Bi-amping the Statements

[Bear]

That's the part I take issue with. It's not same power, it's same voltage. The impedance load changes when you split the woofer and tweeter, so each channel requires less current after you bi-amp.

Power is what causes action (P = V*I = R*I^2). Differences in impedance will cause differences in voltage. See my example, above, about having different impedance across a woofer and tweeter section. The tweeter example has the same power, higher voltage, but lower current draw.

If you did really double the power usage when you bi-amp, where does the extra power go? Heat or sound?

Heat. Always heat. Unless you have one of those cool arc-things. Then you are also concurrently running a radio transmitter.

I'm a Geek, not a gEEk, so there could be a lot of mistakes in my understanding, but let's not throw around the word "misinformation" too casually. :T

[Bear]

Another thing: amps are never 100% efficient so when driving multiple amps ... doesn't the inefficiency add up to create more inefficiency?
Same for the connections: every connection has loss of signal ... doesn't adding more cables and more connections add more loss?

General rule of thumb: Class A is 25% efficient, Class AB is 50% efficient and Class D is 80% efficient. However, each of these will have a base current draw that contributes to the noise floor of the amplifier. The more efficient the amplifier, the lower the base current draw. Some of Nelson Pass's Class A designs basically pull a full-load from the wall all the time and are designed where you could touch the heatsinks for only a few seconds during operation. Some of Bob Carver's designs barely get above room temperature.

Yes, there would be incremental power loss with multiple sets of wires, but it is relatively minimal -- assuming good wire. The loss from multiple balanced or unbalanced interconnects is trivial compared to what's going on in even the most efficient amplifier.

I might be wrong here (I am a newbie after all) but in my job as a telephone and XDSL technician I know that having less cables and less connections is almost always better as it reduces the chance of having problems with noise or conductivity.

A good practice generally for any of a number of reasons, but the specifics are going to be highly implementation-dependent. Personally, I would minimize cabling for its own sake: I don't like messing around with a lot of cable.

[Renovator]

I'm a Geek, not a gEEk, so there could be a lot of mistakes in my understanding, but let's not throw around the word "misinformation" too casually. :T

Thanks Bear.
Nice measured explanation. Too easy for forum discussions to get heated if we let them.

I rapidly run out of knowledge .......

In my experience, people that open discussions with that statement usually know shed loads more than they let on. (or at least are open to learning more) :T

[rhubarb9999]

I bi-amp with tubes on top, solid state on the bottom.

Works for me.

[owdi]

Power is what causes action (P = V*I = R*I^2). Differences in impedance will cause differences in voltage. See my example, above, about having different impedance across a woofer and tweeter section. The tweeter example has the same power, higher voltage, but lower current draw.


Heat. Always heat. Unless you have one of those cool arc-things. Then you are also concurrently running a radio transmitter.

I'm a Geek, not a gEEk, so there could be a lot of mistakes in my understanding, but let's not throw around the word "misinformation" too casually. :T

Heat? Really? ...and where does that heat go? Voltage changes with impedance??? No, that is not what happens.

Let me try again. Think of a two tone signal, of 100hz and 10khz. The amp has to produce voltage peaks and dips that are the sum of both signals. Play this into a two way speaker with a crossover, and measure the voltage across either the tweeter or woofer. You will find the voltage is much lower, because it has been attenuated by the crossover.

Drive this speaker with a single channel, and that channel has to produce enough current for both the woofer and tweeter. Bi-amp this speaker, and each channel only has to supply current to the individual driver. The woofer will draw plenty of current at 100hz, but at 10khz impedance is way up and current is negligible. The tweeter draws lots of current at 10khz, but at 100hz impedance is way up and current is negligible. Add them together and it's exactly the same amount of current as if you had wired the two in parallel and run them off one channel. Same current and same voltage equals same power.

Bi-amping does not lead to a magic 50% decrease in efficiency requiring twice the power for the same output. It does usually give you more headroom, because a typical amp can produce higher voltage when you increase impedance of the load, which you do by splitting up the drivers to bi-amp them.

[Bear]

Sigh. If we are to continue this, then we need to start citing sources so that we all can learn.

Heat? Really? ...and where does that heat go?

Heat is dissipated in three major places in an amplifier/loudspeaker combination: the heatsink attached to the amplifier (hopefully), the crossover components themselves, and the voice coil in the individual drivers. When talking about heat dissipation, you need to consider what's going on in all three. For purposes of this discussion, the crossover and the heatsink for the amplifier are the two most relevant ones. In the specific case of a reactive load, that would be dissipated in the amplifier itself.

Voltage changes with impedance??? No, that is not what happens.

In the spirit of now providing sources, let's start with Ohm's Law, for those playing the home game:

http://en.wikipedia.org/wiki/Ohm%27s_law

Because we are dealing with heavily reactive loads, you need to worry about not just what is going on at the loudspeaker itself, but also what is essentially being sent back up the line to the amplifier.

Let me try again. Think of a two tone signal, of 100hz and 10khz. The amp has to produce voltage peaks and dips that are the sum of both signals. Play this into a two way speaker with a crossover, and measure the voltage across either the tweeter or woofer. You will find the voltage is much lower, because it has been attenuated by the crossover.

I think that there is some additional discussion missing here. "Lower" voltage as compared to what? To what reference load is this referring?

An amplifier will try to reproduce the entire signal sent to it, barring limiter circuitry. That will become important in a moment. If, instead, we are back to thinking that biamping is synonymous with active crossovers, then I give up. The discussion I have been talking about is one where an amplifier sees a (practically) full-range signal as it might get directly off of a preamp.

Drive this speaker with a single channel, and that channel has to produce enough current for both the woofer and tweeter. Bi-amp this speaker, and each channel only has to supply current to the individual driver.

These are not DC circuits and current draw (from the wall) has a heavy dependence on amplifier topology (e.g., current source vs. voltage source) and the nature of the speaker itself. Here is a good read that is not too technical (at least, I can understand it) on amplifier issues:

http://passdiy.com/pdf/leaving_class_a.pdf

Also, the crossover is, itself, electrically-relevant. It will dissipate current under all operating conditions. Also remember, when you split the crossover, the tweeter is no longer essentially acting as a short-circuit for the woofer, and vice-versa. In other words, the full-spectrum signal going to the woofer has to be absorbed by the woofer-amp combination, since the high frequency power is not pushed over to the tweeter where the impedance is much less. The same goes for the tweeter channel. Low frequency signals must be dealt with by the tweeter-amp combination that would have gone to the woofer (this is part of the reason you want a decent-sized capacitor protecting your tweeters).

The woofer will draw plenty of current at 100hz, but at 10khz impedance is way up and current is negligible.

As measured at the driver, perhaps. Negligible is a strong word.

[quote] The tweeter draws lots of current at 10khz, but at 100hz impedance is way up and current is negligible. Add them together and it's exactly the same amount of current as if you had wired the two in parallel and run them off one channel. Same current and same voltage equals same power.[quote]
Time for another link. Since we are not dealing with DC power, we have to take into account not just the naked resistance in the woofer or tweeter circuit, but also the reactive elements. This is where the math gets sticky (beyond my patience to do for an example):

http://en.wikipedia.org/wiki/Ac_power

Of note, here, is that we have to distinguish between real power and apparent power. Simplistically, the inductors and the capacitors in the crossover elements do interesting things to the current required from the amplifier (why capacitor ratings are in Volt-Amps, rather than Watts). What ultimately matters, in terms of economics of operation, is what you are pulling from the wall, and that is where apparent power matters (what a device like a Kill-A-Watt would show, not what a multi-meter would show when measured across the speaker terminals).

Bi-amping does not lead to a magic 50% decrease in efficiency requiring twice the power for the same output. It does usually give you more headroom, because a typical amp can produce higher voltage when you increase impedance of the load, which you do by splitting up the drivers to bi-amp them.

That short-circuit thing, above, is definitely not magic, though the complex math (who came up with the square root of -1, anyway?) can make it seem like it. When you eliminate the ability for the current from the amplifier to follow a path of least resistance from one side of the crossover to the other, you expose the amplifier to two independent loads, and each load covers the entire relevant frequency band (though with a rising impedance the farther outside the passband you go). We agree that the current draw changes as measured at the driver with the change in resistance, but we differ in whether you are holding power constant or voltage constant. Since Power is what actually causes work to occur (work over time), this is what we care about.

Now let's look at my practical speaker example. Let's refer back to how we measure sound (SPL), but to keep things simple, let's also ignore weighting (dBA or dBC) since that only makes the math more complicated, but not the insight.

http://en.wikipedia.org/wiki/Sound_pressure

The simplified equation we care about is 20 * Log10(Power). Ignoring things like Xmax and Xlim, let's assume that we have a speaker with an efficiency of 60dB 1W at 1m. We then apply 100W of power to this speaker, and we get a 40dB gain in SPL. We are now operating at 100 dB across the audio spectrum (flat frequency response -- aren't ideal speakers awesome!). Let's also assume that my crossover is an LR-infinity. In other words, a cliff slope, or square wave, between the woofer and the tweeter.

As mentioned earlier, the two halves of the loudspeaker act like short-circuits when connected together. After you sever them, you still need to produce 100dB out of each half. As a result, you need an amplifier channel that produces 100W of power going into the woofer. Great, but now you have no treble. So, I need another amplifier channel for the tweeter.

In order to level-match the two halves, the SPL from both needs to be equal. Taking 100dB as my starting point and 60dB at 1W-1m for my sensitivity, how much power do I need to apply to the tweeter to equal 100dB? Yep, you guessed it: 100 watts. So, I now have two amplifier channels in operation, each producing 100W of power into their respective loads.

What do you see at the wall? Well, to an extent, that's a testable hypothesis, but it also has a dependency upon the amplifier topology (i.e., it is circuit-specific). For most Class AB and D amplifiers, this will approximate a doubling of the power draw from the wall, setting aside things like base loads (bias) and power supply inefficiencies.

So, I'm happy to be wrong, but let's leave out the magic references. After all, TANSTAAFL and the second law of thermodynamics are on my side.