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) |
Butterworth C1
= 0.1125 / (Rh*f) Linkwitz-Riley C1 = 0.0796 / (Rh*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