Hello,
I hope I'm not totally off this forum with this question - if so please excuse the ignorance.

I recently helped installing a 2 amp - 8 speaker - sound system in a friends restaurant (2 floors with two different sounds).
The system we installed is the following:
2 amps by CREST - models no. cpx 900 (180W, stereo) and cpx 1500 (300W, stereo), the specs can be found at: http://www.crestaudio.com/service/documents/index.cfm

4 speakers for street level room: polk Audio atium 55, 55 W each
4 speakers for first floor room: polk Audio atrium 45, 45 W each

so on each amp Channel there are two speakers connected. Channel A two speakers (left) and Channel B two speakers (right).

Here is my question: if the amp has a load of 8 Ohm per Channel and the speakers are made to take an 8 Ohm load. If y now dougble the number of speakers per channel from 1 to 2, does that double the Ohm to 16.

If that is 'not correct', does this setup (maybe only over time) harm the amps?

Thanks for your inputs!

andyface

In a parallel connection - how you will be connecting your cabinets - the impedance is determined by each transducers resistance divided by the total number of transducers. In your case 2 cabinets at 8 ohms will equal 4 ohms total.

Your chosen amp's nominal impedance is 8 ohms, however they are servicable down to 2 ohms. At 4 ohms your amps provide 500W per channel. So where your concern should be is within the power capabilities of the cabinets. If you keep your levels low, you won't have a problem. But if a dumb employee decides he wants to crank it up...You'll have some blown speakers in your possession.

MrMusicMan!
liked your short comments - thank you. So the additional cabinet devides the load by two - from 8 to 4 Ohms - correct? - Would you happen to know some link with a little more solid input on this issue, as I understand that I'm missing some important information here, this would be of great help.
I'll have an eye out on that barman who is crazy for volume ... !!!
thanks again-

try this link...

http://www.the12volt.com/caraudio/boxcalcs.asp

it will help you calculate the speaker load depending on how you wire them up.

nice little calculator - thank you.
basically its adding up for parallel connections and dividing for serial connections by the amount of cabinets connected. -
You wouldn't happen to know a link for more input on audio electronics? -

whoops - got that wrong, its the other way around, dividing for parallel and adding up for serial -

Feel free to start any new threads concerning audio electronics, I actually quite like the subject and also i feel it is a valuable subject for any audio person to research - particularly regarding signal paths.

In fact this sentiment can be heard in this podcast:

http://media.libsyn.com/media/audio/ac23.mp3
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Recording Studio Suntans

Some good input here and just to summarise.

For series connection you add the ohmage up, therefore:

series = speaker A + speaker B

Series 8 + 8 = 16 Ohms

Here in series the load delivered would be far less than the maximum power transfer the amp is capable of, so VERY safe for the amp AND speakers.

Parallel connection is different, for this you divide.

For formula is:

1/zt = 1/z1 + 1/z2 + 1/z3 + 1/z4 etc ... (z is impedance by the way)

So in your example this would be:

1/zt = 1/8 + 1/8

1/zt = 0.125 =0.125

1/Rt = 0.25

zt = 1/0.25

z total = 4 Ohms

This is probably what your amp will like best, though as many amps do go down to 2 Ohms you need to really make sure they do, because if not they will easily over heat and break!

If you are confused with how to wire this series or parallel config here it is:

SERIES


PARALLEL



And you can have combinations:

SERIES/PARALLEL


Also as your speakers have a wattage rating far below what the amp is capable of, mrmusicman is correct, this is fine, in fact the amp will purrrrrr along nicely with no issues and this is the way to run systems. But do be cautious, might be an idea to limit the output somehow or not allow employees near the gain.
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Recording Studio Suntans