How to use it
- Choose what to find: the frequency, the inductance or the capacitance.
- Enter the two known values and pick units for each (H, mH, µH, nH; µF, nF, pF; Hz, kHz, MHz).
- Choose series or parallel and enter the resistance: the series resistance of the loop, or the resistor across a parallel tank. Leave 0 for an ideal LC circuit.
- Read the result, ω₀, the reactance at resonance, the Q factor and the bandwidth.
Formulas
- Resonant frequency: f₀ = 1 / (2π√(LC)); ω₀ = 2πf₀.
- Solving for L or C: L = 1 / ((2πf₀)² C), C = 1 / ((2πf₀)² L).
- Reactance at resonance: X = 2πf₀L = 1 / (2πf₀C) = √(L/C).
- Q factor: series Q = X / R, parallel Q = R / X.
- Bandwidth: BW = f₀ / Q.
Worked example
L = 10 mH and C = 100 nF: LC = 0.01 × 100 × 10⁻⁹ = 10⁻⁹, √(LC) = 3.1623 × 10⁻⁵ s, so f₀ = 1 ÷ (2π × 3.1623 × 10⁻⁵) = 5.0329 kHz and ω₀ = 31,622.78 rad/s. The reactance is √(0.01 ÷ 10⁻⁷) = 316.2278 Ω. With 10 Ω in series, Q = 31.6228 and the bandwidth is 5,032.9 ÷ 31.6228 = 159.1549 Hz.
To tune 1 kHz with a 100 mH coil you need C = 1 ÷ ((2π × 1,000)² × 0.1) = 253.303 nF.
Tips
- Strays count at high frequency. A few pF of wiring and component capacitance shift a tank tuned with tens of pF; measure, then trim.
- The coil’s own resistance sets Q. Enter the coil’s series resistance in series mode to see the Q a real inductor allows.
- Units: 1 µH = 1,000 nH, 1 nF = 1,000 pF. A slip of a thousand moves f₀ by a factor of about 31.6.