Measuring power consumption of a 230vac sub 1 watt device. - Page 1
Forum section: electronics / projects

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: September 13, 2013, 01:17:57 pm
content: Measuring power consumption of a 230vac sub 1 watt device.HiI just fishing for ideas on how to measure the power draw on a 230V ac powered sub 1 watt device.The device is non linear in its current draw.The current waveform looks like what you would expect from a Switched Mode with no pfc, or for that matter a linear DC psu. ie Caps charging up only near the peak voltages, no draw for the rest of the half cycle.Therefore I can say Vrms x Irms != Power.Obviously I have no equipment specifically designed for this job but I do have a MSO-X 2024A.I cant see any multiplication function on it, which would make things easy.The only other idea I can think of is to capture all the points for V and I simutaneously and do the calcs myself in Excel.Any other suggestions?

author: kripton2035
author_rank: Super Contributor
timestamp: September 13, 2013, 01:28:28 pm
content: it would be a really shame if a msox2024 does not know how to multiply 2 channels ...

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: September 13, 2013, 01:31:25 pm
content: I better get the manual out.edit:There it is on the right hand side, a button marked 'Math'Ok problem solved.Thanks

author: jmcdonald
author_rank: Contributor
timestamp: September 13, 2013, 07:11:27 pm
content: How are you going to hook up the oscilloscope up to 230v? Do you have a current probe?

author: Harvs
author_rank: Super Contributor
timestamp: September 14, 2013, 12:26:57 am
content: Quote from: jmcdonald on September 13, 2013, 07:11:27 pmEasiest way for testing is just to use an isolation transformer for the device.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: September 14, 2013, 01:11:58 am
content: It says it can take 300v rms on the DSO input, after accounting for the x10 probes i guess. But I will use a voltage divider, thanks for the concern.The DUT is on an isolation transformer too, just in case.Trying to learn how to set this thing up to average my VxI function. Lots of new functionality that I haven't used, which is great.

author: johnwa
author_rank: Frequent Contributor
timestamp: September 14, 2013, 09:14:02 am
content: I don't know if this would give you enough accuracy, but have you considered getting one of those cheap plug-in power meters, and replacing the shunt with one ten or a hundred times the original value? The Arlec ones seem to be fairly good on nonlinear waveforms, though I have not tested this extensively. Obviously, if you attempt this modification, you should include a suitably rated fuse.

author: Kjelt
author_rank: Super Contributor
timestamp: September 14, 2013, 09:24:51 am
content: If you want to know the powerconsumption in kWh's you could measure with a kWh meter for one or couple of hours, that averages out all peaks.I wonder if you can use 230VDC and measure the average consumption, it's the integral (waveform surface) you want it shouldn't matter if that is a tiny surface all the time or a few bigger surfaces each 20ms as long as the period you measure is large enough.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: September 14, 2013, 01:29:00 pm
content: But i am thinking that those power meters are made for up to 100A and so wont give any sensible result for what is only about 4mA. Especially the ones with a CT.I will try the DSO directly or the integral-area method, either via the the uploaded csv from the DSO.I will also set up the current sense resistor to be high enough to create and accurately measured voltage, and divide the voltage too, to make it comfortable for the DSO.I've had a couple of beers tonight so the measuring will have to wait.

author: Mechatrommer
author_rank: Super Contributor
timestamp: September 14, 2013, 01:53:42 pm
content: use dso substract function on a shunt resistor. multiply VI in your head.

author: Unknown
author_rank: Unknown
timestamp: September 14, 2013, 03:11:40 pm
content: The limited CMRR may be a problem, depending on the shunt resistance. Measuring 1 Vrms voltage drop with ~100V rms common mode voltage is beyond the CMRR of most scopes, which is usually something like 20:1.

author: Mechatrommer
author_rank: Super Contributor
timestamp: September 14, 2013, 03:17:29 pm
content: verify (or recalibrate if possible) the channels and the shunt resistor if it can do the job beforehand. other diy options will also impose the same problem... accuracy. the more accurate answer to this thread will be... get the proper equipment that is up for the task. YMMV.

author: Unknown
author_rank: Unknown
timestamp: September 14, 2013, 03:29:34 pm
content: This is not about accuracy, this is about your hypothetical 1 Vrms differential signal getting swamped by 5 Vrms of common mode signal.Of course using the isolation transformer to ground the shunt side helps to reduce the common mode signal, but at that point you don't need the differential arrangement anymore. Note that grounding one side of the isolation transformer defeats most of its safety features.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: September 15, 2013, 12:46:28 am
content: Thanks for the inputI've got to admit I didn't even consider CMRR a problem in this case, because of the isolation transformer. Also I didn't know that many scopes have a CMRR of 20! Is that to do with the 10x probe?I don't understand how you can say that I will get a large common mode signal as I will have the ground of the probe connected to the neutral of the DUT. The DUT would otherwise be floating because of the isolation transformer and so it should be effectively shorted to the scopes common which is connected to Earth.So let me know again if I have this wrong and how I can fix it, if it is.Mechatrommer you are right to say I need to allow for the power loss in the shunt resistor, but because it is a linear device it is actually very easy.I just checked the scope with the probe shorted and attached to the isolated DUT in various places. I do get some noise but it is nearer to 100Mhz than 50Hz. No sign of Common mode noise at 50Hz.When I check the divided input voltage I am reading a nice 300mV rms.QuoteYes it is mainly about lowering noise and protection of the DSO.Dont worry I have worked live many times, only when I have to and only carefully.Importantly I have RCDs fitted on the sub circuits of my workshop which should make it very hard for me to damage myself.

author: Unknown
author_rank: Unknown
timestamp: September 15, 2013, 02:13:11 am
content: Check the specs for your scope. CMRR will probabably be something between 20:1 and 100:1 depending on the frequency. This is not due to the probe (although the 10x probe would limit it to less than 100:1), but because matching the channels is tricky.Like I said, you shouldn't have to worry about CMRR if you ground the shunt-side of the isolation transformer. But why bother with the whole differential arrangement then? Just ground it through the probe ground lead and use a single-ended measurement.How is the isolation transformer protecting the scope? One side is grounded, and you hook up the probe to the other side to measure the voltage. So the scope will still see the full mains voltage. A decent 10x probe should be able to handle mains (CAT II 300 V or so), but check the specs, not all can. The scope will also need a 10 V/div mode for a 230 Vrms signal to fit on screen.

author: HackedFridgeMagnet
author_rank: Super Contributor
timestamp: September 15, 2013, 03:13:04 am
content: QuoteThat is what I am doing.QuoteIf during testing I mistakenly to connect the probe ground to some non ground point on the DUT, such as Active, then I would get a short through the probe to Mains Earth. The DSO may be able to handle this but I am not sure what would happen, something would give and it may not be the Circuit Breaker.Mine is rated Cat I 300v rms. 400V peak.

author: Unknown
author_rank: Unknown
timestamp: September 15, 2013, 04:42:49 am
content: As soon as you ground one side of the isolation transformer, then touching the (other) probe ground to any other point will have the same effect. The only advantage may be that the isolation transformer (depending on its size) may limit the short circuit current somewhat. On the other hand it's less likely to trip the breaker. I'm not sure whether an RCD would trip if you would clip the ground leads of both probes on each side of the transformer, I'm guessing it may not.CAT I 300 V is not what I would consider enough for bare mains (which is at least CAT II, and may be CAT III/IV if close to overhead lines or other high-energy circuits). I'm not sure if an isolation transformer provides enough isolation to qualify for CAT I, I guess the small ones would.

author: NiHaoMike
author_rank: Super Contributor
timestamp: September 15, 2013, 05:29:52 am
content: A small transformer (not autotransformer) would in fact be CAT I, but there could be something greater later on if there's a large cap in the circuit.