# Help with my math in this 2nd order audio filter circuit

1. Feb 3, 2016

### INTP_ty

I understand how to use the reactance equations (for capacitor, xc=1/(2PiFC), & for inductor xL=2PiFL when the filter circuit is of first order to get the component values I need, but I don't understand how the math works when the circuit is of second order. The man at the electronics store told me to simply put boxes around the components, treat them as resistors, and solve. Okay, so for the capacitors I did that in figure #2 & or inductors, see figure #3.

That's what I don't get. When resistors are in series, you simply add them. When in parallel, it's 1/Rtotal=1/R1+1/R2 ...

This is a series-parallel circuit though! I watched some Youtube videos titled "series-parallel resistor circuits" & they simply... uh, let's suppose you have one resistor, R1, in series with 3 resistors, R2, R3, R4, that are in parallel. The Youtube mentor ignored the singe resistor R1 in series & focused on the ones in parallel, so R2, R3, & R4. He combined them, drew a big box around them, & called it R2. Then all that was left was R1 and "R2" which were now in series.

Simple, yeah.

But, in my circuit, there is a load present. Wouldn't that matter? So is it really just R1+R2? I can't decide whether or not it's a series circuit. Yeah the two resistors are in series, but not with respect to the load (the speaker)

And how does this all fit into the reactance equation? So for capacitor, you have xc=1/(2PiFC). Would I simply solve for C, then take that figure toss it in the parallel capacitor equation?

Lastly, the instantaneous impedance listed at each speaker, the tweeter 4 ohm, and woofer 4 ohm ...well I lied. Tweeter is more like 3.3 ohm around desired x over point and wooer ...well the manufacturer decided not to include it to keep people like me from getting decent results I guess? How is that going to work?

Now how in the world does that work with the reactance equation? XC is no longer 4 ohms.

2. Feb 3, 2016

### Wee-Lamm

Series/Parallel.
In your image 1, C1 is in series with your tweeter, L1 is parallel to your tweeter, C2 is Parallel to your woofer, L2 is in Series with your woofer, and your Tweeter is Parallel to your woofer. Ignoring C1, C2, L1, L2, the effective impedance would be about 2ohms, or slightly less.
Series impendence: Z = Z1 + Z2
Parallel: 1/Z = 1/Z1 + 1/Z2
Or
Z = Z1 * Z2 / Z1 + Z2

This may help on understanding that, and how to include C & L in the equation.
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imped.html#c4

The biggest difference between a resistor and a Speaker, is that a resistor opposes the flow of D/C current where a speaker opposes the flow of A/C current.
A resistors resistance value will not fluctuate significantly when the frequency does, and the current passing through will also remain constant. The impedance of a speaker 'will' fluctuate as the frequency changes, and as the impedance increases the current flow will decrease proportionately. The voltage across both should remain relatively constant.

The important part here then is that the "4ohm" rating, and the "3.xohm" reading you would measure on an unconnected speaker load, is more relevant to matching the speaker to the amplifier output impedance, but is a good staring point to rough-tune your circuit. For fine tuning the values for your RLC components, checking the impedance at various frequencies would be beneficial. Most Driver manufacturers will provide data sheets that include a frequency response graph, and many post them to their website for download.

3. Feb 4, 2016

### INTP_ty

Let's keep things simple & go back to the first order scenario.

For Xc & XL, am I supposed to use the effective impedance or Zeq? I did not use Zeq, but instead 4 ohms (the "instantaneous" impedance of each speaker at it's intended x-over point). You mentioned later in your post that it was more relevant for matching the speaker to the amplifier's output impedance.

But, if Xc=4 ohms in Xc=1/(2PiFC), then C=10uF whereas if Xc=2 ohms in Xc=1/(2PiFC), then C=20uF.

Huge difference. That's probably why my tweeter is distorting at high-er volumes. I'm only using a 10uF capacitor when I'm supposed to be using a 20uF ...I think

*I am several octaves above the tweeter's resonant frequency.

And,

Since the manufacturer didn't list an impedance plot (it's a no name speaker), could I do it myself with a multi-meter? Say download a 3Khz test tone & feed it into my speaker. I don't understand how they can measure the impedance of a speaker without measuring whatever is driving the speaker too. I assume this is done indirectly somehow?

Last edited: Feb 4, 2016
4. Feb 4, 2016

### Wee-Lamm

It is possible to measure with a multi meter and injecting a set frequency, which can be tedious checking everything from 20 - 20k Hz. TrueRTA may be worth a look, I use it to inject a tone for troubleshooting guitar amplifiers. It's free version does everything I need and more.

I am curious of what you mean by being several octaves "Above" the tweeters range? I would think we would only want to allow the higher frequencies there in a narrower band than the Midrange, and let the Midrange driver do the bulk of the work. It is likely more tolerant of "some" high frequencies than the tweeter would be of even lower midrange. I think the Tweeter would gag immensely on the low end.

5. Feb 4, 2016

### jim hardy

Are you familiar with something called "Polar to Rectangular Conversion" ?

......

Speakers are almost but not quite a pure resistor.
Their DC resistance is about 80% of their nominal 8 or 4 ohm 'impedance'
that is around 6.4 or 3.2 ohms
so most of the power going into them just heats the voice coil.

The rest of the power would ideally make sound and the speaker would appear to be electrically a simple 8 or 4 ohm resistor.
That's not a bad approximation for somebody just starting out.

So you could consider your tweeter a 4 ohm resistor
and it's in parallel with inductor L1
That parallel combination is some new impedance
and note it's a different number of ohms at every different frequency
and it's in series with C1 which is likewise a function of frequency.

So you can solve, for any frequency, what fraction of applied voltage appears across your tweeter and C1.
I'd set up a spreadsheet or Basic program to iterate over the audio frequency range and plot results.

Your 1/2piFC doesn't mention phase angle
if you ignore phase angle it'll not affect your results drastically , you'll still see effect of 2nd vs 1st order crossover. Values at breakpoints will be off a little that's all.

I would make XL and XC both equal to speaker impedance at crossover frequency.
If you set up that spreadsheet or simulator program you can experiment with tweaking L and C.

There's lots of tutorials on speaker building, it's a fun hobby. I enjoy making sawdust more than doing algebra.

Have at it !

6. Feb 4, 2016

### Wee-Lamm

The only things I can comment on here is that the Audio signal will be AC, which explains why the 90 degree phase shift.
Also, it would not be unusual to see the impedance as high as 70ohms during much of the middle region of the band that is referenced by the network. This impedance is not a resistance in that it does not oppose volume but rather is an inductor that resists dramatic changes in volume. This serves to level out the frequency response which reduces the occurrence of noticeable peaks and dips.

7. Feb 4, 2016

### jim hardy

I was hoping to induce INTP_ty to try his hand at simple math

i'm not sure whether he's accustomed to AC and complex arithmetic

so

if he'll calculate a table for about ten frequencies

first column frequency
second column XC
third column XL
fourth column parallel combination of tweeter ohms and XL, call it Ztweeter
fifth column is fraction of applied voltage that appears across Ztweeter, calculated by Ztweeter/( Ztweeter +XC )

he'll see the voltage across tweeter is less at low frequencies

i'd use steps of 1::2::5::10 for frequency
ie 10 hz, 20 hz, 50 hz, 100 hz, 200 hz,,,,,,,,,, 20khz

that was point of my post

if he incudes angles so much the better

same approach will work for the woofer

and i think that adresses his questions

old jim