1
\$\begingroup\$

I have a datalogger that periodically samples current values by measuring the voltage across a power resistor connected in series with a battery and sensor (the load). The datalogger also samples the voltage across the sensor. Let's call the current samples i[n] and voltage samples v[n]. Sampling of both signals is controlled programmatically (i.e., I specify in code how often to sample each parameter).

p(t)=i(t)v(t) and if we had measured the current and voltage signals continuously (as analog signals and not sampling, as I'd done), we could have built a circuit that "continuously multiplied" i(t) and v(t). p(t) would have been a continuous value.

Instead, we have the discretized the signal by sampling it and so we could only compute p[n] as i[n]*v[n].

How representative is p[n] of the actual power being drawn, p(t)? Why might someone (say an engineer who's in charge of designing the power subsystem) be interested in sampling p[n] at a higher frequency?

Also, what can we say about p[n] if the samples weren't simultaneous (i.e., i[1] was measured 1.5 seconds from the start and v[1] was measured 1.525 seconds from the start)?

\$\endgroup\$
  • \$\begingroup\$ Intuition tells me that you only need to sample as fast as needed to resolve the features that you're interested in seeing on the voltage or current signal (e.g., if current or voltage oscillates at startup (causing your sensor to behave erratically) and you want to figure out the cause, you'd need to sample faster than twice the frequency of oscillation). \$\endgroup\$ – Minh Tran Jan 24 at 6:40
  • \$\begingroup\$ If your goal is to calculate energy consumed by the load over time, then this problem can be adequately solved by simply low-pass filtering the V and I signals in the analog domain prior to digital sampling. The low-pass filter does not destroy the information you need. It just computes an average, which is what you want when measuring cumulative energy consumption. \$\endgroup\$ – mkeith Jan 25 at 19:37
2
\$\begingroup\$

How representative is p[n] of the actual power being drawn, p(t)?

Potentially, not at all. For instance, let's say that the applied voltage is a pulse waveform with 1% duty cycle - that is, the voltage is zero for 99% of the time, and (let's say) 10 volts for 1% of the time. The sensor is a 1 ohm resistor. Finally, let's say that the signals are sampled once per voltage cycle, and always during the narrow 1%. It should be obvious that ignoring the actual waveform will produce an estimate which is 100 times too high.

Why might someone (say an engineer who's in charge of designing the power subsystem) be interested in sampling p[n] at a higher frequency?

The previous section should answer that one pretty well.

What you need to do is become familiar with the concept of signal bandwidth. For a signal with a limited bandwidth, if you sample at least twice as fast as the bandwidth you can accurately recreate the signal. Anything less than this will potentially produce inaccurate results.

Also, what can we say about p[n] if the samples weren't simultaneous (i.e., i[1] was measured 1.5 seconds from the start and v[1] was measured 1.525 seconds from the start)?

That is even worse. Depending on the details of the signal, the resulting calculation can bear no resemblance at all to reality.

EDIT - As an example, let's say that the signal is a square wave which has levels of +10 and -10 volts. The load is a 10 ohm resistor. The signal transitions from +10 to -10 at 1.510 seconds. Then the voltage acquired will be 10 volts, the current will be - 1.0 amps, and the computed power will be -10 watts. Just in case you're really new to electronics, this is simply impossible for a resistive system.

As Edgar Brown has commented, if you know enough about the system, the delay between the two measurements can somtimes be compensated for (if the sample rate meets the Nyquist Criterion - but not (in general) otherwise.

\$\endgroup\$
  • \$\begingroup\$ Regarding your last point. If sampling satisfies Nyqvist criterion, that is just a delay operator before multiplying. Not really a problem if you know what you are doing, potentially catastrophic if you don't. \$\endgroup\$ – Edgar Brown Jan 24 at 22:16
  • \$\begingroup\$ I don't fully understand the conclusion of your second example (the computed power will be -10 Watts). A square wave signal that reverses polarity won't change the fact that the 10-ohm absorbs power (Watts > 0). \$\endgroup\$ – Minh Tran Jan 25 at 13:57
  • \$\begingroup\$ @MinhTran - If the samples had been taken at the same time, voltage would be 10 and current 1, for a power of 10 watts. In my example, voltage is measured at 10 volts, but the current polarity has reversed (since the voltage polarity has reversed) and the computed power is -10 watts. Negative apparent powers are common in circuits with capacitance or inductance, which produce phase shifts, but are not physically valid for ideal resistors. And while "the fact that the 10-ohm absorbs power " is true, it does not change the fact that the computed power is negative - ... \$\endgroup\$ – WhatRoughBeast Jan 25 at 21:18
  • \$\begingroup\$ @MinhTran - ...which shows that you can get into a great deal of trouble computing power from non-simultaneous sampling of voltage and current. Which answers your last question in the OP. \$\endgroup\$ – WhatRoughBeast Jan 25 at 21:20
  • \$\begingroup\$ @MinhTran - And actually, "won't change the fact that the 10-ohm absorbs power" is simply wrong. A resistor will emit or dissipate power, not absorb it. Although I do understand what you mean. \$\endgroup\$ – WhatRoughBeast Jan 25 at 21:22
1
\$\begingroup\$

There is a way to calibrate accuracy but you do not have a design spec nor given enough design details that will achieve any reasonable spec. of x % under varied conditions.

Unfortunately you have a lot to learn in power measurements Start reading here. This is only a beginning.


  • p[n] ≠ i[n]*v[n]
  • instead you have p[n] ≠ i[n]*v[n+1]
  • Sample/hold ADC is prone to errors from averaging, power factor and surges when the power is non-sinusoidal or reactive
  • if two consecutive I(t) readings change then is it because of V(t) or the load?
  • What accuracy do you want? expect? under what conditions of load and time duration?

Suggestion

  • choose dual slope integration type ADC with 20ms for 50Hz
  • or dual true multiplier with filter to ADC

FYI

https://www.maximintegrated.com/en/app-notes/index.mvp/id/1041

Line Rejection

One of the most attractive attributes of the DS-ADC is its rejection of unwanted 50/60Hz signals. If the integrate cycle lasts exactly time T , all frequencies of N × 1/T are completely rejected (theoretically). So for T = 100ms, multiples of 10Hz are rejected. The actual limitation of this rejection is due to the finite swing of the integrator (since we don't want it to saturate) and the inevitable "wobble" of the 50/60Hz frequency itself. Over a long period of time, 50/60Hz can be averaged to get extremely accurate time bases. Over a short time however, it jitters by a few Hertz. This will limit the actual line rejection to about 40-60dB.

Added: Explanation with demo

Measuring True RMS power averaged over many cycles is normally done with thermal R sensors or true multipliers with very high bandwidth, anti-aliasing filters high sampling rate to capture all the spectrum.

True 19.25 Arms at source

enter image description here

Precision rectified current sense

enter image description here

4th order LPF rectified current sense.

enter image description here

Vrms * Irms ≠ Prms when the multiplying does not use the exact same sample time even if the interval is the same.

  • e.g. 230Vrms * 19.235 rms = 4.427 kW yet TRUE Prms= 4.04kW ~ 4kW or 10% error
  • It is easy to convert average to RMS for a Sine , but not when it includes PF and spike currents. The errors are unpredictable but often > 10%

  • What is your error spec for crest factor error and this does not even consider PF error !!**

  • There is always a significant accuracy compromise with cost and complexity with with waveform factor conversions using average current or peak converted to RMS.

    • These will contribute significant errors due to the power crest factor and width and phase shift.

Averaging with a short time sample may introduce measurement errors depending on the phase angle and deviation of phase angle between samples. This simulate shows max/min peak values with RMS or Avg for power, voltage and curent. Compare each result to see why there are differences. The true result is displayed at source using V(t)*I(t) averaged near 4kW for an example of mixed loads.

enter image description here

enter image description here

\$\endgroup\$
  • \$\begingroup\$ Can you elaborate why p[n]≠i[n]*v[n]? it seems like reasonable extension of p(t)=i(t)*v(t) when t=nT with sampling interval T. \$\endgroup\$ – Minh Tran Jan 24 at 14:05
  • \$\begingroup\$ If power spectrum contains high slew rate current pulses or harmonics up to N*f then the sample time interval must be line frequency DS-ADC, otherwise 3x the fN harmonic sampling rate with antialiasing filters. Sampling phase within T may/ can introduce errors converting avg or peak to RMS \$\endgroup\$ – Sunnyskyguy EE75 Jan 24 at 18:55
  • \$\begingroup\$ See for yourself tinyurl.com/ycdz7syz crest factor and PF will introduce errors. How much depends on load. typ >10% whereas mechanical power meters have << 1% error. \$\endgroup\$ – Sunnyskyguy EE75 Jan 24 at 20:42
  • \$\begingroup\$ Hi Sunny. I don't mean to sound rude and not entertain your comments but your demo is more advanced than what I currently understand. It'll take me some time to unpack your response but I really appreciate the exposure! \$\endgroup\$ – Minh Tran Jan 25 at 18:51
  • 1
    \$\begingroup\$ If you are trying to design a power meter to >10~25% accuracy ok use any method you like but if you expect 1% accuracy, your method will not work expect for perhaps oven power and light bulbs. and even then you can have significant errors if done wrong. Look up peak to avg to RMS conversion factors for pulse & sine \$\endgroup\$ – Sunnyskyguy EE75 Jan 25 at 18:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.