Calculate crossover frequency, resistance, or capacitance from any two values using the 1/(2πRC) formula in Hz, kHz, or MHz.
Crossover Frequency Formula
The crossover frequency for a simple first-order RC circuit is the frequency where the signal is reduced by 3 dB. This calculator uses resistance, capacitance, and crossover frequency, and lets you select the quantity to solve for and enter the other two values.
- fc = crossover frequency in hertz (Hz)
- R = resistance in ohms (Ω)
- C = capacitance in farads (F)
- π = pi, approximately 3.14159
To solve for resistance, the formula is rearranged as:
To solve for capacitance, the formula is rearranged as:
The calculator converts your selected units to base units before calculating: ohms for resistance, farads for capacitance, and hertz for frequency. After the calculation, it converts the result to the selected result unit.
- Calculate crossover frequency: enter resistance and capacitance to find the RC cutoff or crossover frequency.
- Calculate resistance: enter capacitance and target frequency to find the required resistor value.
- Calculate capacitance: enter resistance and target frequency to find the required capacitor value.
Common RC Crossover Values
These values show approximate crossover frequencies for common resistor and capacitor combinations. They are useful for checking whether your result is in a reasonable range.
| Resistance | Capacitance | Approx. Crossover Frequency |
|---|---|---|
| 1 kΩ | 1 μF | 159.15 Hz |
| 10 kΩ | 100 nF | 159.15 Hz |
| 10 kΩ | 10 nF | 1.591 kHz |
| 1 kΩ | 100 nF | 1.591 kHz |
| 1 kΩ | 10 nF | 15.915 kHz |
Unit Conversions Used
Use these conversions when checking inputs manually. The formula must use ohms, farads, and hertz.
| Quantity | Unit | Base Unit Equivalent |
|---|---|---|
| Resistance | 1 kΩ | 1,000 Ω |
| Resistance | 1 MΩ | 1,000,000 Ω |
| Capacitance | 1 mF | 0.001 F |
| Capacitance | 1 μF | 0.000001 F |
| Capacitance | 1 nF | 0.000000001 F |
| Frequency | 1 kHz | 1,000 Hz |
| Frequency | 1 MHz | 1,000,000 Hz |
Example Calculations
Example 1: Calculate crossover frequency
You have a resistor of 10 kΩ and a capacitor of 100 nF.
- R = 10 kΩ = 10,000 Ω
- C = 100 nF = 0.0000001 F
The result is approximately 159.155 Hz.
Example 2: Calculate capacitance
You want a crossover frequency of 1 kHz using a 10 kΩ resistor.
- fc = 1 kHz = 1,000 Hz
- R = 10 kΩ = 10,000 Ω
The result is approximately 0.0000000159 F, which is 15.915 nF.
Ideal Speaker Crossover Networks
The speaker-network modes calculate ideal passive component values for two or three drivers. Enter each driver’s constant load resistance and the crossover frequency, or lower and upper frequencies for three-way operation. Select first order, second-order Butterworth, or second-order Linkwitz–Riley. The RC calculation above remains a separate first-order circuit calculation.
These are electrical starting values for a zero-impedance voltage source, lossless components and resistive loads. Real loudspeaker impedance, sensitivity and phase vary with frequency. Component values alone do not establish a flat acoustic sum, a safe tweeter operating range or the final acoustic crossover. Measure the drivers and combined response before choosing parts.
Woofer and tweeter branches
Let R be that branch’s load resistance in ohms and ω = 2πf in radians per second. First-order values are L = R/ω and C = 1/(Rω). Use a series inductor for the woofer and a series capacitor for the tweeter.
For second order, L = R/(Qω) and C = Q/(Rω). Butterworth uses Q = 1/√2; Linkwitz–Riley uses Q = 0.5. The woofer uses series L with shunt C across its load. The tweeter uses series C with shunt L across its load. Each branch is connected across the source. At the chosen frequency, ideal Butterworth branches are −3.01 dB and Linkwitz–Riley branches are −6.02 dB relative to their passbands. Ideal two-way LR2 electrical summation requires one branch’s polarity reversed; real acoustic alignment must still be checked.
Three-way midrange branch
The lower frequency must be less than the upper frequency. For the midrange, define B = 2π(f_high − f_low) and ω₀² = (2π)²f_low f_high. The center frequency is √(f_low f_high). The calculator uses a frequency-transformed band-pass branch so its two edges have the selected attenuation; it does not simply cascade independently sized passive filters, which would load each other.
First order per edge: L_series = R/B and C_series = B/(Rω₀²). Connect both components in series with the midrange. This band-pass has two poles total and approaches 6 dB/octave on each side.
Second order per edge: L_series = R/(QB), C_series = QB/(Rω₀²), L_shunt = RB/(Qω₀²), and C_shunt = Q/(RB). Connect the series L and C in series with the midrange; connect the shunt L and C in parallel with each other across the midrange. This band-pass has four poles total and approaches 12 dB/octave on each side.
The woofer branch uses the lower frequency and the tweeter branch uses the upper frequency. Every displayed capacitance is in microfarads (µF) and inductance in millihenries (mH). The stated slopes and edge levels describe individual ideal electrical branches. Arbitrary three-way frequency spacing and real driver responses do not guarantee a flat combined response.
FAQs
What is crossover frequency in an RC circuit?
Crossover frequency, often called cutoff frequency in RC filters, is the frequency where the output level is reduced to about 70.7% of the passband level. This point is also described as the -3 dB frequency. For a simple RC filter, it depends only on resistance and capacitance.
Do I need to enter values in ohms, farads, and hertz?
No. You can select kΩ, MΩ, mF, μF, nF, Hz, kHz, or MHz in the calculator. Internally, the calculation is done using ohms, farads, and hertz because the standard RC formula requires base units.
Why does increasing capacitance or resistance lower the crossover frequency?
The formula places resistance and capacitance in the denominator. When either value increases, the denominator gets larger, so the crossover frequency gets smaller. This is why larger capacitors or larger resistors produce lower cutoff frequencies in a first-order RC circuit.
