Audio Crossover Frequency Calculator
In passive speaker crossover design, a first-order RC (resistor-capacitor) filter divides the audio spectrum at a -3 dB crossover frequency defined by f = 1 / (2 pi R C). Below this frequency a low-pass capacitor passes signal to the woofer; above it a high-pass capacitor blocks lows from the tweeter. The same formula applies to equalizer shelving networks and tone control circuits. Enter resistance in ohms and capacitance in microfarads (uF) to find the crossover frequency, or enter a target frequency to back-calculate the required capacitance.
Crossover frequency formula
f = 1 / (2 π R C)
Where R is in ohms and C is in farads. The result is the -3 dB frequency in Hz. To convert microfarads to farads, divide by 1,000,000.
Design notes
- For a target crossover frequency, solve for C: C = 1 / (2 pi f R)
- With R = 8 ohms and C = 10 uF: f = 1 / (2 x pi x 8 x 0.000010) = 1,989 Hz
- A second-order Butterworth crossover requires two reactive elements and achieves -12 dB/octave rolloff.
- Inductors in series with woofers follow: f = R / (2 pi L)
Frequently asked questions
What is a crossover frequency?
A crossover frequency is the frequency at which a filter transitions between passing and attenuating signals. In speaker systems, a crossover splits audio into bands directed at woofers, midrange drivers, and tweeters.
What is the RC crossover formula?
For a first-order passive RC filter, the -3 dB crossover frequency is f = 1 / (2 pi R C), where R is resistance in ohms and C is capacitance in farads. At this frequency, the output is 3 dB below the passband level and phase-shifted by 45 degrees.
What is the difference between first and second order crossovers?
A first-order crossover rolls off at 6 dB/octave and uses one reactive component (L or C). A second-order rolls off at 12 dB/octave and uses two components. Higher-order crossovers have steeper slopes and are more complex to design.
How do I choose R and C values for a target frequency?
Rearrange the formula: C = 1 / (2 pi R f). For a 1 kHz crossover with a 10 ohm load, C = 1 / (2 x pi x 10 x 1000) = approximately 15.9 microfarads. Choose the nearest standard capacitor value.
Is this the same as calculating inductor-based crossovers?
For an RL filter (inductor low-pass), the cutoff frequency is f = R / (2 pi L). The RC and RL formulas are analogous but use capacitance and inductance respectively.
Official sources
- AES: Audio Engineering Society.
- IEEE: Institute of Electrical and Electronics Engineers.
- OpenStax University Physics Vol. 2, Chapter 15: Alternating-Current Circuits.
Reviewed by the CalculatorHub team, edited by James Graham, 15 June 2026. See our methodology.