I am planning to buy these current transformers which I am wanting to use to measure electricity consumption of a device where household power is 220V, 50Hz. I want to use these in combination with a microcontroller's analog-to-digital converter. I think I will need to rectify and smooth the output of these so the signal is acceptable for an ADC. Here is the circuit I've come up with (current transformer is signified as an AC current source:


simulate this circuit – Schematic created using CircuitLab

I have annoted this to try to convey what I was trying to accomplish. I need to be able to sense current between zero and 18A.

Is this design sane? What should I be thinking/worrying about?

  • \$\begingroup\$ To extend what Nick T has said, because the current transformer has a 1:1000 turns ratio, I expect that, at 18A (which is my expected maximum input) I'll get 0,018A on the other side of the transformer. This will result in a voltage of 1.8V dropped over the resistor \$\endgroup\$ – Matthew Sainsbury Nov 19 '13 at 17:44
  • \$\begingroup\$ Such a circuit would have a fairly high amount of error in it. You would be better off amplifying the current signal with an op-amp or the like. I also might not even bother rectifying the signal; it's a plodding 50 Hz; if you sample even at a leisurely 1 kHz, you shouldn't have much trouble finding the amplitude. It would increase part count and sources for error when you could do it in software. \$\endgroup\$ – Nick T Nov 19 '13 at 17:48
  • \$\begingroup\$ What sort of error will I experience? I was planning on just tying VREF to around 2V to get a wide spread of digital values \$\endgroup\$ – Matthew Sainsbury Nov 19 '13 at 17:51
  • \$\begingroup\$ your diode voltage will vary wildly when you're passing a small amount of current through it. Better to rely components that are as linear as you can manage for measurement (e.g. ceramic rather than electrolytic caps). \$\endgroup\$ – Nick T Nov 19 '13 at 18:26
  • \$\begingroup\$ Right, of course. I should have seen that. Thank you \$\endgroup\$ – Matthew Sainsbury Nov 19 '13 at 18:51

Is this design sane? What should I be thinking/worrying about?

There are two basic things wrong with this proposal: -

  1. The rectified C.T. output will not produce a linear representation of current.
  2. Electricity consumption (aka Power) cannot be measured with a dc signal that represents current.

With a 100 ohm burden and 10 A in the primary, 1 V RMS is produced across the burden - this is a peak voltage of 1.414 volts and the forward volt drop of the diode (if it is schottky i.e. lowest volt drop type) will reduce it to about 1 volt. That's the basic problem because if you only had 3 A flowing, the peak voltage before the diode would be 0.42 volts and you would barely see a few milli volts from after the rectifier.

The second problem is that power, in an AC circuit is NOT simply volts x amps (even if you were measuring actual RMS amps, and it was linear). To measure power effectively you have to multiply the voltage waveform by the current waveform in real time and then average the result. You cannot know the phase angle between voltage waveform and current waveform and even if you did, you cannot use a dc signal representative of RMS current because you will be getting an error in the power calculation due to harmonics in the current.

NB - if the supplied voltage is a sinewave (and it pretty much is in most households), the power is the RMS of fundamental-frequency of the current x RMS of voltage x Power factor.

You cannot determine the RMS of the fundamental frequency of the current if you are rectifying the CT's output then smoothing it.

If you want to do this properly take your CT output and multiply the waveform by a waveform representative of the voltage (sampled I might add simultaneously). You'll probably need to sample at about 1kHz to get it accurate.

Why 1kHz sampling - well more is better but if you have significant harmonics in your current wavefrom, say up to ninth, you'll need at least two samples at 450Hz and that takes you to running at 1kHz.

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  • \$\begingroup\$ Thank you for your detailed answer. I only really have experience with DC so this answer (and the project) really has me feeling out of my depth. Are you saying that, because I can't know the phase difference between the current and voltage, it is not possible to measure power use with just a current transformer? Do you just guess a power factor? How do you measure power? \$\endgroup\$ – Matthew Sainsbury Nov 19 '13 at 18:29
  • \$\begingroup\$ You can measure power with a CT but not if you rectify the output. You have to simultaneously sample the voltage waveform and the current waveform, do instantaneous multiplies and average the result to get true (bill-able) power. \$\endgroup\$ – Andy aka Nov 19 '13 at 18:33
  • \$\begingroup\$ My understanding is that a current transformer acts like a variable current source. Doesn't this mean that the voltage out of the CT is just what is needed to keep the current at whatever the CT is designed to provide? What are we measuring the voltage waveform of? \$\endgroup\$ – Matthew Sainsbury Nov 19 '13 at 18:37
  • \$\begingroup\$ Voltage waveform aka AC voltage waveform. Current waveform aka the signal from your CT. power is average of v.i where v is your mains ac voltage waveform an i is something that accurately represents you current waveform. See this electronics.stackexchange.com/questions/82188/… for some advice on power measurement too. \$\endgroup\$ – Andy aka Nov 19 '13 at 18:43
  • \$\begingroup\$ But how do I get at (measure?) the AC voltage waveform? \$\endgroup\$ – Matthew Sainsbury Nov 19 '13 at 18:46

It might be sane but not very accurate. Since 100 Ohm load translates as 1V per 10A, the diode basically introduces a threshold of 6 to 7A before any current is detected.

Thus a precision Op Amp rectifier is preferred with a smaller cap. If you wish to average readings in software, the accuracy will improve. The RC charge and discharge times could be specified to define what spurious current spikes if you want to ignore, then chose values accordingly.

It won't read true RMS with peak currents, and no power factor measured, but then simple and accurate are tradeoffs.

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May I recommend a transimpedance amplifier

enter image description here

Take for instance you are planning on connecting to a 5V ADC: By setting the feedback resistor (in this case 1k8) to such a level that for the peak secondary current you receive 2V, the 2v5 offset then allows you to measure AC currents.

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  • \$\begingroup\$ This circuit will destroy the electronics if used without a "burden resistor" across the CT secondary. Isec = Iprim/turns_ratio and Murphy and mother-nature will adjust the voltage on the secondary until the needed current flows. \$\endgroup\$ – Russell McMahon Nov 9 '14 at 20:58
  • \$\begingroup\$ Think of utilizing an IC for the above circuit, such as TSM103W, it is excellent choice with built-in reference voltage source 2.5V The circuit above certainly requires input protection (second stage). I would put a transorb, varistor or bipolar zener across OPAMP input. \$\endgroup\$ – user111547 May 30 '16 at 3:46

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