Measuring The Efficiency Of AC-DC SMPS - Page 1
Forum section: electronics / projects

author: arthur92710
author_rank: Newbie
timestamp: March 18, 2014, 08:23:22 pm
content: Hi guys im having some trouble figuring out the efficiency of a switching wall wart power supply. (110-240Vin, 12V 2Aout)The DC side is simple. Connect some load, lets say 36 ohms.Pout = V^2/R, so that's about 4w.Now on the AC side. I can measure the apparent power(VA) with my multimeter.Vin = 118.3V, Iin = 0.067A, so that's 7.92VAThis gives an efficiency(Po/Pi) of about 50%. Pretty poor for a SMPS. But it does not account for the power factor.Next I decided to use my oscilloscope to measure the power. (Both me and the scope are still alive)For the current shunt I used a 1% 0.01ohm resistor.The yellow wave is the voltage across the shunt (=Current/100). The Blue is the input voltage. The pink is the math function of multiplication.But this just shows instantaneous power.I also exported the data in a CSV and calculated the power in excel. The data is discretized so a simple average was easy to find.This gave me 9w.Now it's even worse then 50%. But actually, the real power now becomes larger then the apparent power, which can't happen.So, what am I doing wrong? What is the right way to measure the efficiency of SMPS?

author: Christopher
author_rank: Frequent Contributor
timestamp: March 18, 2014, 08:33:54 pm
content: Load the power supply to maximum using a resistor. Measure the voltage using kelvin connections RIGHT on the output connector, so no current flows effecting your result. That's your output voltage.Measure the output current using a current shunt (or meter). That's your output current.There you have your output power!Now kelvin connections on the input does not matter with such low current (as the voltage will be so high).Measure the voltage and current (again either using a shunt or shunt in meter) using a True RMS meter.Simple! Now you just need a way of measuring all of this at the same time .... Unless you can be certain that your output voltage/input voltage/currents won't change drastically...........EDIT: Don't just assume your output voltages/currents are correct. My line voltage is probably 240V, but if I measure it now it may be 238V.... Also don't assume it's perfect AC.50-60% efficiency is pretty typical for a really crappy wall wart. More expensive ones normally have a much better eff, but no better than 80-85% probably.Also, the efficiency will be much worse at minimal load (obviously depending on how well the beast is designed)

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 18, 2014, 09:56:25 pm
content: Protip: P < VA. Something's wrong.Can you set the scope to measure average(Math)?

author: Phaedrus
author_rank: Frequent Contributor
timestamp: March 18, 2014, 10:12:56 pm
content: I work in a lab dealing with PSUs 24/7. Unfortunately using an oscilloscope to measure PF for efficiency calculations just isn't practical. I've tried. Variations in PF, AC voltage, etc over time throw off the calculation. Unfortunately, you just have to spend some money.A cheap Home Depot meter like the P3 Kill-a-Watt will get you about 2.5% accuracy, which should give you ballpark efficiency. But at low loads especially it's not very accurate, so I wouldn't rely on one. The upside is they only cost $20-$30.Brand Electronics makes some decent entry level power meters that are suitable for an amateur lab, or as a backup in a professional lab. Their 4-1850 model costs $150 and gives you 1% VA accuracy up to 1800W. The 20-1850 model is the same, but adds PF measurement and one or two other features for a total of $250.Note, the last 20-1850 we got read 5% higher than our Extech meter (calibrated last year and fairly stable) out of the box. I used some 1% 20ppm/C 500 ohm resistors and a spare CPU heatsink (total cost <$20) to build a 166.65 ohm dummy load, which let me calibrate the readings to within +/-0.25% of our Extech meter, which is ok for our needs and should be sufficient for yours.

author: Zero999
author_rank: Super Contributor
timestamp: March 18, 2014, 10:40:03 pm
content: Quote from: Christopher on March 18, 2014, 08:33:54 pmThat won't be accurate, even with a true RMS meter, since the phase relationship between the current and voltage is just as important as the wave shape, when power factor is concerned.Try it yourself. Connect a 1µF capacitor across the mains, at 230V 50Hz, the current will be 72mA and a perfect sinewave, giving an apparent power of 16.6VA but a real power of near zero and a near zero power factor.

author: arthur92710
author_rank: Newbie
timestamp: March 19, 2014, 01:44:46 am
content: Quote from: Phaedrus on March 18, 2014, 10:12:56 pm$250 a good price, Ill have to look into it.Quote from: T3sl4co1l on March 18, 2014, 09:56:25 pmYep, that is a problem.My scope can add, subtract, multiply and divide the two channels. In my pic, the pink line is the product of the two channels.Quote from: AcHmed99 on March 19, 2014, 12:29:15 amGood idea, I'll try that tomorrow.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 19, 2014, 02:21:21 am
content: Quote from: arthur92710 on March 19, 2014, 01:44:46 amYeah, I mean, on the Measure menu, select Average, of trace MATH (or whatever it's called, that pink line).Tim

author: arthur92710
author_rank: Newbie
timestamp: March 19, 2014, 02:33:42 am
content: Quote from: T3sl4co1l on March 19, 2014, 02:21:21 amIt cannot do that, sadly.

author: Richard Head
author_rank: Frequent Contributor
timestamp: March 19, 2014, 07:26:58 am
content: Without suitable equipment an accurate measurement is quite difficult on the AC side. The problem is that the current waveshape is non-sinusoidal so you don't get an accurate reading when using simple multimeters. If there was active power factor correction you would get quite accurate results with just a couple of multimeters.I have a Voltech PM100 that makes the measurement a breeze. Dave recently did a teardown of a PM300 (3-Phase unit) that was interesting.I recently tried to measure the efficiency of a high efficiency (95%) power supply and found it surprisingly tricky due to the accuracy required on the AC side.Dick

author: muvideo
author_rank: Frequent Contributor
timestamp: March 19, 2014, 08:20:45 am
content: Quote from: arthur92710 on March 18, 2014, 08:23:22 pmHelloDid you integrate over an even number of power cycles?Also, are you sure you are using a 10mOhm resistor?From the traces it seem to me* you are using a 1ohm shunt for primary side,if you are using a 10mohm and you are seeing that waveforms it could be friedExporting csv from scope and doing the math: integral over a power cycle(better an even number of power cycles) of V * I is fine, just dont expectmuch for the accuracy, it will be probably no better than +-10%.Since you are not working with high-pf stuff, I would follow AcHmed99suggestion to measure directly after the bridge.* ballpark eye work:from the traces and *by eye only* if you have a 1ohm resistor the mostpower is taken around the current peak, if we approximate it to (yellow trace)200mV fo 2ms, we have about 160V during the peak, it repeats every 8.3msso we have P=.2V/1ohm * 160V * 2ms/8.3ms= slightly less that 8WPS:I'm using a bench dmm with THD measurement for my smal psu, but I'mdoing only high PF work, so for me now THD is more important than totalefficiency. Measuring high efficiencies is tricky, an extreme option I'mconsidering when I have more than 90% (I've seen probably 95%sometimes) is a calorimetric approach, but I havent had time to experimet.

author: Conrad Hoffman
author_rank: Super Contributor
timestamp: March 19, 2014, 02:31:04 pm
content: AC power measurements are difficult to do with accuracy. Just for a reality check, when doing some tough high frequency measurements, I put my circuit in a stirred air calorimeter and measured the temperature rise over time. All losses go to heat, and this method, though not 1% accurate, gave me a number I could trust, and even compare to a known resistor dissipating a known amount of power. IMO, your best bet is to find a new or used VAW meter like an old Clarke-Hess 255 or similar.

author: Mechatrommer
author_rank: Super Contributor
timestamp: March 19, 2014, 04:03:36 pm
content: Quote from: Conrad Hoffman on March 19, 2014, 02:31:04 pmhow about checking the meter counter infront of your home that was installed by your electricity provider? that is my reality check i believe, no pun intended.

author: nerdyHippy
author_rank: Contributor
timestamp: March 20, 2014, 05:01:44 am
content: Quote from: Mechatrommer on March 19, 2014, 04:03:36 pmSo in this case you would unplug everything in your entire house except the SMPS under test and measure power from the utility meter?

author: nctnico
author_rank: Super Contributor
timestamp: March 20, 2014, 09:25:18 am
content: How about feeding the SMPS with DC? It would take some investigating on which voltage to apply but every SMPS makes DC from AC first...

author: arthur92710
author_rank: Newbie
timestamp: March 20, 2014, 06:34:41 pm
content: Quote from: nctnico on March 20, 2014, 09:25:18 amIIRC, it would be the peak AC voltage, 170v. This is also a good idea I think. I just don't have a 'HV' DC supply.