Measuring high DC supply voltage with an Arduino

Af · august 31, 2015

For my home-monitoring setup I would like an Arduino to measure the supply voltage it is getting from a DC battery UPS (Uninteruptible Power Supply). Unfortunately (actually by design, but that’s another story), the power supply is 24V, which means it will put out anywhere from 21.3V-29.8V (according to the manufacturer), which is far too much to measure with the Arduino’s 0-5V input range. For simplicity’s sake, lets assume we want to measure a 20-30V voltage. The immediate answer is to use a voltage divider, which will bring a voltage in the 0-30V range into the 0-5V range. The general formula for the resistor divider is:

    \[V_{out} = \frac{R_2}{R_1+R_2} \cdot V_{in}\]

We want V_{in} = 30 to give V_{out} = 5, so

    \[\frac{5}{30} = \frac{R_2}{R_1+R_2}\]

resistordivider Now, just as a sanity check we should calculate the current of the resistor divider, to make sure we’re not converting too much electricity into heat. Ohm’s law gives us

    \[ I = \frac{U}{R}\]

which in this cases gives

    \[ I = \frac{30}{12000} = 0.0025 A = 2.5 mA\]

No problems there. This works okay, but we lose a lot of precision, as only ~1/3 of the Arduino’s range is actually used: the Arduino’s ADC has 1024 different readings between 0-5V, so when reading the 0-30V range the precision is just about 30V / 1024 \approx 0.03 V over the range. If only we could move the lower bound, so that 20V would map to 0V on the Arduino. A wild Zener Diode appears! One use of a Zener diode is as a voltage shifter. voltageshifter Zener diode voltage shifter. This work is licensed under the Creative Commons Attribution 3.0 License, https://en.wikipedia.org/wiki/File:VoltageShifter2.png. The closest Zener diode I could find was an 18V of the BZX79 series. This resulted in the following circuit: zener-voltage-divider which I hacked into my Arduino box. Hacked supply monitoring Now, theoretically the formula for translating an voltage at the Arduino to the supply voltage should be:

    \[Vcc = V_{in} / (4700/(4700+6800)) + 18 = V_{in} \cdot 2.4468 + 18\]

I then did some quick measurements of various input voltages and the resulting voltage at the Arduino pin:
Input voltage Arduino pin
18V 0.32V
20V 1.16V
26V 3.60V
28V 4.41V
29V 4.81V
Plot it into a spreadsheet, create a graph and add a linear regression gives: Now, this formula is a bit different compared to the theoretical one, mainly in the Zener diode drop. However, the datasheet for the BZX79 actually has the 18V C-type (\plusminus 5\%) as between 16.8-19.1V, so this is well within spec. Since this is just a one-off, I’m happy to just use the measured formula, as this will be more accurate. The final precision should be 12V / 1024 = 0.012V. The current should be around I = \frac{U}{R} = 30V/11500 Ohm \cdot 1000 \frac{mA}{A} = 2.6mA, which again is ok.

Gemt under: Extern, HAL9k

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