Help me find/understand voltage equation - Page 1
Forum section: electronics / beginners

author: omglol
author_rank: Regular Contributor
timestamp: April 23, 2017, 10:20:21 am
content: Hi all!I need to measure AC and DC power of a typical automotive alternator:DC power is easy, just DC voltage multiplied by rectified output current.However, I am having problems with calculating AC power (which must not be calculated from output DC current). If a had information about phase voltages (each stator coil between red and green dot), this would be easy. The sum of all phase voltages in this case is always zero. However, I only have information about voltages from phase to rectified ground (red to blue dot). I was told, that in this case, the sum of all phase voltages is actually 1/3 of the mean voltage level. I do not understand this, could you help? Thanks!

author: omglol
author_rank: Regular Contributor
timestamp: April 23, 2017, 10:35:36 am
content: So, just to make it clear, I need to calculate this:PowerAC = 3 * Vphase * IphaseHowever, Vphase is the voltage measured between red and green dot, while I am measuring between red and blue dot.

author: Vtile
author_rank: Super Contributor
timestamp: April 23, 2017, 12:00:50 pm
content: If you would measure between alternators two outputs you would then get the V between Red and Green by dividing that measured value (RMS) with squareroot of 3 (=1.732). Unfortunately I'm not familiar with the car alternators. I assume the starpoint (green) is connected to general ground (chassis) as is the blue, but I'm not sure. In which case the RED-BLUE and RED-GREEN would be indentical.Hopefully someone with more experience with car alternators can give you more precise answer.

author: omglol
author_rank: Regular Contributor
timestamp: April 23, 2017, 12:09:22 pm
content: No, the blue (chassis ground) and green (starpoint) points are not connected, they are not at the same potential. Tested and confirmed by myself.Also, if you scope voltage between blue and red, it is, as expected, AC (positive and negative), while voltage between blue point and red point is always positive.

author: Ian.M
author_rank: Super Contributor
timestamp: April 23, 2017, 12:11:22 pm
content: I don't think you can do it that way. The waveforms are extremely complex due to the three phase bridge and the battery load, and even with a True RMS meter rated for the maximum ripple frequency, are going to be significantly dependent on the rectifier diode parameters, the battery voltage and all the coil and wiring resistances.I've attached an approximate LTspice model of an alternator with fixed constant excitation, that you can probe to see what you are up against. The harmonics are fairly horrible and the power factor certainly isn't unity.

author: Vtile
author_rank: Super Contributor
timestamp: April 23, 2017, 12:44:03 pm
content: Quote from: Ian.M on April 23, 2017, 12:11:22 pmThe line voltage divided by 1.732 gives you the phase voltage in three phase symmetrical system (with phase shift of 120 degrees). It is of course approximation. The output is fairly constant DC compared to heavily pulsed one with the 1phase fullwave rectified output.Everything depends how accurate / delicate system the OP is after.I'm not sure who you were referring.

author: Ian.M
author_rank: Super Contributor
timestamp: April 23, 2017, 01:18:03 pm
content: I was addressing the O.P.If you aren't allowed to use the DC power and subtract the diode losses,, you'd need access to all three phases for voltage, + a current clamp round one phase conductor. If you use three 1K resistors in a Y connected to each phase conductor to derive the neutral potential, and a 1:10 current clamp, you could simply read off 1/3 the power on a Gossen METRAHIT ENERGY multimeter.Edit: I've just noticed the sim gives an unrealistically high battery voltage. Decrease Vbatt to 12V and Vin to 11*sqrt(2) to make it a bit more realistic.

author: bson
author_rank: Unknown
timestamp: April 23, 2017, 08:02:24 pm
content: Maybe put a dummy load on it and measure that; the output is what you care about in the end, isn't it?

author: IanB
author_rank: Super Contributor
timestamp: April 23, 2017, 08:27:09 pm
content: Quote from: omglol on April 23, 2017, 10:35:36 amAlthough you have only shown one "red dot" on the diagram, there are in reality three similar red dots on each of the three phases. Can you not measure the AC voltage between two of the phases (between two red dots)?

author: omglol
author_rank: Regular Contributor
timestamp: April 23, 2017, 09:20:34 pm
content: Quote from: IanB on April 23, 2017, 08:27:09 pmYes, correct. I am measuring all of those three red dots by ADC. But no, I cannot measure voltage between those red dots.I am also measuring all three phase currents, so everything is taken care of about currents ... the only problem are phase voltages.

author: omglol
author_rank: Regular Contributor
timestamp: April 23, 2017, 09:21:35 pm
content: Quote from: bson on April 23, 2017, 08:02:24 pmNo, I need to measure exactly what I asked about.

author: omglol
author_rank: Regular Contributor
timestamp: April 23, 2017, 09:24:50 pm
content: Quote from: Vtile on April 23, 2017, 12:44:03 pmDo you mean to say that voltage between red and blue dot is line voltage?

author: IanB
author_rank: Super Contributor
timestamp: April 23, 2017, 09:27:35 pm
content: Quote from: omglol on April 23, 2017, 09:20:34 pmThis statement is contradictory. If you are measuring all three red dots, you are by definition able to measure the voltage between two red dots.For clarity, I must assume you are measuring the actual time-varying AC voltage waveform, and not just an average value?

author: omglol
author_rank: Regular Contributor
timestamp: April 23, 2017, 09:31:16 pm
content: Quote from: IanB on April 23, 2017, 09:27:35 pmMaybe I need to clarify. Hardware is built in this way, so that red dots are connected (via voltage dividers etc.) to the microcontroller's ADC. Sample frequency is 80x higher than phase frequency. And no, I cannot change the hardware.

author: IanB
author_rank: Super Contributor
timestamp: April 23, 2017, 09:35:02 pm
content: OK, so apparently you cannot measure two red dots at the same time and subtract one measurement from the other?

author: omglol
author_rank: Regular Contributor
timestamp: April 23, 2017, 09:44:56 pm
content: Quote from: IanB on April 23, 2017, 09:35:02 pmYes sure I can, I can measure all three red dots at the same time. However, voltage is measured between red dot and blue dot, as stated many times.What will I get if I subtract voltage between two measurements?

author: IanB
author_rank: Super Contributor
timestamp: April 23, 2017, 10:00:11 pm
content: All voltages in an electric circuit are potential differences between two points in the circuit. Therefore, if the voltage at the first red dot is denoted by Vr1 and the voltage at the blue dot is denoted by Vb, then when you measure the first red dot you are actually measuring (Vr1 - Vb) -- the difference in potential between these two points in the circuit.Therefore, if you measure a first red dot and a second red dot and subtract the two measurements, what you are measuring is this:(Vr1 - Vb) - (Vr2 - Vb) = Vr1 - Vb - Vr2 + Vb = Vr1 - Vr2Therefore you are measuring the voltage between the two red dots. (These must be actual instantaneous voltages, not averages, when doing this calculation.)Now of course, given the voltage between two red dots you can compute the voltage between a red dot and the green dot by means of phase angles and trigonometry, knowing that the phase angle is 120 degrees in a three phase system.

author: Ian.M
author_rank: Super Contributor
timestamp: April 23, 2017, 10:41:48 pm
content: Calling the red dots A, B, and C, to match the phases in the sim I posted, and the green neutral dot N, If you want the instantaneous voltage that would be measured between A and N:V(a,n)=V(a)-(V(a)+(V(b)+V(c))/3If you want to check that, you can add a trace to the sim and paste(V(a)+(V(b)+V(c))/3in as the trace to be added. It will accurately track V(n) no matter which point you choose as your ground reference.Multiply V(a,n) by your instantaneous phase current for A(I(V1) in my sim)and you've got the instantaneous power for that phase. Average over a whole cycle and multiply by three and you are done.Any timing skew between your supposedly simultaneous ADC readings will distort your result.Edit: fixed copy/paste goof on LHS of equation.

author: Vtile
author_rank: Super Contributor
timestamp: April 24, 2017, 08:40:50 am
content: Quote from: omglol on April 23, 2017, 09:24:50 pmNo it would be phase voltage if the Green and Blue would have been connected, but they are not as you stated. The line voltage is the voltage between two red dots.Voltage between Blue-Red is something else, in my mind pretty odd meaurement point. If the measuring points were in both ends of the 6-pulse bridge it would make more sense to me.

author: Ian.M
author_rank: Super Contributor
timestamp: April 24, 2017, 09:15:33 am
content: As you want three times the result, the maths simplifies a bit:3*V(a,n)=2*V(a)-V(b)-V(c)and because the difference between neutral and the reference point cancels out, you can even do that with the raw ADC results, and any offset will drop out as well.

author: omglol
author_rank: Regular Contributor
timestamp: April 24, 2017, 07:57:00 pm
content: Quote from: Ian.M on April 23, 2017, 10:41:48 pmThank you very much! So, if I understand correctly, if I want to calculate power, I just need to solve this equation (voltages a, b and c are taken direct from my ADC):P = 3 * (V(a)-(V(a)+(V(b)+V(c))/3) * I(a)Note that I did not use any simplification just to make the equation more clear. Is this correct?

author: Ian.M
author_rank: Super Contributor
timestamp: April 24, 2017, 10:38:35 pm
content: You are on the right track:Without the3*, that's the instantaneous power from ONE phase. As it swings from 0 to maximum twice every cycle of V(a,n), you *MUST* average it over a whole number of cycles of V(a,n) to get a valid result*.For greatest accuracy, average between zero crossings of V(a,n) as the current and thus the power is practically zero near the zero crossing (due to the three phase bridge rectifier + battery load).* Unless you have current sensors on all phases and can compute the true AC instantaneous power output of the alternator by summing all the phase powers.

author: omglol
author_rank: Regular Contributor
timestamp: April 25, 2017, 06:47:08 am
content: Quote from: Ian.M on April 24, 2017, 10:38:35 pmYes, I indeed have current sensors an all phases (3x). So how do I put all of them in this equation? Thanks.

author: Ian.M
author_rank: Super Contributor
timestamp: April 25, 2017, 07:14:02 am
content: That's simple. First compute your instantaneous neutral voltage V(n).V(n)=(V(a)+(V(b)+V(c))/3Then for each phase, compute the instantaneous voltage with respect to neutral (example phase A):V(a,n)=V(a)-V(n)Once you have all three phase voltages with respect to neutral, the total instantaneous output power is:P=V(a,n)*I(a)+V(b,n)*I(b)+V(c,n)*I(c)That's the trace in purple on the plot in reply#21. It then would need a little averaging to smooth it for display - either average over a whole cycle or use something that doesn't require you to store a whole cycle's worth of samples e.g a software low pass filter.The - in front of the expression in the plot is a SPICE artifact - devices providing power have negative power dissipation. Ordinary lossy components have positive dissipation.N.B. the phase voltage inputs go negative of the ground point (blue dot) by one diode Vf drop. I do hope the existing hardware is designed to allow for that without taking the ADC inputs outside their operating range, otherwise it will clip the negative peaks and if my results are correct, clipping at 0V will cause the calculated power to be about 10% low.

author: omglol
author_rank: Regular Contributor
timestamp: April 25, 2017, 08:11:41 pm
content: Thank you very very much! I am very grateful for your help.Everything makes perfect sense, except the equation for neutral voltage. Could you explain it better? Thank you in advance!Quote from: Ian.M on April 25, 2017, 07:14:02 am

author: omglol
author_rank: Regular Contributor
timestamp: April 25, 2017, 08:13:33 pm
content: I am also interested in this situation with clipping - I understand it, but how did you calculate that 10%? Thanks!Quote from: Ian.M on April 25, 2017, 07:14:02 am

author: Ian.M
author_rank: Super Contributor
timestamp: April 25, 2017, 09:16:28 pm
content: The neutral voltage is the average of the phase voltages.I simmed the ADC input clipping using behavioural voltage sources then plotted the power. Due to the fact that you cant reuse a calculated trace in other trace expressions, I also added a BV source for the calculated neutral. See voltages aa, bb, cc and nn in attached sim which are the clipped version of a,b,c, and n.

author: IanB
author_rank: Super Contributor
timestamp: April 25, 2017, 09:28:05 pm
content: Quote from: omglol on April 25, 2017, 08:11:41 pmQuote from: Ian.M on April 25, 2017, 09:16:28 pmThink of the three phase voltages as points equally spaced around the circumference of a circle, 120 degrees apart. Each phase voltage is then represented by an arrow from the center of the circle to one of those points. The neutral is of course at the center of the circle. Now if you take the average of the coordinates of the three points around the edge of the circle, you will find the center where the neutral is.

author: Ian.M
author_rank: Super Contributor
timestamp: April 26, 2017, 01:52:26 am
content: Yes. I did make the assumption the phases are well balanced. If you want to handle deriving a neutral for unbalanced three phase systems (either phase angle or voltage imbalance) properly it gets a lot more complex, however the simple average gives you a theoretical neutral that is the closest possible to balanced for that set of phase to phase voltages, and if used as the voltage reference point to calculate the power, will give the same result summed over all the phases as calculating with the actual neutral as the reference point. Add some imbalance to my sim and try it. You'll need to add another BV source for the calculated neutral without clipping.

author: omglol
author_rank: Regular Contributor
timestamp: June 24, 2017, 09:43:11 am
content: Quote from: Ian.M on April 25, 2017, 07:14:02 amAdditional question - how can I calculate voltage between two phases without calculating instantaneous neutral voltage?

author: Ian.M
author_rank: Super Contributor
timestamp: June 24, 2017, 10:48:05 am
content: Just take the instantaneous difference.V(a,b)=V(a)-V(b)The voltage of the reference point used for the measurement drops out as its the same for both. However, its subject to the same requirements that the ADC (and the maths in the firmware) must not clipV(a)orV(b), and that you must use instantaneous voltages (i.e. sampleV(a)andV(b)as close to simultaneously as possible and perform the calculation with each pair of samples).