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bitTRL1000 · 10-30-2005, 11:59 PM

See, that's what I'm talking about. I didn't think that maybe the PS could still put out power on the 5 or 3V rails while at its peak on the 12V. Which means that if I add (12 X '12V-rail amp rating') + (5 X '5V-rail amp rating') + (3 X '3V-rail amp rating') that it should be equal, or at least close to the wattage rating of the PS. Also, hopefully it will account for the fact that some PS' have a higher 12V amp rating than another PS while having a higher wattage rating than that same PS. I'll look at some PS ratings on Newegg and see.

I'm well aware of how to use just a resistor across a voltage source to measure internal resistance of that voltage source and the amp draw. The problem is that unless the internal resistance is truly linear then you need a variable resistor with fine resistance control to pick out the load which puts the PS at the cutoff voltage.

10-30-2005, 11:59 PM	#19
bitTRL1000 Member (7 bit) Join Date: Sep 2004 Posts: 70	See, that's what I'm talking about. I didn't think that maybe the PS could still put out power on the 5 or 3V rails while at its peak on the 12V. Which means that if I add (12 X '12V-rail amp rating') + (5 X '5V-rail amp rating') + (3 X '3V-rail amp rating') that it should be equal, or at least close to the wattage rating of the PS. Also, hopefully it will account for the fact that some PS' have a higher 12V amp rating than another PS while having a higher wattage rating than that same PS. I'll look at some PS ratings on Newegg and see. I'm well aware of how to use just a resistor across a voltage source to measure internal resistance of that voltage source and the amp draw. The problem is that unless the internal resistance is truly linear then you need a variable resistor with fine resistance control to pick out the load which puts the PS at the cutoff voltage.