Electronic – arduino – Measuring an Arduino’s power source voltage using Analog In

If you can't connect the circuits an analog optocoupler like the IL300 may be useful:

The input and output remain separated, yet you have the analog value of the battery's level available to the ADC on the other side. The IL300 has an excellent 0.01% servo linearity.

(Vcc and ground left and right of the optocoupler are obviously different.)

transfer function
Opamp U1 will try to make its inverting input equal to \$V_{IN}\$, that's

\$ I_{P1} = \dfrac{V_{IN}}{R1} \$

It controls \$I_{P1}\$ by varying the LED's current \$I_F\$, but we don't need this value in our calculation. Since the photodiodes are matched \$I_{P1} = I_{P2}\$, and the output of U2 is

\$ V_{OUT} = I_{P2} \times R2 \$

so that

\$ V_{OUT} = \dfrac{R2}{R1} \times V_{IN} \$

So, even when R1 is drawn far away from U2 it plays a role for it. The circuit might not work if you choose R2 ten times larger than R1 and your input voltage is 2V.

edit
The short circuit current for the photodiodes is 70 µA. If Vin is for instance 1 V then R1 must be at least 15 kΩ to allow the opamps to get Vb also to 1 V. A value of 100 kΩ for R1 (and R2) will give you an input range of several volts.

An instrumentation amplifier would be able to offset the differential signal measured across the shunt. An AD8221 could run from smaller supplies (+/-2.5 volts) so this saves power immediately and its current consumption is about 1mA.

Your input signal is going to be in the tens of millivolts so no problem with the AD8221's limited common mode input range (on a split supply) and the AD8221 output swing is to within 1.2 volts of either supply rail so, from a +/-5 volt rail you can get an output that is +/-3.8 volts but you need an "offset" to bias it to a standard ADC input range of (say) 0V to 2.5 volts.

The reference pin on the AD8221 can be fed from a 1.25 volt reference to centre the offset from the AD8221 at 1.25 volts and make it suitable for a whole load of low power single supply ADCs. The ADC input would need to be protected from the AD8221 producing a negative voltage in case the signal from the shunt was bigger than expected but this is acheived by a simple series resistor.

Typically the AD7988-5 (a 16 bit ADC that I've used recently) consumes less than 1mW at a sample speed of around 100kSps so this would fit the bill but TI have very similar offerings as do LT. It has a maximum input current of over 100mA so a simple series resistor of about 1kohm is not going cause a problem BUT check the data sheets.