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Old 2nd December 2005, 12:39 PM   #1
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Default speaker wiring vs. handling capacity

I am starting this thread to expand my mind in regards to loudspeaker wiring configurations and how it directly affects power handling capacity.

Let us consider these points. Each cabinet will use full range speakers. There are four identical speakers in each cabinet. The cabinets are identical. The speakers are 8 ohms each.
The power handling capacity of each identical speaker is 100 watts RMS.

In example one, the 4 speakers are all wired in parallel for a total cabinet impedence of 2 ohms. I think most of us would agree that the total handling capacity ( in theory ) would be 400 watts
RMS.

In example two, the 4 speakers are all wired in series ( not series
parallel ) for a total cabinet impedence of 32 ohms. I have always thought that in this wiring configuration ( SERIES ) that the total handling capacity is equal to only one speaker. In other
words, only 100 watts RMS.

Can the series wiring configuration also handling 400 watts RMS
without melting the first speaker that sees the audio signal ?

If there are any fellows out there with plenty of experience that can help me out with this, I would really appreciate your responses to this thread.
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Old 2nd December 2005, 12:46 PM   #2
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barth2102,

>> Can the series wiring configuration also handling 400 watts RMS
>> without melting the first speaker that sees the audio signal ?

Only if you change for speakers handling each 400 watts ...

PS: In series, each speaker see all the current...

Alain.
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Old 2nd December 2005, 12:53 PM   #3
sreten is offline sreten  United Kingdom
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The power handling of the various connections - parallel,
series/parallel and series is the same at 400W total.

Though for the 32 ohm case you would need a 1.6kW into 8 ohm
amplifier to apply enough voltage juice to get to 400W total.

For the 2 ohm case you only need a 100W into 8 ohm amplifier
to get to 400W total, though in this case the amplifier will very
likely give up the ghost below 4 ohms more than 200w load.

/sreten.
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Old 2nd December 2005, 01:06 PM   #4
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I hope for many more replies to this thread. So far one fellow agrees with me, and one does not.
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Old 2nd December 2005, 02:18 PM   #5
sreten is offline sreten  United Kingdom
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Quote:
Originally posted by barth2102
I hope for many more replies to this thread. So far one fellow agrees with me, and one does not.
Except one of us is wrong and the other is right. Personally
I wouldn't post definitively on a subject I wasn't sure about.

A large amount of contradictory replies isn't going to help anyone.
You and Alain are both not seeing the situation correctly.

/sreten.
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Old 2nd December 2005, 02:34 PM   #6
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I'm taking your side sreten. There, now it's 2-2. If we get another who says anything other than what you did, I'll put 'im in the bin
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Old 2nd December 2005, 02:41 PM   #7
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Hey, I hope all these drivers can actually take the power without disintegrating...
Just go for a nice little Apogee Scintilla for your 0.5w SET. A pleasent, easy, 1Ohm load
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Old 2nd December 2005, 03:56 PM   #8
Tweeker is offline Tweeker  United States
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Your speakers powerhandling in terms of wattage remains the same regardless of how they are wired (with some dvc exceptions like Atlas). Power (P) = Volts (V) * Amps (I). P=V*I, V = I/Z, I=V/Z

Lets say you have 4 8watt 8 ohm impedance(Z) speakers, they each can handle 1 amp of current, and it takes 8 Volts to drive this current. For 1 speaker: 8Z/V=1I. V=8. 8V * 1amp = 8watts.

If you wire them all series you get a 32 ohm system, current handling remains the same. It takes 32V of drive to give 1 amp of current here: 32 V *1I = 32watts.

If you wire them all in parallel you get a 2 ohm system. With 4 drivers in parallel you have 4 amps of current handling. It takes 8V to drive 4 amps through 2 ohms. 8V * 4I = 32watts.

If you parallel a pair of 2 drivers in series you get an 8 ohm system that can handle 2 amps of current. It takes 16V to drive 2amps in an 8 ohm system. 16V * 2I = 32 watts.

Powerhandling remains the same throughout, but its alot easier to find 400watt @ 2ohm amplifiers than 400watt @ 32ohm. As sreten mentioned before it would likely mean a 1600watt @ 8ohm PA amp to do 400watts @ 32ohms. Your amp determines what the best arrangment of speakers would be. With some amps 8 ohms would be best, and with others 2 ohms. With a tube amp that had a custom wound transformer for this app 32 ohms.
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Old 2nd December 2005, 09:53 PM   #9
maxro is offline maxro  Canada
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Why not just do a series/ parallel wiring of the four drivers to keep the impedance reasonable? Or, is the question more theoretical and not something which you will actually execute?

Max
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Old 2nd December 2005, 10:31 PM   #10
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Max, the question is more theory at this point. Let us assume that we have a nice PA amp that can deliver 400 watts RMS into a
32 ohm load with no problem. It can also deliver 400 watts RMS
into a 2 ohm load no problem. The question is not about amps or how much power it will take to drive either speaker wiring configuration. The question is regarding how much power the series wiring configuration will handle. We all know that 4 identical resistors wired in parallel will have 25% of the rated impedence of one resistor and a wattage handling capacity of 4 times the amount of one resistor. Everyone agrees with me here,
YES ?

4 - 8 ohm 100 watt resistors wired in parallel will have a 2 ohm load and handle 400 watts, correct ?

Now for the series configuration.

4 - 8 ohm 100 watt resistors wired in SERIES ( not series parallel ) will result in a 32 ohm load. But will it handle 400 watts?

As I mentioned before, I am old school. I built all my guitar cabinets when JBL used real ALNICO for their magnets. The wiring theory I obtained from JBL manuals of yesteryear. I still think that in the series configuration, the first speaker would fry
when it sees a 400 watt RMS signal coming from the amp, cause that speaker is rated for only 100 watts RMS.

So I agree with the Alain, the first reply to my post. Come on fellows , let's hear from you on this. It is not a difficult question, correct ?
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