May 11 2020

2.5A Hall Effect Current Sensor

Category: Uncategorizeddq @ 9:56 PM

 

Problem Statement:

My application requires a 2A or 3A +/- monitor.  The ACS712-05B is very limited on it range for what I need:

  • 05B = 0.185V / 1A
  • 20A = 0.100V / 1A
  • 30A = 0.066V / 1A

 

The quiescent is 2.5V, so the 05B @ 5A will be:

  • 2.5V+(0.185V *5) = 3.425V Max
  • 2.5V-(0.185V *5) = 1.575V Min

3.15 volts are not used, that is a lot of waist for my project.

 

Solution:

I will amplify the output:

  • For 2A MAX = 0.185V/1A to 1.250V/1A (gain of 6.8x)
    • @ 1A = 3.750V (instead of 2.685V)
    • @ 2A = 5.00V (instead of 2.870V)
  • For 3A MAX = 0.185V/1A to 0.833V/1A (gain of 4.5x)
    • @ 1A = 3.333V (instead of 2.685V)
    • @ 2A = 4.166V (instead of 2.870V)
    • @3A = 4.999V (instead of 3.055V)

 

A map will be used in the Arduino code to display the actual Amp value:

  • 2A = map(0,1023,-2.00,2.00)
  • 3A = map(0,1023,-3.00,3.00)

 

I searched the internet for a common solution, the below is close and I will update the values to expand the range.

As drawn, the gain is  3.3X.  Add a 10K (10 turn) resistor between the 100K to be able to zero out the value (This can also be accomplished in code).

Here are the values I need:

  • 2A
    • R3 =6.8K
  • 3A
    • R3=4.5K

R1 and R2 can be as low as 1K and still consume only 2.50mA (12.5 mW).  (Need to change the 10 turn resistor to 100 ohms).

I will use a LM741 instead of the LM321.

 

 

Conclusion:

Need to build and prove this solution has much better resolution.

 

Reference:

http://www.theorycircuit.com/hall-effect-current-sensor-circuit/

https://www.youtube.com/watch?v=RZmGt3-HVlw

https://www.youtube.com/watch?v=DVp9k3xu9IQ

https://www.engineersgarage.com/arduino/acs712-current-sensor-with-arduino/