Impedence and Crossover affects

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A question that I can not figure out is how the crossover affects impedence of a system.

I realize that most speakers are wired in parallel. And 2 8 ohm speakers in parallel = 4 ohms. But if I want a 2-way, with 2 8 ohm speakers, to be an 8 ohm system, how do I accomplish that? I assume it's done in the crossover.

I was looking at a MTM configaration with 2 8 ohm mids, and a 6 ohm tweeter. The overall system was rated at 4 ohms. How does it get there?

Maybe some simple examples would help to clarify.
 
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The best I can tell you is that impedance is a variable spread over the entire frequency range and that if you separate the highs from the lows and provide only one 8 ohm driver to each part of the system, you maintain an 8 ohm load.

The crossover provides a large increase in the resistance to frequencies above where the inductor takes over and the same is true for the the low frequencies and the capacitors.

You have separated the spectrum into 2 halves, you now provide an 8 ohm driver for each half. Think of them as two separate parts, like you would for left and right, only this is for high and low.

This same theory applies for three way XO's etcetera.

Hope this helps,

Cal
 
Impedance of a speaker is nearly always determined by bass units,
the crossover cannot change the impedance of the speaker.

A 16 ohm speaker has a 16 ohm bass or 2 8 ohm bass in series.

An 8 ohm speaker has two 16 ohm bass in parallel,
or an 8 ohm bass, or 2 4 ohm bass in series.

An 4 ohm speaker has two 8 ohm bass in parallel,
or a 4 ohm bass, or 2 2 ohm bass in series.

Basically you can ignore higher frequency units crossed over
to unless they have an impedance lower than the bass units.


:) sreten.
 
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Originally posted by squidbait
I realize that most speakers are wired in parallel. And 2 8 ohm speakers in parallel = 4 ohms. But if I want a 2-way, with 2 8 ohm speakers, to be an 8 ohm system, how do I accomplish that? I assume it's done in the crossover.

I was looking at a MTM configaration with 2 8 ohm mids, and a 6 ohm tweeter. The overall system was rated at 4 ohms. How does it get there?

I wonder of you are asking about two 8 ohm drivers carrying the same frequencies. Parallel gives you 4 ohms and series gives you 16, as you already know. If you want to end up with 8, then you either add resistors (works for mids and highs, but not really for woofers) or you rethink your project.

Cal
 
Any time two drivers play in the same freq range their impedences are parallel. So your two M's play the same range that means 8ohms in parallel with 8ohms = 4ohms. The tweeter is in a different freq range so it stands alone at 6 ohms. This is a gross oversimplification but you get the idea. If you had a woofer, mid, and tweeter there would be no overlap of drivers(there is some but dont worry) so each would be it own inpedence without any parallel interaction(there is some but dont worry, again).
 
squidbait said:
So in the MTM system I described above, the tweeter impedence lends nothing to the overall system inpedence, it's the two mid's that make it a 4 ohm system?
A different way to think about it: the way crossovers work is by adding frequency-dependent variable impedance.

If we postulate simple first-order filters (a cap on the tweeter, a coil on the woofers), we find that at low frequencies the cap presents a very high impedance and so the combination of the tweeter and cap need not be considered in terms of overall system impedance. At high frequencies, the coil presents a very high impedance so the coil/woofer combo drops from the equation. Somewhere around the crossover point, the impedance is approximately the same for each branch and so the overall impedance is a "shared" figure.

That said, most of the energy in music and movie sound is in the the bass and mid, so one can generalize that the tweeter is unimportant in terms of impedance, most of the time.
 
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