Quality Factor (RLC)
Q = (1/R) × √(L/C)
Quick answer
This calculator solves Q = (1/R) × √(L/C) for Q. With Resistance 0.47 kΩ, Inductance 100 µH, Capacitance 100 pF, the result is 2.128.
Calculator
Formula
Description
This form calculates Q directly from the component values of a series RLC circuit. The term √(L/C) has units of ohms and is called the characteristic impedance of the LC network. Q is then the ratio of this characteristic impedance to the series resistance. Lower resistance means less energy loss per cycle and higher Q. This formula reveals why component selection matters: for a given resonant frequency, you can trade off L and C values to achieve the desired Q while keeping f₀ = 1/(2π√(LC)) constant.
Variables
- Q — Quality factor (dimensionless)
- R — Series resistance in ohms (Ω)
- L — Inductance in henries (H)
- C — Capacitance in farads (F)
Practical Notes
For a parallel RLC circuit, the formula inverts to Q = R × √(C/L), because the parallel resistance damps the circuit differently. In real inductors, the series resistance (DCR) is the dominant loss mechanism that limits Q, which is why air-core inductors achieve higher Q than ferrite-core ones at high frequencies.
Symbols and units
| Symbol | Meaning | Default unit | Accepted units |
|---|---|---|---|
| R | Resistance | Ω | mΩ, Ω, kΩ, MΩ |
| L | Inductance | µH | nH, µH, mH, H |
| C | Capacitance | pF | pF, nF, µF, mF, F |
Step by step
- 1. What you know: Resistance 0.47 kΩ, Inductance 100 µH, Capacitance 100 pF
- 2. The relation: Q = (1/R) × √(L/C)
- 3. Substitute: R = 0.47 kΩ, L = 100 µH, C = 100 pF
- 4. Result: Q = 2.128
Worked examples
Minimum range
Given: Resistance 1 Ω, Inductance 100 nH, Capacitance 1 pF
Result: Q = 316.2
Low range
Given: Resistance 10 Ω, Inductance 1 µH, Capacitance 10 pF
Result: Q = 31.62
Typical
Given: Resistance 0.47 kΩ, Inductance 100 µH, Capacitance 100 pF
Result: Q = 2.128
High range
Given: Resistance 4.7 kΩ, Inductance 0.47 mH, Capacitance 1 nF
Result: Q = 0.1459
Maximum range
Given: Resistance 47 kΩ, Inductance 4.7 mH, Capacitance 10 nF
Result: Q = 0.01459
How Q changes with Resistance
| Resistance (input) | Q (result) |
|---|---|
| 47 Ω | 21.28 |
| 117.5 Ω | 8.511 |
| 0.235 kΩ | 4.255 |
| 0.3525 kΩ | 2.837 |
| 0.47 kΩ | 2.128 |
| 0.705 kΩ | 1.418 |
| 0.94 kΩ | 1.064 |
| 1.88 kΩ | 0.5319 |
| 3.29 kΩ | 0.304 |
| 4.7 kΩ | 0.2128 |
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<p><a href="https://vbatpower.com/formulas/quality-factor-rlc/">Quality Factor (RLC)</a> Calculator by vbatpower.com</p>Frequently asked questions
How do you calculate Quality Factor (RLC)?
Use Q = (1/R) × √(L/C). With Resistance 0.47 kΩ, Inductance 100 µH, Capacitance 100 pF, Q works out to 2.128.
What is the formula for Quality Factor (RLC)?
The formula is Q = (1/R) × √(L/C). This form calculates Q directly from the component values of a series RLC circuit.
Related Concepts
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