Inductor Quality Factor
Q = 2πfL / Rdc
Quick answer
This calculator solves Q = 2πfL / Rdc for Quality Factor. With Frequency 10 MHz, Inductance 100 µH, DC Resistance 0.47 kΩ, the result is 13.37. It solves in 4 directions, so you can enter the values you have and get the one you need.
Calculator
Formula
Description
The quality factor of an inductor measures how effectively it stores energy relative to the energy it dissipates per cycle. Higher Q means lower losses and sharper frequency selectivity in resonant circuits. Q increases with frequency up to a point, then decreases as skin effect and core losses become significant. Air-core inductors achieve the highest Q values but require larger physical size. Ferrite-core inductors are compact but have lower Q due to core losses, especially at high frequencies.
Variables
- Q — Quality factor (dimensionless)
- f — Operating frequency (Hz)
- L — Inductance (H)
- Rdc — DC winding resistance (Ω)
Practical Notes
Typical Q values: power inductors 20-50, RF inductors 50-200, air-core inductors 100-500. At high frequencies, replace Rdc with the AC resistance including skin effect. The self-resonant frequency (SRF) of an inductor is where Q drops to zero because parasitic capacitance resonates with the inductance. Always operate well below the SRF.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| Q | Quality Factor | — | — |
| f | Frequency | MHz | Hz, kHz, MHz, GHz |
| L | Inductance | µH | nH, µH, mH, H |
| Rdc | DC Resistance | Ω | mΩ, Ω, kΩ, MΩ |
Step by step
- 1. What you know: Frequency 10 MHz, Inductance 100 µH, DC Resistance 0.47 kΩ
- 2. The relation: Q = 2πfL / Rdc
- 3. Substitute: f = 10 MHz, L = 100 µH, Rdc = 0.47 kΩ
- 4. Result: Quality Factor = 13.37
Worked examples
Minimum range
Given: Frequency 100 kHz, Inductance 100 nH, DC Resistance 1 Ω
Result: Quality Factor = 0.06283
Low range
Given: Frequency 1 MHz, Inductance 1 µH, DC Resistance 10 Ω
Result: Quality Factor = 0.6283
Typical
Given: Frequency 10 MHz, Inductance 100 µH, DC Resistance 0.47 kΩ
Result: Quality Factor = 13.37
High range
Given: Frequency 100 MHz, Inductance 0.47 mH, DC Resistance 4.7 kΩ
Result: Quality Factor = 62.83
Maximum range
Given: Frequency 1 GHz, Inductance 4.7 mH, DC Resistance 47 kΩ
Result: Quality Factor = 628.3
Solving for each variable
Solve for Quality Factor
Frequency 100 kHz, Inductance 100 nH, DC Resistance 1 Ω → 0.06283
Solve for Frequency
Quality Factor 0.025, Inductance 100 nH, DC Resistance 1 Ω → 39.79 kHz
Solve for Inductance
Quality Factor 0.025, Frequency 100 kHz, DC Resistance 1 Ω → 39.79 nH
Solve for DC Resistance
Quality Factor 0.025, Frequency 100 kHz, Inductance 100 nH → 2.513 Ω
How Quality Factor changes with Frequency
| Frequency (input) | Quality Factor (result) |
|---|---|
| 1 MHz | 1.337 |
| 2.5 MHz | 3.342 |
| 5 MHz | 6.684 |
| 7.5 MHz | 10.03 |
| 10 MHz | 13.37 |
| 15 MHz | 20.05 |
| 20 MHz | 26.74 |
| 40 MHz | 53.47 |
| 70 MHz | 93.58 |
| 100 MHz | 133.7 |
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<p><a href="https://vbatpower.com/formulas/inductor-q-factor/">Inductor Quality Factor</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Inductor Quality Factor?
Use Q = 2πfL / Rdc. With Frequency 10 MHz, Inductance 100 µH, DC Resistance 0.47 kΩ, Quality Factor works out to 13.37.
What is the formula for Inductor Quality Factor?
The formula is Q = 2πfL / Rdc. The quality factor of an inductor measures how effectively it stores energy relative to the energy it dissipates per cycle.
How do you find Frequency when you know the other values?
Rearrange for Frequency: with Quality Factor 0.025, Inductance 100 nH, DC Resistance 1 Ω, Frequency is 39.79 kHz.
How do you find Inductance when you know the other values?
Rearrange for Inductance: with Quality Factor 0.025, Frequency 100 kHz, DC Resistance 1 Ω, Inductance is 39.79 nH.
Related Concepts
All Inductors formulas →Need more features?
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