"Buck Turgidson" said:
OK. I'll make up a supply spec. There is no supply brand mentioned
in the advert. This is a 535W power supply.
+3.3V@32A, +5V@32A, +12V1@18A, +12V2@18A,
[email protected],
[email protected]
265.5W combined , 408W (34A) combined, 9.6W , 12.5W
<--------------- 512W ------------------>
4200+ 89W/12V*(1/0.90) = 8.24A from 12V2 (90% efficient Vcore conversion)
Geforce 6600 12V at 4 A (Xbitlabs.com has measured these)
Fans 12V at 0.5A
Hard drive 12V at 0.6A (spinup completed, sitting idle)
Hard drive 12V at 0.6A (spinup completed, sitting idle)
Hard drive 12V at 0.6A (spinup completed, sitting idle)
DVDROM 12V at 1.5A (no real info available, probably less
DL-DVDRW 12V at (1.5A) (when gaming, only one drive has game CD)
Total 12V1 at 7.8A
Many of the other items, are 1W chips.
Memory is powered by a lower voltage rail. 3.3 is good for making 2.5V,
or 5V can be used as well. Combined 3.3V and 5V consumption should
be less than 100W or so (enough to completely run some older computers).
This calculation is not completely accurate, as there should be
a calculation done under two conditions. When the computer is
first turned on, each hard drive draws 2A while it is accelerating
the hard drive spindle. That operation completes in about 10 seconds.
If you have enough hard drives doing that, it is theoretically possible
to trip out on overcurrent. With the above supply as an example,
if we had 12A of 12V1 to spare, then that would be enough to spin
up six hard drives at the same time. With three drives in the system,
there should be room to spare. When doing the spinup calculation,
assume the processor is running at half power, and the video card
would be roughly half power (as it is running in 2D mode). If you
had more hard drives, then a second calculation and column of numbers
would be needed.
To summarize, 12V1 at 7.8A, 12V2 at 8.24A, 3.3/5V probably 100W or so.
About 300W total roughly. If you inserted another 6600 video card,
then 12V1 rises by another 4A, and there is room for that.
HTH,
Paul