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Old 14th July 2009, 07:11 PM   #1
wpc is offline wpc  United States
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Default Component speakers and Impedance values

I noticed most separate component speakers say they are 4 ohm speakers. But that's connected to the crossover and both speakers connected to the crossover.

What is the impedance of the tweeter and mid separate? Would that make each of them 2 ohm run in parallel or 8 ohm in series? Or does the passive crossover do some magic and the speakers are really 4 ohm each?
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Old 14th July 2009, 09:26 PM   #2
claff is offline claff  United Kingdom
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Get the dmm out and test them

otherwise they could be loads of different combinations that would make the amp see 4 ohm. 2 lots of 8ohm would do the trick, also 5 ohm and 20 ohm etc. etc.
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Old 14th July 2009, 11:23 PM   #3
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4 ohms each. The crossover isn't magic, it's purely physics. Mostly Ohm's Law.

Internally, a two-way M/T crossover will have separate crossovers for the midrange and tweeter. The inputs to the individual sections are tied together in parallel.

Let's look at a simple first order (6dB) HP tweeter crossover:

It's just a capacitor in series with the tweeter...

The value of the capacitor is chosen so that it presents an impedance (Xc) equal to the tweeter's at the desired crossover (-3dB) frequency. Here's the formula: C=1/(2*pi*f*Xc)

Let's say the tweeter is indeed 4 ohms, and you want to crossover at 4000 Hz.

C=1/(2*3.14*4000Hz*4ohms) = 0.00000995 Farads or 9.95uF

Now, lets look at what's happening with a 4000 Hz sine wave running through that system... You have a 4 ohm speaker in series with a 4 ohm capacitor. So that's an 8 ohm load at 4kHz as far as the amp is concerned. The tweeter is only "getting" half (4/8) of the crossover's input voltage, and output would be 3 dB less than it would be without the capacitor.

Turn the equation around to find the capacitor's impedance (actually called reactance) at 8000 Hz:
Xc=1/(2*3.14*f*C)
Xc = 1/(2*3.14*8000Hz*9.95e-6uF) = 2 ohms

So, at 8kHz the capacitor is 2 ohms. The cap and tweeter in series present a 6 ohm load to the amp, and the tweeter "gets" (4/6) 2/3 of the crossover's input voltage, more than half. The capacitor's reactance continues to fall until it is effectively out of circuit.

OK, let's go down below the crossover f3 to 2000Hz and see what's happening.
Again, Xc=1/(2*3.14*2000Hz*9.95e-6) = 8 ohms

So the capacitor is 8 ohms at 2kHz. The cap and tweeter in series present a 12 ohm load to the amp, and the tweeter "gets" (4/12) only one third of the crossover's input voltage. The capacitor's reactance increases as frequency goes down until the tweeter is effectively out of circuit.


Exactly the opposite happens in the low pass circuit, and the two circuits in parallel are a reasonable equivalent to the load of the individual speakers in the region where they are effectively working.

Higher order crossovers simply add more components to increase the rate that the output voltage changes with frequency.

Hope that helps.
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Old 20th July 2009, 09:42 PM   #4
wpc is offline wpc  United States
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Wait, are you saying that each are 4 ohms but the impedance presented to the amp is even higher than 8 ohms because of the crossover?
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Old 21st July 2009, 12:46 AM   #5
ppia600 is offline ppia600  United States
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The crossover can make the amp see an average of 4ohms even though both of the components (woofer and tweeter) may be 4 ohms each. The capacitors, chokes and resistors work to supply each component with frequencies within their useable range.


Something like that...


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Old 21st July 2009, 01:19 AM   #6
infinia is offline infinia  United States
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Hi
The crossover circuit helps define the range of frequencies for each driver and outside that range, the specific drivers impedance will rise due to the crossover, so as not to load each other too much in parallel. So generally the midbass impedance will rise due to added series inductance and the tweeter impedance will rise due to series capacitance. So the net impedance seen by the amp at the crossovers input is kept around the range of 3-6 ohms. Is that clear as mud?
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Old 21st July 2009, 11:55 AM   #7
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Quote:
Originally posted by wpc
Wait, are you saying that each are 4 ohms but the impedance presented to the amp is even higher than 8 ohms because of the crossover?

If you are considering just the woofer or just the tweeter with it's crossover section, yes it can be.

To make matters even more counter-intuitive, the individual speaker's impedance (without the crossover) will also rise above 8 ohms as frequency rises. Impedance, by definition, isn't constant.
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Old 21st July 2009, 12:02 PM   #8
AndrewT is offline AndrewT  Scotland
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if you are building a 4ohm speaker then select 4ohm drivers and select a crossover for 4ohm drivers.

If you mix driver impedances by choosing an 8ohm and a 4ohm driver or by paralleling a pair of 8ohm drivers and a single 8ohm driver then your crossover becomes more complicated. Each half of the crossover must be designed for the load that you attach to it.
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