When Bi-amping is there change in sensitivity


I am thinking of Bi-amping my speakers with a 80 wpc tube amplifier driving mid & high frequencies and a 500 wpc solid state amp to drive the LF driver. I was wondering if the tube amp will be able to keep up with the same volume levels as the solid state amp. I assume that it would be possible if there were difference in sensitivities for the LF and mid/hi frequencies. Does anyone know if the sensitivities change when bi-amping or if it stays the same because of the crossovers?
thanks
gago1101

Showing 4 responses by bombaywalla

03-20-12: Ngjockey
This is why I don't turn the lights off. Don't want the electricity to build up in the wire like water behind a dam.:)
LOL!!
sheesh! don't confuse Manitunc any more......
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.