Cannot get a good reading from ACS711 current sensor


I bought three ACS711 current sensors about a month ago and now I’m starting to test them.

I want to measure the starting current (as well as the nominal one) of a three phase motor. This motor is been powered from a variable source (50 volts AC) so there will not be an issue with the 100 V limitation of the sensor. I supply power to the sensor with a 5.1 volt source (DC) and have the recommended bolts just like in the product picture.

When I turn on the motor, it start’s spinning; when I turn on the sensor it gives me 2.5674 volts at the output. But if I increase the voltage, the current does not increase, or if I decrease it, it does not decrease.

Strange… if the sensor is not working, it shouldn’t output the value for 0 Amps; if the sensor is not working, the motor shouldn’t spin since it is in series for the phase to work.

Will post more info tomorrow (a picture maybe).

PS: the motor has a nominal current of 4 amps.


I am sorry to hear you are having trouble with your ACS711 current sensors. Do you have the -12.5 to +12.5 A version or the -25 to +25 A version?

What are you using to measure the sensor’s output voltage? Do you observe that the output changes by a very small amount or that it does not seem to change at all?

Do you have a multimeter or something else you can use to get an independent measurement of the current actually flowing through your motor when it is running? I am not sure what “nominal current” means, but if it is the current that the motor is expected to draw under its maximum load or at peak efficiency, then it is probably not a useful indicator of the motor’s actual current when it is free-running.

Please go ahead and post a picture of your setup.

- Kevin

hi , im having trouble with ACS711 current sensor .after connecting (the schematic is attached),the circuit is being shorted.Is my connection correct?Do help me out here,thanks!


No, that connection is not right; you are just shorting out your power supply through the sensor (the resistance between IP+ and IP- is only 1.2 mΩ). The way you use the sensor is to put it in series with your current path so that the current flowing through your load also flows through the current sensor from IP+ to IP-. One way to do this is to connect VIN to IP+ and then connect IP- to the rest of your circuit where you currently have VIN connecting. Hopefully, nothing has been damaged by your incorrect setup.

- Ben

Hi,thank you so much.It worked.I have received from OUT and GND around 100mV.How do I use this formula VOUT = (0.11 V/A * i + 1.65 V) * Vcc / 3.3 V to calculate the I?thanks alot :slight_smile:

You need to rearrange that formula to solve for i.

- Ben

Thanks Ben.Im wondering what is the meaning of the formula (0.11 V/A ).What is the V and the A?

Those are units. V is volts and A is amps.

- Ben

Hi Ben,

Thanks again it works fine now. I use a free running motor which is rated at 250mA .Putting 0.25A as i into the equation

VOUT=(0.11 x 0.25 + 1.65 ) (5.5 / 3.3)

I get about 2.5V on the multimeter which is fine. However there is a problem in the equation when making i the subject :

i = ( (VOUT x 3.3 / Vcc ) -1.65) / 0.11
= ( (2.5 x 3.3 / 5.5) -1.65) / 0.11
= - 1.363 A

Am I doing this right? Appreciate your help.

Yeah, it looks like you have the equation right. Notice that if you plug in 2.8V for VOUT, you get that i = 0.27 A.

- Ben

Thanks Ben for everything.Works well.Thanks again.