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Current measurement with arduino Answered


I try to measure current with hall-sensor amploc 25A, it sense magnetic-fields, offset voltage is 2.5V and I use arduino uno.
I get 519 value from analog pin and 37mV/A is hall sensor sensitivity.

My formula is outputValue = ((((sensorValue*5000.0)/1023.0)-2500)/37

multimeter shows 0.112amps when I put load but I dont gett correct value on LCD

Is my formula OK?




4 years ago

OK correct me if i'm reading the datasheet wrong, but at peak rated current of 25A the sensor outputs about 0.9V. So the sensor outputs about .036V per A. The arduino registers about 0.0049V per unit. So 1A will give you an analog read of about 7. The value of 519 you got was most likely noise from static electricity in the air. In short, this sensor isn't usable with the arduino on it's own. You'll need to put an amp between the sensor and the arduino and factor that into your calculation.

In answer to your question. No the calculation isn't correct.


Reply 4 years ago

Best option may be to find a different sensor. Unless you have a need to measure up to 25A. You shouldn't need to measure more then 5A to 10A in any hobby application. The lower the max rating of the sensor is the more accurate your reading will be using an arduino.


4 years ago

Sorry but I can not follow you here....

Current is measured as a voltage drop on a shunt resistor.

A hall sensor is used for magnetic field measurements.

We can try guesswork here but if you still need help it would be good to inculde some infos.