How to calculate resonant frequency in LC circuits
Use f0 = 1 ÷ (2π √(LC)). A 10 µH inductor with a 1 nF capacitor has √(LC) = 100 ns, so f0 = 1.592 MHz. At that frequency the inductor and capacitor each have a reactance of √(L/C) = 100 Ω; they cancel, so a series LC looks like a short and a parallel LC like an open circuit.
LC resonant frequency formula
An inductor’s reactance rises with frequency and a capacitor’s falls. Resonance is the one frequency where they are equal, so ω0 L = 1 ÷ (ω0 C), and solving that gives the resonant frequency equation.
f0 = 1 ÷ (2π √(L × C))
- f0
- resonant frequency in Hz (ω0 = 2πf0 in rad/s)
- L
- inductance in H
- C
- capacitance in F (include stray capacitance across a parallel tank)
Rearranged for the part you are choosing:
L = 1 ÷ ((2πf0)² × C) C = 1 ÷ ((2πf0)² × L)
The characteristic impedance is the reactance of either part at resonance:
Z0 = √(L ÷ C) = 2πf0 L = 1 ÷ (2πf0 C)
- Z0
- characteristic impedance in Ω
Worked examples
Example 1: a 100 mH and 10 µF tank
LC = 0.1 × 10⁻⁵ = 10⁻⁶ s², so √(LC) = 1 ms and f0 = 1 ÷ (2π × 1 ms) = 159.155 Hz. Z0 = √(0.1 ÷ 10⁻⁵) = 100 Ω: at resonance both parts have 100 Ω of reactance. This is the example tank circuit in Lessons in Electric Circuits, which gives 159.155 Hz.
Example 2: a 10.7 MHz tank with a 1 µH coil
C = 1 ÷ ((2π × 10.7 MHz)² × 1 µH) = 221 pF. The nearest E12 capacitor, 220 pF, gives 10.73 MHz. Add 5 pF of board and coil capacitance and the same parts resonate at 10.61 MHz, so the capacitor to fit is really 216 pF. At these values, a few picofarads matter; a trimmer capacitor or an adjustable coil takes up the difference.
Example 3: a buck converter output filter, 10 µH and 22 µF
f0 = 1 ÷ (2π √(10 µH × 22 µF)) = 10.73 kHz, with Z0 = 674 mΩ. In a voltage-mode buck converter this is the output filter’s double pole, which the loop compensation has to account for. With such a low Z0, the capacitor’s ESR and the load resistance decide how much the filter peaks; the RLC circuit calculator shows the damping.
LC circuit resonant frequency for common L and C values
f0 = 1 / (2π √(LC)) for decade values of L and C. Each step of 100× in L or C moves f0 by 10×.
| L \ C | 10 pF | 100 pF | 1 nF | 10 nF | 100 nF | 1 µF |
|---|
| 1 µH | 50.33 MHz | 15.92 MHz | 5.033 MHz | 1.592 MHz | 503.3 kHz | 159.2 kHz |
| 10 µH | 15.92 MHz | 5.033 MHz | 1.592 MHz | 503.3 kHz | 159.2 kHz | 50.33 kHz |
| 100 µH | 5.033 MHz | 1.592 MHz | 503.3 kHz | 159.2 kHz | 50.33 kHz | 15.92 kHz |
| 1 mH | 1.592 MHz | 503.3 kHz | 159.2 kHz | 50.33 kHz | 15.92 kHz | 5.033 kHz |
| 10 mH | 503.3 kHz | 159.2 kHz | 50.33 kHz | 15.92 kHz | 5.033 kHz | 1.592 kHz |
Capacitance to resonate an inductor at common frequencies
C = 1 / ((2πf)² L), before stray capacitance. Subtract the board and device capacitance from the value you fit.
| Frequency | L = 100 nH | L = 1 µH | L = 10 µH | L = 100 µH |
|---|
| 100 kHz | 25.3 µF | 2.53 µF | 253 nF | 25.3 nF |
| 455 kHz (AM IF) | 1.22 µF | 122 nF | 12.2 nF | 1.22 nF |
| 1 MHz | 253 nF | 25.3 nF | 2.53 nF | 253 pF |
| 10.7 MHz (FM IF) | 2.21 nF | 221 pF | 22.1 pF | 2.21 pF |
| 13.56 MHz (ISM) | 1.38 nF | 138 pF | 13.8 pF | 1.38 pF |
| 27.12 MHz (ISM) | 344 pF | 34.4 pF | 3.44 pF | under 1 pF |
For the inductance instead, L = 1 / ((2πf)² C): for example 122 µH resonates with 1 nF at 455 kHz.
Real inductors and capacitors
The formula assumes ideal parts. A real inductor has winding resistance, which sets the Q and bandwidth of the resonance, and winding capacitance, which makes it self-resonate; choose one whose self-resonant frequency is well above f0. A real capacitor has ESR and ESL and self-resonates too, at the minimum of its impedance curve. For tuned circuits use C0G/NP0 ceramic, mica or film capacitors: X7R and X5R ceramics lose capacitance with DC bias and temperature, which moves f0. Below about 100 pF, include the stray capacitance of the layout, the coil and any connected input. Iron and ferrite cores change inductance with current and temperature, so check the inductor’s saturation current against the peak current in the tank.
Glossary
- Resonant frequency (f0)
- The frequency where the inductive and capacitive reactances are equal and cancel.
- Tank circuit
- An inductor and capacitor in parallel, which stores energy that swings between the two at f0.
- Characteristic impedance (Z0)
- √(L/C), the reactance of either part at resonance.
- Reactance (X)
- The frequency-dependent opposition of an inductor (2πfL) or capacitor (1 / (2πfC)) to AC, in ohms.
- Self-resonant frequency (SRF)
- The frequency where a real inductor or capacitor resonates with its own parasitics; above it the part behaves as the opposite type.
- Stray capacitance
- Unintended capacitance from pads, traces, windings and device inputs that adds to the tank capacitance.
Frequently asked questions
What is the formula for the resonant frequency of an LC circuit?
The resonance frequency formula is f0 = 1 / (2π √(LC)), with L in henries, C in farads and f0 in hertz. It is the frequency where the inductive reactance 2πfL equals the capacitive reactance 1 / (2πfC).
What is the resonant frequency of 10 µH and 100 nF?
159.2 kHz. LC = 10⁻⁵ × 10⁻⁷ = 10⁻¹² s², √(LC) = 1 µs, and 1 ÷ (2π × 1 µs) = 159.2 kHz.
Is the resonant frequency the same for a series and a parallel LC circuit?
For ideal parts, yes: both resonate at 1 / (2π √(LC)). A series LC has its minimum impedance there and a parallel LC its maximum. With resistance in the branches, the parallel circuit’s impedance peak and zero-phase point move away from that value, by less as Q rises.
What is the resonant frequency of an RLC circuit?
The ideal value is the same 1 / (2π √(LC)), and resistance sets the Q and bandwidth of the resonance. With resistance in the branches of a parallel circuit, the impedance peak and zero-phase point move away from that value, by less as Q rises; the RLC circuit calculator works out Q, bandwidth and damping.
What is the characteristic impedance of an LC tank?
√(L/C), in ohms. It equals the reactance of the inductor and of the capacitor at resonance, and it sets how the tank interacts with resistance: in a series circuit Q = √(L/C) ÷ R, in a parallel one Q = R ÷ √(L/C).
How do I choose L and C for a given resonant frequency?
Pick the impedance level Z0 first, then L = Z0 ÷ (2πf) and C = 1 ÷ (2πf × Z0). For 1 MHz at Z0 = 50 Ω that is 7.96 µH and 3.18 nF. Then round one part to a standard value and solve for the other.
How much do component tolerances move the resonant frequency?
f0 scales with 1/√(LC), so a fractional change in L or C moves f0 by about half as much. If both L and C are 5 % high, f0 drops by 4.76 %; if both are 5 % low, it rises by 5.26 %. Trim with a variable capacitor or a tuned coil when the frequency must be exact.
Why does my LC circuit resonate lower than calculated?
Usually extra capacitance: pads, traces, the coil’s own winding capacitance and the input capacitance of whatever is connected add a few picofarads, which matters when C is small. Class 2 ceramics (X7R, X5R) also change value with DC bias and temperature, so use C0G/NP0 for tuned circuits.
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