To calculate LC resonant frequency, use the formula fr = 1 / (2π√(L × C)), where fr is the resonant frequency in Hertz, L is inductance in Henrys, and C is capacitance in Farads. This formula defines the exact point where inductive reactance (XL) and capacitive reactance (XC) cancel each other out, leaving only the parasitic resistance of the circuit.
The Core Formula and Symbol Definitions
The fundamental equation for the resonant frequency of an ideal LC tank circuit (applicable to both series and parallel configurations) is derived from setting XL = XC. Below is the strict definition of every variable in the equation.
| Symbol | Parameter | Base SI Unit | Common Practical Units |
|---|---|---|---|
| fr | Resonant Frequency | Hertz (Hz) | kHz, MHz, GHz |
| L | Inductance | Henry (H) | mH, μH, nH |
| C | Capacitance | Farad (F) | μF, nF, pF |
| π | Archimedes' Constant | Dimensionless | ~3.14159265 |
Rearranged Forms: Solving for L or C
In practical bench work, you rarely solve for frequency from scratch; you usually have a target frequency and need to find the missing component value to pair with a standard part you already have in your bin. Here are the algebraically rearranged forms of the master equation.
- To solve for Inductance (L):
L = 1 / ( (2πfr)2 × C ) - To solve for Capacitance (C):
C = 1 / ( (2πfr)2 × L )
Notice that (2πfr) is the angular frequency (ω). Therefore, the equations can also be written in shorthand as L = 1 / (ω2C) and C = 1 / (ω2L). Always square the entire (2πfr) term before multiplying by the known component value.
Worked Examples with Strict Unit Tracking
The most common point of failure in resonance calculations is unit mismanagement. You must convert all practical units (milli, micro, nano, pico) into base SI units (Henrys and Farads) before plugging them into the formula. Below are two step-by-step derivations.
Example 1: Audio Crossover Network (Low Frequency)
Given: An inductor L = 10 mH and a capacitor C = 4.7 μF.
Goal: Calculate fr.
- Convert to base units:
L = 10 × 10-3 H = 0.01 H
C = 4.7 × 10-6 F = 0.0000047 F - Multiply L and C:
L × C = 0.01 × 0.0000047 = 4.7 × 10-8 - Take the square root:
√(4.7 × 10-8) = 0.00021679 - Multiply by 2π:
2 × 3.14159 × 0.00021679 = 0.0013621 - Invert to find frequency:
fr = 1 / 0.0013621 = 734.16 Hz
Example 2: RF Bandpass Filter (High Frequency)
Given: An RF choke L = 47 nH and a tuning capacitor C = 2.2 pF.
Goal: Calculate fr.
- Convert to base units:
L = 47 × 10-9 H
C = 2.2 × 10-12 F - Multiply L and C:
(47 × 10-9) × (2.2 × 10-12) = 103.4 × 10-21 = 1.034 × 10-19 - Take the square root:
√(1.034 × 10-19) = 3.2156 × 10-10 - Multiply by 2π:
2 × 3.14159 × 3.2156 × 10-10 = 2.0204 × 10-9 - Invert to find frequency:
fr = 1 / (2.0204 × 10-9) = 494,951,494 Hz ≈ 495 MHz
Common Unit Traps and Realistic Magnitude Ranges
If your calculated answer is off by a factor of 1,000 or 1,000,000, you have fallen into a unit conversion trap. The most frequent error is confusing microfarads (μF, 10-6) with picofarads (pF, 10-12). A 100 pF capacitor is not 0.1 μF; it is 0.0001 μF. Always write out the scientific notation (e.g., 10-12) on your scratchpad.
Furthermore, use this magnitude table to sanity-check your final answer. If you are designing an audio circuit and your math yields 45 MHz, you missed a decimal point. For a comprehensive overview of AC circuit behaviors across these domains, see the Electronics Tutorials AC resonance guide.
| Application Domain | Typical fr Range | Typical L Values | Typical C Values |
|---|---|---|---|
| Mains / Power Filtering | 50 Hz - 120 Hz | 10 mH - 1 H | 100 μF - 10,000 μF |
| Audio Crossovers | 20 Hz - 20 kHz | 1 mH - 50 mH | 1 μF - 100 μF |
| Switching Power Supplies (Buck/Boost) | 10 kHz - 2 MHz | 1 μH - 100 μH | 100 nF - 47 μF |
| RF ISM Bands (Sub-GHz, NFC) | 13.56 MHz - 915 MHz | 2 nH - 500 nH | 1 pF - 50 pF |
| Microwave / Radar | 2.4 GHz - 60 GHz | 0.5 nH - 5 nH | 0.1 pF - 2 pF |
Decision Path: Sizing an LC Tank for a 433 MHz ISM Band Filter
Let's apply the rearranged formula to a real-world design constraint: building a parallel LC bandpass filter for a 433 MHz RF transmitter module. We need to select physical, purchasable surface-mount components.
| Design Condition | Required Action | Component Selection Rule |
|---|---|---|
| Target fr > 100 MHz | Avoid wirewound ferrite cores due to high core losses and low SRF. | Select air-core, ceramic-core, or multilayer SMD inductors. |
| Requires tight frequency tolerance (±2%) | Avoid standard X7R/Y5V capacitors due to voltage/temperature drift. | Select C0G/NP0 dielectric ceramic capacitors. |
| Target fr = 433 MHz | Fix L to a standard high-Q value, solve for C. | Pick L = 15 nH. Calculate C. |
Executing the Final Calculation
We have chosen a standard 15 nH inductor. Now we solve for the exact capacitance required to hit 433 MHz.
- Target: fr = 433,000,000 Hz
- Known: L = 15 × 10-9 H
- Equation: C = 1 / ( (2π × 433 × 106)2 × 15 × 10-9 )
- Step 1 (Angular Frequency): 2π × 433 × 106 = 2.7206 × 109 rad/s
- Step 2 (Square it): (2.7206 × 109)2 = 7.401 × 1018
- Step 3 (Multiply by L): 7.401 × 1018 × 15 × 10-9 = 1.110 × 1011
- Step 4 (Invert): 1 / 1.110 × 1011 = 9.009 × 10-12 Farads
- Result: 9.0 pF
Final Bill of Materials (BOM) Recommendation
Do not settle for generic 'high-frequency' labels; specify exact part numbers with verified SRF and dielectric data. For high-Q RF applications, consult the Murata MLCC RF specifications to ensure low ESR at your target frequency.
The Concrete Pick:
For your 433 MHz LC tank, purchase the Coilcraft 0603CS-15N (15nH, 0603 package, ±5% tolerance, SRF > 3 GHz) paired with the Murata GJM1555C1H9R0 (9.0pF, 0402 package, C0G/NP0 dielectric, ±0.25pF tolerance). This specific pairing guarantees a resonant peak within 1% of 433 MHz, avoiding the insertion loss and detuning typical of off-brand X7R capacitors and low-SRF ferrite chokes.
