Two-way cross-over basics

by: Jonas Holmgren


In audio, there are a couple of different passive cross-over designs/theories. A passive cross-over is a one using only passive components to filter out signals. These passive components are capacitors and coils.

The two most common design/theories are Butterworth and Linkwitz/Riley. By constructing cross-over networks using a number of coils and capacitors, we can achieve different cross-over slopes. The slope is at what rate we filter out the signal. The most common slopes are 6, 12, 18 or 24dBs/octave. A high slope means the signal is quickly reduced in amplitude and therefore will be more quickly inaudible.

For two speakers that will both need filtered signals, we design two-way cross-overs. A two-way crossover will split one full range signal into two signals, one that contains low frequencies with filtered out high frequencies; we call it low-pass signal. And the other contains higher frequencies with filtered out low frequencies, called high-pass signal. (Well, in fact, nothing can be filtered out, it can only be reduced in amplitude at a selected ratio/slope)

The following table includes formulas for the 6dB/octave Butterworth filter and for 12dB/octave Butterworth and Linkwitz-Riley filters. In the formulas the following terms are used:

Rh: driver impedance receiving the high-pass signal (could for example be a tweeter),
Rl: driver impedance receiving the low-pass signal (could be a bass/midrange-unit)
f: selected cross-over frequency in Hz.


C1 is capacitor value in Farads (multiply with 1.000.000 to get uF)
L1 is coil value in Henry (multiply with 1.000 to get mH)

 



Design/formula

Network diagrams

Butterworth

C1 = 0.159 / (Rh*f)
L1 = Rl / (6.28*f)




Butterworth

C1 = 0.1125 / (Rh*f)
L1 = (0.2251*Rh) / f
C2 = 0.1125 / (Rl*f)
L2 = (0.2251*Rl) / f

Linkwitz-Riley

C1 = 0.0796 / (Rh*f)
L1 = (0.3183*Rh) / f
C2 = 0.0796 / (Rl*f)
L2 = (0.3183*Rl) / f



Now it's time for some examples. Let's calculate coil and capacitor values for a 6dB filter, cross-over point at 2.000 Hz (2kHz) with the midrange having an impedance of 4 ohms, and a tweeter that has an impedance of 6 ohms.

C1 = 0.159 / (Rh*f) = 0.159 / (6*2000) = 0.0000132 F = 13.2 uF

L1 = Rl / (6.28*f) = 4 / (6.28*2000) = 0.0003184 H = 0.32 mH

 

For a 12dB filter of Linkwitz-Riley type, with the same drivers and cross-over point we get:

C1 = 0.0796 / (Rh*f) = 0.0796 / (6*2000) = 0.0000066 F = 6.6 uF
L1
= (0.3183*Rh) / f = (0.3183*6) / 2000 = 0.0009549 H = 0.95 mH
C2
= 0.0796 / (Rl*f) = 0.0796 / (4*2000) = 0.0000099 F = 9.9 uF
L2 = (0.3183*Rl) / f = (0.3183*4) / 2000 = 0.0006366 H = 0.64 mH