I'm working on a project at the moment to measure power through a current sensor.

I posted a similar question before, and received very helpful advice which I have tried to heed.

I have improved the design (I hope) by incorporating a precision rectifier.

I still need to get at the voltage waveform of the household supply. My thinking is that the phase will be constant at any point in the household circuit (assuming the house is on a single phase) and so I was thinking of getting a transformer from something like a cellphone charger and plugging it into a wall socket, and measuring that in the same way as I am measuring the current. Here is a diagram:

enter image description here

Is this a good way to do it? What factors do I need to keep in mind? Is there a better way?


R4 is not needed - short it out. R5 is wrong - remove it and connect the bottom of XFMR1's secondary to the op-amp's 0V signal rail. Apart from the fact that using ADC1 and ADC2 doesn't give you real power but RMS(volts) x RMS(amps) then it should work. Power is not measured this way in an AC circuit.

The AC phase should be almost guaranteeable anywhere in the house but there will be a small phase shift through XFMR1. However, as you are trying to calculate "power" using RMS steady state values this is of zero consequence.

Here's what a power waveform looks like at various phase angles between V and I: -

enter image description here

With V and I in phase, power is at maximum. In scenario 2, I lags V by 60 degrees and power is half of scenario 1 because cos(60deg) = 0.5 i.e. power factor = 0.5. Scenario 3 is when V and I are 90 degrees to each other - there is still the same amplitude power waveform but average (real kW) power is zero. Scenario 4 shifts the current to lagging at 120 degrees and cos(120) = -0.5 hence average power is negative.

To complete this answer it's probably necessary to think about what happens when the current has third (or higher) order harmonics present: -

enter image description here

The voltage waveform multiplied by the 3rd harmonic (green in top diagram but strangely(?) blue in lower diagram) produces an average power of absolutely zero watts. This is why measuring RMS current may be a very poor indicator of real power. Despite the RMS of current being increased by the 3rd harmonic being present, this has no bearing on power consumed whatsoever - this is why, these days, we are urged to use devices with power factor correction circuits in - not only do they align the current and voltage but they reduce harmonics of currents - what gets sent down the power cables to our houses is therefore greener and makes the power companies much happier because they don't have to "over-rate" the cables they use to carry useless harmonic currents.

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  • \$\begingroup\$ wouldn't I get a half-wave representation of the actual wave out of the rectifiers? I wanted to calculate the phase and amplitude of the current and voltage waveforms so I can multiply them in software \$\endgroup\$ – Matthew Sainsbury Dec 13 '13 at 17:05
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    \$\begingroup\$ @Matt - why go to all the trouble of rectifying the signals - just bias the signals so that the resultant ac waveform becomes centred on your ADC's midpoint and do it this way. This would be the proper way to do it and you get the best of all worlds - you make me happier (LOL) and you use less hardware. \$\endgroup\$ – Andy aka Dec 13 '13 at 17:09
  • \$\begingroup\$ Something as simple as a voltage divider? \$\endgroup\$ – Matthew Sainsbury Dec 13 '13 at 17:33
  • \$\begingroup\$ Yep that's do it - if you put two resistors in series across the transformer output and connect the centre of those two resistors to (say) 2.5V you've biased up either of the transformer outputs suitable for a 5V ref ADC. \$\endgroup\$ – Andy aka Dec 13 '13 at 17:51
  • \$\begingroup\$ Like this ? I think I might have misunderstood your last comment... No point seems to yield biased AC when simulated \$\endgroup\$ – Matthew Sainsbury Dec 13 '13 at 18:17

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