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Power Supply efficiency

Posted by: Ytuloyal on
Hi PI Engineer, How can I calculate my power supply efficiency. I designed a power supply with tny275pn. Input 85-265VAC and 2 outputs 12V/0.25A and 5V/0.25A. Thanks,

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Submitted by prasun on 08/22/2008
Hi mr sadik, if u r using POWER meter, then input rms voltage multiplied by input rms current is input power. Now connect both the loads, Measure, output voltage 1 , output current 1, output voltage 2, output current 2. Now put into following equation: [[ {output voltage 1 * output current 1}+ { output voltage 2 + output current 2}]/[ input voltage * input current]]. after this multiply answer by 100. this gives you efficiency. BUT IF YOU DO NOT HAVE POWER METER TO MEASURE INPUT VOLTAGE RMS & INPUT CURRENT RMS, then try this! Cut the positive path after the bridge rectifier & before the input capacitor, Measure the current at this point by connecting Multi meter in I dc mode. Also, measure the voltage across input bulk capacitor. Now.This voltage & current are DC. there use this formula to measure efficiency in this case:[[ {output voltage 1 * output current 1}+ { output voltage 2 + output current 2}/ [ {input voltage * input current}* Power factor] now as you may not know what if Power factor, so if you can accept rough figure, consider 0.5 or 0.6 for rough idea. Any problem further, mail me at prasunkuls@hotmail.com
Submitted by PI-Chekov on 08/22/2008

Hello Sadik Algul

Prasun's approach does work if you don't have an AC power meter - we have a video in the works showing this. There is a video completed already showing how to correctly measure the output of the supply to avoid errors due to cable voltage drops (http://www.powerint.com/en/design-support/appstv#6).

I've also attached a sketch of the description Prasun gave. Note that this power measurement doesn't include losses in input bridge or input filter. For the input rectifier use VF x 2 x IDC where VF is the voltage drop of bridge diode and IDC is the DC current you measured into the bulk cap. The factor of two accounts for two diode conducting - I recommend using a voltage of 1.1 V for the VF. For the input inductor or common mode measure the DC resistance (for common mode choke multiply by two to account for both winding). Estimate the power using R X IDC^2. Ideally we would use RMS currents but the DC approach results in an efficiency estimate within 2-3% of the actual value on the high side ie actual efficiency is slightly lower.

Note that because you are measuring DC you do not have to account for power factor (this is why measuring on the AC side and multiplying AC current and AC voltage doesn't work).

Cheers

PI-Chekov

Submitted by micheal on 11/19/2009

A power supply unit (PSU) is the component that supplies power to the other components in a computer. More specifically, a power supply unit is typically designed to convert general-purpose alternating current (AC) electric power from the mains (100-127V in North America, parts of South America, Japan, and Taiwan; 220-240V in most of the rest of the world) to usable low-voltage DC power for the internal components of the computer. Some power supplies have a switch to change between 230 V and 115 V. Other models have automatic sensors that switch input voltage automatically, or are able to accept any voltage between those limits mortgage calculator.

The most common computer power supplies are built to conform to the ATX form factor. This enables different power supplies to be interchangeable with different components inside the computer. ATX power supplies also are designed to turn on and off using a signal from the motherboard, and provide support for modern functions such as the standby mode available in many computers. The most recent specification of the ATX standard PSU as of mid-2008 is version 2.31 credit card.

Note that some manufacturers, most notably Compaq and Dell, have produced power supplies using the same connectors as ATX but with different voltages on different pins travel insurance; mismatching such PSUs and motherboards can result in damage to either or both.