Toroid Core Inductance
L = AL × N²
Quick answer
This calculator solves L = AL × N² (AL in nH/turn²) for Inductance. With AL Value 0.5, Number of Turns 1.1, the result is 0.605 nH. It solves in 3 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
The inductance of a toroidal inductor is the product of the core AL value and the square of the number of turns. The AL value (inductance factor) is a property of the core material and geometry, specified in nH per turn squared. Toroidal cores provide excellent magnetic shielding since the flux is largely contained within the core, minimizing radiated EMI. This makes them the preferred choice for EMI filters, common-mode chokes, and power inductors in noise-sensitive applications.
Variables
- L — Inductance (H)
- AL — Core inductance factor (nH/turn²)
- N — Number of turns (dimensionless)
Practical Notes
AL values are specified at low flux density and decrease as the core approaches saturation. Common toroid cores: T37-2 (iron powder, AL = 4.0), T68-2 (AL = 5.7), FT37-43 (ferrite, AL = 885). The nH/turn² unit means the result of AL × N² is in nanohenries, which is internally converted to henries. Doubling the turns quadruples the inductance but also increases DCR and may require a larger core.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| L | Inductance | µH | nH, µH, mH, H |
| AL | AL Value | — | — |
| N | Number of Turns | — | — |
Step by step
- 1. What you know: AL Value 0.5, Number of Turns 1.1
- 2. The relation: L = AL × N² (AL in nH/turn²)
- 3. Substitute: AL = 0.5, N = 1.1
- 4. Result: Inductance = 0.605 nH
Nearest standard component values
| Series | Nearest value | Error |
|---|---|---|
| E12 | 0.56 nH | 7.4% |
| E24 | 0.62 nH | 2.5% |
Worked examples
Minimum range
Given: AL Value 0.025, Number of Turns 0.055
Result: Inductance = 7.56 × 10⁻⁵ nH
Low range
Given: AL Value 0.25, Number of Turns 0.55
Result: Inductance = 0.07563 nH
Typical
Given: AL Value 0.5, Number of Turns 1.1
Result: Inductance = 0.605 nH
High range
Given: AL Value 0.8, Number of Turns 1.76
Result: Inductance = 2.478 nH
Maximum range
Given: AL Value 8, Number of Turns 17.6
Result: Inductance = 2.478 µH
Solving for each variable
Solve for Inductance
AL Value 0.025, Number of Turns 0.055 → 7.56 × 10⁻⁵ nH
Solve for AL Value
Inductance 100 nH, Number of Turns 0.055 → 33060
Solve for Number of Turns
Inductance 100 nH, AL Value 0.025 → 63.25
How Inductance changes with AL Value
| AL Value (input) | Inductance (result) |
|---|---|
| 0.05 | 0.0605 nH |
| 0.125 | 0.1513 nH |
| 0.25 | 0.3025 nH |
| 0.375 | 0.4538 nH |
| 0.5 | 0.605 nH |
| 0.75 | 0.9075 nH |
| 1 | 1.21 nH |
| 2 | 2.42 nH |
| 3.5 | 4.235 nH |
| 5 | 6.05 nH |
The same result in other units
| Unit | Value |
|---|---|
| nH | 0.605 |
| µH | 0.000605 |
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<p><a href="https://vbatpower.com/formulas/toroid-inductance/">Toroid Core Inductance</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Toroid Core Inductance?
Use L = AL × N² (AL in nH/turn²). With AL Value 0.5, Number of Turns 1.1, Inductance works out to 0.605 nH.
What is the formula for Toroid Core Inductance?
The formula is L = AL × N² (AL in nH/turn²). The inductance of a toroidal inductor is the product of the core AL value and the square of the number of turns.
How do you find AL Value when you know the other values?
Rearrange for AL Value: with Inductance 100 nH, Number of Turns 0.055, AL Value is 33060.
How do you find Number of Turns when you know the other values?
Rearrange for Number of Turns: with Inductance 100 nH, AL Value 0.025, Number of Turns is 63.25.
What units does Toroid Core Inductance use?
Inductance is shown in nH, µH. Values are held in SI base units internally, so you can mix prefixes without converting by hand.
Related Concepts
All Inductors formulas →Need more features?
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