Skip to main content
Arduino

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

3
  • Thank you for the quick reply! The Arduino will never receive the full 12v because i'm powering the sensor separately from the Arduino. The Arduino is only connected to the sensor via the analog pin and ground. The most voltage it would ever get into its analog pin would be 8v if there was a catastrophic failure with the sensor and it sent the full supply. That may still be enough to kill the pin, but probably not the whole Arduino, as far as I know. So you're recommending that I test the system using 10K for R1 and 24K for R2? Just want to clarify. Commented May 8, 2017 at 14:31
  • 1
    @Andres: If you know for sure that the input voltage will never be more than 8 V, then you could use 10 kΩ between the sensor and the Arduino and 15 kΩ between the Arduino and ground. The Arduino will then read no more than 4.8 V. Commented May 8, 2017 at 15:24
  • 1
    Honeywell has also Current Sensors for 5V (the CSNX series). If you are only reading DC currents, then you can add a little capacitor parallel to the 15k resistor of about 10nF (1nF to 100nF) to reduce some noise. Do you have a modern 3.3V Arduino Due or do you really have a very old Duemilanove ? For the Duemilanove it is allowed to pull or push 1mA into a Arduino pin. That means with 10k and 15k the Arduino is protected for voltages from -10V to 19V. A voltage divider is accurate and works very well, anything else (for example OpAmps) would only introduce more inaccuracy. Commented May 9, 2017 at 6:36