CF-4 realistic power handling?

[itsmyforte]

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On 7/10/2003 5:39:55 AM djk wrote:

"its simple one amp feeds highs to the mid-tweet, and the other amp feeds the woofer. That is bi-amping. it doesn't matter if they have passive or active xovers. passive xovers eat power, unless there is no resistance in the networks."

If you don't understand the issues it might be best to keep quiet, otherwise you risk looking like the ID10T that you seem to be.

I wish everyone had an oscilloscope and an electronic crossover so they could see and hear the difference.

I've been doing this so long (30+ years) I sometimes forget there was a point in time that I didn't get it either.

obviously we can't agree on this, i am just definiting bi-amping according to text book definition. or should i say the definiton i have always went by. bi meaning 2, amping, meaning amplifiers. so 2 amps on 1 loudspeaker where the xover is elecronically separated between woofer and tweeter is bi-amping. also bi-amping, can be done with electronic xovers before amplification. thats my understanding as i have used for some time now.

[Ray Garrison]

djk,

I've spent quite a bit of time poking around the web looking for the ways people use the term "bi-amping", and you are the only person I can find who is defining the term the way you are defining it.

This is not to say that you are wrong and everyone else is right. You may very well be correct in everything you are saying.

If you look at the PS Audio website, the Arcam website, the Manley website, the Cary website, the B&W website, the Pass Labs website, the archives at Positive Feedback, Stereophile, EnjoyTheMusic.com, the rec.audio.high-end newsgroup, or anyplace else I've looked, the consistant definition of bi-amping is using two amps driven by one preamp, with one amp driving a pair of binding posts connected to the mid / high drivers and the other amp connected to the woofers.

Whatever you think this *really* is, this is what everyone else in the industry calls bi-amping. What you are referring to is usually called "active bi-amping".

I still say that one big advantage of this technique ("fool's biamping" if you will) is that by separating the amps driving the high and low frequencies, if you clip one (trying to play the Telarc 1812 cannons and clipping the bass amp) you don't clip the other amp, thus reducing the distortion that you would otherwise have in the midrange and treble.

Do you think that's a valid point?

[Andy W]

I'm with djk on this one. Bi-amping without an active crossover offers no real benefit, and in contrast to Ray, I've not heard of bi-amping being done without the active crossover; that would be silly.

djk is also correct in his signal analysis. If you have two frequencies, say 100Hz and 1kHz, both having a peak voltage of 10V, the amplifier will need to be capable of producing 20V peaks to cleanly reproduce the composite signal. When you are bi-amping with an active crossover, the amplifier only needs to be capable of 10V peaks.

-Andy W

[Ray Garrison]

This is getting silly.

Here is a link to the B&W wesbsite, specifically the manual for their top-of-the-line speaker, the Nautilus 800. On page three of this manual, they describe connection options.

What are they describing? (cut and paste from Acrobat multi-column pages is a pain, or I'd've done that to save the trouble of going to the link. )

Whether you like B&W speakers or not, I think they qualify as something of an authority in the field.

[Andy W]

They even say it themselves! Without the active crossover, the voltage output of the amplifier has to be high enough to prevent distortion on the musical peaks.

"Bear in mind that, even though midrange and, even

more so, tweeter drivers can (and only need to)

handle less continuous power than bass drivers,

the amplifier feeding them needs to have an

adequate voltage swing in order to supply the

shortterm highfrequency peaks in music without

distortion. A high voltage capability implies high

power, so it is not particularly desirable to have

a lower power amplifier feeding the midrange

and tweeter than is used for the bass drivers."

However, IF you use an active crossover, you CAN use a lower power/voltage amplifier for the mid and tweeter frequencies, because the high frequencies are not riding on the large (in voltage) low frequency signals.

Again, if you want a performance benefit, you HAVE to use an active crossover.

-Andy W

[Ray Garrison]

(sigh)

There is no question that using an active crossover in front of two amps is a better solution to bi-amping than driving two amps full signal that then feed a passive crossover. No body is arguing that.

What I am saying is that there *IS* and advantage to the passive biamp approach. Not as *BIG* and advantage as in active bi-amping, but an advantage nontheless.

If a 100 watt amp is inadequate to drive my speakers to the volume level I want when I'm playing something with demanding bass, and I clip the amp, the resulting high frequency distortion (the upper harmonics of the square wave we get when the amp is clipped) will be blocked by the low pass network feeding the woofer, but will be passed by the tweeter network, and will hit the high frequency driver. This will, at a mimimum, distort the high frequency sound, and might very well damage the tweeter by hitting it with large amounts of high frequency hash. If I replace the single 100 watt amp with two 50 watt amps, passively biamped, and play the same music at the same attempted volume level, what happens? Well, I still clip the amp driving the woofers (and I'll clip tham at a 3dB lower level than I clipped the 100 watt amp, assuming the high frequency portion of the signal wasn't demanding a lot of power from the 100 watt amp), but I will *NOT* be feeding a lot of garbage into the high frequency drivers. The other amp will continue unmolested, driving the high frequencies with an undistorted signal, while the poor woofer amp is being thrashed and clipped. As long as the high pass and low pass components of the passive crossover are electrically isolated (as they are in the CF4), the tweeter will remain blissfully unaware of the clipping in the bass.

Right?

[Andy W]

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On 7/11/2003 9:42:11 AM Ray Garrison wrote:

The other amp will continue unmolested, driving the high frequencies with an undistorted signal, while the poor woofer amp is being thrashed and clipped. As long as the high pass and low pass components of the passive crossover are electrically isolated (as they are in the CF4), the tweeter will remain blissfully unaware of the clipping in the bass.

Right?

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No.

Excuse the lame drawing, and pretend that the low frequency would be sent to the woofer by the crossover, and the high frequency would be sent to the tweeter. Also pretend that they have the same peak voltage. If both signals are present at the same time, say on a particular CD that you enjoy, you would get a waveform that looks like the summed signal.

A) If you are not bi-amping, and this signal is to be reproduced cleanly, the amplifier need to be capable of an output voltage equivalent to the sum of the peaks of the two discreet frequency signals. If not, the signal is clipped, and the tweeter will get the signal shown.

If you bi-amp using an active crossover, the low frequency will be filtered out before it goes to the tweeter amplifier, and that amplifier needs only be capable of 1/2 of the voltage of the amplifier in example A.

C) If you bi-amp WITHOUT an active crossover, then the amplifier has to be capable of the same output voltage as in example A or the same clipped signal at the tweeter will result.

D) Done with an active crossover a pair of 50W amplifiers can sound like a 200W amplifier. Done without an active crossover a pair of 50W amplifiers sounds a like a single 50W amplifier.

-Andy W

P.S. For Example D to be completely true the crossover frequency would have to be in the neighborhood of 300Hz, too low for most consumer 2- and 3-way speakers, but a very real proposition in the world of pro audio.

[Ray Garrison]

Okay, if I understand what you are saying, then this would be true.

Start with the CF4 connected to a single amp. The speaker wires terminate at the upper binding posts (connected to treble horn). The straps between the upper posts and the lower posts are in place. Play a pink noise track on a test CD. Increase the gain until the amp just begins to clip. Pause the CD player. Remove the straps, leaving the amp connected to just the upper binding posts. Resume playing without touching gain control. The amp will still clip.

Is that right? Seems counter-intuitive.

[Andy W]

That's right.

The amplifier takes the voltage signal at its input and makes a larger version of the signal at it's output -- up to the point at which the amplifier clips. If you remove the straps, the voltage output is still the same (ignoring power supply droop). The amplifier will see the change in load and no current* will flow at low frequecies, and therefore no power will be dissipated in the amplifier at low frequencies. The voltage is still there, but it has nowhere to go (because of the high impedence at low frequencies) and it eats up all the headroom in the amplifier. If you filter out the low frequencies in an active crossover this doesn't happen.

(* actually there will be a _small_ amount of current and therefore power at low frequecies since the impedence at low frequencies is finite)

Bi-amping is difficult to accomplish for the end consumer in the real world without intimate knowledge of the drivers in question. One way is to remove the passive crossover and create an active network. The other (better?) way is to determine the crossover frequency, leave the passive network in place, and then create an active network at least an octave out of band for the high and low pass section. This way the original sound of the system has a better chance of surviving.

-Andy W

[Ray Garrison]

uh, first...

I am not arguing with you, or claiming that I'm right and you're wrong, or anything... I'm really just trying to understand this. If I need to go back and reread a basic electrical engineering text before I'm going to get this, please feel free to say so. (last time I cracked one open was circa 1977).

Having said that, I still don't get it.

When the straps are in place, I am driving a network with high pass and low pass crossovers connected in parallel. When I play pink noise, my impedence is X. I don't know how to calculate X - in a DC circuit it's simply the reciprocal of the sum of the inverses of the DC resistance of the high pass and low pass sections, but in a reactive environment I'm way over my head. Anyway, whatever it is, it is X. Now, when I remove the straps and the low pass segment of the crossover is no longer there, our total impedence would (I think) *HAVE* to be greater than X. As the total impedence has increased, then by Ohms law our power output has decreased (assuming voltage is held constant), and the amp is putting out fewer watts, and so will not clip.

Help help help.

Ray

[Andy W]

Basically and amplifier clips when it runs out of voltage, not when it reaches a certain power level. For instance a 50W into 8 Ohm amp will clip at 25W when driving 16 Ohms. If the power supply is adequate and the output is not limited (either by the output transistors or current limiting) it will deliver 100W into 4 Ohms.

Speaker impedance (which varies with frequency) and the available amplifier output voltage determine the output power.

Assuming that the amplifier is operating within its design parameters (i.e. not driving a low impedence and/or triggering any current limiting circuitry), clipping is a function of power supply voltage and output signal voltage. Even if there is no load on the amplifier (speaker disconnected) it will still clip. Clipping usually occurs within a couple volts of the power supply rails. Under heavy loads the rail voltage will sag and clip at a still lower voltage, partly due to the lower rail voltage and partly due to the increased voltage drop across the output transistors (current gain drops as collector current increases). How much the power supply rails sag depend on the power supply.

-Andy W

[WMcD]

I'm gonna put my big toe into this pool of very deep water.

Part of the problem is how additional frequencies add, and what is the amplifier "rating". I had to think it through. I invite corrections.

Let's do the last first.

When I set a test tone for speaker testing, I use an RMS value of 2.828 volts out of the amp. This makes sense because the power equation is P = V x V / R. We're assuming an 8 ohm resistive load.

Now, we know that the RMS value of the sine wave really means that the peak positive is 1.414 times the RMS value and the negative peak (valley) is negative 1.414 of that RMS.

So the peaks are 1.414 times 2.828 or 3 time square root of 2 or 4.242 volts. In other words, the peak to peak (should be peak to negative peak) is 8.484 volts.

Suffice to say that the positive and negative supply rails of the amp has to be 4.242 volts. We'll therefore conclude that an amp capable of making a (note well, one single sine wave) 1 watt sine wave at the output has have that 4.242 volts available.

We'll assume that is the only limitation on the amp and its power supply. But it is the hard limit above which output clipping occurs.

The other issue is how sine waves add and how it effects the limit set by the supply rail voltages (clipping).

We'll assume that amplifier gain being what it is, it takes a 1 volt peak (2 volts P-P) sine wave input to drive the output to the 1 watt limit set above. Anything higher in input is going to drive it to clipping.

To examine this we have to appreciate the initial question or conundrum and do a bit of a thought experiment.

Lets assume we put in a 1 volt peak single sine wave at 50 Hz into the amp at a near clipping level; the output is 1 watt. We can turn that off and then put in a sine wave of 1000 Hz sine at the same one volt level. Again the output is 1 watt. Seems simple enough.

Now . . . big issue. If we put the two sine waves in at the same time; don't we get 1 watt at 50 Hz and 1 watt at 1000 Hz? It would be nice but we know it can't be true. If it was, we could put in sine waves at all individual frequencies and have thousand of watts coming out.

The reason this doesn't work is that the two input sine waves add. At some points they are both positive (or both negative) and will get above the "one volt limit" to drive the amp into clipping. I suppose I could work up the wave form on a spreadsheet, but you get the idea.

If on the other hand, you're sending the two sine waves each to individual amps, the amps, individually, don't "see" an addition at their input and are not driven to clipping. Their two loads will each get 1 watt. You should get 2 watt out to the combined, individual loads total.

- - - - -

You may have seen I covered myself in the sine wave addition issue by saying that at some points the two will overdrive the amp to clipping. At some points there individual values don't add up to the limit because their not positive enough or one is negative while the other is positive.

So, the question comes up whether there might be some combination of frequencies and levels where everything adds up to less than the limit all the time in the cycle.

The one I know of is the Fourier type addition in which weighted odd harmonics of a fundamental are added to make up a square wave. The resulting square wave has a level of 1 volt. However, the higher harmonics are reduced in level to make it all come out right. And you need all the upper harmonics at the calculated decreasing levels.

- - - - -

Behind the issue may be a question of whether the use of two amps with split frequency inputs make the system more "powerful". Yes, that is what the numbers say. But it is only power doubling which is 3 dB. It also assumes you're driving the system 3 dB from clipping (not good), and it only works, in my analysis if the problem frequencies are being divided. Those are fairly large assumptions.

Please also note the split amp use requires that there be a crossover network at the inputs of the amps, not at the outputs.

Gil