Have you ever wondered how the tweeter and wooferhave their responses combined in a speaker system, or what the crossovernetwork is and how it works? If so, thisarticle will shed some light on the least seen and perhaps most undervaluedpart of the speaker system, the crossover network. Unlike the electronic world where the filtershave gone from using tubes, to analog computers to digital signal processors,modern day passive crossovers, a pair offrequency selective filters designed to work together as a single network, arebasically the same as they were about 100 years ago when the speaker businesswas in its infancy. While crossovercomponents and materials and construction have changed, the underlying theoryand practice has not.
What does the Crossover Do?
Like yourown brain, the crossover network is a director.In addition to its primary function, the passive crossover in practiceis expected to do more than simply split the frequency band in two.
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1) The crossover allows you to placetwo 8 ohm speakers in parallel, one for the highs and one for the lows, andstill present an 8 ohm, not a 4 ohm load to the amplifier (which is what theamp would see without the network)
2) The crossover allows you to matchtwo drive units of different efficiencies so that the combination of the twodoes not result in a large shelf in the frequency response.
3) The crossover allows you to equalizethe frequency response of either the woofer or tweeter, making the compositeresponse smoother than the unfiltered curves of either driver (woofer ortweeter) would seem to indicate is possible. Wine for mac apps.
Let’s getback to the basic job of the crossover, and see if we can understand how itaccomplishes this by learning a few new but simple concepts. The basic job is to send the high frequencyinformation to the tweeter, and eliminate this same band of high frequenciesfrom the woofer. Next, send the lowfrequency information to the woofer, and eliminate this same band of lowfrequencies from the tweeter. Acrossover is in essence two frequency filter sections working in parallel. A high pass section and a low passsection. These terms are selfexplanatory. The high pass passes highfrequencies, the low pass passes low frequencies. The high pass is placed in series with thetweeter, and the low pass is in series with the woofer. A normal two way crossover will have sixterminals. Two in, and four out. Usually the two sections (high-pass andlow-pass) are in parallel and both sections ONLY work properly when the wooferand tweeter are both attached and operational. To understand how the crossoverdoes its job, we must first understand the concept of impedance.
The threemain passive elements used in all crossovers are resistors, inductors, andcapacitors. Impedance is a measure ofhow much the resistor, capacitor or inductor impedes the flow of electrons atany given frequency. Resistors are thesimplest components, and the easiest to understand. The have a constant resistance to the flow ofelectrons which does not vary with the frequency of the signal. Resistors do not change the phase of the signal, the time relationshipbetween voltage (electrical pressure) and current (flow of electrons). Theyonly consume power and convert it to heat.That is all they do. In a perfectspeaker system there is no need for resistors at all. Of course, while there are no perfectspeakers, all the basic crossover theory's we use start with that premise.
The secondelement we shall consider is a capacitor.The capacitor impedes the flow of electrons through it in a waywhich is NOT independent of frequency. (It would not be technically correct touse the word resistance, but you can certainly think of it as such toconceptualize this). The capacitor has an impedance which is inverselyrelated to frequency. This is to saywhen you double the frequency of a signal applied to the capacitors terminals,its impedes that electrical flow by half as much as it did at the lower frequency. If you halve the frequency applied, thecapacitors impedance (How much it impedes the flow of electrons) isdoubled. This is a very usefulcharacteristic which we use to create networks which are frequency selective(like a crossover).
The Maths
Theimpedance of a capacitor is determined by the equation: Z = 1/j(2piF*C)
Where Z =Impedance magnitude,
J = thesquare root of (-1) ('Imaginary' component to describe the phase)
pi = 3.1414etc
C = Valueof Capacitance in Farads; 1 Farad = 1,000,000 ufd
I think thefirst time I read about the square root of (-1) I had an allergicreaction. So, for those of us who arefinding it difficult to comprehend phase for the first time, we can pretend itis not there, and to simplify the discussion we will admit to this equationbeing somewhat incomplete without the “j”, but for the determination of thesteady state and not transient value of the impedance, we are OK.
When acapacitor is put in series (shares one electrical terminal) with a resistor,the capacitor acts as a high pass filter. At some frequency it has the sameimpedance as the resistor, and at a much higher frequency it is like a short,just a piece of wire with little to no resistance (at least in a perfectcapacitor). At very low frequencies theimpedance of the capacitor becomes so high, that essentially nothing can passthrough it. It is this characteristicwhich is used in endless electronic circuits to allow AC signals to passthrough yet completely block DC voltages from one part of the circuit to another.
Tabs in word for mac. Figure 1
Lets say wehave a 10 ufd capacitor and put in in series with an 8 ohm resistor. At some frequency, the capacitors impedancemagnitude will be equal to the 8 ohm impedance of the resistor. (Which ofcourse is still 8 ohms at any frequency from DC to daylight.)
Lets do the math.
Z = 8 =1/{(2)*(3.14)*(F)*(10)*(10-6)}
The capacitor is MICRO Farads, which is millionths of a Farad, so wemust multiply the 10 by 0.000001( 10-6) to convert microfarads intoFarads, so we can solve for the frequency (F) in Hz. By cross multiplying (multiplying both sidesby F and dividing both sides by 8) we get:
F =1/(2*3.14*8*10*10-6)) = 1990 Hz.
What thistells us is that at 1990 Hz (2 kHz if you are among friends) the impedance ofthe capacitor is equal to that of the resistor, so at this point the power intothe resistor is half what it would be without the capacitor. Put another way, at 1990 Hz, the power intothe resistor is 3 db less and this is your crossover point. Above this frequency the attenuation(reduction in power to the resistor) is less, and as we go lower in frequencythe capacitors impedance grows larger, while the resistor is unchanged, so theattenuation of signal to the resistor increases. At a frequency of 995 Hz, the attenuationwill be 6 db. At 498 Hz, it will be 12db. We call this a first order or 6 dbper octave crossover. That is, in anutshell, how this works. It grows incomplexity when more parts are used, but the basic function is the same. Now if your speaker is a perfectly resistiveload, your 10 ufd series capacitor will be a perfect passive 6 db per octaveHigh Pass (HP) filter. Don't worry,life is not that simple.
The otherimportant element used in a crossover network is the Inductor; sometimes calleda choke, and sometimes called a coil.They are all refer to the same thing.The Inductor is in many ways the inverse of the capacitor. It too changes impedance with frequency, butin a way entirely opposite to the capacitor.The inductors impedance grows LARGER with increasing frequency, notsmaller like the capacitor. Itsimpedance grows larger at the same rate as the capacitor grows smaller, and thesimilarity ends there. So if we replaceour capacitor with an inductor in the first case we have a first order low passfilter. (As shown in figure # 1)
Note thatin higher order networks, second order and above, both inductors and capacitorsare used in the same circuit to provide high-pass (HP) or low-pass (LP)filtering. In the HP filter, capacitorsare put in series with the load while inductors are put in parallel. (A resistor represents the load in figure 1,and the speakers represent the load in figure 2). In the LP filter the opposite is true, theinductors are in series and the capacitors in parallel with the load (orspeaker if you are making a crossover network).Hopefully this makes sense if you keep in mind the direction of theimpedance changes of the parts with increasing or decreasing frequency.
Figure 2
In thenetwork pictured in figure 2, the HP blocking capacitor, C1, increases inimpedance with decreasing frequency, while at the same time the shortinginductor L1, decreases in impedance with decreasing frequencies, hence shortingout the low frequencies while the capacitor is increasingly blocking them. The inverse is true of the LP filter. L2 is blocking highs and passes lows, whileC2 is shorting out the highs and passing the lows. This is how the LP filter blocks highs andpasses lows while the HP filter is performing the inverse function. It is this dual action that allows this kindof network to create twice the blocking action (12 db) per octave of frequencythan the simpler 6 db/octave network. Ifyou did not understand this on the first read, that is OK, you have a lot ofcompany. Think about how the capacitorand inductor change their impedance with frequency, and take another look atthis circuit diagram. If you try, youcan figure out what is going on as the frequencies change.
Back to thesimplest network, the 6 db per octave. If I have an 8 ohm resistor, and a 0.001Henry (1.0 millihenry inductor), at what frequency do I have a 3 db attenuationacross my resistor? Let's calculate thefrequency where the inductor equals 8 ohms.
The valueof the impedance of an inductor is determined by the formula
Z ={(2pi)*(J)*(F)*(L)} where F= Frequencyin Hz, L = Inductance in Henries Z= Impedance in Ohm
A Crossover Song Is
Like we didfor the capacitor, we shall ignore the imaginary 'J' term and pushahead.
8 =(2pi)(F)(0.001H) Therefore: F = 8/{(2pi)(0.001)= 1273 Hz
A Crossover Trial
If we wantto set the Low Passfrequency equal to the HighPass frequency of 1990Hz, what inductor do we need? We set F =1990 and then we can solve for L.
L =8/{(2pi)(F)} = 6.4 + 10-4 Henries =0.64 millihenries
We have calculated the parts needed for our first crossover network for our 8ohm tweeter and 8 ohm woofer with a crossover frequency of 1990 Hz Network below:
A Crossover Vehicle
FIGURE 3