Stupid line array question number 3 [message #23583] |
Wed, 21 March 2007 18:14 ![Go to next message Go to next message](/forum/theme/AudioRoundTable/images/down.png) |
Elvis
Messages: 6 Registered: May 2009
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Esquire |
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I'm assuming that when you wire speakers in series they look like a line with the positive wired to the negative and then the negative wired to the positive etc. Groups of series and parallel look like a ladder. The bottom of the ladder are the connections to the amp: one positive, and one negative. The rungs are the groups of series wired speakers. IF my calculations say that I need 2 four speaker groups wired in parallel, then I would have 2 sets of 4 speakers in series. Each of these 2 sets would be the rungs of the ladder. My stupid question is: is the last rung, one of the two sets of 4 speakers in series, or do the sides of the ladder continue and connect with a solid wire effectively joining the positive and negative wires in one more rung above the speakers which is a solid wire, or does it not really matter? Elvis
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Re: Stupid line array question number 3 [message #23584 is a reply to message #23583] |
Thu, 22 March 2007 08:58 ![Go to previous message Go to previous message](/forum/theme/AudioRoundTable/images/up.png) ![Go to next message Go to next message](/forum/theme/AudioRoundTable/images/down.png) |
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Wayne Parham
Messages: 18748 Registered: January 2001
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Illuminati (33rd Degree) |
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If you need two parallel blocks, and each is connected in series, this is what it would look like: (+) | | | (+)------speaker-----(-) | | (+)------speaker-----(-) | | (+)------speaker-----(-) | | (+)------speaker-----(-) | | | (-)------speaker-----(+) | | (-)------speaker-----(+) | | (-)------speaker-----(+) | | (-)------speaker-----(+) | | | (-)
Total impedance of the array will be half the impedance of a single driver.
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Re: Stupid line array question number 3 [message #23586 is a reply to message #23583] |
Thu, 22 March 2007 10:20 ![Go to previous message Go to previous message](/forum/theme/AudioRoundTable/images/up.png) ![Go to next message Go to next message](/forum/theme/AudioRoundTable/images/down.png) |
cfranz
Messages: 23 Registered: May 2009
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Chancellor |
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Wire the two 4-speaker series separately; Identically but separately. Each should have a + and - line back to the crossover. Yes, you could also run a from the + input to the + of each parallel group (and the same for -) but if your confused already, why bother? Crossover  Crossover (-) (+) | | | | (-)speaker(+)----(-)speaker(+)----(-)speaker(+)----(-)speaker(+) | | (-)speaker(+)----(-)speaker(+)----(-)speaker(+)----(-)speaker(+)
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Re: Stupid line array question number 3 [message #23590 is a reply to message #23583] |
Fri, 23 March 2007 05:28 ![Go to previous message Go to previous message](/forum/theme/AudioRoundTable/images/up.png) |
FredT
Messages: 704 Registered: May 2009
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Illuminati (1st Degree) |
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It's not a stupid question at all. I've wired several arrays and I still get so confused that I use different color wiring for the positive leads (white), negative leads (black), and the leads that go from one speaker's positive terminal to the next speaker's negative (gray). See the link below for a wiring diagram of two groups of four drivers each, where the drivers in each group are series-wired, and the two groups are parallel-wired. This scheme results in an overall impedance that's twice the impedance of each individual driver.
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