Linearteam WinISD Pro
Passive crossovers are installed in series with the speakers in your box, and serve to "funnel" the output frequencies from your amp to the appropriate speaker in your box. Sending low frequency notes to your tweeter will smoke it in no time flat. Sending high frequency notes to your woofer wastes high frequency power that would be better utilized by your tweeter. Your midrange fits in the middle there, and is subject to both scenarios.
Passive crossovers are constructed from capacitors and inductors:
Capacitors (caps) block low frequencies and pass high frequencies. Here's the electronic symbol for a capacitor:
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This is an accurate symbol in that it does a pretty good job of representing how a capacitor is actually built. Instead of getting into plates and dielectrics and all the rest, we'll just discuss how the thing functions. It's important to note that no electrical current flow actually passes THROUGH a capacitor. Think of a capacitor like this...imagine a horizontal pipe a foot long and about as big around as a coke can. Now, imagine that in the middle of the pipe is a rubber diaphragm (if you were to look into the pipe, you couldn't see all the way through..rather you'd see the rubber diaphragm in the middle of the pipe). Now, you couldn't pass anything (such as a smaller diameter pole, for instance) THROUGH this pipe from one end to the other. BUT, you could give the illusion of passing a pole BACK and FORTH through the pipe if your were to alternately push and pull on two different poles that actually met in the center of the pipe. As you push one way, the diaphragm will strecth a little, and then as you push the other way, it would stretch back to the other side. To an audience that didn't know any better, they'd swear that the pipe is indeed hollow!
This is how a capacitor affects electrical current. A direct current (DC) is completely blocked, while an alternating current (AC) is able to pass. Since DC is essentially ZERO Hertz, then it follows that maximum impedance is presented to LOW frequencies. The impedance drops lower and lower as the frequency increases (the faster you push/pull on the poles, the easier it becomes). Remember, a capacitor offers a "free ride" to HIGH frequencies.
Inductors are the exact opposites of capacitors. An Inductor is a winding of wire which serves to block high frequencies, while giving the low frequencies their "free ride"! The electrical symbol for an inductor is:
__UUUUUU__ <---pretend like that is a non-broken line.
Any time current flows in a wire, an electric field is produced around the wire. When the current is then removed from the wire, the field "collapses". If you take a wire and wind it into a coil, things get interesting. As you initially begin to pass current through the coil, a field tries to build up around each winding in the coil (after all, it is in reality still just one length of wire). Trouble is, that since the coils are all so close to each other, they actually begin to interfere with each others progress! BUT, once a steady current flow is established in the wire, the fields around the coils quit fighting. As long as that current is kept constant, everbody is happy. When the current begins to change again, though (such as when the current is increased or removed), they get to fighting again, and continue to fight until the current quits changing. For this reason, inductors OPPOSE any CHANGE in current.
As far as DC and low frequencies are concerned, the inductor isn't much of a big deal. But for high frequencies which are always changing (and rapidly at that), the inductor is a bit of a pain in the ass. Remember, the inductor offers a "free ride" to low frequencies.
Ok, we know that the impedance of caps and inductors change with corresponding changes in input frequencies. The impedance that caps offer is called Capacitive Reactance (Xc), and is measured in ohms, given by the formula:
Xc = 1/(2pi * F * C)
Where,
Xc = Capacitive Reactance (ohms)
2pi = 6.28
F = Input Frequency
C = Capacitance value of the capacitor itself (in Farads)
The impedance that inductors offer is called (surprise) Inductive Reactance (Xl), and is also
measured in
ohms, given by the formula:
Xl = 2pi * F * L
Where,
Xl = Inductive Reactance (ohms)
2pi = 6.28
F = Input Frequency
L = Inductance value of the inductor itself (in Henries)
To put this in perspective, take a look at the following diagram of a typical 2 way passive
crossover
design (2- way means a woofer and tweeter only, no midrange driver):
Here, you see that the inductor is in SERIES with the woofer (marked "lo"), and the cap is in SERIES with the tweeter (marked "hi"). In keeping with what we discussed above, notice that the inductor (who gives the free ride to low freqs) is indeed connected to the woofer. This allows all the low, bass notes to travel to the woofer, and also blocks the high freqs from reaching the woofer. Likewise, the cap is connected to the tweeter, which gives the free ride to the high frequencies, while blocking those deadly low frequencies from reaching the tweeter.
Naturally, the next logical question is "Yeah, but what size cap and inductor do I use?". Good question. The first thing to determine, of course is what frequencies do you want to chop off?
If you want to stop all frequencies above, say, 1500 Hz from reaching the woofer, then you need to use an inductor that will offer the same impedance that the woofer is rated at, at 1500 Hz. For example, if you're using an 8 ohm woofer, and your cut-off frequency is to be 1500 Hz., then you need to choose an inductor whose impedance is 8 ohms at 1500 Hz!
Recall the formula for Inductive Reactance:
Xl = 2pi * F * L
Well, we know that we want the inductive reactance to be 8 ohms at an input frequency of 1500
Hz. So, plugging that in, we get:
8 = 2pi * 1500 * L
We need to get that L alone, so we divide both sides by (2pi * 1500):
8/(2pi * 1500) = (2pi * 1500 * L)/(2pi * 1500)
The (2pi * 1500) on the right side of the equation cancels itself out:
8/(2pi * 1500) = L
Ah, that's better. Breaking out the trusty calculator (or use the built in calculator in
Windows by clicking START, RUN, and type in CALC, and hit ENTER), we get:
L = 0.0008493 Henries.
Move the decimal 3 places to the right to get the answer in the more popular milli-Henries
(mH):
L = 0.849 mH
Now, suppose we want the same cutoff frequency for our tweeter. The same rule applies..we
need a cap that will offer 8 ohms (provided the tweeter is an 8 ohm tweeter, of course) at
1500 Hz. Looking at the formula for capacitive reactance (Xc) again:
Xc = 1/(2pi * F * C)
We get:
8 = 1/(2pi * 1500 * C)
We need to get that C alone, so begin by taking the inverse of each side:
1/8 = (2pi * 1500 * C)/1 <---the 1 cancels itself since any number divided by 1 is
itself.
Which gives us:
1/8 = (2pi * 1500 * C)
Dividing both sides by (2pi * 1500) frees up the C:
(1/8)/(2pi * 1500) = C
Ah. Now for the calculator:
C = 0.00001327 Farads
Moving the decimal 6 places to the right gives us the answer in the popular micro-Farads
(uF):
C = 13.27 uF
Ok, so our cap is gonna need to be 13.3 uF, and our inductor needs to be 0.849mH in the
crossover. Look again at the diagram:
At 1500 Hz input frequency (from the + and - terminals in the diagram), our inductor, L, will offer 8 ohms of reactance (resistance). Since the woofer is also rated at 8 ohms, this means that the voltage being applied to the woofer is half of what it would be if the inductor weren't there. This results in a 6 dB loss from the woofer at 1500 Hz. And remember, inductors hate high frequencies! So that means that when a higher frequency comes along, the reactance offered by the inductor will be even HIGHER than 8 ohms. In fact, for every octave (doubling of) increase in frequency, the inductor will lower, or attenuate, the output from the woofer by an additional 6 dB! This is referred to as a "6 dB/Octave" crossover (plug in other frequency values into the inductive reactance formula and you can see how much reactance the inductor offers to the circuit at those frequencies...remember to use 0.000849 as the value for H)....
Likewise, at 1500 Hz, the cap also offers 8 ohms of reactance, which cuts the output from the tweeter by 6dB as well. And, since caps block low frequencies, each octave (doubling) of frequency drop will cause a corresponding 6 dB of attenuation to the cap as well. Again, plug in different values to the capacitive reactance formula and look at the different impedances offered into the circuit by the cap at different freqs (remember to use 0.0000133 as the value for F).
You can use these formulas to figure out which size caps or inductors are needed to cutoff any speaker at any frequency.
Sometimes, cutting off a speaker at a rate of 6 dB/Octave isn't enough. At 6 dB/Octave, especially in high power systems, alot of lows are still powerful enough to get through and damage the tweeters. In this case, a 12 dB/Octave crossover can be used to "steepen the curve" of attenuation. In this case, the actual cutoff frequencies are still the same, but for each corresponding octave change in frequency, the input to the respective driver is reduced by 12 dB, instead of only 6 dB. How? Look at the following diagram:
In this design, BOTH the woofer and the tweeter are connected to an inductor AND a cap! We'll stay away from the long math (just let WinISD figure it out for you). However, a quick explanation will clear it up for you.
First, let's look at the woofer. Inductor L1 is in SERIES with the woofer. This means that any and all frequencies that pass through the woofer MUST also pass through L1. Well, L1 works just like he did before. He blocks the highs from reaching the woofer. But, since L1 can only block 6 dB/Octave's worth of high freqs, there are still some high freqs that sneak through. Here's where cap C1 comes into play. Notice that C1 is wired in PARALLEL to the woofer. This means that the cap can shunt, or bypass, frequencies around the woofer. Remember, capacitors give high freqs a "free ride". So, any high freqs that sneak through L1 are given a shortcut by C1. Why should the high freqs bother going through the woofer when C1 is giving them a "free ride" straight over to the other side? What it all boils down to is that L1 does his 6 dB/Octave's worth of work, and C1 does an additional 6 dB/Octave's worth of work. Together, they combine to yield a net of 12 dB/Octave of attenuation!
Capacitor C2 and Inductor L2 work exactly the same routine on the tweeter. C2 blocks as many lows as he can, and any lows that sneak through are given a free ride by L2.
Inductors and capacitors only come in certain values. Our circuit called for a 13.3 uF cap. You're not going to find any 13.3 uF capacitors. However, 10 uF caps are quite common, and using a slightly different value won't kill you. Also, the same is true for inductors. In fact, you don't have to put a crossover network on your woofers...
12 dB/Octave passive crossovers can be dangerous to your amp. Look again at the 12 dB/Octave crossover:
Notice that in each branch, the inductors and capacitors are in SERIES with each other (L1 is in series with C1, and C2 is in series with L2). In each branch, the inductor and capacitor are each chosen so as to offer 8 ohms of reactance at 1500 Hz. What this means is that at 1500 Hz, the cap is offering 8 ohms of capacitive reactance, and the inductor is offering 8 ohms of inductive reactance. Capacitive reactance and Inductive reactance are exact opposites, and cancel each other out! If you have 23 ohms of capacitive reactance, and 18 ohms of inductive reactance in a circuit, the net reactance is 5 ohms (the difference of the two). If the woofer or tweeter are removed from the circuit (become unplugged, or electrically "open" during play), then what you have is 8 ohms of capacitive reactance, and 8 ohms of inductive reactance in that branch. Since the two reactances cancel each other out, then the amplifier effectively sees a grand total of ZERO ohms at 1500 Hz! This will result in a damaged amp. Keep this in mind when considering implementing a 12 dB/Octave passive crossover into your system. Amps with built in short circuit protection will most likely survive this scenario, however...
Active crossovers are used mainly on multi-amp systems. Active crossovers are small, electronic circuits which serve to separate frequencies at the "line-level", and send those frequencies to separate amps. Each amp then plays the frequencies it's given. Each amp then powers it's own special set of speakers (one amp for woofers, another amp for tweeters, etc.)
WinISD will figure out the values you need to construct active crossovers, however, the in's and out's of the design of these devices is way beyond the scope of this Help file. If you can't look at the diagrams and immediately recongnize what components you need to buy and solder together, then don't even attempt it. Go buy one from the store!
A good source for active filter design is:
Active Filter Cookbook (2nd Edition)
Don Lancaster
ISBN 0 7506 2986 X (Newnes)
(c) 1996