How to measure negative voltage on ADC? - Page 1
Forum section: electronics / projects

author: hal9001
author_rank: Regular Contributor
timestamp: October 28, 2021, 11:33:34 am
content: I have a charge pump that outputs -10 V. Theres no other negative power supply on the board. How can I monitor the -10 V supply with an ADC?I can send the -10 V through an inverting opamp with voltage divider but the negative supply of the op amp will have to be powered by the same -10 V charge pump and put extra load on it. Is there a better way?Cheers!

author: sicco
author_rank: Regular Contributor
timestamp: October 28, 2021, 11:41:44 am
content: Check out the latest ‘over the top’ from Analog Devices.https://eu.mouser.com/datasheet/2/609/ada4099_1-2238218.pdfpage 24.

author: voltsandjolts
author_rank: Unknown
timestamp: October 28, 2021, 11:47:09 am
content: The op-amp can still be powered as normal from the positive rail, it doesn't need to be connected to the negative rail.Bias the op-amp input appropriately for the value of resistor used to connect to the -10V.It's similar to using a resistor divider to measure, say, +48V rail using an op-amp running from 3v3 rail, with some extra biasing from a volt reference or such.Or use an ADC that has a negative input range like the ADS8665 (which uses this input biasing technique).

author: David Hess
author_rank: Super Contributor
timestamp: October 28, 2021, 12:07:10 pm
content: Quote from: hal9001 on October 28, 2021, 11:33:34 amThe common mode input voltage of an inverting operational amplifier is the same as the voltage at the non-inverting input, which could be ground, so there is no need for a negative supply. Inverting amplifiers are sometimes used to do exactly what you need without a negative supply.

author: mikerj
author_rank: Super Contributor
timestamp: October 28, 2021, 01:07:48 pm
content: You could simply use a potential divider with the top connected to your reference voltage, and bottom connected to your -10v rail and the ADC in the centre. Scale the resistor values appropriately so that 0v on the ADC represents your minimum negative voltage (i.e. full scale). You then just need to invert and scale the value in your software.Watch out for the divider current pulling the -10v rail up past 0v if the rail could ever be in a high impedance state.

author: RoGeorge
author_rank: Super Contributor
timestamp: October 28, 2021, 01:25:54 pm
content: Quote from: hal9001 on October 28, 2021, 11:33:34 amYes, there is. Since you only want to measure that, you do not need an opamp, a resistor divider will suffice. So all you need is two series resistors, with one end tied to the -10V you want to monitor, and the other end tied to a positive (stabilised) voltage you may have. Calculate the two resistor in such a way that the resulting voltage divided (for the ADC) is always positive.For example, if you have a +5V reference, the two resistors could be R+= 1 k\$\Omega\$ and R-= 5 k\$\Omega\$. The resulting voltage for the ADC will be +2.5V, and the exact measured value will depend of the -10V you want to monitor.VADC= V-+ (V+- V-) * R-/(R++ R-)From that, V-can be calculated by measuring VADCand knowing R+and R-and V+.The attached files are for QucsStudio 4.2.2.

author: voltsandjolts
author_rank: Unknown
timestamp: October 28, 2021, 01:31:47 pm
content: Random example fromherejust because a picture is helpful.

author: Caliaxy
author_rank: Frequent Contributor
timestamp: October 28, 2021, 01:57:16 pm
content: Or just use another charge pump to invert the voltage again, since you already know how to do it and might have a few spare chipsSome chips have more than one charge pump inside, so chances are you might have an unused one (if you’re lucky)

author: hal9001
author_rank: Regular Contributor
timestamp: October 30, 2021, 11:11:03 am
content: Thanks all!Quote from: Caliaxy on October 28, 2021, 01:57:16 pmA voltage divider without op amp is very simple but the failure mode you pointed to is something to think over.Quote from: mikerj on October 28, 2021, 01:07:48 pmQuote from: David Hess on October 28, 2021, 12:07:10 pmCheers. Does this require a positive reference as some mentioned? Or is it as simple as making Ri=Rf*3 to divide the input -10 V to -3.3 V for an op amp with +5 V positive rail to work with?

author: magic
author_rank: Super Contributor
timestamp: October 30, 2021, 06:09:43 pm
content: Solution can be as simple as grounding IN+ and taking Ri=3·Rf to turn -10V into+3.3V, yes. Use a single supply opamp like LM358, give it 5V power and it will be happy.The solution that requires a precise positive voltage is to put two resistors from, say, +5V to the -10V rail in, say, 1:5 proportions, so that you get +2.5V when the rail is exactly -10V. No opamp necessary.

author: Zero999
author_rank: Super Contributor
timestamp: October 30, 2021, 06:55:23 pm
content: A passive divider will do for most applications. If the input goes too negative, the MCU's internal ESD protection diodes will conduct and clamp the voltage. Just use high enough values to ensure the current isn't too high, or add a Schottky diode between the input and 0V, if you're really paranoid.

author: TimFox
author_rank: Super Contributor
timestamp: October 30, 2021, 08:28:46 pm
content: If you use the simple inverting amplifier (with single positive supply rail) in Reply #8 above, using an op amp that can tolerate the inputs at the negative rail, the output will be positive with a negative input, and will therefore be within the usable range of the op amp, so long as the input voltage into the first resistor is reasonably negative. "Rail-to-rail" opamps can usually tolerate inputs at the negative rail, but cannot take their output literally to the negative rail (although pretty close).Proportion the two resistors so that the expected range of negative voltage gets mapped into a positive voltage within the op amp limits that is suitable for your ADC.