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MultiWay Conventional loudspeakers with crossovers 

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11th May 2013, 02:04 PM  #1 
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Join Date: Jul 2010
Location: Western China

Extremely Basic Low Pass Filter Question
I want to create a low pass filter and just understand this point better.
Reactive Capacitance is: XC= 1 / ( 2πfc) Where c is in farads and f is in hertz. I want to calculate the rolloff and cutoff frequency for a given value of capacitance  let's assume 4.7uf  at a given frequency. How do I calculate this rolloff/cutoff frequency of a low pass filter with only one capacitor in the circuit? Example a woofer connected to a tweeter with a 4.7uf wired across the woofer terminals. What's the rolloff/crossover frequency? TIA! 
11th May 2013, 07:20 PM  #2 
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Join Date: Mar 2013

First, lets go over some basics. A single capacitor in series with the tweeter is a 6db/oct high pass. A single inductor in series with the woofer is a 6db/oct low pass. I don't know of a situation that you would put a lone capacitor across the woofer.
Second, your formula does not look correct. For speaker applications the formula will take into account the impedence of the driver. 
12th May 2013, 02:40 AM  #3  
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Join Date: Jul 2010
Location: Western China

Quote:
So how to calculate the xo frequency for that 6dB/octave high pass filter using only one capacitor? I obtained that formula from several difference sites. What is the correct formula. 

12th May 2013, 02:54 AM  #4 
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Location: San Antonio

Think of it a different way... imagine the capacitor as a (frequencydependent) resistor. Picture the circuit with that resistor, rather than a capacitor. And keep in mind that current flows through the path of least resistance.
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12th May 2013, 03:05 AM  #5 
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Right, that is how I think of it. So given that it is a frequency dependent resistor, what would be the way to calculate the crossover response of said capacitor with a given driver? I wish to calculate where to cut off the high pass at a given frequency.
"For speaker applications the formula will take into account the impedance of the driver. " Ok. So this formula would be...what? Last edited by miragem3i; 12th May 2013 at 03:09 AM. 
12th May 2013, 03:23 AM  #6 
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Join Date: Apr 2011

Not an expert or even experienced, but these pages might help:
Passive Crossover Networks Passive Crossovers, Capacitor and Coil Calculator 
12th May 2013, 03:33 AM  #7 
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Join Date: Jun 2011

The 3dB (half power) point (generally quoted as the nominal frequency of the network) of a 2 component network is where Z1 = Z2.
So in the case of an RC network (one resistor and one capacitor), the power will be split 50/50 across the 2 components when R = Xc, or R = 1/(2*pi()*f*C). You can jiggle this around to get f = 1/(2*pi()*R*C).
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12th May 2013, 03:34 AM  #8 
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Join Date: Nov 2005
Location: San Antonio

C = 1/2πfR
where R is the driver impedance or f = 1/2πCR
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12th May 2013, 03:50 AM  #9 
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First Order Formulas
Note, top of both formulas, the 159,155 has a comma, the second one is a decimal point.  Rick........ Last edited by Richard Ellis; 12th May 2013 at 03:55 AM. Reason: Maybe can't see it. 
12th May 2013, 12:50 PM  #10 
diyAudio Moderator

This calculator is also very good and easy to use.
2Way Crossover Designer / Calculator Instructions are here: 3 Way Crossover
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