ohm question?

Rick Tyson · From: Ohio

I have herd one can reduce the ohm of a speaker by placing a resistor across the (+) & (-) on the speaker.
If this is true,,what size resistor would one connect this way to reduce an 8ohm speaker to a 4ohm speaker?? & does the direction of the resistor make a difference??
Thanks in advance

Rick

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Joe E · From: Houston Texas

I have also asked this question. It seems to me that I was told that this would not work. The resiter would lower the Ohms but would have no way to release the energy, thus not giving you the results you were looking for.

I hope someone could explain it better for both of us.

Bill Crook · From: Goodlettsville, TN , Spending my kid's inheritance

Basic AC OHMS law....

I(Current) = E(Voltage in volts) over(divided by) R(Impedance in ohms).

I really don't know what your trying to accomplish here but, really all that is going to happen is the sound output will be reduced by a factor of X.

This could possible be used to connect a transducer (read this as a speaker) of 50 watts capability to an amp that has a 100 Watt output.

Now this is where it gets a bit harrie, Lets say for instance, you have a 8 ohm, 50 watt speaker, (It's been a few years sence I done any of this, so if anyone feels free to correct me, Please do so.) and you wish to use a amp of 8 ohm,100 watt output.

O.K. Now we wish to keep the impedance at 8 ohm but reduce the power(in watts) to the speaker, we need to bleed 1/2 the power off toward the resistor. so, if the speaker is of 50 watt capable, oned needs for the resistor to be capable of 50 (or more)watts too. (Remember we are trying to sink 100 watts here.) 1/2 thru the speaker and 1/2 thru the resistor. The value of the resistor isn't going to be 4 ohms as we would normaly think, as this would result in more voltage/current being dropped across the resistor than the speaker, thus reduceing the signal to the speaker, again by a factor of X.

The desired results is to keep the impedance at 8 ohms, yet not suck any power away from the speaker. (at least up to 50 watts) so the resistor value will have to be of a number that will still mantaine the total desired impedance ohmage. (in our case, the value of the resistor WILL be more than 4,or 8 ohms. Your amp and speaker setup here will be the determining factor here.

Hope this helps.

Bill

Rick Tyson · From: Ohio

Bill
Thanks so much for the Info.
I guess I should tell you why I asked this question to start with.
I have a super pair of 18" Black Widows (4ohm) & wanted to install them in a PA cabinet along with a horn driver. Seems I cant find any 4ohm drivers for the horn though they are all 8ohm??
Any siggestions

Thanks Rick

Joe E · From: Houston Texas

If you are going to use a screw on type driver for the horns, you should be able to find a 4 ohm unit. If you are using an 18" for the speaker, you will deffinately want a midrange horn and not a mid/high or high freq horn. Most likly a horn that will x-over at 500 or 600hz. I was using a set up of 2x15 bass bins, 1x15 mids and mid range horns that went from 500-8000hz. Still have the horns if your interested. (sell them cheap too).

Also what type of x-over will you use. If your bi-amping it wouldn't matter if the horns were 8 ohms. If your using a passive crossover it would.

Joe

mike j · From: newberry springs ,calif

series circuits 2 + 2 = 4 ohms

parralel circuits 2+2
--- = 2 ohms
2

Paul Stertz · From: Pittsburgh,Pa. USA

Rick, you would use an 8 ohm to reduce it to 4 ohms. Just make sure you get a resistor with enough of aa power rating to handle what you're putting out. Another option is to hook another 8 ohm speaker in parallel with the first speaker (+ to +,- to -). This will also give you 4 ohms, plus with the "twin" speakers, you can get a good bit more volume out of the setup. To answer the other part of your question, there is no polarity on a resistor. Polarity is noted on a speaker to keep proper phasing (when the + terminal is positive with respect to the - terminal, the speaker cone will move outward). Hope this helps.

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Paul Stertz
Blue laq. Mullen D10 8+5
Blue laq. Yamaha 302S

Terry Downs · From: Wylie, TX US

It is very impractical to connect resistors in parallel or series with speakers. The power in the resistor develops heat. You usually want sound pressure levels not heat.

You can never efficiently use horns and bass speakers in a bi-amp or passive crossover cabinet that are not the same impedance. More power will be dissipated in the horn vs. the woofer.

As you begin to lower the impedance of the speaker load to an amplifier, the power output increases...up to a point. This is because the current is higher for a given amplifier output voltage swing. Current x Voltage = Power. Unfortunately, the amplifier has a finite output resistance. Otherwise we could make speakers extremely low impedance and get lots of power!! The amplifier output impedance is usually very low and even the speaker cables will be a significant contributor to the output impedance. Amplifiers are rated by damping factor. This is the ratio of the load impedance to the output impedance. If an amplifier has a damping factor of 100, the amplifier output impedance with an 8 ohm speaker load is 0.08 ohms. If you connected this amplifier to a speaker with a cable having 1.02 ohms of two way resistance, you would have an effective output impedance of 1.1 ohms. If your total speaker load was 1.1 ohms, you would get half of the max power into the speaker and the other half would be dissipated in heat in the amplifier and cable. When the cable impedance plus the output impedance of the amp equals the load impedance, the power delivered to the speaker is half what is dissipated in the amp and cables.

In addition, no modern day amplifier should require a fixed load. The only desirable application of a resistor load is to over load a tube amp to get intentional distortion that would otherwise require high sound pressure levels to acquire.

A correction to the resistor calculations shown earlier are as follows:

Resistors in series
Rtotal = R1 + R2 + R3 + ………Rn

Resistors in parallel if all resistors are the same value
Rtotal = R/n where n = number of resistors

2 resistors in parallel of different values
Rtotal = (R1 * R2)/(R1 + R2)

N resistors in parallel of any value
1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ……..1/Rn

I hope this helps.

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Terry Downs
http://nightshift.net
terry@nightshift.net

Bill Crook · From: Goodlettsville, TN , Spending my kid's inheritance

Terry.....

Thanks for the clarfication on the resistor formulars. Like I said, It's been a few years back for me in doing this kind of stuff. I guess I need to go back through the books again.

The other post are right too. If the transducers (speakers) are of different types, (tweeters,midrang,and/or NOT directly across the output of the amp,That is,having a coupling cap in SERIES with output of the driver) these calculations go right out the window, almost.

All this could really get into a specialized area, which I am not nearly trained enought to say anything is bullet-proof.

The one thing I have found tho is :
Most modern transistorized ampifiers do not have a output transformer drivin' the speakers, so, after more than a number of tests, I have concluded, that as long as the total impedance of the transducer (speaker/speakers) do not fall below 4 ohms, the ampifier will not be damaged.

C Dixon · From: Duluth, GA USA

Bill, you wrote:

"This could possible be used to connect a transducer (read this as a speaker) of 50 watts capability to an amp that has a 100 Watt output."

Very respectfully, if you have an amp capable of delivering 100 watts of power at 8 ohms to an 8 ohm load, only capable of handling 50 watts of power, one cannot reduce the power impressed upon that load and keep the impedance the same! (As seen by the amp).

You are correct in what you can do. Just not keep the impedance the same. So here is what happens. Two scenarios.

1. Install a resistor of say 8 ohms in series with the speaker. Make sure the resistor can handle twice the maximum power delivered to it. In this case 100 watts (50 watts twice). So buy, if you can an 8 ohm, 100 watt resistor. Put it in series with the speaker. You now have a 16 ohm load.

However, the total watts will be divided equally across both the speaker and the resistor. With this caveat. No longer will the amp be putting out 100 watts! Reason is; only when a perfect match exists between an amp's output impedance and a load's impedance, will the amp deliver its maximum rated power (watts). It will always be somewhat less.

2. Put the 8 ohm 100 watt resistor in parallel with the speaker. The wattage division is identical as in the first case. With this one difference: The load the amp sees now is not 8 ohms, it is 4 ohms. And if the amp is rated at 8 ohms, this is a NO NO!

Never go below an amp's rated output impedance. Always safe to go above. NEVER below.

But again because the impedance is not matched, you will have somewhat less than maximum power deliverd to the combined load.

Hope this helps to clarify a small part of the very ununderstandable world of electronics. When I used to teach electronics, most technicians never understood the above. And unfortunately, they never will.

God bless them, and all of you,

carl

Michael Brebes · From: Northridge CA

In most PA speakers you will actually find that the horn is or can be a higher impedance. This works because of the efficiency of the horn, it usually being much louder than the woofer. If you are using an active crossover, who cares what the resistance is. If it is a passive crossover, it will still have a smaller effect than though because usually the primary crossover frequency components are a capacitor and inductor (f=1/6.28LC). If you are using a simple capacitor on the horn then you have to remember that the crossover frequency will be cut in half compared to a 4 ohm load (f=1/6.28RC, where f is frequency,R is resistance, and C is capacitance). Hope that helps out mor than it confuses.

Rick Tyson · From: Ohio

I sure would love to have the knowledge that would allow me to understand the above replies

A sincere thanks goes out to all of you though

If I may, Ill ask this:
1. I have a power amp that will deliver power at 8,4 or 2 ohms loads.

2. I have a 4 ohm speaker.

3. I have a 8 ohm horn

What kind of x-over & resistors would I need to make these components work as a PA cabinet & how would I wire it up, if it will work at all??

Thanks again fellas.

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Bill Crook · From: Goodlettsville, TN , Spending my kid's inheritance

hey,Rick .....

I'm going to make your life easier here,

You stated:

quote:
Bill I guess I should tell you why I asked this question to start with.
I have a super pair of 18" Black Widows (4ohm) & wanted to install them in a PA cabinet along with a horn driver. Seems I cant find any 4ohm drivers for the horn though they are all 8ohm??
Any siggestions
Thanks Rick

O.K. this may not have the exact desired output results your lookin' for, but it will not hurt your amp or speakers.

connect the 18" BW as you normally would.
+ of amp to + of speaker
- of amp to - of speaker

now, you stated that both speaker and the horn will installed in a singular cabinet, so

1) Bring a wire off the - side of the speaker and connect it to the - side of the horn.

2) from the + side of the speaker, connect one leg of a non-polarlized capacitor (NP) with a value of beween 10 and 47mfd, (The exact value here isn't really important as it's function here is to pass only hi -freq through to the horn) it dosen't make any difference which end of cap you connect to the speaker as it isn't polarity sensitive.

3) connect the other end of the cap to the + side of the horn.

I have used this method in several home entertainment systems and in our PA equipment. works well and has lasted for a number of years with no negative effects.

Rick Tyson · From: Ohio

Bill
Thanks my friend, thats exactly what I was looking for.
Sorry I bugged everyone so much.
Thanks again Bill & all who responded.

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