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post #21 of 27
Quote:
Originally Posted by NrGx
But where does that powerloss come from? Heat! Why is heat created? Powerdissipation because of a voltage drop. WHy do you get voltage drops? RESISTANCE. The wires and transformer have resistance.

Correct me if I'm wrong...

Here's the thing. He discused how transformers have a ratio on them (like 10:1 for 12V rail) but what he didn't say is that that's a ratio, not how many times the wire is wrapped around the cores. The more times you wrap around the cores, the better quality the output. This creates a catch-22 though. As you increase the number of wraps, you increase the resistance through the wire (Because you get more and more wire). Typically when the current travles through transistors, it'll move through multiples of football fields in length of wire (I pulled one apart..... it wasn't pretty).

So, there's a big reason for drop in power in the unit. Also, you must know that transformers are not perfect. NrGx talks about the "Electromagnetic induction" of the transformer, but he doesn't explain that since the magnetism has to travel through a medium (usually steel), it looses some of it's magnitude/power (For those of you in physics, the E Field looses some strength because it has to travel a bit). This also causes a drop in the effeciency of the PSU.

Hope all this helps

[edit]:stupid thumb not working
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post #22 of 27
and 220/240 in many other countries

Also, you should mention something about efficiency. Due to power loss, 'Power in' = 'Power out' is incorrect.

When an efficiency of 80% is spec-ed for a PSU, that means that a 500W PSU (power always is listed on the 'out' side of the equation, thus the total max power delivered on the several rails), will draw 500/0.80 = 625W from the wall outlet. This also changes the power use, and thus the electrical cost of the PSU.

Also, a PSU rarely runs at full power, let alone continually.
    
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post #23 of 27
Quote:
Originally Posted by Chozart
Also, you should mention something about efficiency. Due to power loss, 'Power in' = 'Power out' is incorrect.

ideally it's correct, but in the realworld, that never happens, so you're both right.... but something does need to be said for effeciency.
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post #24 of 27
Most of the heat comes from the MOSFETs, which is why they are attached to those large aluminum heatsinks. In order to regulate the voltage, they use internal resistance, which results in heat being produced and consequently power loss.
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post #25 of 27
While the original post was a good start, as others have mentioned, it's lacking quite a bit of material. The other's made mention of some very key concepts like MOSFETS and rectifiers. You don't need to go in depth, just explain that they are in the PSU and what their basic function is.

You may also want to mention power factor correction. PFC is basically how PSU's efficiently convert energy. You can read my FAQ on the concept here.
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post #26 of 27
Hey nice FAQ, I really didn't know what was happening in my PC's PSU.
WELL I would like to correct something at the beginning, if u don't mind. When u say that P=V*I u are right but this is true for the direct current (active power). If u use this formula in alternative power(the power at our plugs at home is alternative) u get the effective power which is measured in VA(volt-amperes). to convert VA(volt-amperes) into W(watt) simply multiply VA by 0.7. Example
Power in VA is 400VA
power in W will be 400*0.7=280W.
I have done this at school.(I am studying electronics). If I am wrong I would be really happy about any suggestions. I am just willing to make this FAQ better.
Always go on like this NrGx.
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post #27 of 27
Thus, the formula P = V * I is valid 12V, 3.3V, 5V and all the other rails of the PSU, since this is all DC.

On the AC side of things, it gets more complex, and I found some horrid formulas. I assume your 0.7 is an estimate to avoid these formulas and is a workable number.

For the discussion about PSUs, it does not matter that much (although the initial example on the AC side should be corrected).
    
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