When Bi-amping is there change in sensitivity

03-19-12: Manitunc
In a passive bi-amp situation, where a preamp is feeding two power amps and the signal from those power amps are feeding a bi-ampable speaker where the connection between the bass and midrange/treble is removed, where does the excess signal go? The preamp is feeding the full frequency range to the amplifier, which is amplifying that full frequency range and feeding it to a speaker's woofer, for instance. What happens to the part that is above that woofer's frequency cut off range. Doesnt it just get absorbed in the crossover and turn into heat. If so, how does that result in a decreased load to the amplifier? ....

That's a logical question. The answer is that the excess power doesn't get absorbed or turned into heat because it is never generated in the first place.

Keep in mind that power equals voltage times current (or less, if the load is not purely resistive). The crossover circuit that is in the mid/hi section of the speaker prevents low frequency currents from being supplied by the mid/hi amp and flowing into that section of the speaker. The near zero current means that the amplifier is delivering near zero power at low frequencies, even though its output voltage corresponds to the full-range signal.

Likewise, the crossover circuit in the low frequency section of the speaker prevents mid/hi frequency currents from having to be supplied by the low frequency amp, resulting in near zero power being supplied by that amp at mid/hi frequencies.

Another way to look at it is that the crossover networks result in the impedance looking into the mid/hi section of the speaker being very high at low frequencies, and the impedance looking into the low frequency section being very high at mid/hi frequencies. For a purely resistive load (i.e., impedance and resistance are the same), power equals the square of voltage divided by impedance, so at frequencies for which impedance is high power is low.

Re Drew's comment, I agree that in general there will not be a great deal of improvement in the clipping point or headroom of each amplifier, although it may be marginally significant in some cases. However, as I understand it a major rationale for passive biamping is the POTENTIAL for the sonics of the amplifiers to improve as a result of their being less heavily loaded. Secondarily, there may be a modest but in some cases significant increase in the total power that is available and that can be utilized, depending on the power ratings of the two amplifiers, on the crossover point, and on whether the maximum power capability of each amplifier, in combination with the impedance characteristics of the particular speaker, is limited by the onset of clipping or by current or thermal limitations. Finally, along the lines of Manitunc's comment, having tubes on top and solid state for the lows is POTENTIALLY and hopefully a way of combining the best of both worlds, although that trades off against possible loss of coherence, especially in the crossover region.

Regards,
-- Al

The preamp is feeding the full frequency range to the amplifier, which is amplifying that full frequency range and feeding it to a speaker's woofer, for instance. What happens to the part that is above that woofer's frequency cut off range.

Manitunc. Almarg has explained the situation quite well. Some additional comments:
look at this audio x-over plot/graph right of the paragraph "Overview":
http://en.wikipedia.org/wiki/Audio_crossover

starting from the left side of graph note that the blue curve (or even the red curve) starts at 0dB & is flat (ie. 0dB) to about 1 rad/sec (this plot is in normalized freq but don't worry about this as it does not detract from the basic explanation) and then begins to roll-off. This is a low-pass filter that is used as the x-over for a bass driver. At, say 2rad/sec frequency, the blue or red curve is about -12dB compared to the amplitude at 0rad/sec. In terms of non-dB numbers this means that the 2 rad/sec frequency is approx 1/16 the power of the 0rad/sec frequency.
If the amplifier is providing an amplified full range 20Hz-20KHz audio signal, the frequency above the x-over point (in this example the 2 rad/sec frequency) is taking only 1/16 the power i.e. means 1/16th the current from the amplifier.
So, just like Almarg wrote - there is hardly any power generated by the power amplifier for the frequencies above the cut-off frequency.
So, what happens to the part that is above the woofer's cut-off frequency? Nothing happens to it 'cuz power is not generated in those frequencies. No power generated thus no power dissipated.
hopefully, Almarg's + the explanation above also answer:

Applying the OP's situation, where he wants to use an 80 watt tube amp on top and a 500 SS amp on the bottom, where does the 80 amps bass power go to?

, it would seem that the only way you could take advantage of the SS amps increased power is to direct frequencies before they get to the amp, so the tube amp never sees the bass frequencies.

NOW, you're talking.......
real biamping!! You're absolutely correct - that's how real biamping is done & that's why you find several members here write that real biamping is not for the faint-hearted. In my earlier post in this very thread I have posted a link to a well-explained article on biamping. I will not repeat that link again as all you have to do is look at my prev post of 03/19/12.

Reading the posts above has got me to thinking about this, and my prior approach using passive biamping makes little sense.

glad that it hit you finally. better late than never, they say..... ;-)

A correction: in the Wikipedia x-over plot, the y-axis is dB Gain (i.e. voltage), so the 2 rad/sec point is -12dB or 1/4th the ampltude at 0 rad/sec.
All-the-same the point remains unchanged - freq above the x-over cause generation of very small amounts of power from the amplifier.

Almarg,
"Keep in mind that power equals voltage times current (or less, if the load is not purely resistive). The crossover circuit that is in the mid/hi section of the speaker prevents low frequency currents from being supplied by the mid/hi amp and flowing into that section of the speaker. The near zero current means that the amplifier is delivering near zero power at low frequencies, even though its output voltage corresponds to the full-range signal.

Likewise, the crossover circuit in the low frequency section of the speaker prevents mid/hi frequency currents from having to be supplied by the low frequency amp, resulting in near zero power being supplied by that amp at mid/hi frequencies."

I dont quite understand how the crossover circuit prevents the low frequency currents from being supplied by the mid/hi frequency amplifier since the amp is being fed the full range and only after it reaches the speaker does it get split off by the coil used in the crossover. How does that coil draw off the low frequencies if they never get there in the first place? That is why I dont believe that biamping with the same amplifiers changes the sound other than perhaps some additional headroom for the mid/hi amp.

Since the amp is being fed the entire spectrum and amplifying it to the speaker where it is diverted by the crossover, where does that diverted energy go?

Manitunc, I'm not sure I understand what you are not understanding. Energy is proportional to power (factored by time). Power is proportional to voltage times current. As you undoubtedly realize, in a passive biamp configuration there is no connection between the mid/hi amp and the low frequency section of the speaker, and there is no connection between the low frequency amp and the mid/high frequency section of the speaker. As a result of the high impedance that is presented by each section of the speaker at frequencies that it is not intended to reproduce, there will be little or no current flow at those frequencies, hence little or no power will be generated or delivered at those frequencies, hence there will be little or no energy to be diverted, absorbed, dissipated, or consumed at those frequencies.

I would draw an analogy with turning on a light fixture via a switch on the wall. When the switch is in the off position it presents a high (essentially infinite) impedance to the AC that is supplied through the house wiring. Therefore the light fixture draws no current and consumes no power or energy. Similarly, the high impedance of the mid/hi crossover at low frequencies prevents any current, power, or energy from being drawn from the mid/hi amp in response to the low frequency content (i.e., the low frequency spectral components) of the output voltage of that amp. Similarly, the high impedance that the low frequency crossover has at high frequencies prevents any current, power, or energy from being drawn from the low frequency amp in response to the mid/high frequency content of the output voltage of that amp.

Think of the output voltage of each amp as being a summation of many different frequencies. The amount of current that is drawn from the amp at each of those frequencies depends on the impedance of the speaker at each of those frequencies.

Regards,
-- Al