3-way active crossover network

Rohit Balkishan Dubla

Description:

This is a 12db/octave active crossover. It is based on 2nd order butterworth filters. The resistors Ra and Rf set the gain of each filter to 1.582, which is slightly less than the required value of 1.586. This value of gain (Ao) follows from the formula k = 3 - Ao where k = 1 / (Q-factor of the filter). For a 2nd order filter, the value of k can be obtained from the butterworth circle for n = 2. It turns out that k = 1 / cos(x), where x = PI / (2 * n) which is PI / 4 in this case. Thus for a butterworth response, the Q-factor turns out to be 0.707. Please see references (1) for more details. Increasing Q beyond this results in peaking at the cut-off frequency for each individual filter. Likewise, reducing Q makes the filter response more and more gradual. Thus a value of 0.707 for the Q-factor gives the flattest pass-band gain & sharpest roll-off at cut-off frequency. Note that the gain of each individual filter must be less than 3, otherwise the circuit will oscillate. To get an even sharper roll-off the order of the filters must be increased (keeping Q = 0.707). In the above schematic are shown the low-mid range, mid range & mid-high range filters. The crossover frequencies chosen are 300 Hz and 3000 Hz. Thus the low-mid range filter has a cut-off frequency of 300 Hz, the mid range has a lower cut-off at 300 Hz & an upper cut-off at 3000 Hz, and the mid-high range has a cut-off frequency of 3000 Hz. Please see references (2) for the crossover frequencies.

The calculations for the crossover are as follows:

Low-mid range: Low pass filter, fh = 300 Hz.

fh = 1 / (2 * PI * R1 * C1), assuming R1 = R2 & C1 = C2 = 10nF. This yields R1 = R2 = 53K (used 56K).

Mid range: Low pass filter, fh = 3000 Hz, followed by a high pass filter fl = 300 Hz.

Assuming R1 = R2, R3 = R4, C1 = C2 = C3 = C4 = 10nF.

For the low pass, fh = 1 / (2 * PI * R1 * C1), yields R1 = R2 = 5.3K (used 5.6K).

For the high pass, fl = 1 / (2 * PI * R3 * C3), yields R3 = R4 = 53K (used 56K).

Mid-high range: High pass filter, fl = 3000 Hz.

fl = 1 / (2 * PI * R3 * C3), assuming R3 = R4 & C3 = C4 = 10nF. This yields, R3 = R4 = 5.3K (used 5.6K).

Note that the mid-range filter is preceded by an inverting amplifier. This is needed for 2 reasons - Firstly, the gain of the mid-range is (1.582 * 1.582) which must be brought back to the level of the low-mid & mid-high ranges (1.582). Secondly (and more importantly), a 2nd order filter has an inherent property of shifting the phase of any signal passing through it depending on the signal's frequency so that the signal is 90° out of phase with the input (direction of shift depends on whether the filter is high-pass or low-pass). Thus the low-mid range filter has shifted the signal by 90° and the mid range has also done the same (but in an opposite direction). Hence, the signals appearing at the low-mid range & mid range outputs are going to be 180° out of phase with each other & will cancel out (electrically or acoustically). The same happens to the mid range & mid-high range filters at. The inverter, with a gain of -0.63 serves to solve both the problems. Using a 4th order filter (assuming it to be a cascade of two 2nd order butterworth types) in place of the ones shown will not have the phase-reversal at crossover problem, but you will still need to bring down the mid range filter's gain (this equates to 2 inverters). The 0.1µF capacitor (Cin) is used with R5 to obtain a lower 3dB frequency of about 15 Hz. With the arrangement shown, the overall magnitude response exhibits peaks at the crossover frequencies when the 3 outputs are combined, either electrically or acoustically. This ordinarily does not pose a problem, since the speaker deficiencies themselves will tend to hide (rather veil) the peaks, but with really good speaker systems, the peaking could become evident. The op-amps used must preferably have a high slew rate and all resistors must be of 1/4 W, 1% metal film type. For the crossover that I have made, the op-amps used are TL074 quad devices. Any other op-amps of your choice can be substituted in place of these. All inputs & outputs must use fully shielded cables (as short as possible). If the circuit is to be assembled on a general purpose board, then try to keep all component leads and wiring as short as possible to avoid possible pick up of radio signals (mine did, but only on touching certain resistor leads).

It may be noted that changing the gain of the amplifiers in each filter to unity (all Rf's = 0, and remove all Ra's) and increasing Rf1 in the inverter to 100K yields a crossover whose summed output is flat, with each filter crossing over at -6db as against -3db for the current configuration. This crossover behaviour is much more preferable to the butterworth crossover and constitutes a Linkwitz-Riley crossover (www.linkwitzlab.com). After the suggested modifications, each filter will be operating with an LR2 response (Q = 0.5).

That concludes the description of the crossover. I hope that the reader will find the material presented here to be of some help in understanding electronics.

References:

1) Integrated Electronics by Millman & Halkias, McGraw-Hill (ISBN 0-07-Y85493-9)

2) Bi-Amplification (not quite magic, but close) - ESP