I noticed something today. I am powering the Arduino Nano from a battery and a 5v step up converter. That works just fine and I'm getting a stable 5v output. I have a second cable coming out from the battery which plugs right into analog input A0 set as input so I can measure the battery voltage. However when I disconnect the battery from the usb plug and leave the voltage sensing cable connected to the analog input the power led stays on. I want it to shut down and be disconnected from the power source and have the voltage sensing cable still connected. How can I fix this?

2 Answers 2


Power is flowing from the analog pin, through the internal clamping diodes, to the Vcc line. Those diodes are however not designed to handle that much current, so don't continue doing this.

Either also disconnect the ground wire from the battery, or simply add a resistor between the battery and the analog pin. Something like 10k. That way less than a mA of current will be clamped, and the arduino will stay off.

  • But how high of a voltage drop will I have then? I can't have any if I'm measuring battery voltage which needs to be precise
    – qwertz
    Commented Aug 23, 2015 at 11:54
  • Negligible. The analog pins are high impedance inputs. The ADC is optimized for analog signals with an output impedance of approximately 10k or less.
    – Gerben
    Commented Aug 23, 2015 at 11:58
  • Thank you. Another question: Do you know any transistors/FETs with a very low voltage drop which I need for switching off a sensor?
    – qwertz
    Commented Aug 23, 2015 at 11:59
  • @qwertz That is a completely different topic. Don't ask it in a comment. Instead that question is best asked in electronics.stackexchange.com since it has nothing to do with Arduino.
    – Majenko
    Commented Aug 23, 2015 at 12:02
  • Ok so I'll create a new topic for that, thanks
    – qwertz
    Commented Aug 23, 2015 at 12:02

This addresses a subset of your question which was covered in your added comment-question. I've addressed use of a 20V full scale input range to make the answer more general. The results would be "even better" for your 5V input case. Note that is Vin ~= 5V you would want a measurenet range somewhat above 5V max so that you can handle typical voltage variations.

Do you know any transistors/FETs with a very low voltage drop which I need for switching off a sensor?

If you used a say 10k resistor in the upper leg of a voltage divider and added a MOSFET switch with an on resistance (Rdson) of 2 Ohms (a terrible spec) between Vin and the divider then the error introduced would be about 2/10k = 1:5000. If measuring a say 12V battery and using a Vinmax voltage of 20V the error in voltage measurement would be in the order of 20V/5000 = 4 mV. If using an Arduino Nano with a 10 but ADC the resolution (not accuracy) is 1 in 2^10 = 1:1024 or 20V/1024 = about 19 mV for a 20V input range. ie the resistor introduces an error of about 1/5th of a bit of available measuring accuracy. Worse (or better) - if you use 1% resistors they will swamp errors caused by ADC resolution. Even use of say 0.1% resistors or calibration is liable to produce a result where ADC resolution is not the main limiting factor.

Even with no resistive divider, getting accuracies that are doinated by the 1:1024 ADC resolution and not other factors takes care and effort. At 5V in ax ADC resolution of 1:1024 means that a variation of 1 part in 2048 could cause an erroneous reading. Input offset voltages, thermal effects, board and connector resistances and ground voltage dropsm and offsets can easily provide greater error magnitufes.

Test: Take a working design. Set meter to lowest voltage range such that 1 mV can easily be seen. (Ideally 200 mV range. 2V range marginally OK). Measure ground voltages between various "ground" points at various locations in the system. Differences ofm mV to 10's of mV can often be found. Differences are poptentially higher if power current are present - say 10's of mA or more.

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