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Efficient Power Supply Contest
A reader writes: "In the June (paper) issue of Scientific American, there is a mini-article descibing the energy being wasted by power supplies in computers. Those things are only 60-70% efficient in converting line-voltage AC to low-voltage DC, and there are so many millions of them out there that a modest efficiency increase could trim $1billion or more from the annual energy costs of the USA. Well, various governmental agencies are seeking to get improved power-supply efficiency into the marketplace. The central "clearinghouse" site is at efficientpowersupplies.org, and details of their contest are in this PDF."
You can also check out power supply reviews on Silent PC Review. They concern themselves with efficiency since an efficient power supply can be quieter and produce less heat.
The site also has a lot of other good info.
Switching supplies can approach 90% efficiency if they are carefully built. Such supplies will cost more, naturally, but an improvement from 60% to 90% efficiency will save you the extra cost over the course of a year or so. And, of course, you can feel better that you are contributing slightly less to carbon dioxide emissions.
by switching from energy guzzling CRTs to cool power efficient flat screens. I went from a 19" CRT at 350w to a 19" flat screen at 50w quite painlessly.
I doubt you could achieve that kind of savings no matter how power efficient you made the PS.
[wallwarts with the load unplugged] are still converting even though it's more efficient than normal since there is smaller load.
Actually, they're LESS efficient than normal. With no load, ALL the power they consume is wasted - efficience is 0%. B-)
Now the total AMOUNT of waste IS typically lower. But it's not trivial. Even the lowest tech wallwart burns power heating copper in the transformer and making up leakage in the capacitors. If it has a switching regulator it's also burning a bunch of power keeping that alive. And a voltage-flattening/capacitor-discharging resistor actually INCREASES the amount of power wasted in the wart when the load is gone (by eating some of the power that WOULD have gone into the load).
So why waste ANY by leaving the wart plugged in?
You can guesstimate the power by feeling the wart when it's been sitting there with no load for a while. The hotter, the more waste.
Bantam Dominique roosters crow a four-note song. Once you've heard it as "Happy BIRTHday" you can't NOT hear it that way
DC power supplies are usually distributed because resistive (heat) losses in wires are proportional to current^2. Since power supplies consume a relatively contstant amount of power=voltage*current, a higher voltage will result in a lower current, which means less power given off as heat; if DC was produced in the basement, thick (and expensive) copper wires/busses would be needed to distribute it. In fact, the reason AC was chosen over DC for the power grid was because AC could be stepped up to higher voltages and therefore produced at a far away central location.
--- You shall know the truth, and the truth shall make you mad- Neal (not Cowboy) Boortz
Switching supplies can approach 90% efficiency if they are carefully built.
A downside of high efficiency is that the energy lost to heating is a tiny fraction of the energy handled. When certain components start to fail they can increase their losses - and this increases the heating. The higher the overall efficiency, the greater the extra heating is as a percentage of the NORMAL heating.
If this is not taken into account in the design of the supply (and its cooling budget), the supply may be prone to thermal runaway and catastrophic failures as components age.
Bantam Dominique roosters crow a four-note song. Once you've heard it as "Happy BIRTHday" you can't NOT hear it that way
That maybe the case, but it doesn't change the basic logic. If a 500W is 70% effecient, then it is pulling in 715W. If 500W is what you need, then at 90%, you now only need a PS that pulls 555W. Dropping almost 200W from your input, decreases your heat, decreases your fan requirement, decreases your output (and therefore input) requirement. See?
1) Because the whole electronics industry has already been built up on electronics based on DC supplies, all chips, the circuits learned in EE class for common functions, etc.
2) The semiconductor technology that 98% of our electronics know-how is based on operates on low voltages, so you'd have to convert the higher 120-220-400 line and transmission voltages to low voltages anyways.
3) Most electronic active components in our current technology (semiconductors, even tubes), are asymmetric with regards to polarity and do not have "friendly" characteristics with truly bipolar (AC) signals and supplies.
4) Much of electronics can be viewed as tasks in signal processing, particularly signals that vary in time. AC power is itself electrical power that varies in time (e.g. 50-60hz). Therefore using AC as a supply into circuit would inherent introduce a LARGE signal on top of any signals you were actually interested in.
5) Batteries are inherently DC sources, so making circuits that can run of both batteries and an AC power source would be more complicated if the circuit required AC to run (you'd have to build the equivalent of a DC->AC inverter which is considerably more difficult than a AC->DC power supply, and doing so would waste battery power (inefficiencies in conversion), which is much more precious in most applications than wasting power originating from an AC powerline source.
No, your question and your understanding was valid. The power rating on a power supply states what maximum power the supply can deliver to its load. The actual power consumed *from* the power supply is solely a function of the load attached to it (i.e. the "computer" components it runs). The actual power consumed *from* the wall outlet is the sum of the power consumed by the power supply's load (i.e. the computer components) plus the extra power consumed by the power supply (i.e. the waste) which is directly proportional to the power supply's efficiency.
WarriorPoet42 got it right the second time around - but this did not make your question "stupid."
BY THE WAY: Just because you have a 400W power supply in your PC does NOT mean you are consuming 400W of power from the AC outlet. If you put an older (slower) CPU/mobo with no expansion cards, and run, say, a modern low-power hard drive, etc., the LOAD presented to the 400W power supply will be much lower. Think about it. Small form factor PCs are often built with 150W power supplies. This means that the components NEVER consume more than 150W, and probably seldom if ever hit that peak.
A side-effect of this is that the power supply efficiency does not necessarily always *waste* its ratedpower-minus-(1-minus-efficiency).
(whaatt??) Let's say:
R is the power supply's rated power.
E is its efficiency expressed as a fraction of 1 (i.e. 90% efficiency is expressed as 0.9)
So, a 400W (R=400) power supply with 80% (E=0.8) efficiency will *waste* 400*(1.0 - 0.8) 80 watts of power. But ONLY if the LOAD is drawing the full 400 watts of power!
Now let's say we have a 400W power supply with 80% efficiency, but the computer components only draw 180W of power. Let's use C to represent the power draw of the computer, so C=180. Now, just substitute C for R and you get:
C*(1-E) = 180*(1.0 - 0.8) = 36W. This is what you are REALLY losing due to power supply inefficiency.
Note: A switching power supply will have some minimal losses even if there is NO load attached to it. These are small compared to the efficiency losses in normal operation, so for practical purposes may be ignored. You could add a constant (say, K) to the equations above to account for this static power loss in the power supply, but K would be small, when compared to C, so has little effect on the math....