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What I want to do is make a low pass filter using a simple RC circuit (first order), or possibly an LC circuit (which would be second order). Right now, I am in the process of determining the values for the resistance of the resistor and capacitance of the capacitor (I'm currently going with the RC circuit idea). As the equation is f=1/(2*piRC), I understand that if I want a crossover frequency of 200Hz, I will have theoretically infinite choices or R and C. Which values are better to choose? I've heard I should choose C first, because capacitors are more difficult to find in the right denomination. Still I am left with a few hundred possibilities.

Also, is there a best choice of type of capacitor? (electrolytic, etc)

My application, if needed, is that I want to filter out high frequencies of sound from entering my car sub woofer (powered by a 70 W amp), so I'm going to put the low pass filter between the amp and sub woofer (I would put it upstream of the amp, but in this case, the amp and head unit are one in the same device).

Thoughts?

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    \$\begingroup\$ An RL filter is first order also. An LC or RLC filter is 2nd order. \$\endgroup\$ – John D Jul 3 '18 at 22:24
  • \$\begingroup\$ So the second comment on this forum is incorrect? \$\endgroup\$ – RJP Jul 3 '18 at 22:27
  • \$\begingroup\$ Reed, the choice of balancing R and C is about impedance. A large R and a small C will load down the source less but can't drive much of anything at the other end, either. \$\endgroup\$ – jonk Jul 3 '18 at 22:27
  • \$\begingroup\$ My sub woofer has 3.3 ohm impedance, as recorded with my multimeter. Will this help? \$\endgroup\$ – RJP Jul 3 '18 at 22:28
  • \$\begingroup\$ @ReedPetersen The comment I see says, "an inductor and capacitor make a second order filter", which is correct. This does not say that an RL is second order. It says that LC is second order. \$\endgroup\$ – jonk Jul 3 '18 at 22:28
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I understand that if I want a crossover frequency of 200Hz, I will have theoretically infinite choices or R and C. Which values are better to choose?

Think about what the input impedance will be to the RC low pass filter when the input signal frequency is much, much higher than 200 Hz - the capacitor can be regarded as a short circuit and this means that the input impedance is simply "R" and, if you decide that you cannot live with an R value lower than (say) 50 ohms, your capacitor has to be no greater than 16 uF.

Next think about what the output impedance will be at frequencies much lower than 200 Hz - clearly, the capacitor is not part of this analysis so, if you chose 50 ohms for "R" previously, the output impedance becomes 50 ohms. Can you live with this impedance? Maybe you could live with an output impedance of 1 kohm? In which case the capacitor value is no greater than 80 nF.

So, decide on input and output impedances and constrain your design to these requirements.

I'm going to put the low pass filter between the amp and sub woofer

It's likely that a simple RC will not do the job because your sub-woofer will have a loading effect of maybe 8 ohms and this means the resistor value will need to be at least one-tenth of this value. With R at 0.8 ohms, you will have problems when the input frequencies are high because the input impedance becomes 0.8 ohms.

Using an LC circuit is what you need because as input frequencies rise, the inductor impedance rises with them and doesn't create a stupidly low input impedance that your amplifier has to try and drive into. Try this type of circuit: -

enter image description here

With R at 8 ohms (your sub woofer impedance), L at 8.2 mH and C at 82 uF you get this response: -

enter image description here

Source of inter-active tool.

The cut-off frequency (3 dB point) is 216 Hz and has values that are not too far from specific calculators for cross-over units you can find on-line like this one: -

enter image description here

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  • \$\begingroup\$ Thanks for the detailed response! What happens if my sub's impedance is 4 ohms? As I said, I measured it with my voltmeter, and I think this is the case. Do the above values still work, or should I plug them back into the online resource you cited? \$\endgroup\$ – RJP Jul 5 '18 at 13:05
  • \$\begingroup\$ Use the interactive tool and put values in and try it out but the short answer is that the flatness of the response up to 200 Hz won't be as good BUT this can be fixed by altering the L/C ratio but keeping the product of L and C about the same. In other words if you increase L by 50% you should decrease C by 50%. @ReedPetersen \$\endgroup\$ – Andy aka Jul 5 '18 at 13:16
  • \$\begingroup\$ I reckon 4700 uH and 150 uF would be suitable with 4 ohm loads. \$\endgroup\$ – Andy aka Jul 5 '18 at 13:22
  • \$\begingroup\$ Do this inductor and this capacitor look like good choices? I'm not certain what other specs I am looking for, but I found two components in the right range, anyways. \$\endgroup\$ – RJP Jul 5 '18 at 20:39
  • \$\begingroup\$ Firstly the inductor - it has a DC resistance of over 1.5 ohms and this would be a serious problem. It's also only rated for peaks of 1 amp and this is also a problem if you are expecting tens of volts RMS across your 4/8 ohm load. I'd be thinking about a DC resistance that is sub 0.2 ohms and if your peak voltage across the 4 ohm load is (say) 20 volts (implying a 50 W sinusoidal power to the speaker), the current would peak to 5 amps. Inductors are always trickier.... \$\endgroup\$ – Andy aka Jul 5 '18 at 20:46
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If you need to place it before the speaker, go for LC, any R (be it RL or RC) will give you voltage drop and useless power dissipation, not to mention less attenuation. Also they'll have to have a (much) lower value than the load, so in the range of \$\Omega\$, which means the C will turn out very large, and the resistor will have to be of great power to accomodate the losses, which also means it'll probably be wire type, which means it will have large parasitic inductance, which means you're better off using an inductor.

I assume the driving is done by a typical audio amplifier, thus voltage driven, thus low output impedance(?), so you could calculate your filter based on the data you have. Since the measured resistance is 3.3\$\Omega\$, you can deduce that the impedance is 4\$\Omega\$ (edited formula):

$$Z=4\Omega$$ $$L=\frac{Z\sqrt{2}}{2\pi f}, C=\frac{1}{2\pi f Z\sqrt{2}}$$

which will give you a Butterworth type of filter (also prefered for its flatness). If you need some cross-over, then Linkwitz-Riley will be preferred, but not mandatory, and then replace \$\sqrt{2}\$ with 2. Your choice(s) are, basically, any all-pole 2nd order filter is up for picking, not necessarily the best for the job, though.

The inductor will have to be made without a core (air core), to avoid nonlinearities, and with a wire that supports the maximum RMS current (it may turn out bulky), while the capacitor needs to be nonpolarized -- these tend to be more expensive so you can "cheat" by placing two polarized caps in series, with common + pin. You may also need an additional, small valued resistor (but higher power) to make it a lowpass RLC, for damping purposes, or for load matching, unless you get the LC values right, in case it's needed.

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