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How to test power supply efficiency?

A

AgentJean

Banned
How would I go about testing the efficiency of my PSU? I have a cheap(by cheap I mean 30 bucks + shipping) 600 watt(might be a 500, I forget) and I'm wondering if I'd be better of buying something a little better that would save me money on my power bill in the long run.

Any suggestions on how to do this?
 
The efficiency is probably around 60%. I don't think you can test the efficiency without expensive equipment.
 
Check Howard (longtime AT'er) for his list of bad PSUs and recommended while you're at it.

Finding a efficient and reliable PSU is easy nowadays. The work has been done for you. jonnyguru.com has tested many PSUs and all you have to do is read the results.
 
WARNING:
This can be dangerous if you are not sure what you are doing.

You set up the PSU so that it is only connected to a variable resistive load. The power rating of such a resistor must be higher than the highest output power you intend to test. You can have multiple resistors to load the 12V rail as well as the 5V rail.

Use an AC power measurement device like Kill-A-Watt to measure the input power to the PSU (Pin) in Watts.
http://www.p3international.com/products/special/P4400/P4400-HG.html

Use a digital multimeter to measure the output voltage on each rail just to be sure since the voltages may not be exactly 12V and 5V.
http://www.sears.com/sr/javasr/subcat.d...ers+%26+Accessories&BV_UseBVCookie=Yes

Use a multimeter to measure the current from each rail.

For each rail, multiply the measured voltage (DC V) by the measured current (DC A). The result is power in Watts.
Add the power of all the rails. That is the output power of the PSU (Pout).

Efficiency is Pout/Pin.

I answered the question to the best of my knowledge.
However, these measurements have been done by others who have reported their measurement results. Why not use those results? This is not a simple measurement, and could be dangerous.
 
It would likely cost you less money to buy a good PSU than purchase the equipment you would need to test such a large load. Navid's answer is correct though - power out divided by power in.
 
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