hall effect ic for current sensing

[robismyname]

I found an ic that can sense the current of my battery. I want to take my 12V battery and interface it to a microcontroller via the attached hall effect sensor so that my microcontroller can tell me how much current my battery is providing to the load.

**broken link removed**

The ic has pin 1-2 as +IP and 3-4 as -IP. So would I take the positive lead of my battery and connect it to pins 1-2 and 3-4 go to ground?

The data sheet says "the Hall IC converts into a proportional voltage" So does that mean that if my current is 5 amps Vout will be 5 Volts?

Is there possibly a simpler solution to measure current from my battery with a micro?

 

[walkura]

Pin 1&2 go to your battery pin 3&4 go to your load (not directly to ground but the current go's first to the load connected)
This sensor will give (in case of the 5 A type) 185 mV x the amount of ampere's.
0 Ampere means You will have half the powersupply voltage on your output (2,5 Volt out with 5 volt Vcc)
5 Ampere will make this voltage go up to 3,425 Volt.
(or with "negative" amperes it would go the same amount down)

To be honest, i seriously doubt if You could find a more easy method.

 

[guangjin]

Hi, i am using a similar IC ACS172 but i am applying 230V ac to it. and the output is an AC wave. is there a ways to make the AC waveform in to a DC waveform ?? As my ADC can only take in DC value.

 

[Mehrshad74]

Hi, i am using a similar IC ACS172 but i am applying 230V ac to it. and the output is an AC wave. is there a ways to make the AC waveform in to a DC waveform ?? As my ADC can only take in DC value.

Don't worry!
the output of the acs712 is biased to VCC/2 it means that at zero current its output is 2.5v and with +30A its 4.5v and with -30A its 0.5v
So you can read its output directly and you just need to subtract 512 from the ADC result register .
you can apply any function like ABS or RMS on the current in your code.
(assuming 5V supply and 5V ADC ref voltage)