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Topic: Carvin BX-500 speaker ohms question |
Nathan Guilford
From:
Oklahoma City
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Posted 28 Jun 2015 5:28 pm |
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I really really like the combination of my Carvin BX-500 and my Telonics 15", but I've got an outdoor gig coming up without microphones for the amps and I wanna run two cabs. The second has a peavey loudspeaker 15" (scorpion plus). It can handle the power just fine but it is 8 ohms. If I run both out of the back of the amp (4ohm and 8 ohm) Am I gonna cause a problem with the load? The amp has a switch to select minimum load 2 or 4 ohms. What do you think, electronics whizzes? And thanks in advance.
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'02 Carter S-12 uni
‘76 MSA D-12
www.toothbrushers.com |
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Lane Gray
From:
Topeka, KS
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Posted 28 Jun 2015 5:42 pm
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I'd select 2. A 4 and an 8 come out to 2 2/3 (32÷12)
_________________
2 pedal steels, a lapStrat, and an 8-string Dobro (and 3 ukes)
More amps than guitars, and not many effects |
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Bill Moore
From:
Manchester, Michigan
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Posted 29 Jun 2015 8:09 am
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| Lane, two 8 ohm speakers, parallel, equal 4 ohms. A 4 and an 8 ohm speaker, parallel, equals 6 ohms. That said, I use an 8 ohm speaker with the BX500, I also leave the selector at 2 ohms all the time, no problems. I have plugged in various other speakers, in addition to the 8 ohm, both 4 and 8 ohms, without problems. Still kept the ohm selector at 2 ohms. |
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Lane Gray
From:
Topeka, KS
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Posted 29 Jun 2015 8:27 am
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No. Parallel impedances and resistances are products divided by sums.
Look it up. A 4 and an 8 in parallel is 2.667
(R1*R2)÷(R1+R2)
| Code: |
(8*8)÷(8+8)=64÷16=4
(8*4)÷(8+4)=32÷12=2.667 |
_________________
2 pedal steels, a lapStrat, and an 8-string Dobro (and 3 ukes)
More amps than guitars, and not many effects |
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Bill Moore
From:
Manchester, Michigan
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Posted 29 Jun 2015 10:08 am
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| Hey, I've been wrong before, I'll still leave the selector set at 2 ohms. |
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Nathan Guilford
From:
Oklahoma City
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Posted 29 Jun 2015 2:48 pm Fun with numbers
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I did look it up. Thanks guys for the insight. You are BOTH right! Check it out
http://www.speakerimpedance.co.uk/?act=two_parallel&page=calculator
Two speakers in parallel will follow the formula that Lane suggested. So Bill you're right that 8ohm and 8 ohm load = 4 ohms
(8*8 )/(8+8 )= total ohms
(64)/(16)= 4 ohms
but as Lane pointed out
(8*4)/(8+4)= total ohms
(32)/(12) = 2.67 ohms
So - both right - put the switch on 2ohms and should be good. Thanks for the lesson guys.
_________________
'02 Carter S-12 uni
‘76 MSA D-12
www.toothbrushers.com |
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