AC Current Measuremnt Strategy - Page 1
Forum section: electronics / projects

author: mmitton
author_rank: Newbie
timestamp: April 02, 2023, 12:44:55 pm
content: Hi All,We are busy with a project that takes measurements from a Current Transformer (CT) to track current usage in attached appliances. The CT is connected through an amplifier circuit to a micro-controller ADC pin. The AC wave (positive cycle only) is sampled at 100us intervals (there are a few channels to sample) and a snapshot of the positive half of a wave is recorded. From this snapshot we hope to calculate the current usage of the attached load.We understand that in order to get an accurate power reading of the attached load we need to get an RMS current value from the CT, an RMS voltage value of the supply and the power factor (measure the phase angle difference between the supply voltage and load current).Step 1 is to convert the ADC reading from the CT to a current value that is consistent with a Digital Multimeter (DMM) or within a small margin of error.We believe that it should be possible to take snapshots (ADC values) of various types of loads with similar current ratings and the calculation employed should yield values that correspond to DMM readings. ie. If the DMM measures 2 different types of 15W loads and sees a current of 69mA and 58mA respectively (ratio of 1.19), then our calculation using ADC values should yield a similar ratio between the 2 loads.Here are the values recorded for each load, 15W Incandescent and 15W LED lighting. (see graph)The various measurement strategies yield the following results and ratios: (see table)Where we calculate these values using the following formulas:ADC sum = sum of all valuesADC Ave = average of all valuesADC RMS = peak * 0.707ADC True RMS = Square root of the average of the sum of the squaresAs you can see the sum/ave have a similar ratio but in the opposite direction and the RMS / True RMS calculations are way off.Is the strategy to get the AC current employed here correct or valid?What is the DMM using (Doesn't say it's a True RMS meter)?Is the phase angle difference the issue here and do we need it for an accurate current measurement?What are we missing?Thank you for any insights.Mark

author: Kleinstein
author_rank: Super Contributor
timestamp: April 02, 2023, 01:16:54 pm
content: The LEDs are a potentially nonlinear load. So the current is not necessary sine waveform. So one can not rely and the simple relative between the average and peak and RMS values valid for a sine wave.Ideally one would have the current transformer with an offset, so that the µC internal ADC can sample the whole waveform and not just 1 polarity. One could than use the average to subtract the DC offset and than do the RMS calculation for the current.For the current measurement alone one does not need the phase, as with just there current there is nothing to have a phase to.With nonlinear loads like the LEDs the power calculation is no longer working OK with RMS current times RMS voltage times cosine of phase shift, as with a non sine waveform there is more than just a phase shift to describe the power factor. The way to calculate the real power is than to use the average of voltage * current for each moment in time. The ADC would switch fast between voltage and current signal and likely use interpolation for the voltage to get values for the same time. So no mode need to also calculate the RMS voltage or current, but directly get the power. The same method can also be used with linear loads, so no need for seprate voltage , current and phase.

author: mmitton
author_rank: Newbie
timestamp: April 05, 2023, 12:15:19 pm
content: Thank you for your reply, we are going to look full wave rectification to flip the negative side of the wave positive, as well as move the zero point higher away from the adc minimum and noise floor.

author: jonpaul
author_rank: Super Contributor
timestamp: April 05, 2023, 01:09:39 pm
content: simply use analogy TRMS IC,current>>CT>>r>>TRMS IC>>DC V>>ADCJ

author: Kleinstein
author_rank: Super Contributor
timestamp: April 05, 2023, 03:59:12 pm
content: Quote from: mmitton on April 05, 2023, 12:15:19 pmPrecise full wave rectification can be tricky. Directly sampling the AC waveform may be more accurate and for the instant power calculation one would need the full voltage anyway.

author: Terry Bites
author_rank: Super Contributor
timestamp: April 05, 2023, 04:44:23 pm
content: RMS and power factor are hard to measure with non sinusoidal and complex impedance loads.RMS meters won't come to the rescue in these situations.You need a measure of crest factor as well.Luckily, you can do this in software.Measure the Peak and RMS, CF = XPEAK ÷ XRMS , this applies to both the voltage and current waveforms./www.electricalvolt.com/2020/03/what-is-crest-factor-or-peak-factor/"If the peak current is too high, the metering CT inputs can clip, causing inaccurate readings.This means that when measuring loads with high current crest factors, CT current rating needs to be selected considering crest factor of the load. For example, if your load draws 15 amps RMS, but has a crest factor of 4.0, then the peak current is 60 amps. If you use a 20 amp CT, the meter will not be able to accurately measure the 30 amp peak current."

author: cscaccetti
author_rank: Newbie
timestamp: April 13, 2023, 07:24:39 pm
content: Have you tested the CT and calculated its RCF (Ratio Correction Factor) and taken that into account. Also what is the CT rated for? Are you ensuring that you are saturating the core and getting past the knee point?