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12th March 2009, 12:53 PM  #11 
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Join Date: Dec 2006

if you want W/ch, drive both channels full power into whatever load (8r for instance). the formula is E^2/R so if you get 28.3V rms, it's (28.3*28.3)/8 which should give 100.111 watts. total power would then be 200.222 watts, since you're driving both channels to full power. i have no isea where Peavey came up with the 58% figure, unless it was a calculation for heat dissipation or something, or maybe it's a correction factor for watts per channel if you test the power with only one channel driven to full power,
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12th March 2009, 04:18 PM  #12 
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Join Date: Oct 2007

PV probbaly came up with the 58% figure from efficiency  the CS800 should be about 58% efficient at full power. If it's putting out 400 wpc, it will draw 1379 watts. Leaving a power dissipation of 579 watts. It will draw more than 1379 voltamps because of the nonsinusiodal AC current draw so you can't use that to calculate thermal dissipation.
Measuring output power should be straightforward in principle. Volts (RMS) squared divided by ohms. The trouble is twofold. First, most modern amps will throw a fit and shut down when asked to dissipate 579 watts of heat. Compare the heat sinks in that old boat anchor vs. its newer bretheren, vs. a "comparable" QSC  RMX1450. No comparison. That figure grows substantially with amps "bigger" than a CS800. Second, 1379 watts can be 2000 or more voltamps. The transformers can probably take this for an hour or more, but the mains breaker will trip in a few seconds. The test needs to be run quickly enough that the thermals aren't excessive, and the average current draw well under 20A. "Burst testing" is sometimes done  but if the burst is short enough you end up with inflated power numbers. 10 or 50 milliseconds won't drain the filter caps. One second on full sine, 5 seconds off is a good compromise. And a good amp should be able to take that for what would be the duration of a gig. 
12th March 2009, 07:30 PM  #13 
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Join Date: Mar 2003
Location: Locked Up In The Amp Rack

My confusion lies with the Crown PSA 2 & Macrotech 2400 amplifier.
The PSA 2 offers 75volt rails, two Power Transformers each, having its own 20amp fuse. The MA 2400 offers 57volt rails, one power Transformer for each channel in addition to a 15amp circuit breaker for each transformer. The PSA 2 is 275 watts per channel @ 8 ohms, 460 watts in a 4ohm load. The MA 2400 is 515 watts per channel @ 8 ohms, 710 watts in a 4ohm load. When I calculate use the Volts times Volts divided by 4 ohms, I get a sum of 812.25 watts for the MA 2400. When I calculate the PSA 2, I get 1406. 25 watts. Can anyone explain what is happening here? Cheers!
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12th March 2009, 09:10 PM  #14 
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Join Date: Oct 2007

First off, the MA2400 is a bridge configuration with floating supplies for each channel. The effective rail voltage is +/114 using a normal topology.
And it isn't anything special like class G or H to reduce current consumption or dissipation  it is a hog. Now in order to get stated performance for the 2400: Unloaded supply = +/114V Assume 75% regulation: Vcc = 85.5V Two sets of saturated transistor banks @ 5V each: Vpeak = 75.5V Output voltage = 53.4V RMS, 712W/4R. At 65% efficiency and 65% power factor the AC current draw would be 3370VA. Now to get stated performance for the PSA2: Unloaded supply = +/75V Assume 88% regulation: Vcc = 66V One set of output transistors in saturation: Vpeak = 61V Output voltage 43.1V RMS, 465W/4R. At 65% efficiency and 65% power factor, the AC current draw would be 2200VA. I would say that the PSA2 is more conservatively designed  that is, designed to be pushed harder or run at higher duty cycles, based on the overcurrent protection. 
13th March 2009, 01:59 PM  #15 
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Join Date: Mar 2003
Location: Locked Up In The Amp Rack

Cheers wg_ski!
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