Who actually knows this for sure....??

[86svo 10+ year member]

I've read so many conflicting things and have yet to find a definitive answer on the voltage output when you bridge a multi-channel amp!

Going by JL-audios web-site, you are supposed to take the ohm rating of the speaker, divide that in half(we'll use 4-ohm speakers: 4/2=2) then multiply by your bridged channel rms value(I'm using a Alpine F-250, so bridged it's 100rms: 2*100=200) then you take the sqrt(sqrt200= 14.1) then finally you take your voltage and multiply by two(so, 14.1*2= 28.2). This is the way JL audio describes how to do it, but I'm not sure if that is only for their amps or for all amps.

Going by the good old normal method I come out with 20v (sqrt(100*4)=20). Now, this is a fairly significant difference and I'm not too good with electrical math/knowledge to know the reasons why JL says to do what they show.

Can anyone give any strong/reasonable input as to one way or the other?? For now I'm sticking with the 20v just to be on the safe side, but if I can go more, I would like to.

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[ThatChevyGuy 5,000+ posts]

I just use my ears, and years of tuning amps.

 

[Atimm693 10+ year member]

What exactly are you trying to figure out? Voltage for gain setting?

As far as I know you take Volts*Ohms and sqrt the total. But JL's amps can't be much different then all of the other amps.

 

[86svo 10+ year member]

What exactly are you trying to figure out? Voltage for gain setting?
As far as I know you take Volts*Ohms and sqrt the total. But JL's amps can't be much different then all of the other amps.

Trying to determine the correct method for finding the gain setting on a bridged amp.

 

[squeak9798 5,000+ posts]

I've read so many conflicting things and have yet to find a definitive answer on the voltage output when you bridge a multi-channel amp!
Going by JL-audios web-site, you are supposed to take the ohm rating of the speaker, divide that in half(we'll use 4-ohm speakers: 4/2=2) then multiply by your bridged channel rms value(I'm using a Alpine F-250, so bridged it's 100rms: 2*100=200) then you take the sqrt(sqrt200= 14.1) then finally you take your voltage and multiply by two(so, 14.1*2= 28.2). This is the way JL audio describes how to do it, but I'm not sure if that is only for their amps or for all amps.

Going by the good old normal method I come out with 20v (sqrt(100*4)=20). Now, this is a fairly significant difference and I'm not too good with electrical math/knowledge to know the reasons why JL says to do what they show.

Can anyone give any strong/reasonable input as to one way or the other?? For now I'm sticking with the 20v just to be on the safe side, but if I can go more, I would like to.

Where are you getting this formula from JL ??

JL's tutorial is worked out using a single channel on their amplifiers. In their example, the 300/2 is 17.3V per channel @ 2ohm, so for a bridged 4ohm load (which = 2ohm per channel) you would take the single channel 2ohm voltage and double it.

In your example, if you have 100w @ 4ohm bridged, I'm going to guess you have 50w per channel @ 2ohm. So, using the same "method" as in the JL tutorial for an apples to apples comparison, you would find the voltage of a single channel @ 2ohm sqrt(50*2) = 10V, then double that voltage to find out what the bridged @ 4ohm voltage would be. 10V * 2 = 20V

Or, bypassing all of the BS and just using ohms law straight from the get go sqrt(100*4) = 20V

Both methods result in the same answer.....JL's tutorial just takes the long route because they skip the actual math & just give you the per-channel voltage. So to find a bridged voltage with their charts, you need to find out what voltage would be per channel at half the impedance and double it.

 

[86svo 10+ year member]

Cool, that clarifies a lot then. I had not seen where it showed to cut the rms rating in half as well and I believe that is where a lot of people, myself included, were getting confused.

It makes sense the way you described it. Thanks for the clarification!