Measuring power draw of more complex loads such as switching PSU

Question

I'm trying to create an extension cord capable of measuring the power draw of devices connected to it. My approach is using a microcontroller with a non-invasive HWTC current sensor. I am using an ADS1015 12-bit ADC which is capable of converting up to 3300 samples per second. The microcontroller is programmed so that it finds the highest voltage peak flowing through the sensor (as I am measuring AC) and then I am calculating the RMS current.

This is a simplified code running on my MCU:

/// Gets the peak voltage over the next 500 ms
/// \returns Peak voltage in mV
float getPeakVoltage(ADS1015& adc) {
    Int16 maximum = 0;
    const Ulong start = millis();
    // Measure for 500 ms
    while ((millis() - start) < 500) {
        const Int16 value = adc.readDifferential();
        if (value > maximum) {
            maximum = value;
        }
    }
    // 1 bit from ADC = 3mV
    return static_cast<float>(maximum) * 3.0f;
}

const float peakVoltage = getPeakVoltage(adc);
// Using Ohm's law
static constexpr float resistorValue = 220.0;
const float peakCurrent = peakVoltage / resistorValue; // Sensor current ratio is 1000:1
const float powerDrawRms = 230.0 * peakCurrent * 0.707f; // in Watts

This approach works flawlessly for simple loads such as a Wolfram light bulb, however, when I connect, for example, a phone charger, it yields garbage. And by garbage, I mean way too high current draw.

Now I, unfortunately, don't have an oscilloscope to see what the waveform looks like, nonetheless, why my approach fails with different loads attached to the sensor?

Answer 1

I believe your problem is that you are assuming the current waveform is sinusoidal since you are taking only the peak value and then calculating the RMS value. In practice, since many devices use switching power supplies, the current waveform will not be sunusoidal and may have large peaks. Since your ADC can sample fast enough, you should take many more samples over one or more cycles of the input current waveform (either 50 or 60 Hz) and calculate the RMS value from them.

Answer 2

If you change your algorithm to do a simple average to RMS conversion you will see less error. But, doing true RMS calculations is more accurate but consumes most of the uC time at 3200 samples per second.

Pulse Vpk to Vrms, Vpk/√d at d=5% Vrms=0.22Vp
- but using Vpk/√2=0.707Vp thus result is 0.707/0.22= 3.2x true rms
- but using rectified Vavg, \$V_{RMS}=\frac{2}{\pi}V_{AVG}= 0.637 V_{AVG}\$
- then on the same pulse Vavg=Vpk*d=Vrms
- but if using the same conversion the error Vrms=0.637*Vavg
- your d=5% pulse error is -36.3% instead of +320%

This is just another approach and not necessarily the best which requires your error assumptions for power factor PF and crest factor (Pk/rms) with measurement error tolerance.

LPF method

Another approach is to apply a low pass filter on the rectified current waveform with -20dB at 4f. This steep filter is accomplished with an 8th order Butterworth LPF at 3f where f is the fundamental line frequency (50 or 60) This reduces the crest factor of noise pulses but can be implemented in one quad OpAmp with gain. and is 8x better than integration since it is 8th order with only 0.3% attenuation at f instead of -3dB (-31%). But then the same filter must be applied to voltage so that power factor angle difference does not change.

Best is a true RMS multiplier function or IC for RMS power.