rating the power of your set up?

[nads]

just been mulling this over. how do work out what power you are putting into your speakers if you are Bi-amping? do you take the power of each channel and add? and then the Ohms of the speaker does it stay the same? half? double? for figures 2 amps at 100 watts per channel and 8 ohm speakers. are you putting 200watts into each 8Ohms speaker? or..... i get lost with this ***....

 

[Anonymous]

I suppose it would depend on the sonic filters on each driver about what impedence you are pushing. Im not sure wether it works the same as resistors. If so you have two resistors in parralel when they are jumpered, to create the 8 Ohm load, so each must be providing a 16 Ohm load.

 

[Anonymous]

i thought if you bi amped you did not get any more power. i think its only if you use 2 power amps with an integrated or pre amp. I bi amp and i dont get any more power from my set up, just a much clearer more detailed more open sound. A friend did use 2 power amps with his integrated and he got a lot more power. i think he mono'd them. I could be wrong though about not getting more power from 1 power amp and if so perhaps someoene could enlighten me.

 

[nads]

[quote user="richardjlarby"]I suppose it would depend on the sonic filters on each driver about what impedence you are pushing. Im not sure wether it works the same as resistors. If so you have two resistors in parralel when they are jumpered, to create the 8 Ohm load, so each must be providing a 16 Ohm load.[/quote]

Rich. you see where i am coming from.....

so still looking for an answer?

 

[Anonymous]

I do recall being told by a tech at car entertainment gods, Alpine, that I couldnt use a resistor to correct the issue I once had with my subwoofer being of too low impedance for the amp I wnated to drive it, because the speaker doesnt work like a resistor.

 

[Anonymous]

Yes, i believe you just add the watts together according to the ohms rating. Although, i believe with speakers the resistance may be different on HF and LF, certainly it varies with frequency.
It doesn't matter if you are biamping with two amps that have different power ratings, as long as they have the same gain and "sound".

 

[Anonymous]

Okay, this is all going to be a little simplistic, but hopefully usefull.

(By the way, Thaiman, just log out now, this is a biggy

I'll only consider the tweeter for now, as the woofer is basically the same but the other way around for frequency.

The tweeter has a high-pass filter between it and the amplifier that stops any low frequency signal going to it. If it didn't it would zing merrily past your ear the first time somebody hit a bass drum, probably before. What this filter does, regardless of whether or not you're bi-amped, is ramp up the impedance that the amplifier sees for low frequencies to that driver. With one amplifier driving the whole speaker, the amplifier still sees a normal impedance lower down because of the low frequency part of the speaker. In a bi-amped set up it doesn't. The impedance for high frequencies hasn't changed, but that for low frequencies has.

So what does this do for the power output?

First a question - what does an amplifier actually do? - it provides voltage gain - i.e. for a given input voltage the output voltage is multiplied up by some figure, limited by a maximum output voltage which is the output from thepower supply. This hasn't changed at all - the amplifier is still increasing the voltage by exactly the same ratio it was before, but the impedance it sees has changed for the low part of the frequency range - it's gotten much much bigger. So the output power, which is basically dependent upon the voltage and the impedance of the load is the same for the treble part and much lower for the bass part. The other amplifier has the opposite effect. Overall you do add the two outputs together, but they're not the same outputs you see on the spec sheet anymore because of the changed impedance that the amplifiers see. And the net result is more or less the same as before you started.

At least this is true for a passively bi-amped system. For an active system, you remove the crossover from the speaker altogether and put it in place between the pre-amp and the power-amp. So each amplifier sees a different input signal - a lower input voltage because it's for only part of the whole signal. In this case the headroom for each amplifier is increased because it doesn't see a large part of the signal and the output is ... well ... bigger. It's not doubled (not in the real world) because the bass signal is probably the higher amplitude part and that will clip well before the treble does, but you'll definitely get more by doing this.

In a passive system you will actually get more too, but for a different reason and to a much lesser extent - amplifiers have two limiting factors - the maximum voltage they can supply and ditto for current (a bit more woolly this though). When the power supply in the amplifier is trying to provide too much current the voltage that it puts out will droop slightly, more so the more current is being drawn - this equates, with non-regulated or poorly regulated power supplies (and plausibly with regulated ones though only if the voltage drops quite a lot) to non-linear amplifier behaviour since the output is essentially the input * max output / max-input and max output is the output of the power supply. So it will "clip", but not in the hard-clipping sense of the input voltage exceeding the maximum allowed that you would see with an infinite load impedance. I've referred to this a "current-clipping" before and been totally slated for it, but I don't know of a proper term. Possibly "gain-clipping" might be a better term? I don't know and would welcome advice if available.

The above isn't very clear is it?

Basically - bi-amped the amplifier is being asked to provide less current, which makes it's life easier and so it can work better. You might, at the limit, get more power out because of this. (How often is the amplifier at its limit?)

But going active is the proper way to do it.

Oh, and in the real world lots of more modest speakers don't actually provide a low-pass filter for the bass unit, they just use the natural electrical and acoustic properties of the driver to do the job for them and just put a capacitor across the tweeter. It's not really an issue, but thought I'd mention it anyway

I hope this helps at least a little, and if you think it's rubbish feel free to provide a better explanation

 

[nads]

Jim, thanks for the essay but i think getting the multi meyer out might actually give me the answers i an trying to find. Or at least let me see the Questions a bit more clearly.

 

[Anonymous]

Thanks for reading it.

As for taking measurements, well, it's all AC and with programme music (as opposed to test tones) you should see - exactly the same voltage as you do for a normal amp, but reduced current draw from each. It won't be symmetrical however. The voltage being the same is the reason you don't get more power since each amplifier will still clip at the same input and output voltage and you've not changed anything to affect that by bi-amping.

 

[nads]

Jim i was looking at the Speakers rather than the Amps.

So if you have a nominal 8Ohm speaker and then you in effect make it 2 speakers so what is the Ohms of them?

Most people are aware that any given amp will provide more power to a speaker with a lower rating... but that is not what i am looking for.

 

[Anonymous]

Let's simplify it a bit.

Take two full range speakers wired in parallel to make a bigger speaker. Both are 8 ohm. The amplifier attached to them sees a 4 ohm load and delivers double the power it would into either of the speakers individually... So if you connect each speaker directly to a single amplifier - you get - the same power - half from each amplifier that's now driving an 8 ohm load.

Each amplifier is working less hard and so your top limit is probably higher ,but you are always limited by both the voltage and current capacity that your amplifier(s) can support.

It's the same thing for bi-amping, but the actual impedances in question aren't quite so simple.

 

[nads]

[quote user="jimwall"]
Let's simplify it a bit.

Take two full range speakers wired in parallel to make a bigger speaker. Both are 8 ohm. The amplifier attached to them sees a 4 ohm load and delivers double the power it would into either of the speakers individually... [/quote]

right thats the bit that i was looking for.

so in theory the bi-amped system will be seeing 16Ohms at the end of each speaker cable.

taking it really simply. (Ok as you said earlier it is not that simple).

 

[Anonymous]

I don't quite know how to put this, but it's really not quite so simple. The example I gave is I think in retrospect, too simple, but the thought was there...

The point I maybe should have made was the reason the impedance changes in that example - the amplifier provides an output voltage, the current that comes out (and thus the power) is a consequence of that voltage. The voltage hasn't changed. For the two parallel speakers, the voltage across each is the same whether bi-amped or single amped. Therefore the current going through each is the same in both cases. Bi-amped the current comes from one amplifier per speaker, single amped from one amplifier for both speakers. So bi-amped each amplifier puts out half the current of the single amp. Each amplifier sees the whole load from a single driver. single-amped it sees both loads in parallel and puts current out for each.

But... it really isn't quite that simple - what I was trying to explain in the long post to start with was...

The impedance that you see is frequency dependent - that's how the crossover actually does it's job. Pondering that thought, go back and read the first post again and maybe you'll see why I answered it that way. And I hope this helps

 

[nads]

Jim i thank you again, and i remember why i did not take electronics to far at Uni.

Give me a car engine or anything mechanical no probs.... But as above just confuses me.

anyway thanks again. and thankfully i do not need to understand it to enjoy it.