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:
![\[V_{out} = \frac{R_2}{R_1+R_2} \cdot V_{in}\]](http://blog.smartere.dk/wp-content/ql-cache/quicklatex.com-ee1f44dbd85a1cd2218265ee49ce5255_l3.png "Rendered by QuickLaTeX.com")
We want
to give
, so
![\[\frac{5}{30} = \frac{R_2}{R_1+R_2}\]](http://blog.smartere.dk/wp-content/ql-cache/quicklatex.com-f02f447533b604db627794c4ebd6140b_l3.png "Rendered by QuickLaTeX.com")
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
![\[ I = \frac{U}{R}\]](http://blog.smartere.dk/wp-content/ql-cache/quicklatex.com-93017fbe68496cbd35e880ea8ee1c4b5_l3.png "Rendered by QuickLaTeX.com")
which in this cases gives
![\[ I = \frac{30}{12000} = 0.0025 A = 2.5 mA\]](http://blog.smartere.dk/wp-content/ql-cache/quicklatex.com-6631b393a242838c08feb147241b12dd_l3.png "Rendered by QuickLaTeX.com")
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
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.
The closest Zener diode I could find was an 18V of the
BZX79 series. This resulted in the following circuit:
which I hacked into my Arduino box.
Now, theoretically the formula for translating an voltage at the Arduino to the supply voltage should be:
![\[Vcc = V_{in} / (4700/(4700+6800)) + 18 = V_{in} \cdot 2.4468 + 18\]](http://blog.smartere.dk/wp-content/ql-cache/quicklatex.com-2bf301ad44061de9fd4492ef666f33ad_l3.png "Rendered by QuickLaTeX.com")
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 (
) 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
. The current should be around
, which again is ok.
10 sep
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