Unbridged values vs bridged backwards?

[Immacomputer 10+ year member]

No, it wasn't frustrating because I can read what you type.

If it would have looked like this, it would have been frustrating:

"ok so leik guyz im havn truble w/ wiring up my system b/c i dunt no how 2 brige my amp .....ahhhhh i am confsued on who 2 do it cuz i dun no wat or how u guys r doin it ur ur ur random bad...."

 

[iriejp1 10+ year member]

One last comment and I'll let this dead horse lie. I think this all started when a shop showed me that if I simply hook up my front and rear speakers up to the same 2 channels on an amp I could get more power to the speakers. I just wouldn't be able to use my fader on my HU. I assumed he was a professional and went along with it. This, I believed, must be bridging. I ran the setup across my electronics theory and it seemed to make sense. Speakers in parallel, less load to the amp, more current...this makes sense. Was this setup dangerous to my equipment?? Is this done ever??

 

[Immacomputer 10+ year member]

It depends on the load of the speakers and what the amp was able to handle. There is no way to know without model numbers.

 

[iriejp1 10+ year member]

oh ok. But this is not common right?

 

[Immacomputer 10+ year member]

For mids/highs? No, not really unless it's a 4 channel amp and you're only powering 2 speakers with the front and rear channels bridged.

 

[iriejp1 10+ year member]

It was a long time ago...but I think it was a 3 channel and he was putting the front and rears on 2 (L/R) channels and my sub on one. It was all garbage equipment from top to bottom..but it actually worked for quite a while before I finally replaced it all. Anyways thanks...

 

[Thnking 10+ year member]

Actually,

When the amplifier is bridged, each channel sees the entire load. Current doubles because it's cumulative for each output.

So running a 4ohm speaker bridged off the amp, each channel outputs current to drive a 4ohm load.

 

[thch 10+ year member]

this is incorrect for previously stated reasons.

I have plenty of easy explainations for the "each channel see's half" suffice to say, the ohm's law one is the easiest typed version and its already in this thread. there is a circuit equivilence explanation that also works well, but i don't feel like making/uploading the pic.