Filtering Out frecuencies

Well, I have a Sony Stereo. Pretty old, but has a CD player. Well, the thing is, I have added 2 small speakers (about 2-3" in diameter). My puzzle is , Which kind of capacitor do I have to add in order to filter out bass(20-250hz)? My stereo now has a large speaker and the small speakers I have added on each channel. Therefore, now I have 4 speakers. I want to filter out bass from the small ones because they could get damaged and I don't have time to go get new ones. Thanks for the help!

you dont want a filter, you need a cross-over have a search for cross-overs

Sorry,

What is cross over? I am a complete newbie. I heard of it , but I don't know what it is....

Re: Sorry,

juan123 said:

What is cross over? I am a complete newbie. I heard of it , but I don't know what it is....

Click to expand...

Look here for every thing you want to know https://colomar.com/Shavano/crossover6db.html.

Styx said:

you dont want a filter, you need a cross-over ...

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Actually, a crossover is a type of filter. Usually in speaker systems with multiple drivers (speaker elements) the amplifier is connected directly to the bass speaker ("woofer"). Since the woofer doesn't reproduce high frequencies very well, due to it's physical size, you won't usually hear the high freqs from it. The audio signal is also sent thru a high-pass filter (the crossover) to remove the low freqs and then sent to the "tweeter" or high-frequency speaker. In some systems, you will also find a "midrange" or mid-frequency speaker. In this case, the crossover will also have a band-pass filter, which is designed to filter out the low and high freqs, while leaving the middle freqs (the mid-band) to be sent to the midrange speaker.
JB

true but a search at "google" for filters will give lots-o-hits
but a search for "cross-over" will get you what you want very quickly

Let's say you want to attenuate the power by 50% at a given freq of 80Hz. If the amp's output resistance is 4Ohm, adding another 4Ohm impedance in series at 80Hz will accomplish this. Frequencies below 80Hz will be attenuated even more and above 80Hz less

So Capacitive reactance(Xc) = 1/(2*pi*f*C), where f=Hz, C=Farad

C works out to be about 497uF, for 8 Ohm speaker it will be 248uF

Because you need a non-polarized capacitor for audio (which is AC) and non-polarized caps of this value will be expensive if you can get it, you may use 2 electrolytic capacitors each of 1000uF (for 8 Ohm 2 470uF)connected in series (back to back), positives to positives or neg's to neg's to create a non-polarized cap. Voltage rating of each cap should be about 25v each.

TheOne said:

Because you need a non-polarized capacitor for audio (which is AC) and non-polarized caps of this value will be expensive if you can get it, you may use 2 electrolytic capacitors each of 1000uF (for 8 ohm 2 470uF)connected in series (back to back), positives to positives or neg's to neg's to create a non-polarized cap. Voltage rating of each cap should be about 25v each.

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Say what??? How on earth does this create a nonpolarized cap?? It's not going to change the fact that one cap is always backwards!

Oznog said:

TheOne said:

Because you need a non-polarized capacitor for audio (which is AC) and non-polarized caps of this value will be expensive if you can get it, you may use 2 electrolytic capacitors each of 1000uF (for 8 <a href="#">ohm</a> 2 470uF)connected in series (back to back), positives to positives or neg's to neg's to create a non-polarized cap. Voltage rating of each cap should be about 25v each.

Click to expand...

Say what??? How on earth does this create a nonpolarized cap?? It's not going to change the fact that one cap is always backwards!

Click to expand...

i remember that i have read about this before. but i didnt understand how this would be true.

Oznog said:

Say what??? How on earth does this create a nonpolarized cap?? It's not going to change the fact that one cap is always backwards!

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It's crude and nasty, but it does work - it's often been used in crossovers for many years.

TheOne said:

Let's say you want to attenuate the power by 50% at a given freq of 80Hz. If the amp's output resistance is 4Ohm, adding another 4Ohm impedance in series at 80Hz will accomplish this.

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Hi TheOne,
Actually, an amplifier's output resistance is extremely low, about 0.08 ohms or less, unless it is an old tube (valve?) amp with a resistive output transformer. If an amplifier had a 4 ohm or more output resistance it would provide very poor damping of the speakers and they would resonate all over the place.
The crossover cap must match the load (speaker's impedance), not the source.

Sorry, I meant to say, an amplifier designed to drive a 4 or 8 ohm load. :wink: Cause damping factor= Rload/Rout

Interestingly damping factor has very little influence on loudspeaker performance-Ref Douglas Self "Audio Power Amplifier Design Handbook"

audioguru said:

If an amplifier had a 4 ohm or more output resistance it would provide very poor damping of the speakers and they would resonate all over the place.

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Also it would give very poor efficiency, the theoretical maximum efficiency would only be 50%, so probably only 30-40% in practice.

The link I posted above gives ready calculated values for various frequencies and various speaker impedances, and even gives the required formulas as well.

But considering the problems finding odd sized capacitors, plus their wide tolerance, you just need to look at the listing and pick a capacitor that gives you a crossover frequency somewhere near where you want.

Looking at the listing, 99.4uF gives 200Hz, or 49.7uF gives 400Hz for 8 ohm speakers - obviously 100uF or 50uF non-polarised would be the obvious choices - or something in betwenn (if you can find anything).

Don't forget, we're not talking a precision filter here - only preventing a tweeter from blowing up!.