PSU efficiency calculation - Page 1
Forum section: electronics / beginners

author: aheid
author_rank: Regular Contributor
timestamp: September 12, 2020, 11:43:13 pm
content: So I bought this 12V 100W PSU on AliExpress, mostly just to discover what I'd get. A surprise self-gift of sorts.I wanted to measure it's efficiency to see how it fares. I hooked up a multimeter on the primary side, a current clamp meter around one of the leads on the primary. The multimeters measures RMS voltage/current respectively though not True RMS but AFAIK this should be good enough for this, no? I then connected the PSU output to my electronic load and made some measurements.For a 50W load the input voltage was 238V(rms), input current 0.34A(rms), output voltage was 12.1V and output current 4.14A. So am I correct in saying the efficiency factor is 50.1W/80.9W = ~0.62?The PSU is based on the OB2283 controller, and as far as I can see follows the reference design from the datasheet very closely.Not that I'm going to modify it right away, but for curiosity where would one start to look if one wanted to improve the efficiency of this board? I noticed the output diodes got quite hot so possibly replacing those with something better might be a start I guess. But from my calculations and looking up the diodes "only" ~5W should be wasted there, though I need to verify this as who knows what those diodes are.I attached some images of the board for reference.:https://datasheetspdf.com/pdf/897595/On-BrightElectronics/OB2283/1

author: bob91343
author_rank: Super Contributor
timestamp: September 13, 2020, 12:47:23 am
content: Yes the warmest parts are dissipaing the most energy and so either should be replaced with better components or the circuit design changed to reduce parasitic losses.Losses can be grouped into areas. There are conduction losses, namely volts times amps in a device or perhaps current squared times resistance. These can be due to saturation resistance of semiconductors or, less of a problem, leakage currents while shut off.There are switching losses due to finite time to go between conduction and off states. There is a spike of heat generation between off and on.Transformers have losses due to iron and copper and radiation loss. The first can be divided into hysteresis and eddy currents and radiation. The second is due to the effective resistance of the windings multiplied by current squared. Effective resistance must take into account skin efffect.In your circuit there is also a snubber. This minimizes stress on the semiconductor but additionally dissipates heat.So the answer to your question depends on where the most power is wasted. Once you learn that, you can address how to reduce it.

author: Le_Bassiste
author_rank: Frequent Contributor
timestamp: September 13, 2020, 10:27:53 am
content: Quote from: aheid on September 12, 2020, 11:43:13 pmnope, this is not the way you evaluate efficiency of a PSU. while your DC-side measurement setup is correct, your AC-side setup is missing out on one essential information: phase between current and voltage. what you are actually measuring isapparentpower input, but what you need to measure isrealpower input:Papparent = VRMS x IRMSPreal = VRMS x IRMS x cos(phi), where phi is the phase angle between voltage and current.so, to get the proper real input power value, you also have to measure the phase betweeen voltage and current.this can be easily achieved if both voltage and current are sinusoidal, but will become pretty difficult, if the PSU doesn't have a good PFC on the input and hence terribly distorts the current waveform (which is unfortunately exactly the case with PSU in your schematic).therefore, if you want to really dig into it, the better way to measure the real power input is by sampling V(t) and I(t), multiply them to get P(t) and then average the result. this way, you get the most accurate result for real input power.

author: aheid
author_rank: Regular Contributor
timestamp: September 13, 2020, 10:38:41 am
content: Quote from: Le_Bassiste on September 13, 2020, 10:27:53 amAh yes, that makes more sense. I had a niggling suspicion but I got confused by some websites I read.

author: aheid
author_rank: Regular Contributor
timestamp: September 13, 2020, 10:46:58 am
content: Quote from: Le_Bassiste on September 13, 2020, 10:27:53 amAs "luck" would have it, I seem to already have ordered some isolated line voltage monitoring module as well as a current transformer some time ago, so I guess it shouldn't be too hard to set up a small test rig that'll sample them both and compute true power.

author: David Hess
author_rank: Super Contributor
timestamp: September 13, 2020, 10:57:48 am
content: The current and voltage are probably in phase but there will be harmonic currents and a high crest factor unless the power supply has power factor correction.

author: bob91343
author_rank: Super Contributor
timestamp: September 13, 2020, 05:59:57 pm
content: It gets more complex. When the current is not a sine, the harmonics will of course read on any meter you use but will not contribute to input power. Unless the input voltage isn't sinusoidal. You can't assume it is, if in the real world the source has a finite impedance and thus is no longer sinusoidal due to harmonic currents being drawn.One way is to use a dynamometer wattmeter to measure the input power. Due to its construction, it indeed only measures real power although its frequency response might be a concern.You also need to decide whether you want to include output ripple power in your calculation of output power. If the ripple is 1% and causes 1% current, there is some output power that needs to be accounted for, although in general it's not useful power.We are taught the basics but without regard for the nuances. Nuances can ruin you unless you are aware of them.Here is another nuance: If a diode is wasting power due to its forward drop and its switching losses and its leakage current, that loss changes with temperature so that if you let the setup get warm its temperature changes and so do its losses.One way to get around all this is to submerge the power supply in a calorimeter and measure the heat generated. This gives a more accurate measure of efficiency if the efficiency is high. However one must consider the change in performance of the unit due to the environment of the calorimeter.The devil is in the details.