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
No, the sensitivities will not change. You will have to provide some means of compensating for the difference in the gain of the two amps. Otherwise a tonal imbalance will result.

Most likely the 500W amp will have higher gain than the 80W amp, so if the 500W amp does not provide a means of reducing its gain (i.e., a volume control), and assuming that you are referring to passive biamping (i.e., no active crossover ahead of the amplifiers) you would have to place some sort of passive attenuator in series with the input of that amp.

Also, again assuming that you are referring to passive biamping, since the equalized gains will result in both amps having to output essentially the same voltage at any given instant, corresponding to the full frequency range of the signal, you will probably not be able to utilize a lot of the power capability of the 500W amp. How high you will be able to turn up the volume control on your preamp, without distortion, will be limited by the clipping point of the lower powered amp.

Regards,
-- Al
Thanks for your response, Al. The tube amp in question does have level control to try to match the gains in case of bi-amping. You mentioned active crossover, would that be a more preferable approach over the passive bi-amping? In that case I would be able to adjust the gains from the crossover itself. And if I end up doing this, would you suggest removing the crossover from the speakers?

Another question I had was if the transients in the mid-high frequencies require as much headroom as the transients in the bass region. Would a 80wpc amp be able to give enough power to provide clean high volume sound (mid-high only) with a moderately affecient speaker, say 89db sensitivity?
I'm not sure that the level control on the tube amp will be able to do the job. My suspicion would be that even with it set to its maximum setting the 500W amp would still have a higher gain. If so, you would still have to provide a means of reducing the gain of the 500W amp, such as a passive attenuator in series with its input.

Yes, an active crossover would presumably provide the necessary gain adjustment, and would also keep low frequencies out of the high frequency amp, resolving the problem I mentioned of not being able to utilize all of the power capability of the high powered amp.

I have no experience with active (or passive) biamping, but those who do generally recommend removing the crossover from the speakers, which would seem logical. However before proceeding down that path you would want to obtain detailed information on the technical characteristics of the crossover in the speakers, and make sure that the active crossover you choose would be able to emulate (match) those characteristics. Some speakers have complex crossover characteristics that may not be matchable by most or all active crossovers.
Another question I had was if the transients in the mid-high frequencies require as much headroom as the transients in the bass region. Would a 80wpc amp be able to give enough power to provide clean high volume sound (mid-high only) with a moderately affecient speaker, say 89db sensitivity?
It depends on the frequency of the crossover point. It has been mentioned in past threads here that a crossover frequency of 350 Hz will typically result in an equal division of power between the low frequency and mid/hi frequency sections. The higher the crossover point, the lower the amount of power that will be required for the mid/hi section.

Regards,
-- Al
>03-18-12: Gago1101
>Another question I had was if the transients in the mid-high frequencies require as much headroom as the transients in the bass region.

Yes provided that you are passive bi-amping (although I'd argue active bi-wiringis a more useful description of what's going on). Both amplifiers see the same input signal and (assuming you don't change the tap on an ampliier with transformer coupled outputs) will clip almost exactly where they would have when running the entire speaker full-range.

Active (real) bi-amplification lacks this drawback but takes a line level cross-over which in hi-fi applications must be designed for the drivers and enclosures in question and means it isn't a viable DIY proposition for most people.

>Would a 80wpc amp be able to give enough power to provide clean high volume sound (mid-high only) with a moderately affecient speaker, say 89db sensitivity?

Probably.
here's the earlier thread on bi-amping that Almarg was alluding to:
http://forum.audiogon.com/cgi-bin/fr.pl?cspkr&1331402108&&&/Bi-amp-question

and here is a really good article on real bi-amping:
http://sound.westhost.com/bi-amp.htm
Issues about wether it's actually beneficial or not aside, I am intrigued by a couple of the technical matters discussed so far. Great info from Al (as always) and Drew, BTW.

Is it not true that of great importance is each amp's input sensitivity? And is there a correlation between an amp's power rating and it's input sensitivity as Al suggests? That has not been my experience.

Also, while I understand that each amp would see a full range signal at it's input, I don't understand why the amp would not still benefit from not having to DRIVE a full range signal. What am I missing?
Thanks, Frogman.
Is it not true that of great importance is each amp's input sensitivity? And is there a correlation between an amp's power rating and it's input sensitivity as Al suggests? That has not been my experience.
In referring to sensitivity I was referring to the sensitivity of the speaker, which is what I thought the OP was asking about. Now that you mention it, though, I'm not completely certain that I was interpreting the question as it was intended.

In any event, I certainly agree that in general there will not be a high degree of correlation between an amp's power rating and its input sensitivity. The GAIN of the amp, though (the relation between output voltage and input voltage), will have a high degree of correlation with the relation between its output power capability and its input sensitivity. As I indicated, and as I'm sure you realize, the gains of the two amps must be closely matched. The sensitivity of the amps (the input voltage that will drive the amps to their maximum power capability) is indirectly relevant, because of its relation to gain.
Also, while I understand that each amp would see a full range signal at it's input, I don't understand why the amp would not still benefit from not having to DRIVE a full range signal. What am I missing?
That is all correct, and you are not missing anything. I was not implying anything to the contrary. In a passive biamp arrangement, each amp will benefit from a significant reduction in the amount of CURRENT and POWER it has to deliver. The point I was addressing, though, is that since the VOLTAGE that will be generated at the output of BOTH of the amps will correspond to the complete full frequency range signal, if there is a large disparity in the maximum power capability of the two amps it is likely that the lower powered amp will be driven into clipping at a volume level at which the higher powered amp is delivering far less power than it is capable of. And of course the volume level cannot be turned up beyond the point at which EITHER amp is clipping, or severe distortion will be heard. Therefore a substantial fraction of the power capability of the higher powered amp will not be able to be utilized.

Best regards,
-- Al
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?

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? If it still has to produce that power, just to get absorbed in crossover, then your speaker is limited to what your lowest power amplifier can produce at the bass frequencies, even though the speaker is not producing that power from that amplifier.

Usually, the point of using a lower power tube am with a higher power ss amp is to get the sweetness of tubes in the midrange and up, while getting the power and control of SS in the bass. Since bass notes require considerably more power for the same percieved volume, 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. Then the tube amp can play very loud without clipping because the mid/hi range takes much less power than the bass.
Reading the posts above has got me to thinking about this, and my prior approach using passive biamping makes little sense.
Thank you for clarifying. It all makes sense.
>03-19-12: Frogman
>Also, while I understand that each amp would see a full range signal at it's input, I don't understand why the amp would not still benefit from not having to DRIVE a full range signal. What am I missing?

There's "a benefit" but it's not significant.

It's like what happens to your car when you neglect to eat breakfast. Assuming the two of you together weigh 3700 pounds the half pound you don't eat before work reduces the weight reduces rolling resistance and kinetic energy at a given speed by 0.01%. With most of your power going into overcoming aerodynamic drag your gas mileage increases will be even less substantial.

You'll get a little less power supply sag but aren't going to net a full dB of headroom. If you weren't clipping before you'll still be fine, and if you were clipping you'll probably still be and assuming you keep the same passive speakers and cross-overs need amplifiers with at least 2-4X the power rating to avoid that.
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
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?
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.
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
>03-20-12: Manitunc
>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.

You're confusing voltage (potential, like the 80 PSI or whatever that your pipes are at, or the voltage across the amplifier output terminals) and current (what's flowing - crack the faucet so there's high resistance to flow and only a trickle comes out. Open it all the way creating a low impedance and you get wet. Little energy flows into a high impedance load and lots goes into a low impedance).

Reactive components have inductive (coils) and capacitive (capacitors) impedance that varies with frequency.

A capacitor's impedance magnitude is 1 / 2 pi f C with f in Hz and capacitance C in farads.

As f approaches 0 impedance becomes infinite. With current flow voltage divided by impedance current and therefore power (V * I) approach 0 as you get down to DC.

An inductor's impedance magnitude is 2 pi f L with f in Hz and inductance L in Henries.

Impedance is proportional to frequency; so as frequency goes up, less current flows, and less power is delivered.

The voltage dropped across impedances in series is proportional to them. IOW, put 3 volts into 1 and 2 Ohms in series and you'll have 1A flowing with 1V dropped across the first resistor and 2V across the second resistor.

The simplest possible cross-over circuit (first order electrical) puts a capacitor in series with the tweeter and inductor in series with the woofer.

At the highest frequencies the capacitive impedance becomes negligible, current flow is as high as it can get due to driver impedance, and all the voltage goes into the tweeter. The inductive impedance is large, essentially no current flows, and the voltage drop is almost entirely across the inductor in the low frequency circuit.

At the lowest frequencies the inductive impedance becomes negligible, current flow is as high as it can get due to driver impedance, and all the voltage goes into the woofer. The capacitive impedance is large, essentially no current flows, and the voltage drop is almost entirely across the capacitor in the high frequency circuit.

Assuming a purely resistive text book driver load (which only exists where the capacitance of the moving mass equals the inductance from the suspension compliance and voice coil) with resistance R at the cross-over frequency of 1 / 2 pi R C about 29% of the voltage is being dropped across inductor and capacitor and the two drivers are each seeing about 71% of the total.
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.:)
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......
Get it now.