amp troubles

ohms is the amount of resistance in a circuit, the less resistance a circuit has the power(amps/watts) you will get.  For your amps power, it depends on the speakers.  The speakers voice coils are what an ohm applies to.  most commonly speakers are 4 ohms.  But many subs can be 1, 2, 4, or 8. It also depends on how many voice coils your sub has.  Let's say you have a sub with dual 4 ohm voice coils, you can run this sub in 2 ohms or 8 ohms depending on how its hooked up.  In a dual voice coil sub, you will have 4 ports for speaker wires, 2 positive, 2 negative.  Hooked up in parallel you'll get a 2 ohm load, in series you'll get an 8 ohm load. for 2 ohms, you run one channel to both ports, positive to both positive's on sub and same for negative wires. this will get you 2 ohms. Don't bridge your amp to a 2 ohm load unless it's rated to do so.  Let me know if this clears things, up or if you want to know more.  Understanding resistance is very important to car audio setups.

That makes things a bit more clear. Now what still throws me out a bit. So if a two channel is running at 4ohms. It means its 4 ohm stable, right? Now say the subs i connect is 2ohm, how does that affect the sub?

OK. Here we go.

It all has to do with what the amplifier "sees" as an effective load.

Your amp is stable to 2 ohms per channel, that means the maximum load (the minimum impedance) you can put on each channel, simultaneously, is 2 ohms. Anything below that impedance, and the amplifier will overheat (potentially catastrophically) or, if designed with a good protection circuit, not turn on at all.

::::::NOTE::::::
You'll notice I never referred to it as "ohm load", and you shouldn't either... It's not a correct phrase. Those are both words used to describe the same thing, but used oppositely. Low OHMS = high LOAD
::::END NOTE::::

Bridging is different altogether. When you bridge an amplifier, you still have the same load that your woofer is presenting, i.e. 8, 4, 2, 1 ohms... whatever the woofer is rated, however, whenever you put that (let's just say you have a 4 ohm woofer, to keep it simple) 4 ohm woofer on an amplifer in bridged mode, you are presenting half the impedance (twice the load) to each channel, in the case of a 4 ohm woofer, the amplifer will be effectively loaded to 2 ohms per channel.

It's really easy, actually. All you would do, is take the total load, presented by the woofer(s) and divide by 2, if connecting that load to a bridged amplifier. If connecting to a mono amplifier then you don't do anything - whatever the total ohms is, that's the load the amplifier will see.

::::::!!!!WARNING!!!!::::::: MISTER WIZARD TIME AHEAD
Turn off you computers NOW, if you think you might get bleary-eyed at another long, drawn-out haemphyst scientific explanation!

Why does it do this? Why does 4 ohms not equal 4 ohms in bridge mode? Because you are doubling the voltage across the load, the power supply will be required to make twice the current. The current production is what a power supply sees, when trying to provide a voltage to drive a load. I'll try to explain a little better.

Let's say that you have a stereo amplifier with an 8 ohm load attached to it. The amplifier is making 100 watts into that 8 ohm load. For that to happen, the amplifer power supply must be making (as in "HAS TO BE making") 28.28 volts peak to peak. The formula for figuring power is this: peak to peak voltage, squared, divided by the load. So:

28.28*28.28=800/8=100

Now, there is also a current requirement here. Voltage cannot flow without current. Voltage without current is called "static" electricity. For the amplifer to make 100 watts into an 8 ohm load, we have already determined that it is making 28.28 volts, but the current will be figured this way: voltage, divided by resistance equals current

28.28/8=3.535A

Now, a perfect amplifier will not care what the load is, it will ALWAYS make 28.28 volts, no matter WHAT speaker you connect to it. It does this by continuing to increase the current capacity on the outputs of the amplifier.

Let's say you replace that 8 ohm woofer with a 4 ohm woofer. Will the power output remain the same? With a cheezy amplifier with a weak power supply, it could, but most likely, you will always get a little more power. Let's use my perfect amplifier example, the one that can maintain that 28.28 volts, no matter what speaker you connect to it, and plug the numbers, OK?

28.28*28.28=800/4=200

Wait... what happened? Just by reducing the load, we got more power? wha...? What happened to our current?

28.28/4=7.07A

The current DOUBLED! That's why we can halve the load and get twice the power.

What happens if we double the voltage, but keep the same 8 ohm load?

56.56*56.56=3200/8=400

Wait... wait... WAIT!!! We only doubled the voltage, but we got 4 TIMES the power? Why? Well, it's in the current side, again.

56.56/8=7.07A

Herein lies the answers to the power outputs involved with bridging, and the loads seen by the power supply, in a bridged amplifier AS WELL AS the reason the amplifier "sees" twice the load (or half) of a given impedance. It's because of the additional current demand.

When you bridge an amplifer, you "invert" one channel, electrically. What this means is that if one channel is making a 60Hz wave, the other channel is making that same 60Hz wave, but 180 degrees out of phase. If you had two woofers, one connected to each channel, you would have very little to no overall output, due to the cancellation. Stereo amplifers all have two voltage rails (the positive and the negative rail) and a "neutral", or return, just like in your house, and the woofer, in stero mode is attached to the return and only one rail at any given moment.

But if we connect a woofer to a bridged amplifier, we suddenly present twice the voltage, (one rail, running at +14.14 volts, and the other rail running at -14.14 volts, rather than 14.14 volts and a neutral, or 0 volt rail) which will mandate twice the current. Now 2*2=4, so twice the voltage, twice the current, equals 4 times the power. What is actually happening, is the woofer is now no longer connected to a voltage rail and the neutral, but rather across the two voltage rails, providing twice the voltage, across the same load. This is why better quality amplifiers will quadruple their outputs into a bridged load, over a stereo load of the same impedance.

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It all reminds me of something that Molière once said to Guy de Maupassant at a café in Vienna: "That's nice. You should write it down."

Uh Yeah, what he said

Thanks guys, It realy clears up a few things.