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Old 11th May 2013, 01:04 PM   #1
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Default 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 roll-off/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!
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Old 11th May 2013, 06:20 PM   #2
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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.
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Old 12th May 2013, 01:40 AM   #3
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Quote:
Originally Posted by jjrenman View Post
I don't know of a situation that you would put a lone capacitor across the woofer..
My mistake.
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.
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Old 12th May 2013, 01:54 AM   #4
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Think of it a different way... imagine the capacitor as a (frequency-dependent) 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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Old 12th May 2013, 02:05 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 02:09 AM.
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Old 12th May 2013, 02:23 AM   #6
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Not an expert or even experienced, but these pages might help:

Passive Crossover Networks

Passive Crossovers, Capacitor and Coil Calculator
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Old 12th May 2013, 02:33 AM   #7
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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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Old 12th May 2013, 02:34 AM   #8
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C = 1/2πfR
where R is the driver impedance
or f = 1/2πCR
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Old 12th May 2013, 02:50 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.


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----------------------------------------------------Rick........
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Last edited by Richard Ellis; 12th May 2013 at 02:55 AM. Reason: Maybe can't see it.
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Old 12th May 2013, 11:50 AM   #10
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This calculator is also very good and easy to use.
2-Way Crossover Designer / Calculator
Instructions are here:
3 Way Crossover
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