Determining effective power consumption by integration, not only for one second. - Page 1
Forum section: electronics / testgear

author: miggarc
author_rank: Newbie
timestamp: October 28, 2018, 12:28:55 am
content: Hi...I'm trying to determine effective power consumption by integration, over a complete cycle at 50Hz,and not over 1 second (as it is in the majority, if not all, oscilloscopes) .The video "How & Why to use Integration on an Oscilloscope" from Davewas very elucidate but there is something I'm still missing.Scope screenshot is at:photobucket.com/gallery/user/miguelgpbs/media/bWVkaWFJZDoxMzAzNTk3NTQ=/?ref=anyway, the image is also attached here...I'm using a Rigol DS1054z* Yellow (CH1) : Mains sine wave voltage, 230Vac rms 50Hz* Cyan (CH2) : Current, measured through a 1 ohm series resistor.* Purple : Voltage (Yellow CH1) times Current (Cyan CH2) math function to compute power* Red vertical lines are the imposed area of measurementAs the current is an irregular waveform, consequently the power is also a irregular one.To integrate this irregular power waveform over a complete cycle, I used the area measurement function of this oscilloscope,confined between the red lines (this gives me immediately the final integrated value) .The result value for the integration (area calculation) was 152mWs (green bordered at bottom right side).Question is:The 152mWs, as the unit of measurement itself states (mWs), is valid for a time of 1 second.So, as long as I understood, this result is not the effective power for a 50Hz frequency.Given this, how to calculate the effective power consumption, from the result 152mWs, for thepower waveform (Purple) over a complete 20ms cycle (50Hz) ?Thanks in advance,﻿Miguel Garcia

author: PTR_1275
author_rank: Frequent Contributor
timestamp: October 28, 2018, 01:10:49 am
content: Is the scope automatically allowing for that figure to be 1second sample time?What happens if you increase to 2 cycles, or 10, or a whole second. Does that figure change by the order of magnitude you expect?

author: MrW0lf
author_rank: Frequent Contributor
timestamp: October 28, 2018, 09:53:11 am
content: Quote from: miggarc on October 28, 2018, 12:28:55 amNope, it's just equivalent amount of energy compared to instantaneous power of 152mW during 1 second. 1J = 1Ws. Think kWh. Does it matter in what amount of time you spent 1kWh? You could have used 100W device during 10h, or 1kW device during 1h. Bill is just the same.You could switch on actual integral function trace on CH1*CH2 to cross check area calculator.

author: miggarc
author_rank: Newbie
timestamp: October 28, 2018, 10:38:27 am
content: Quote from: PTR_1275 on October 28, 2018, 01:10:49 amI will try to increase number of cycles for integration and then let you know.Many Thanks for your answer

author: miggarc
author_rank: Newbie
timestamp: October 28, 2018, 10:41:50 am
content: Quote from: MrW0lf on October 28, 2018, 09:53:11 amHiBut... Dave Jones said on his video, at 11m 02s, we must not confuse this units because it is not an energy unit.Could he be wrong?https://youtu.be/Dh0xYu8YvaE?t=662In fact, if I convert this result from energy to watt, it gives me an absolute erroneous value.Try it at:https://www.rapidtables.com/calc/electric/kwh-to-watt-calculator.htmlInputting those values:kWh : 0.000152(152mW=0.000152)h : 0.000278(1s = 0.000278h)Result is 546.7 Wwhich is out of question because the circuit related to the scope measurementshas an absolute maximum power of 25W .Thanks

author: MrW0lf
author_rank: Frequent Contributor
timestamp: October 28, 2018, 10:50:15 am
content: Quote from: miggarc on October 28, 2018, 10:41:50 amHe is talking about Vs. You are measuring Ws.If it is of any help here is some post about power measurements:https://www.eevblog.com/forum/testgear/what-you-do-with-math-channels-on-oscilloscope/msg1855935/#msg1855935But beware this talk is mostly around sinusoidals and terminology might be not fully correct since I'm not native english speaker, however you might find some general approach how to "take apart" such problems for better understanding.Quote from: miggarc on October 28, 2018, 10:41:50 am0.152Ws/0.02s=7.6W

author: miggarc
author_rank: Newbie
timestamp: October 28, 2018, 11:06:26 am
content: OK MrW0lfI will take a look there.Thanks again

author: David Hess
author_rank: Super Contributor
timestamp: October 31, 2018, 05:22:53 pm
content: Quote from: miggarc on October 28, 2018, 12:28:55 amNote that integration of power produces a measurement of energy and *not* power hence milliwatt-seconds instead of just milliwatts. 1 watt-second is 1 joule of energy.Quote152 milliwatt-seconds is the energy over one powerline cycle. Converting watt-seconds to watts means dividing by time or in this case, 1/50 seconds which is the same as multiplying by 50. So 152 milliwatt-seconds divided by 1/50 seconds yields 7600 milliwatts or 7.6 watts.

author: miggarc
author_rank: Newbie
timestamp: October 31, 2018, 09:56:55 pm
content: Hi David HessAlthough I've already figured out how to obtain power over an entire cycle from the 152 mWs energy result,many thanks for your answer...

author: macboy
author_rank: Super Contributor
timestamp: November 01, 2018, 06:56:39 pm
content: Quote from: miggarc on October 28, 2018, 10:41:50 amThe calculator asks for kWh (kilo Watt hours) and you entered kWs (kilo Watt seconds), which resulting in the wrong answer.To convert mWs to kWh, divide by 1e6 mW/kW (mW units cancel and kW remain), then divide by 3600 s/h (s units cancel, h remains). So 152 mWs is in fact 4.22e-8 kWh or 0.0000000422 kWh.

author: bson
author_rank: Unknown
timestamp: November 01, 2018, 08:06:10 pm
content: Did you really run 230VAC mains through a 1Ω resistor?

author: miggarc
author_rank: Newbie
timestamp: November 01, 2018, 11:29:12 pm
content: HiYes, that's the idea ...

author: Brumby
author_rank: Unknown
timestamp: November 02, 2018, 01:31:03 am
content: 52,900W?!! That's the RMS power you would get for each quarter cycle. That's going to need aseriouslylow impedance source as well as one helluva resistor or some very short ON times..... or is this a purely theoretical exercise?

author: macboy
author_rank: Super Contributor
timestamp: November 02, 2018, 03:43:54 pm
content: Quote from: Brumby on November 02, 2018, 01:31:03 amYou have missed the fact that the 1 ohm resistor is the current sense shunt, not the load.

author: miggarc
author_rank: Newbie
timestamp: November 02, 2018, 03:49:11 pm
content: I forgot the measurement schematics.Here it is.PWRS is obviously a power supply, a switch mode one (SMPS) .At is of interest here is all to the left of the PWRS, including itself:This is the side of the circuit to witch all values are related to.I'm measuring the PWRS efficiency.

author: Brumby
author_rank: Unknown
timestamp: November 03, 2018, 07:56:58 am
content: Quote from: macboy on November 02, 2018, 03:43:54 pmExcuse me, I have to go find something.... Ah here's one: