ENGINEERING CALCULATOR · FORMULAS INCLUDED
LC Filter Calculator
Enter inductance, capacitance and load resistance to calculate the −3 dB cutoff, natural resonant frequency, Q factor and gain of a loaded LC low-pass or high-pass filter.
Input parameters
Calculated results
Inputs changed. Calculate again to refresh the result.
| −3 dB cutoff frequency fc | 245.539656 kHz |
|---|---|
| Natural resonant frequency f₀ | 159.154943 kHz |
| Quality factor Q | 5 |
| Characteristic impedance √(L/C) | 10 Ω |
| Gain at evaluation frequency | 4.178485 dB |
| Voltage magnitude ratio |Vout/Vin| | 1.617798 |
Results apply to: LC low-pass · Evaluation frequency 100 kHz
Ideal voltage source with zero source impedance; ideal L and C, resistive load R. Peaking occurs when Q > 1/√2.
Frequency response reference
Relative to the unity passband gain of the circuit shown; dB = 20 log₁₀|Vout/Vin|.
| Frequency | f/f₀ | |Vout/Vin| | Gain (dB) |
|---|---|---|---|
| 15.9154943 kHz | 0.1 | 1.009895 | 0.08552402 |
| 79.5774715 kHz | 0.5 | 1.321637 | 2.422245 |
| 159.154943 kHz | 1 | 5 | 13.9794 |
| 318.309886 kHz | 2 | 0.3304093 | -9.618955 |
| 1.59154943 MHz | 10 | 0.01009895 | -39.91448 |
| 245.539656 kHz (fc) | 1.542771 | 0.7071068 | −3.0103 |
Formulas and circuit assumptions
f₀ = 1 / (2π√LC) · Z₀ = √(L/C) · Q = R / Z₀
For x = f/f₀, low-pass magnitude is: |HLP| = 1 / √[(1−x²)² + (x/Q)²]
High-pass magnitude is: |HHP| = x² / √[(1−x²)² + (x/Q)²]
Set a = 2 − 1/Q² and y = (a + √(a² + 4))/2. Then fc = f₀√y for low-pass and fc = f₀/√y for high-pass. The half-power point is conventionally called −3 dB; its exact value is −3.0103 dB.
In these equations, L is in H, C in F and R in Ω. Source impedance, winding resistance, capacitor ESR, self-resonance and nonlinear loads are excluded. This is not a complete model of a pi filter, T filter or common-mode choke network.
Worked example
With L = 10 μH, C = 100 nF and R = 50 Ω: f₀ = 159.155 kHz, Z₀ = 10 Ω and Q = 5. The low-pass cutoff is about 245.540 kHz, while the high-pass cutoff is about 103.162 kHz. Both responses peak because Q is high; f₀ is not their −3 dB cutoff.
Frequently asked questions
When does f₀ equal the cutoff frequency?
At Q = 1/√2 the model has a second-order Butterworth response, and fc equals f₀. Changing load resistance changes Q and cutoff, while f₀ remains fixed for the same L and C.
Why can a passive filter have voltage gain above one?
Energy exchange between L and C can produce resonant voltage magnification. This does not create extra energy. Check real losses, source capability and component ratings.
Related tools and references
Coilcraft: second-order filter transfer function · Analog Devices: A Filter Primer