine the inductance of single-layer air core coils, which are commonly used in radio frequency applications, filters, and oscillator circuits.

Applications and Importance:

  • RF circuit design and antenna tuning
  • Filter design for audio and radio applications
  • Oscillator circuit calculations
  • Impedance matching in transmission lines
  • Educational purposes in electronics courses

Formula and Variable Definitions

Wheeler's formula for single-layer air core inductors provides accurate results for coils where the length is greater than 0.8 times the diameter:

$$L = \frac{D^2 N^2}{18D + 40l}$$

Variable Definitions:

Variable Description Unit Range
L Inductance microhenries (µH) 0.1 - 10,000 µH
D Coil diameter millimeters (mm) 1 - 1000 mm
N Number of turns dimensionless 1 - 10,000 turns
l Coil length millimeters (mm) 1 - 1000 mm

Formula Assumptions:

  • Single-layer winding
  • Air core (no ferromagnetic material)
  • Uniform turn spacing
  • Cylindrical coil geometry
  • Coil length > 0.8 × diameter
  • Wire diameter << coil diameter

Step-by-Step Calculation Process

Step 1: Formula Display and Theoretical Foundation

Wheeler's formula is derived from electromagnetic field theory and provides an empirical approximation for inductance calculation. The formula relates the physical dimensions of the coil to its electrical properties.

$$L = \frac{D^2 N^2}{18D + 40l} \text{ (in microhenries)}$$

Step 2: Example Problem

Let's calculate the inductance for a coil with the following specifications:

  • Coil diameter (D) = 20 mm
  • Coil length (l) = 50 mm
  • Number of turns (N) = 100

Step 3: Value Substitution Process

Substitute the given values into Wheeler's formula:

$$L = \frac{(20)^2 \times (100)^2}{18 \times 20 + 40 \times 50}$$

Step 4: Step-by-Step Calculation Demonstration

Calculate the numerator:

$$D^2 \times N^2 = 20^2 \times 100^2 = 400 \times 10,000 = 4,000,000$$

Calculate the denominator:

$$18D + 40l = 18 \times 20 + 40 \times 50 = 360 + 2,000 = 2,360$$

Final division:

$$L = \frac{4,000,000}{2,360} = 1,694.915... \approx 1,694.92 \text{ µH}$$

Step 5: Result Verification and Interpretation

The calculated inductance of 1,694.92 µH is reasonable for a coil of this size. This value can be verified by:

  • Checking dimensional analysis (all units cancel correctly)
  • Comparing with typical inductance values for similar coils
  • Verifying that the coil meets Wheeler's formula assumptions

Visual Representation

The diagram below illustrates how to interpret the calculation results and understand the relationship between coil geometry and inductance:

Inductance Calculation Visualization Physical Coil l = 50 mm D = 20 mm N = 100 turns Wheeler\'s Formula L = D²N² / (18D + 40l) L = (20²×100²) / (18×20 + 40×50) L = 4,000,000 / 2,360 = 1,694.92 µH Result Interpretation • Inductance increases with D² and N² • Inductance decreases with length • Suitable for RF applications • Typical range: 1-2 mH for this size Key Relationships 1. Doubling the diameter (D) increases inductance by 4× (quadratic relationship) 2. Doubling the turns (N) increases inductance by 4× (quadratic relationship) 3. Doubling the length (l) decreases inductance (inverse relationship) 4. Formula accuracy: ±5% for l/D > 0.8, ±2% for l/D > 2