Quote:
Originally posted by gmat
Whoops. That's a common misconception, BB2K.
Heat is produced by inefficiencies, aka losses internal to the PSU. Usual PC switching PSUs have about 70% efficiency. The 30% left are mostly heat.
Higher rater PSUs have simply fatter transistor and lines to sustain heavier current. But as power increases, losses do as well.
So in brief:
- fatter PSU + same setup = same heat
- fatter PSU + heavier setup = more heat
Also, higher wattage PSUs have often lower efficiency (to maintain decent costs)... = even more heat !
|
The efficiency of a powersupply is not fixed for all operating conditions. The manufacturer's efficiency specification will most likely be measured in a "maximally loaded" situation. (How much of the load is on +5V and how much is on +12V is unknown unless you can get the manufacturer to cough that data up.)
A major portion of the losses is going to be "I^2*R" losses due to currents flowing through resistances. Assume a single 12V output supply. If you gradually increase the current draw from zero to maximum:
The output power will go up directly proportionally to I. (P = 12V * I)
The resistive power supply losses will go up proportional to I^2. (P= I^2 * R)
Therefore the supply will most likely be more efficient at output powers below its maximum ratings. A "fatter" supply with an equal efficiency rating (i.e. One with lower resistance in the current paths.) should be beneficial. (There are other losses in addition to the resistive losses and the maximum efficiency will peak somewhere between zero output and max output.)
Whether this hypothetical equally efficient fatter supply can be found I don't know. I haven't done much looking, but most of the supplies seem to be rated around 70% efficient.