# Extremely Basic Low Pass Filter Question

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#### miragem3i

##### Member
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!

#### jjrenman

##### Member
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.

#### miragem3i

##### Member
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.

#### sofaspud

##### Member
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.

#### miragem3i

##### Member
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?

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#### counter culture

##### Member
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).

#### sofaspud

##### Member
C = 1/2πfR
where R is the driver impedance
or f = 1/2πCR

#### Richard Ellis

##### Member
First Order Formulas

Note, top of both formulas, the 159,155 has a comma, the second one is a decimal point.

----
----------------------------------------------------Rick........

#### Attachments

• CrossOver.JPG
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#### miragem3i

##### Member
Those calculators are great, but the OP was about finding the solution for a 1-cap, no coil xo and the 1st Order BW has a coil in it.

I am trying to find the formula WITHOUT an inductor or resistor, just two drivers and a cap. Reverse engineering my S.0 to figure out what they were doing.

High Pass filters allow high frequencies above a selected crossover frequency to pass, filtering out all frequencies below it. In a first order (6dB per octave) filter, this consist of a capacitor in series with a loudspeaker. Just above the crossover frequency, the capacitor begins to add resistance to the circuit. At the crossover frequency enough resistance has been added to equal the resistance of the loudspeaker and reduce the power by 3 dB or 50 %. One octave below the crossover frequency, power has been reduced by 6 dB or 75%. Each octave lower reduces the power by an additional 6 dB. The size of the capacitor will be determined by the impedance of the loudspeaker(s) and the desired crossover point. The smaller the size or value of a capacitor (microfarads, µfd or mfd) is, not physical size, the higher the high pass frequency will be.

This is very clear and I understand it, but there is no formula to calculate values? I understand that the smaller the value of the cap the higher the pass frequency, but HOW DO I CALCULATE THIS when there is no coil in the "crossover"?
What is the formula? Something like Fxo = K * [1 / (c)] where c is in farads and K is a constant.

":Just above the crossover frequency, the capacitor begins to add resistance to the circuit."

How do I calculate this xo frequency in a one capacitor filter with no coil or inductor or resistor in the circuit?

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#### miragem3i

##### Member
C = 1/2πfR
where R is the driver impedance
or f = 1/2πCR

Which driver impedance? There are two drivers in the circuit. The tweeter impedance, the mid impedance or the combined series impedance of both?

I assume you intend this:
f = 1 / [2πCR]

The original set up was one 4.7 uF capacitor, a very common part, two drivers as DC resistance measured below, and no other parts. No coil no resistors.
The value of the tweeter impedance *seems* to be the value to use, but since this circuit includes a coil and mine had no coil I cannot be certain.

I used the calculators above at http://www.diyaudioandvideo.com/Calculator/XOver/

1st Order Butterworth
xo frequency: 5551 Hertz [trial and error using goal seek]
6.1 Ohm Tweeter / 5.4 Ohm Woofer
Parts List
Capacitors
C1 = 4.7 uF
Inductors
L1 = 0.15 mH

But the xo had NO COIL INDUCTOR. How much error would be in the calculations? Or did they just fudge it and include the inductance of the speaker coils and wire in the circuit and call it a day?

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#### Pano

Paid Member
Those calculators are great, but the OP was about finding the solution for a 1-cap, no coil xo and the 1st Order BW has a coil in it.
Did you look carefully at the calculator I linked?
2-Way Crossover Designer / Calculator

You'll see that the 1st order low pass section is just a cap. No coil. The coil is for the low pass section. Please look again. The formula is on the help page, if you want to do it that way.
2-Way Crossover Design / Calculator Help

#### Jerrym303

##### Member
The single capacitor is the high pass tot the tweet. The inductor is the low pass to the woofer. I am not aware that you can create both high and low pass with one capacitor. You will note that the calculator that Pano linked asked for the impedence of each driver.

It may be that your system was running the woofer full range.

#### AndrewT

##### R.I.P.
A 2way crossover is TWO filters combined,
One Filter is a Low Pass that feeds the Bass driver.
The other Filter is a High Pass that feeds the Treble driver.

If you have ONE driver and want a high pass filter then add a capacitor in series.
If you want a low pass filter then add an inductor in series.

A Capacitor in series with a driver can only be a High Pass Filter.

The standard formula applies.
F-3dB = 1 / 2 / Pi / C / Z where Pi = 3.14159, C = Capacitance in Farads, Z = Driver impedance AT THE CROSSOVER FREQUENCY. This impedance may not be the same as the "nominal driver impedance"

#### Michael Bean

##### Member
Hi miragem3i,

One of my favorite all-time quotes is from fairly wise individual named Albert Einstine; "Everything should be made as simple as possible, but not simpler." What you're trying to do is "to simple" and it is going to be extremely difficult to nearly impossible to get it done right. You can't apply any math without some well done measurements of the drivers, and if you don't happen to have a lot of expensive (lab grade) test equipment and maybe an anechoic chamber handy, it's going to be difficult to say the least.
In order to build a 2-way speaker with only one cap for the tweeter high pass will require the bass driver to have a very smooth upper roll-off with minimal cone break up, finding one of those is going to be hard, they are rare. Next you'll have to measure the frequency response and impedance curve of the bass driver to determine the cap value for the tweeter high pass. And the two drivers selected have to be matched for SPL if you insist on "no resistors", yet another constraint on driver selection. And after going through all of that you're still faced with the reality of driver offset causing frequency response problems in the cross over region.
If you want to build a well performing speaker system, I suggest you avoid "to simple" and stick with a more traditional approach. Check out 2nd order linkwitz, they are relatively easy to design and build, and give pretty good results.

Mike

#### weltersys

##### Member
I am trying to find the formula WITHOUT an inductor or resistor, just two drivers and a cap. Reverse engineering my S.0 to figure out what they were doing.
The formulas and calculators are almost worthless, as they assume the speaker is a fixed impedance, like a resistor.

A speaker is not like a simple resistor, it has impedance and reactance that vary with frequency.

Look at the example below, the calculator says an 18 uf capacitor will provide a 550 Hz 1st order Butterworth crossover.

The chart shows the actual measurement of an 18 uf capacitor in series with a 16 ohm horn driver, and the same unit without the capacitor.

As you can see, the response with the capacitor looks nothing like the calculator suggests, in fact the capacitor causes more attenuation above 550 Hz than some frequencies below. The single capacitor is useless as a crossover with this speaker, though it might have the desired effect with another driver that has a different impedance curve.

Calculation of what the capacitor in your S.0 does is impossible unless you had an impedance chart of the driver.

#### Pano

Paid Member
You'll see that the 1st order low pass section is just a cap.
Oops, sorry. In the above I meant HIGH pass. High pass is a single cap. Sorry.

#### miragem3i

##### Member
Did you look carefully at the calculator I linked?
2-Way Crossover Designer / Calculator

You'll see that the 1st order low pass section is just a cap. No coil. The coil is for the low pass section. Please look again. The formula is on the help page, if you want to do it that way.
2-Way Crossover Design / Calculator Help

Why would the calculations be identical if there is a coil removed from the circuit? The tweeter 'sees' the coil across the terminals of the mid as part of that driver.

The inductors (coils) in a LPF have resistance. This resistance affects the impedance of the entire circuit.
If you are claiming that it's too small to matter or close enough or the mH are overwhelmed by the larger circuit impedance, then just say so.

Otherwise why does the omission of this component not matter?::
Inductors
L1 = 0.15 mH

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#### sofaspud

##### Member
Why would the calculations be identical if there is a coil removed from the circuit? The tweeter 'sees' the coil across the terminals of the mid as part of that driver.
It doesn't really see the coil, since it is a parallel circuit (and therefore voltage is the same in each branch). The cap blocks lows to the tweeters, while the coil blocks highs to the woofer. And the cap will block those lows whether the coil is there or not, so the calculations are identical.

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