Bridging question

[lumisonic]

A 2 ohm stable amplifier is capable of 900 watts bridged mono into 8 ohms.

4 - 8 ohm speakers are connected in a way that the combined sum is still 8 ohms.

900 x 8 = 7200
Square root of 7200 = 84.85 rounded

Ignoring the need for limiting (not that I would), and inherent circuit limitations, would all 4 speakers be driven a 84.85 volts each, or does 84.85 need to be divided by 4 (1/4 of 84.85 for each speaker)?

Do 4 - 8 ohm speakers (wired to still = 8 ohms) load an amplifier the same as 1 - 8 ohm speaker?

Or am I not even close with any of this?

[Bill Fitzmaurice]

If you're not really an expert at this don't screw around with bridging. Run dual mono, two 8 ohm cabs per channel.

[lumisonic]

I am trying to learn as much as I can about bridging.
I have been reading everything I can find. Which unfortunately has not been enough.
Is there more to it than just following ohms law?

I don’t think dual mono will give me the voltage swing I would like to achieve with the amps that I have and I am unable to replace them at this time.

Even if I don’t bridge my amps, I still want to learn about it.

I know you have advised on other posts not to mess with bridging if you do not know what you are doing, but I want to learn and I have to start somewhere.

Are there any suggustions for reference material?

[Haysus]

Lumisonic
If you running out of power get an amp to meet your needs without bridging. You haven't told us what speakers you have or what amp, so how can we help you with not enough info.

[doncolga]

All I've ever known about it is whats described in manuals for amps I've had...that's probably the case for lots of us. The extent of my understanding is that it takes two amps and treats it like one big one with double the power. Obviously there's a more technical side that I'm not I'm not familiar with. My experience with bridging was with a big Mackie M2600 bridged into a 4 ohm load, two front loaded 18's. My 3012 LF loaded 18" wide T30's running on one mono channel of a 600 watt amp run rings around them, so based on what I've picked up from here, I don't bridge anymore either. I can measure well over 50 volts of swing of that one channel on that amp, so I still have a whole channel sitting unused for two future T30's...and I don't trip circuit breakers anymore.

[lumisonic]

I have been spending some time lately trying to educate myself on the subject of amplifier bridging. Everything I read dealt with voltage and impedance. I finally found something that included one piece of the puzzle that was missing.

CURRENT. (man, I have been away from my electronics hobby for too long)

It doesn't answer all of my questions, but since it helped me understand things a bit better maybe it will help someone else so I would like to share it.
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Here's some of the science and math behind bridging an amp:

A standard solid-state amp usually has a power supply that is +/- with respect to ground. For this example, we'll use plus and minus 40V, a total swing of 80V possible. During normal operation, one leg of the speaker is connected to ground (common), and the other is connected to the output transistors. The output transistors put out a voltage that is between the "top" voltage (+40 V) and the "bottom voltage" (-40V), depending on the instantaneous value of the input signal, they pull the output voltage up or down. So the most voltage the speaker can see across it is magnitude 40 volts, either positive or negative.

When you bridge a stereo (2-channel) amp, you hook one side of the speaker voice-coil to the transistor output of one channel, and the other side of the speaker voice coil to the transistor output of the other channel. Special circuitry then drives one channel 180 degrees out of phase with the other, so when one output is going up towards +40V, the other output is going down towards -40V. This allows all 80 volts of the power supply to be used as the voltage swing for the speaker.

But the output transistors have a current limit, and this current limit still holds true even in bridging. So even though the voltage has doubled, the current remains the same. So Power, which is voltage times current, is doubled, since the voltage across the speaker is doubled. But to have the same current at 2x the voltage, the impedance has to double as well.

On the amp above,

40volts peak is equal to 28.28 volts RMS, so If the amp is rated for 4 Ohms per side, then
Vsquared / R = power or 800 / 4 = 200Watts
Current=7.07Amps RMS (28.28 / 4 = 7.07)

So now doubling the voltage to the speaker, 28.28 * 2 = 56.56 V RMS, times the same 7.07 amps, gives 400 Watts, as expected. V=I*R, so if V=56.56, and I=7.07, then R has to equal 8 Ohms. If you have a single 4 Ohm speaker bridged from ch1 positive to ch2 positive, the amp may fry, since it will try to put out too much current.
Over-current is one of the main things that kills transistors.

It makes sense, since you are electrically "stacking" the two channels of the amp, then you need to electrically "stack" the 2 speakers they each want to see. 4 ohms + 4 ohms = 8 ohms. 200 Watts + 200 Watts = 400 Watts.

This is just a general explanation, specific cases are slightly different, depending on the amp in question.

[Dave Non-Zero]

And from your explaination, it shows why most of the time you dont want to bridge. The last part of your example shows using 2 cabs to double the impedence, getting a total of 400w, wihich is 200 per cab, which is the same as the single cab per channel.

That also helps answer the OP example, wherehe wants to bridge into mutliple cabs...

If you wire the cabs to make 8ohms, you are series/parrallelling them, so reducing the voltage swing back down to what it would be in a stereo situation.

The parralled pairs get the same voltage across them, but the series connection halves the voltage. So every speaker gets half that 80V.

[jjohnson]

It makes sense, since you are electrically "stacking" the two channels of the amp, then you need to electrically "stack" the 2 speakers they each want to see. 4 ohms + 4 ohms = 8 ohms. 200 Watts + 200 Watts = 400 Watts.

By "stacking" I am assuming you mean in series and not in parallel. I personally don't like series because of the extra work in wiring the whole thing. There are some tricks to do this but it does mean either a prepared breakout box or special wiring in the speaker jacks.

In either case, whether the amp is bridged or not - you are simply presenting the same problem: With the given 8 ohm speaker load, what amount of voltage is needed to drive the speakers? You are not working out how the bridge works - just what putting speakers in series will do to the voltage requirements to power the given speakers. Once bridged the amp is simply a different amp than what you started with, with a new set of specs - don't think of bridged or not bridged - just the numbers.

Example of a 25 Watt @ 4 ohm (10 Volt) amplifier powering a 4 ohm speaker:
Current = Voltage / Resistance = 10 / 4 = 2.5 amps
or
Power = Voltage^2 / Resistance = 10^2 / 4 = 25 watts

Powering two 4 ohm speakers in parallel:
Current = Voltage / Resistance = 10 / 2 = 5 amps
Power = 50 watts

Powering two 4 ohm speakers in series:
I = V / R = 10 / 8 = 1.25 amps
P = 12.5 watts


Using the above equations (to the best of my ability) you would need 120 volts across your 8 ohm load to generate 900 watts/speaker. You'll need to ask what effect series has on the max power/voltage handling of a speaker.

Edited to correct my math error.

[Bill Fitzmaurice]

The extent of my understanding is that it takes two amps and treats it like one big one with double the power. .

That's what the manufacturer's ad copy would lead you to believe, but it's not the case. What bridging does is to double the voltage swing. That's OK if you have a smallish amp and a higher impedance cab, 8 ohms or more. With an amp of adequate power and less than an 8 ohm load per channel all that doubled voltage swing does is make it easier to blow your drivers.

[lumisonic]

Thank you everyone. I finally get this.
It never ceases to amaze me how much I can learn on this forum.

I realize in my first question I forgot to factor the series part of the equation.
Running dual mono the way Bill described would give the same results without bridging.
And I now see why bridging is not a good idea.

It gets scary how not using something you learned a long time ago for an extended period of time, and the addition of more and more “senior moments” can cloud ones ability to reason.

[doncolga]

Thank you everyone. I finally get this.
It never ceases to amaze me how much I can learn on this forum.

+1. I've learned a ton here too, and I've not even had the pleasure of hearing my soon to be finished OT12's in action yet.