Sometime in the late seventies, when I put together my first couple of passive crossovers, the life of a DIYer was fairly simple. Speakers had a nominal impedance, you would use
the usual formulae for determining capacitor and inductor values, round them to the nearest manufactured values, and go to the electronic parts store with a shopping list. If you were a little advanced, you would use a scientific pocket calculator in the process, especially if it could do π with 7 decimal places.
Sure, the undaunted amateur had access to a breadth and depth of knowledge already then, but the measuring equipment to apply this knowledge was way too expensive. Combine this with the sheer computational overhead, and mostly anything really advanced was reserved for the pros. I had to make do with a simple analog multi meter (V, A, and Ω only), a TI-30 pocket calculator, and a Weller
It was around the time of the Apple ][ and VisiCalc, but it would be more than twenty years until I started to use a computer in the process of speaker building. Around the year 2000 I discovered WinISD which taught me what I should have done with a woofer I used in a speaker project in the eighties. At the push of a button it invalidated my former project's enclosure size, along with the dimensions of the port for tuning the woofer/enclosure system. So that is what I should have done? I just had to try this out; hence I bought a table saw and started to make some saw dust. After all, my first degree is in Experimental Physics.
Soon after discovering WinISD, I advanced to LspCAD and got to simulate crossovers. Every capacitor, inductor, and resistor turned into a knob on a big control panel, and twiddling any of these knobs instantly produced a new frequency response curve. I'm still proud of how flat the actual frequency response looked on my RTA once this project was done—although I could not fully explain why.
Along with making it easy to visualize the effects of the filters in a crossover, LspCAD also made things a lot more complicated. Speakers no longer had a nominal impedance of 8 Ω. Now they needed
resonance trap circuits. Not only did the software catch up with my hobby, I effectively had to catch up with the software!
In the process of one of these learning steps, while filling pages and pages of quad ruled paper, reviving calculus of complex numbers I had long relegated to the dusty corners of my brain, I eventually concluded this was getting silly. It was not exactly rocket science anymore what I was doing at that point, merely expanding a chain of series and parallel impedances of the forms R, j·ω·L, and 1/(j·ω·C), and all I wanted to know was the impedance for a particular ω. Couldn't I get Excel to crunch some complex numbers instead of me doing the boring algebra?
Turns out I could—after googling the question, finding the original Office CD, and installing some
Add Ins... Apart from skipping the algebra, yet confirming my understanding, I now had my own crude form of knobs on a control panel. Change a cell, and the spreadsheet instantly updates all the other cells. It was as if the complex algebra had come alive!
When I receive a speaker I've won on eBay (such as a cone transducer, a compression driver, or a ring radiator), I do a series of simple tests. After visual inspection and—for cone transducers—testing if the cone can be moved mechanically without voice coil rubbing, I use a multi meter to measure Re (DC resistance of the voice coil). If for any reason the speaker is shorted, then I obviously don't want to risk any damage to my amplifier.
If the speaker is
safe, I proceed to a quick listening test. Using an SACD, preferably containing a well recorded piano, I try to hear any obvious defects. The setup depends on the transducer: Woofers (or larger cone midranges) are connected directly to the amplifier. For tweeters (ring radiators, compression drivers) I try to find a suitable capacitor. Connected in series with the tweeter, this constitutes a simple high-pass filter. It gives the tweeter some level of protection from the low frequency content of the SACD (at this stage, I don't want to confirm the maximum power handling):
A simple high-pass filter
Capacitors have an impedance ZC (AC resistance) which is inversely proportional to the applied frequency ω = 2·π·f:
ZC = 1/(j·ω·C)
For ω → 0, ZC → ∞, corresponding to an open circuit or perfect insulator (as if the tweeter were disconnected altogether). Conversely, for ω → ∞, ZC → 0, corresponding to a short circuit or perfect conductor (as if the tweeter were connected without capacitor).
Connected in series with the transducer, and ignoring the transducer's voice coil inductivity Le for now, this lets the transducer tap into a voltage divider formed by ZC and Re:
Vout = Vin·Re/(Re + ZC)
Defining gain G(ω) as
G(ω) = |Vout|/|Vin|
and substituting ZC yields
G(ω) = ω·Re·C/(ω²·Re²·C² + 1)½
and, after normalizing with x = ω·Re·C,
G(x) = x/(x² + 1)½
which looks simple enough to see what G(x) looks like without actually seeing it:
un-normalizingx = ω·Re·C and solving for C:
C = 1/(ωxo·Re)
or, after expanding ω = 2·π·f,
C = 1/(2·π·fxo·Re).
Here is what this really looks like, plotting 20·log10(G(f)) in dB against frequency f in Hz.
A simple high-pass filter (fxo = 632 Hz)
Accordingly, when I set about testing a mint pair of JBL 077
slots (vintage version of the JBL 2405H), with a nominal impedance of 8 Ω and crossed over at around 10 kHz, I needed a capacitor of about 2 μF. I had two 1 μF in my parts bin, which adds up to 2 μF when wired in parallel, and thus I could protect the expensive tweeters during my listening test.
At another time the device under test was a vintage JBL 2482 midrange compression driver. Per its specifications, this driver can be crossed over as low as 300 Hz, which is what I wanted to hear. Unfortunately, I did not have a loose capacitor anywhere near the 33 μF required for a 16 Ω high-pass at 300 Hz.
The nearest I had was 82 μF, stuck in an old crossover dating from the eighties. Together with a 10 mH inductor it formed a nominal 8 Ω 12 dB/oct high-pass at 176 Hz. I could take out both capacitors, wire them in series, for a capacity of 82 μF/2 = 41 μF which would be close enough for now, especially considering that the actual impedance of the 2482 is less than 16 Ω. Or I could be lazy and simply connect the 2482 to the midrange branch while leaving the branches for the woofer and tweeter open. After all, if it's open, it shouldn't create a problem for the amplifier, right?
I guess I was lucky and—considering the compression driver's efficiency—conservative with the volume knob for my listening test. Had I read the warning on Lenard Audio beforehand, I definitely would not have been lazy! Here is why:
A 12 dB/oct high-pass filter without tweeter
Without tweeter, the high-pass branch is reduced to C, RL (the DC resistance of the inductor), and L wired in series. The amplifier
sees the impedance
Z = 1/(j·ω·C) + RL + j·ω·L
which is the sum of the respective impedances. Rearranging real and imaginary parts and remembering that 1/j = −j yields
Z = RL + j·(ω²·L·C − 1)/(ω·C).
To decide if this may create a problem for the amplifier, we'll have to look at the magnitude of the impedance
|Z| = (Z*·Z)½
where Z* denotes the complex conjugate of Z, and determine the minimum of |Z|. To do so, I could proceed with some basic calculus by solving
∂|Z(ω)|/∂ω = 0
for ω, which gets pretty complicated, and pretty quickly so: It starts by taking the derivative of a square root, followed by the chain rule applied to the derivative of a quotient, and subsequent applications of the chain rule. In the end, it yields a lengthy expression which handily obfuscates where this is going—even before solving the chain of derivatives for ω.
The alternative is to think before working (
constructive laziness). In the above expression for the impedance Z, only the imaginary part depends on ω, the real part stays a constant RL. Accordingly, whatever the sign of the imaginary part, the magnitude will always be at least RL. In turn, the imaginary part reduces to 0 for
ω²·L·C = 1
ω0 = 1/(L·C)½
or, after expanding ω = 2·π·f,
f0 = 1/(2·π·(L·C)½).
Without the tweeter, at f0 the amplifier sees a minimum impedance
Zmin = RL.
All that impedes current at f0 is the DC resistance RL of the inductor—essentially a lengthy copper wire.
Ouch! I consulted the manual of my amplifier for its maximum output current, and while this figure is remarkably generous, the price of a new amplifier does not encourage further experiments along those lines. Instead, I had LspCAD simulate the fault condition of a missing tweeter. LspCAD wouldn't let me disconnect the tweeter from the tweeter branch of the crossover, but it did let me pretend a tweeter with Re = 1000 Ω. This will have to do for an
open tweeter. Here is the impedance response with the component values taken from the old crossover:
Impedance of crossover's tweeter branch without tweeter as seen by the amplifier
Like I said, I was lucky that my amplifier can handle generous output currents, that the compression driver didn't need much power for a listening test, and last but not least, that RL was a relatively high (!) 0.4 Ω (as specified on my inductor). Depending on the position in the crossover, such as in series with a woofer, it may make sense to use an inductor with thicker wires for a lower RL and thereby smaller insertion losses.
REMEMBER: It doesn't always take a shorted speaker to destroy an amplifier. An open branch of a crossover may do this just as well!!!
The popularity of esoteric speaker wires and associated myths provides me with an ongoing source of entertainment. One such myth is that the wires to both speakers of your stereo (or however many speakers you may have in your home theater) should be of the exact same length. If they are not, the differences in wire lengths are claimed to be audible as some kind of detriment to the sonic experience.
Without understanding the inner workings of speaker wire, it may seem like a wise decision to opt for equal lengths: If the wire did anything objectionable, at least it would do the same to all speakers. Accordingly, unequal wire lengths might throw the holographic stereo or multi-channel image out of balance. So you go for the salesperson's recommendation and buy equal lengths.
At home, it may not always be practical to place the amplifier exactly halfway between the two speakers of the stereo, or even more impractically so, in the center of the surround system. That spot is best occupied by the sofa. But what to do with the extra lengths? Just curl them up in nice bobbins behind the amplifier, like your garden hose in the backyard?
In the following considerations, I'll assume that for some practical reasons one speaker is 3 m (10 ft) further away from the amplifier than the other. This gives me an excuse to save an extra 10 ft of speaker wire for the other speaker. After all, I'll be using expensive 12 AWG (3.3 mm²) speaker wires. If your situation varies, feel free to reach for that pocket calculator and crunch your own numbers.
Electricity travels through copper wires at approximately the speed of light, or roughly c = 3·108 m/s. To travel the extra distance x = 3 m, this delays one of the speakers by an amount of time t
t = x/c
or 10-8 s = 10 ns (nanoseconds).
If you think you can hear this, you will have to compensate it by delaying the sound of the other speaker. To do so simply recess it accordingly: Sound travels at the rate of c = 343 m/s, hence the recess x, away from the listening position, is
x = c·t
or 3.43 µm (micrometers or microns). That would be 0.135 mil if this makes it any easier to grasp. Adjust the position of the respective speaker, but keep your exact same (!) listening position.
If you are a salesman of expensive speaker wires, you may point out that the real difference is the
loss of Ohms in the extra length of wire, which will affect the stereo image.
To get an idea of the potential impact, recall that I was going to use 12 AWG speaker wires. 12 AWG has a resistance of approximately 1.6 mΩ/ft (cf. AWG table here), hence 3 m (10 ft) of it adds up to about 16 mΩ (milliohms). Connected in series with the speaker, and assuming that the expensive speaker is a perfectly resistive load, this represents a speaker attenuator formed by Re (the speaker's resistance) and R (the speaker wire's extra resistance).
A simple speaker attenuator
Its gain G is
G = Re/(Re + R)
or, in dB,
GdB = 20·log10(G).
For the above 16 mΩ wire and a 4 or 8 Ω speaker, this evaluates to -0.035 or -0.017 dB, respectively, which is a very small attenuation for either speaker.
If you think you can hear this, tell me to what tolerance your tweeters were matched before they were installed. The industry standard for driver matching is ±3 dB. If you got them from a reputable source, you may get ±1.5 dB or even ±0.5 dB, but this is still a lot higher than 0.017 dB. To put this into perspective, using 12 AWG speaker wire to attenuate an 8 Ω speaker by 0.5 dB would take an extra length of about 91 m (300 ft).
Alas, if you already have the speaker wire neatly curled up, consider an alternative layout.
Arranging wire in neatly stacked curls forms an inductor. In turn, analogous to the simple high-pass filter, an inductor connected in series with the speaker (woofer) constitutes a simple low-pass filter. It
protects the woofer from high frequency content:
A simple low-pass filter
Inductors have an impedance ZL (AC resistance) which is proportional to the applied frequency ω = 2·π·f:
ZL = j·ω·L
For ω → 0, ZL → 0, corresponding to a short circuit or perfect conductor (as if the woofer were connected without inductor). Conversely, for ω → ∞, ZL → ∞, corresponding to an open circuit or perfect insulator (as if the woofer were disconnected altogether).
Connected in series with the transducer, and ignoring the transducer's voice coil inductivity Le as with the simple high-pass filter, this lets the transducer tap into a voltage divider formed by ZL and Re:
Vout = Vin·Re/(Re + ZL)
Defining gain G(ω) as
G(ω) = |Vout|/|Vin|
and substituting ZL yields
G(ω) = 1/(1 + ω²·L²/Re²)½
and, after normalizing with x = ω·L/Re,
G(x) = 1/(1 + x²)½
which looks again simple enough to see what G(x) looks like without actually seeing it:
un-normalizingx = ω·L/Re and solving for L:
L = Re/ωxo
or, after expanding ω = 2·π·f,
L = Re/(2·π·fxo).
Here is what this really looks like, again plotting 20·log10(G(f)) in dB against frequency f in Hz.
A simple low-pass filter (fxo = 632 Hz)
Except, of course, that we need fxo in terms of L, and moreover, we need the actual L of the extra length of wire nicely stacked up in curls. The former is trivial, while latter gets me deeper into physics than I'm comfortable with right now. Textbook to the rescue!
In vacuum, a cylindrical coil with cross-section A, length l, and N turns of wire has an inductivity L of
L = μ0·N²·A/l
wherein μ0 = 4·π·10−7 is the permeability in vacuum and A = π·r² is the area of a circle. Hence
L = 4·π²·r²·N²·10−7/l.
We are getting closer. Let's curl the extra 10 ft of speaker wire into 10 loops. This leaves 1 ft or 12" per turn, or turns with a radius r = 12"/(2·π) = 2". Let's assume this piles up to a stack of about 1" (12 AWG wire is fairly thick, especially with an expensive looking transparent insulation around it). Thus, in SI units
r = 0.05 m
N = 10
l = 0.025 m
which gets me an approximate value for the
inductor we made out of the extra length of speaker wire:
L = 40 µH.
Granted, this is just a rough estimate. The textbook formula is meant for vacuum (whereas I have actual air in my listening room) and it is valid for lengthy coils only (whereas ours is shorter than wide). But it will have to do for now because my new digital multi meter cannot measure such small inductors.
Armed with the
ballpark figure of L = 40 µH, we can now determine the corresponding crossover frequency of the low-pass filter that the curled up speaker wire has caused, to see if this is of any concern:
fxo = Re/(2·π·L).
Depending on the nominal impedance Re of the speaker, I get the following values:
fxo(8 Ω) = 32 kHz
fxo(4 Ω) = 16 kHz
In other words, 10 turns of extra speaker wire, nicely curled up in 4" curls, stacked 1" high, should cause the response of a 4 Ω speaker to be 3 dB down at 16 kHz.
This can be heard!
The inductor page on Wikipedia quotes a formula for a short cylindrical coil as follows:
L = r²·N²/(9·r + 10·l)
with r and l in inches, and L in µH. Running the numbers with the above formula, I get
L = 14.3 µH
and the following nominal cutoff frequencies
fxo(8 Ω) = 89 kHz
fxo(4 Ω) = 45 kHz
which would be a lot harder to hear. At 20 kHz, a 4 or 8 Ω speaker would be 0.8 or 0.2 dB down, respectively. Can you hear this?
If not the last word, the above theoretical considerations should at least give an idea why to avoid curling up extra speaker wire. In particular, avoid iron bobbins and similar materials with a high permeability (essentially any material attracted by a magnet). Also remember that my numbers are merely examples. Yours may be different. If you're so inclined, have another look at either formula above and think about the consequences of changing the number of turns, the radius, and/or the length of the coil.
In the end, physics has it that every circuit is an inductor, albeit a very small one with a single turn only. In this sense, another post on TrueAudio explores the consequences of the straight speaker cable being a single turn of wire that runs to the speaker and back once. The post mentions a formula for L per foot:
L/ft = 0.281·log10(d/r)
with L in µH, d the center-to-center distance of the two wires, and r the radius of the wire. The units of d and r can be inches or mm, as long as they're both the same, since we're just looking at their ratio.
Luckily, I have a snippet of my expensive speaker wire left over in the parts bin, and measure
d = 10 mm
r = 1 mm
L = 2.81 µH
for 10 ft (3 m). This pushes the nominal cutoff frequencies way above 20 kHz. At 20 kHz, a 4 or 8 Ω speaker would be a mere 0.033 dB or 0.0085 dB down, respectively.
CONCLUSION: Do try this at home and use whatever length of speaker wire it takes, but avoid making ornamental curls with your wire.
Next to esoteric speaker wires, bi-wiring is another popular myth that I yet have to understand (or bust). Bi-wiring designates the concept of running two pairs of wires to each speaker, one directly to the woofer's low-pass filter, the other one to the tweeter's high-pass filter. In other words, instead of splitting a single pair of wires at the
entrance of the speaker, it is split at the
exit of the amplifier.
Googling the subject, I did not find a conclusive answer—certainly not a simple one that would have convinced me with a few bullet points or a single diagram. Nevertheless, several trustworthy websites brought up
back emf (emf is short for electromotive force, symbol ε).
Back emf is a concept that I should have been familiar with, yet I don't recall ever associating it with speakers—a potentially critical oversight in my quest for sonic bliss.
Let's start with the electromotive force. Simply put, the signal sent from the amplifier to (and through) the speaker's coil creates a force that interacts with the speaker's magnet. The coil is attached to the speaker's cone, which in turn is more or less flexibly attached to the speaker's basket. This is the electromotive force at work. It makes the speaker's cone move back and forth along with the received signal. The electrical signal is transduced into cone vibrations. These vibrations alternately compress and rarefy air, which we perceive as sound.
But as the cone is vibrating, it also acts as a generator that creates electricity. Indeed, a speaker can be used as a microphone—a fact which I found out already as a kid while goofing around with spare speakers and an old table radio. Just connect the speaker to a suitable input (
aux, or, if the amp has it,
mic input). Not every speaker is equally suited for the purpose. Some would simply be way too heavy to handhold. But if you have ever looked inside an old telephone receiver, you will have noticed that there is not much of a difference between what's inside the mouthpiece (the microphone), and what's inside the earpiece (the speaker).
What escaped me so far is what happens to the electricity generated by the speaker (the
back emf) when the speaker is connected to the amplifier's output—in other words, during normal operation of the speaker. Wouldn't emf and
back emf be constantly
fighting each other? And what happens once the input signal has stopped? The speaker still vibrates, and while the emf is gone, the
back emf is continuing its
fight until the vibrations have died down.
Two articles had most of the answers. The first article, Solid state amplifiers by Lenard Audio explains how a modern solid state (or transistor) amplifier deals with
back emf. The second article, Active vs Passive Crossovers by Elliott Sound Products explains what happens with
back emf once you put a (passive) crossover network between the amplifier and the speaker. This was the startling part for me, but first things first.
As far as I gather, a modern transistor amplifier acts as a voltage source. To a speaker, the amplifier looks like a kind of a battery—albeit with alternating voltage. In theory, this voltage is independent of the current drawn from this source (ideal voltage source). Whatever the current, the voltage is unaffected. In practice, of course, this is not possible. The current is limited by the design of the particular voltage source or amplifier.
This limitation can be expressed in terms of the
internal resistance (German: Innenwiderstand, English: output impedance) of the voltage source. The following illustration shows a real voltage source as an ideal voltage source with the
limiting resistor Ri in series:
A real voltage source with a real output impedance
In practice, this means that the voltage will drop in relation to the
internal resistance and the
load resistance drawing a current. This voltage drop can be calculated by duly applying Kirchhoff's Loop Rule.
Modern amplifier design strives to keep the output impedance as low as possible. This creates the desirable property of a high damping factor. The damping factor D is defined as the ratio between the load impedance (ideally the DC resistance Re of the speaker) and the output impedance (ideally Ri):
D = Re/Ri
The lower the output impedance of the amplifier, the higher the damping factor gets. In turn, as far as I understand, the higher the damping factor, the better is the ability of the amplifier to
Think of it this way: The amplifier and the speaker form a closed electrical circuit. Both the amplifier and the speaker are some form of
generators. The amplifier is the active generator (the emf), while the speaker generates voltage in response to the amplifier (the
back emf). The emf faces a relatively high resistance Re drawing a small current. This will not cause the generator's voltage to drop all that much. Conversely, the
back emf faces a really low resistance Ri. This causes a violent voltage drop approximating a short circuit. The result? The amplifier
wins. The vibrating speaker cone and coil, the source of the
back emf, come to a premature halt:
Tug-of-war between the amplifier and the speaker
Notice that if the speaker's
back emf didn't have the amplifier's output impedance as
antagonist, the speaker would continue vibrating until its energy is somehow
absorbed mechanically, by a combination of the volume of air trapped in the speaker's enclosure and the properties of the material that attaches the cone to the speaker's basket. In this context, the amplifier's low output impedance acts as an electrical brake, and hence the notion of prematurely halting the vibrations of the speaker's cone. To the degree that the amplifier's output impedance can approximate 0 Ω, it simply
The efficacy of this electrical brake gets reduced once there is a passive crossover between the amplifier and the speaker, such as a 12 dB/oct low-pass filter between the amplifier and the woofer:
A 12 dB/oct low-pass filter between the amplifier and the woofer
In the case of the above crossover, the woofer as a generator of
sees the low-pass filter as a capacitor C wired in parallel with an inductor L, its DC resistance RL, and the
internal resistance Ri of the amplifier:
The same low-pass filter as
seen by the woofer generating
read the diagram from right to left)
The impedance of this circuit is
Z = ZC·ZL/(ZC + ZL)
ZC = 1/(j·ω·C)
ZL = R + j·ω·L
where I have combined the sum of the amplifier's
internal resistance Ri and the inductor's DC resistance RL into a single variable R. Substituting ZC and ZL into Z yields
Z = (j·ω·L + R)/(1 − ω2·L·C + j·ω·C·R).
Compared to the Crossovers as seen by the Amplifier, where we looked at the minimum impedance of a high-pass without tweeter, this time around we want to find out the maximum value of the impedance Z as such would reduce the efficacy of the previously discussed electrical brake. On suspicion that it may be at
ω0 = 1/(L·C)½
I'll substitute ω0 into Z and determine the magnitude of the impedance at that particular value of ω. Thus
|Z(ω0)| = |(j·ω0·L + R)/(j·ω0·C·R)| = (L2/(C2·R2) + L/C)½.
To get an idea of what this could mean in practice, we need some numbers. I'll assume a 12 dB/oct Butterworth low-pass filter for a nominal speaker impedance Rnom of 8 Ω and a nominal crossover frequency of 200 Hz, similar to the previously mentioned old crossover dating from the eighties. For a 12 dB/oct Butterworth filter the ratio L/C equals 2·Rnom2. Furthermore, I'll assume the crossover uses a really good inductor, with a low RL of 0.4 Ω. Last but not least I'll assume a somewhat realistic amplifier output impedance of 0.1 Ω, which at the same time turns out to be the lowest value I can readily simulate.
With these numerical values we can simplify the above expression for |Z(ω0)|. With L/C = 2·Rnom2 and Rnom = 8 Ω, L/C = 128 Ω2. Hence, compared to the quadratic term in L/C, the linear term can be
|Z(ω0)| = (L2/(C2·R2) + L/C)½ ≅ L/(C·R).
Substituting L/C = 128 Ω2 and R = Ri + RL = 0.1 Ω + 0.4 Ω = 0.5 Ω, I get
|Z(ω0)| ≅ 128 Ω2/0.5 Ω = 256 Ω.
In other words, at the nominal crossover frequency of the low-pass filter the speaker sees an impedance of 256 Ω. This is a far cry from the assumed amplifier output impedance of 0.1Ω—it is going to take quite a bit longer to brake that speaker!
With a bit of
sideways thinking I managed to simulate the crossover as seen by the transducer in LspCAD. First, I created a 12 dB/oct Butterworth low-pass filter for the above numbers. Then I reversed the order in which the filter components appear as series and shunt branches. Finally I pretended a
driver (transducer) with a very low DC resistance Re = 0.1 Ω (this is the lowest LspCAD would let me) and no inductivity Le of its own. This
driver served to simulate the
internal resistance of the amplifier. Here is the resulting impedance response:
Impedance of crossover as seen by the transducer
Notice the peak of about 250 Ω at 200 Hz. The simulation confirms the above calculations. One octave below and above the crossover frequency, at 100 Hz and 400 Hz respectively, the impedance seen by the transducer drops below 8 Ω. This is still way higher than the
internal resistance of the amplifier. With Rnom = 8 Ω this barely pushes the damping factor above 1. Notice that for a
lossier inductor with a higher DC resistance, the impedance peak would be lower—at the expense of a
lossier inductor. This doesn't sound like a good alternative.
Somehow this seems unfair. I've just paid for an expensive amplifier with a high damping factor, and now I'm about to negate the benefits of said damping factor with a passive crossover between the amplifier and the transducer.
Now, while these numbers look pretty scary, I do not yet know just how important this is in the grand scheme of things. I imagine that the frequency dependent loss of the amplifier's ability to act as an electrical brake could mask soft passages in music reproduction when the transducer keeps vibrating impetuously after a loud passage. I will have to find out the magnitude of this problem.
In closing, it's funny how this endeavor turned out. I merely wanted to learn more about bi-wiring, but instead I learned about
back emf, and that what looks like a filter to the amplifier looks like a resonant circuit to the transducer. But I still don't know much about bi-wiring.
One thing is for sure, though: Bi-wiring won't be of much help braking the transducer, because the (passive) crossover remains
in the way. To get the amplifier's full damping factor back, I would need to tri-amp my speakers, i.e. use an individual amplifier per driver (low, mid, high) and channel (left, right), for a total of 6 channels, and implement the crossover at the pre-amp stage.
Stay tuned! I will find out—eventually, anyway.
perfect pair matching) was an extra $30 fee I didn't consider at the time.
Page 8: Solid state vs Valve amps.
Active vs Passive Crossoversthen
2.1 The Crossover—a Different View.Elliott Sound Products ask that other websites do not link to individual pages of their site, which I'll be happy to comply with, and hence these instructions for finding what I am referring to.