WmAx said:
Power factor denotes the actual energy delivered to a terminated load(the speakers in this case). From what I can tell, power factor as discussed here has no important bearing so far as the maximum current draw the amplifier will exhibit becuase the amplifier has finite output abilities. For example: the amplifier is not going to produce more unclipped power into a reactive load than it would into a resistive one. The effective power delivered to the load is reduced, but the maximum voltage and current produced by the amplifer remains the same(or may be reduced if the amplifier has trouble with reactive loads). So, let's give an example where we have an AB class 100 x 2 amplifier that is 50% efficient. It will require 400 watts to produce it's maximum unclipped 100 x 2 watts into a resistive load(power factor=1). If the power factor of the terminated load = 0.7, the maximum power delivered to the final load will be 70 watts x 2. But the amplifier is not going to be able to increase it's power output beyond this without clipping. The amplifier is going to consume 400 watts with maximum from it's power source(outlet) with unclipped continuous output at 100 watts x 2 @ PF=1.0 or 70 watts x 2 @ PF=0.7. The power factor here denotes the actual output power delivered to a load by the amplifier.
Now, if you are talking about the power factor of the outlet vs. amplifier load on the outlet, then please provide data on the dynamic behaviour of this interface as power factor of the amplifier vs. loudspeaker load varies so that this data can be analyzed.
-Chris
Chris, we are talking about power/current draws from the wall outlet in this thread, right?
Take the Denon 3805 as an example, it specifies 120V 7.1A. That's great, no issues, so it draws 7.1A under whatever conditions Denon defined (but not published).
Now take that Adcom 7605 that Zumbo mentioned, 1680W, under condition "maximum" in this case. So at 1680W, if you assume a p.f. of 0.7, it will draw
(1680/120)/0.7 = 20A. If you ignore p.f. and use what people constantly quoting the famous expression Watts=Volts*Amps, then the draw will be
1680/120 or 14A. This is enough difference to be noted. My point is, if you know the consumption in amps or VA (Sony, and I believe Yamaha use VA, then its fine to estimate how much it draws from the outlet, but for those who specify consumption in Watts (HK, Adcom, etc.), then you need to at least assume a nominal p.f. as an approx correction factor when trying to calculate its power draw from the outlet.
I don't disagree that for the amp/speaker circuits, p.f. is not going to be constant, but it would be safe to say that in most cases it won't average above 0.8.
Another point is, people often take the volts and amps, or VA numbers (again Denon and others tend to specify amps or VA), and interpret them as though they were Watts. For example, the 3805 would be interpreted as consuming 120X7.1=852 Watts. If you assume an overall efficiency of 50%, its output under that consumption, would be 426 Watts when it fact it is at least 20% less when you take p.f. into account.
Using your example of 100X2X0.7 (@0.7 p.f.)=140W, the power delivered to the load is reduced but the draw from the outlet is as you mentioned, still 400 (base on 50% efficiency as you mentioned), except that it is now 400VA, not 400W. The limiting factor here is the current, not the useful power. In other word, a 15A circuit can deliver 120X15=1800W only to a load with unity p.f. For a load averages at p.f. of 0.7, the 15A circuit can only support 120X15X0.7=1260W. At a given voltage, the power source, from the generator to the power line, cables etc., are ultimately limited by their current carrying capability.