Advanced Audio help.

I have two sets of Speakers running off of one 2 channel amp and i noticed that one set is louder then the other and i was wondering if any one knew if i could put a resistor between the amp and the speak of the louder set. I just want to know if this will work and if it doe i'll figure it out my self. The louder pair pull an rms of about 80 and the quiter and 90rms. Im' not shure if that has any thign to do with the loudness, but im' guessing it does.
Thanks

It's because of efficiency differences between the two sets. You can use a resistor in series with the louder set if you want, just remember to use a resistor strong enough where it won't blow. These are 4ohm speakers right? The amount you attenuate them (in dB) will be:

Att=20*log(4/(4+Rr))
where Rr is the resistance you use

So if you want to attenuate them 3dB you would use 1.6 ohm resistors, if you want to attenuate them 6dB you would use 4ohm resistors, etc.

This is per speaker, so you would find the resistance you want, and put one of them in series with each side to attenuate them equally.

They'll each need to be able to dissipate:
Pr=(P*Rr*R)/(R+4)^2
That's in watts, where Rr is the resistance you use, R is 2ohm, and P is the power your amp puts out per channel at 2ohm.

To get the right resistor values and power requirements, you can wire multiple resistors together in series or parallel. So 2 4ohm 10w resistors in parallel would make a single 2ohm 20w resistor for example.

I"M in an electricity and electronics class in school and wel haven't gotten very far yet. I understood most of what you just said but some of it didn't register, if you could help me a little more i'd like that.
The amp that i have puts out 230 watts per channel @ 2 ohms my speakers are 90wRMS for the 6x9 and my 6.5 comps are the same(thought they were different) the difference in sensitivity is 3db.
I get the first equation (mabe calculus was worth while) on how to figure out the resistance. But i don't understand the second equation you sent me. Also i was thinking that if you resist the speaker then it would mess with the output of the amp making it less right? Or would this not make much of a noticeable difference?

The first equation just came from combining 2 equations.

The first is a normal voltage divider, this is what you get when you wire 2 resistances in series. Some of the voltage drops across the first resistance, the rest across the second resistance. The equation goes:
Va=V*Ra/(Ra+Rb)

where Va is the voltage across Ra, V is the total voltage applied, and Rb is the other resistance you're using. Since we're interested in the voltage dropped across the speaker, we use 4 for Ra and the resistor's value for Rb.

The second part is the equation to convert voltage changes to the decibel scale:
20*log(V2/V1)

In our case, V1 is the total voltage V, and V2 is the new voltage you get after you put the resistor in there and cause a voltage divider (equation above). When you divide them, you're just left with:
20*log(4/(4+Rr))



As for the other one, that's a few different equations molded into one. First you need to find the voltage your amp puts out. To do that you need to know that:
P=V^2/R

A little rearranging gives:
V=sqrt(P*R)

Now, this is the total voltage being applied to your new voltage divider. To find the power being dissipated in the resistor, you need to find the voltage being dropped across the resistor, which is:
Vr=V*Rr/(Rr+4) -same equation as before, but for the resistor instead of the speaker this time

Then, the total power being delivered to the resistor is:
Pr=Vr^2/Rr

Plug in, and you get:
Pr=(P*Rr*R)/(R+4)^2
Where Pr is the power being delivered to the resistor and Rr is the value of the resistor



If the speakers have a real efficiency difference of 3dB (keep in mind, some companies overrate the efficiency by "cheating", so it might actually be a 6-9dB difference if one of them is rated that way), then if you dropped the more efficient speakers by 3dB they should match each other. You could always drop it a little more or less depending on what you're looking for, 3dB isn't very much to our ears.


As for the resistor making the amp put out less power, that's true, which is kind of what you're going for. Right now you have 2 4ohm speakers wired in parallel so the amp is seeing 2ohm and putting out 230 watts, 115 to each speaker. If you put a 4ohm resistor in series with one of those speakers, you would be wiring a 4ohm resistance in parallel with an 8ohm resistance which would give 2.67ohm and your amp would put out roughly 172.5 watts per channel. 115 of that would go to the untouched speaker, and the other 57.5 would go to the resistor/speaker combo. Since it's a 4ohm speaker in series with a 4ohm resistor, power would be split evenly between them and each would get 28.75 watts. Now work backwards, by putting that resistor in series with the speaker you cut the power being delivered to the speaker from 115 to 28.75, you cut it by a factor of 4, which is exactly -6dB, the same result you get by using a 4ohm resistance in that first equation to find the drop in output.

Oh, that 230x2 @ 2ohm is rms power right? That's what matters for all of this stuff, not peak power.

Wow i get it, Yea the speaker companies are audiobahn for the 6x9 and cdt for the comps. Yes that is 230x2w RMS for the amplifier its about 820w Max, but of corse max doesn't mean a whole lot.

I'm kinda confused on this section
As for the other one, that's a few different equations molded into one. First you need to find the voltage your amp puts out. To do that you need to know that:
P=V^2/R

A little rearranging gives:
V=sqrt(P*R)

Now, this is the total voltage being applied to your new voltage divider. To find the power being dissipated in the resistor, you need to find the voltage being dropped across the resistor, which is:
Vr=V*R/(R+4) -same equation as before, but for the resistor instead of the speaker this time

Then, the total power being delivered to the resistor is:
Pr=Vr^2/R

Plug in, and you get:
Pr=(2*P*R)/(R+4)^2
Where Pr is the power being delivered to the resistor.

What does the P stand for in this equation P=V^2/R, is it the power in watts?

I also only wnat to cut power to the louder speaker (not shure if you understood that).
So i'm getting that if i wire a 1.6Ω resistor in with one speaker per channel the 3db soud drop would result?

Thanks for helping you have helped me understad a lot of information.

Actually I just noticed that I made a mistake, I was calling both the impedance you're attaching to the amp and the resistance you're using for attenuation 'R'. For calculating how much power the resistor needs to dissipate it's actually:
Pr=(P*Rr*R)/(Rr+4)^2
where Pr is the power burned off in the resistor
P is the power output of the amp at 2ohm
Rr is the resistance of your resistor
and R is 2ohm

yes 'P' is the amplifier's output power in watts

If you want to drop output by 3dB then you would want to wire a 1.6ohm resistor (25 watt or larger) in series with each of the 2 speakers.

And yes, this is for attenuating whichever speakers you want. If you want to attenuate just the front L 3dB then put a 1.6ohm resistor in series with the front left speaker, if you want to attenuate both of the front speakers 3dB then put a resistor in series with each, etc.

Hey thanks man it all makes sense. I"m gonna try this out but last night i just realized that an easier way to do this would just be to buy a 4ch amp instead of the 2 that i have but i'm gonna try this just incase i can't sell my 2ch that i have now. Thanks again.
BTW Nessian rules