DOUBT: what voltage use to calculate shunt current - Page 1
Forum section: electronics / beginners

author: chinoPR
author_rank: Newbie
timestamp: November 03, 2020, 03:40:37 pm
content: hi!, I have measured the voltage on a resistance in series with the power supply of my board .. with it I must calculate the current with ohm's law. My question is the following, attach the images of the oscillogram, what voltage should I take to calculate the current? the RMS value of a single pulse? the RMS value of the all pulses?

author: thinkfat
author_rank: Unknown
timestamp: November 03, 2020, 03:45:47 pm
content: That depends entirely on what you want to know. Are you interested in peak power? Then use the largest voltage peak. If you want to know average current, use Vrms.

author: chinoPR
author_rank: Newbie
timestamp: November 03, 2020, 03:53:44 pm
content: I need to know the consumption of the board, it is powered by 24 V. at first I calculated it with the average RMS value of all the pulses and calculated the current with ohm's law (finally P = V * I), although I don't know if it is the correct way to estimate consumption

author: Renate
author_rank: Super Contributor
timestamp: November 03, 2020, 08:05:14 pm
content: Well, you're measuring (indirectly) current.And power is proportional to current squared.(With the presumption that the load resistance is constant.)So the root of the mean of the current squared is what you want.(If the load resistance is constant.)That's the RMS of the current.EditPower is proportional to current. (With the presumption that voltage is constant.)So the average of the current is what you want in this case.

author: S. Petrukhin
author_rank: Super Contributor
timestamp: November 03, 2020, 11:30:52 pm
content: Quote from: chinoPR on November 03, 2020, 03:53:44 pmIt's not clear what you want to measure.If you want to specify the current consumption of your board, you must specify the maximum current consumption-the peak value. It is used to select the power of the source.If you want to specify the energy consumed (for using the battery charge, for example), you need to calculate the integral.RMS is a bit different, it indicates an equivalent DC voltage doing the same job as AC voltage.

author: Le_Bassiste
author_rank: Frequent Contributor
timestamp: November 06, 2020, 08:00:19 am
content: all above answers are (at least partly) wrong.given the supply voltage VSUP of the board (of which you want to know the power consumption) is basically constantand the measured supply current (as per your scope) ISUP(t) = VSHUNT(t) / RSHUNT,the power consumptionof the boardis PSUP = VSUP xAVG(ISUP(t)), whilethe power consumptionin the shunt resistoris PSHUNT = (RMS(ISUP(t)))^2 x RSHUNT. ( or, PSHUNT = (RMS(VSHUNT(t)))^2 / RSHUNT)hth

author: perieanuo
author_rank: Frequent Contributor
timestamp: November 06, 2020, 10:31:36 am
content: hi,just calculate rms of that voltage (if you can !!!), than you have your current. no more complicated than this. bot for such waveform is nearly impossible in real world without quick sampling, that's the correct engineering answer.As I doubt you can do this within 5% error, you may estimate.see as an examplehttp://www.eletrica.ufpr.br/edu/artigos/CIL22-012_final_gerson.pdfeverybody here thinks it's easy, it's not

author: thinkfat
author_rank: Unknown
timestamp: November 06, 2020, 10:53:23 am
content: The scope can do the Vrms calculation well enough. But you need to be careful about setting it up. For one, set the Y gain such that the amplitude is maximum without clipping to sample with as many bits as possible. Two, set the timebase so that you get enough cycles into the calculation but not sample too coarsely. Deep memory scopes will do better because you can have more cycles in memory with higher sample rate. It depends on how much energy you expect to be consumed in transients.If you have a scope with statistics functions, might be better to set the timebase to display one cycle and then let the scope average over many waveforms. This will give better fidelity due to increased temporal resolution.

author: Le_Bassiste
author_rank: Frequent Contributor
timestamp: November 06, 2020, 11:25:22 am
content: Quote from: perieanuo on November 06, 2020, 10:31:36 amaaand wrong again with respect to what the TO was asking for.

author: Le_Bassiste
author_rank: Frequent Contributor
timestamp: November 06, 2020, 11:29:48 am
content: Quote from: thinkfat on November 06, 2020, 10:53:23 amall correct what you say about the scope. but again, TO needs the average current, not the RMS current.on top of it, there is a much more precise way to get the _average_ value of the voltage correctly: it seems to be periodic, so he could zoom in on one full period (not many periods) of the signal and calculate the _average_ value via the peak value, the waveform of a single voltage pulse, and the duty cycle.

author: thinkfat
author_rank: Unknown
timestamp: November 06, 2020, 12:26:16 pm
content: The scope can do average calculation, too, I guess ;-)The waveform may be periodic but it's likely not identical cycle-by-cycle. So it's better to use more than one cycle.

author: Renate
author_rank: Super Contributor
timestamp: November 06, 2020, 02:12:23 pm
content: Quote from: Le_Bassiste on November 06, 2020, 08:00:19 amYou have a good point about using average of the measurement if the supply voltage (which we're not simultaneously measuring) is presumed to be a constant and that the perturbations in current are caused by the effective resistance of the load changing.If the resistance of the load is (presumed) constant and the perturbations in the current are caused by perturbations in the supply voltage then RMS is what you want to use.

author: Siwastaja
author_rank: Super Contributor
timestamp: November 06, 2020, 02:27:20 pm
content: Seeing the pulses repeat at around 75kHz, you should add capacitance to supply such short and frequent pulses, getting mostly rid of them, and as a side effect, getting rid of your problem about what to measure and whether the supply can deal with such pulses or not.

author: m k
author_rank: Super Contributor
timestamp: November 06, 2020, 03:25:19 pm
content: Quote from: chinoPR on November 03, 2020, 03:53:44 pmIt's finally up to you what you want to see in that number.Your info is anyway as old as the time to the user of that info.If you get one pulse only then your small digits can also be uncomfortably unstable.Power companies has averages like two hours and so.In case you're making a wattage meter that's not very practical eighter.Test it.Make an average of few pulses first and then widen it, means that you make a new calculation after every pulse but are averaging many old pulses in to it.You can also leave the old value as a base and average it with the new value and so make a sort of an integrator.Difference is that time vice the first one is a flat filter when with the latter you can shape that flatness a bit.For long time accuracy you just add shortest possible values.Means that then you are not thinking anything of those original values, you just use smallest practical time window.