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I'm trying to measure an AC current on a wire using Rogowski coil based current transducer. the CT I have at home has output voltage of 5v and output current of 1A as specified in the datasheet here. I'm looking for an electronic component that will enable me to decrease the output current such that it won't burn my Arduino on one hand, but won't influence the output voltages since those are the measurement results that I'm interested in.

I read about current transformer but wasn't sure it's the right component. can anyone give me advice about which component should I use?

Thank you

E. Ginzburg
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2 Answers2

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I don't think it has an output CURRENT of 1A...

Its output is a voltage that is somewhat proportional to the current passing through the loop. That's it. No problem of current whatsoever.

Looking at the datasheet you power it with 5V. It consumes around 17.5mA and it outputs a voltage up to 5V.

If you Arduino has inputs that accepts 5V then you are fine and can connect it directly to your Arduino.

Blup1980
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  • Thank you for your response, I'm using Arduino MEGA so I guess there should be no problem to connect it directly. In the spec sheet it was mentioned that the Primary nominal residual current rms is 1A, I guess I misinterpreted what that means. perhaps you could clarify this point to me so that I won't get confused in the future? and in any case, do you have any suggestions for which component can I use to decrease current without changing the voltage, I'd be happy to know in any case. thank you once again. – E. Ginzburg Aug 17 '21 at 14:35
  • You MUST provide voltage clamping on the A/D input of the Arduino. You will easily kill the Arduino otherwise. – Jack Creasey Aug 17 '21 at 14:44
  • @JackCreasey as commented in your answer, the module is no a bare tranformer. It already contains everything to ensure that no voltage spike will occur. But it's a good advice to check for the worst cars before plugging something directly into an IO. – Blup1980 Aug 18 '21 at 06:57
  • @E.Ginzburg Well, it's not easy to explain. It's written in the datasheet that the device will output a voltage. Think of the output signal of the module as a power supply or a battery whose voltage is proportional of the current that is passing through the measurement loop of the device. As for voltage power supplies, you can't have "too much output current". Have a look to that answer it may clarify things : https://electronics.stackexchange.com/questions/34745/choosing-power-supply-how-to-get-the-voltage-and-current-ratings – Blup1980 Aug 18 '21 at 07:02
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The Rogowski coil is a voltage output, BUT you must protect the A/D input of the Arduino from potential overvoltage. You need to add a voltage limited OP amp (or some other active clamping circuit) to the input and connect the sensor to that.

This will work:

schematic

simulate this circuit – Schematic created using CircuitLab

The TLV6001 is rated for rail-rail operation and so allows the full range of MCU or A/D input ...in addition it is rated to carry 10mA in the input protection diodes. This would allow the configuration above to withstand voltages of +/-200 V on the input resistor R2 (providing your resistors are rated for this voltage).

Since the TLV600 is powered by the MCU +5 V supply it cannot produce an output voltage above 5 V or below 0 V, so the input is accurately clamped WITHOUT impacting the A/D range at all. You must make sure that your MCU solution is always drawing a minimum current greater than your expected protection current (this is only an issue if you put things into a sleep state). You can select the input resistance needed by simply changing R3. This should work for both 3.5V and 5V Arduino variants.

Jack Creasey
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