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This must be a basic question, which I am struggling to find and understand suitable circuits to be able to read AC current using a micro-controller.

Most of the other examples are calculating DC current, while related questions such as this one, attempt to measure AC current on mains voltage which is not what I am after.

So far I have managed to calculate and measure the RMS voltage from a +-1.5 V sine wave. I am now looking to measure the current by simply using a shunt resistor in series with a larger load resistor (10 K ohm), and an op-amp, without any specifically designed IC's. Vref for my ADC is 3.3 V.

DO the circuits that measure DC current, such as the one below, still apply for AC voltage? I cannot seem to understand this basic concept.

enter image description here

EDIT: Can my AC source replace the above DC supply and still get correct readings, or do I need to make changes to the circuit?

Also if the circuit above was applicable for an AC voltage source, I am thinking need to get the RMS reading at Vs. Is this correct?

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A dc power supply feeding a conventional looking load (as shown) will produce a supply current that may have cyclic changes but all the current will be in one direction. Hence, Vs will always be positive even if it does reflect the cyclic changes in a cyclic load current taken through the unknown load.

If you want to discard the DC content of the current you can do this inside the micro-controller (digitally) by performing an average measurement and subtracting this from the instantaneous values that you need to take to calculate RMS.

Also remember that you need to sample at high enough of a rate to avoid aliasing problems.

However, if RL is just a 10 kohm resistor the DC average current is fundamentally the RMS current (ignoring resistor noise that should be tiny). The op-amp you should use should be able to work properly with its inputs at the power supply negative rail.

If the supply voltage is AC then you need to bias Vs so that it never goes below 0 volts. This can be done several ways.

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  • \$\begingroup\$ What I'm after is if my AC source can replace the above DC supply and still get correct readings. \$\endgroup\$ – Rrz0 Jan 14 '18 at 17:25
  • \$\begingroup\$ I've added to the answer. \$\endgroup\$ – Andy aka Jan 14 '18 at 17:34
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    \$\begingroup\$ Ah okay, that explains it. I have already used a potential divider to bias my small AC signal, I guess I could use this again here... \$\endgroup\$ – Rrz0 Jan 14 '18 at 17:35

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