Resonant frequency calculator (LC and RLC)
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Result
At resonance a series RLC circuit has its smallest impedance, Z = R, so the current is largest.
Find the resonant frequency of an LC circuit, or the inductance or capacitance needed for a given frequency. For series and parallel RLC circuits the calculator also gives the quality factor Q, the bandwidth and the two half-power (−3 dB) frequencies. Choose units from nH to H, pF to µF and Hz to MHz; every conversion is shown in the steps.
Step-by-step solution
Units
Resonance: the reactances are equal
Reactance at resonance
Quality factor and bandwidth
Key formulas
- Natural frequency
- ω0 = 1/√(LC), f0 = 1/(2π√(LC))
- Series RLC
- Q = (1/R)√(L/C), BW = R/(2πL)
- Parallel RLC
- Q = R√(C/L), BW = 1/(2πRC)
- Quality factor
- Q = 2π·(energy stored)/(energy lost per cycle)
- Bandwidth
- BW = f2 − f1 = f0/Q, f0 = √(f1f2)
How to calculate the resonant frequency
- Convert L to henry and C to farad.
- At resonance the reactances cancel: X_L = X_C, so ωL = 1/(ωC) and ω₀ = 1/√(LC).
- Convert to hertz: f₀ = ω₀/(2π) = 1/(2π√(LC)).
- Quality factor: Q = (1/R)·√(L/C) for a series RLC circuit and Q = R·√(C/L) for a parallel one.
- Bandwidth: BW = f₀/Q. The half-power frequencies are f₀·(√(1 + 1/(4Q²)) ∓ 1/(2Q)).
Questions and answers
Does the resistance change the resonant frequency?
Not in the ideal series RLC circuit or with R, L and C all in parallel: both resonate at f₀ = 1/(2π√(LC)). It does change in a tank circuit where the resistance is in series with the coil: there f₀ = (1/2π)·√(1/(LC) − R²/L²).
What happens at resonance in a series circuit?
The voltages across the inductor and the capacitor are equal and opposite, the impedance drops to its minimum Z = R and the current is largest. The voltage across L or C can be Q times the supply voltage.
What happens at resonance in a parallel circuit?
The currents in the L and C branches cancel, the impedance rises to its maximum Z = R and the circuit draws the least current from the source. This is how a parallel tank circuit selects one frequency.
What is the bandwidth?
The range of frequencies between the two half-power points, where the power is half its value at resonance and the current or voltage is 1/√2 of its peak (−3 dB). BW = f₀/Q, so a circuit with a high Q is more selective.
Why are the half-power frequencies not symmetric around f₀?
They are symmetric on a logarithmic scale: f₁·f₂ = f₀². For a high Q they are almost exactly f₀ ± BW/2, but for a low Q the lower one is clearly closer to f₀.
Practise it yourself
Practise LC resonance, Q factor & bandwidth
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