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Old 8th October 2001, 07:54 AM   #1
Fossil is offline Fossil  Singapore
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If it so happens that I were to change to a 2-way spkr without biwire terminals, can I still use the amp which has already split the signals?
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Old 8th October 2001, 12:45 PM   #2
JBL is offline JBL  Canada
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I do not really understand your question. I think you want to biamp with one amp. For biamp you need 2 amplifier nad an active crossover. Is this what you ask for?
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Old 8th October 2001, 02:12 PM   #3
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I'm not positive either, but I think his amplifier is set up for bi amping, but the speakers he's looking at only have a single set of terminals.

Fossil, if you used an active crossover before the amplifier, this will have to be removed. If you want to continue using this amplifier, simply don't use one of the sets of binding posts. I don't know of any amps that <b>require</b> you to biamplify with them.

Is this two way speaker you mentioned a DIY project or a commercially available speaker? I'm assuming its a commercially availble speaker. If you're willing to tinker with the crossover a little bit, odds are you can make it bi-wirable.

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Old 9th October 2001, 02:25 AM   #4
Fossil is offline Fossil  Singapore
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Default Sorry...din make myself clear.

Had a long day and din't manage to explain my situation.

I am planning a DIY amp with in-built active crossover and with 4-ch output. This will allow me to have individual amp driving a pair of biwirable 2-way spkrs, right?

However, if my spkr has only 1 pair of terminals, that is, non-biwirable, can I combine the outputs of the LF & HF and plug into the spkr? WIll I kill the amp?
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Old 9th October 2001, 04:25 AM   #5
paulb is offline paulb  Canada
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It will cause problems because the two amplifiers will fight one another. Even if it did work, you've defeated the purpose of bi-amping by combining the two channels back together.
Can you take apart the speaker and remove the crossover inside, then wire the speakers individually?
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Old 9th October 2001, 06:16 AM   #6
Fossil is offline Fossil  Singapore
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Thanx!
Suspected it might not work and wanted an opinion becoz I have another pair which in non-biwirable.

Just out of curiosity, for amps with several parallel output devices to boost wattage, how is it that the output trans don't see each other as 'loads'?
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Old 9th October 2001, 02:14 PM   #7
grataku is offline grataku  United States
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fossil
even if the speaker had two pairs of connectors you would still need to rip out the passive xover from the speaker cabinets unless you want to filter twice which would be redundant. You you have three choices: remove the passive xover and add a couple of terminals to the box then do the active xover thing, or modify the passive xover and separate the LF from the HF connections then just do biamp without active xover, or, finally, leave everything as it is.
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Old 9th October 2001, 07:30 PM   #8
paulb is offline paulb  Canada
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> Just out of curiosity, for amps with several parallel output devices to boost wattage, how is it that the output trans don't see each other as 'loads'?

The output transistors are driven with the same inputs, and can only pass current in one direction, so they can't see each other as loads. They generally have load-sharing resistors on their outputs, and may even be matched to produce identical outputs.
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Old 9th October 2001, 09:03 PM   #9
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There's a third option that few people use and I think is the "cleanest one": use two power amps with a passive filter at both inputs. A low pass and a high pass.

That way you are not adding active circuits in between.

The only problem of this solution is if the amps are low power. You will also have to adjust the levels someway, by ear or using instruments.


Carlos
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Old 10th October 2001, 02:10 AM   #10
Fossil is offline Fossil  Singapore
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Quote:
Originally posted by grataku
fossil
even if the speaker had two pairs of connectors you would still need to rip out the passive xover from the speaker cabinets unless you want to filter twice which would be redundant.
That would be ideal. But I'm quite unprepared to rip out my spkr's x-over at this point.
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