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Topic: Monitor high voltage solution (Read 276 times) previous topic - next topic

Domnguyen

Dear all.
I have a simple task as below. I want to acquire the monitor voltage from an equipment using Arduino and then, record the data. The point is, the monitor voltage is up to tens of volts (or in the range of -20 to +20 V). I am looking for a converter or "de-amplifcation" module which can reduce the input voltage at a ratio of input:output = 10:1.
Any can advise me or share your own experience?
Thanks.

wvmarle

Resistor voltage divider - 22k & 220k would be a good ratio. Or 10k and 100k. Brings it down to just under 4V total, which allows for a bit of overvoltage.
If you can connect Arduino GND to the -20V line you can just connect the mid point of the divider to your analog in.
For more accurate measurements use 2k2 & 100k to bring down the voltage to <1V and use the internal 1.1V ADC reference to measure.
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

Domnguyen

Great suggestion. Thanks for help.

Koepel

If the signal can be 20 V above and 20 V below GND, then I would use a three-resistor voltage divider.
Two resistors to put the analog input in the middle, and with one resistor to the high voltage.
It requires some calculation: https://electronics.stackexchange.com/questions/266786/arduino-negative-voltage-reading-with-positive-ground.

With the 5V as reference, these values can be used:
Ri = 47k or 56k
Rp = 12k
Rg = 18k
I tried to calculate it, but I got lost :smiley-confuse: Is there an online calculator for this ?

If your 5V is reliable, then you can use the 5V as reference. If the Arduino is powered with a USB cable, then the 5V is not reliable.
The 1.1V internal reference is better to measure voltages.

outsider

#4
Mar 01, 2018, 04:59 pm Last Edit: Mar 01, 2018, 04:59 pm by outsider
Something like this?

wvmarle

Check the schematic (and calculations!) in the accepted answer of the stackexchange question linked above.
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

outsider

Yes, I can see why my suggestion would not work correctly. Why can't ANYTHING be easy?  :P 

wvmarle

It's quite simple, you just try to make it too complex :-)
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

Koepel

#8
Mar 01, 2018, 07:01 pm Last Edit: Mar 01, 2018, 07:38 pm by Koepel
All that is needed now is the math to calculate Vo. I have seen the formula and it was a simple formula, but I can't find it right now.



Perhaps this?
Vo = (Vi*Rp*Rg + Vp*Ri*Rg + Vg*Ri*Rp) / (Ri*Rp + Ri*Rg + Rp*Rg)
The 47k for Ri was too low, I had to change it to 56k.
With: Ri = 56k, Rp = 12k, Rg = 18k
Vi = -20V, Vo = 0.38V
Vi = +20V, Vo = 4.94V

wvmarle

Formulas were also given in that Stackexchange answer.

Better increase Ri a bit more, that 4.94V is really close to the 5V maximum. Resistor tolerances may push it over 5V, as can an increase in input voltage to just over +20V. Changing the values of Rp and Rg would allow to to centre the mid-point a bit better.

Also you do indeed need a very reliable 5V supply, which means you have to use a regulator (either the on-board one, or an external one - not the USB power supply).
Quality of answers is related to the quality of questions. Good questions will get good answers. Useless answers are a sign of a poor question.

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