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This question already has an answer here:

I want to monitor a 12V lead-acid battery using an ADS1110 ADC.

This ADC has a reference voltage of 2.048V. Thus, I (think) need to condition my input from, say a 10V-15V level range to 2 volts (0.4 gain) and then add a DC offset to bring it down to 0V-2V for my ADC.

Can anyone recommend a low current (<1mA), low component solution for me?

Many thanks.

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marked as duplicate by Elliot Alderson, Warren Hill, Edgar Brown, Sparky256, Bimpelrekkie Feb 12 at 14:55

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  • \$\begingroup\$ since that part provides a differential reading, why do you need to kill the offset? \$\endgroup\$ – dandavis Feb 11 at 20:57
  • \$\begingroup\$ Thank you Dan, Good point. I guess I'm looking at a non-inverting op-amp with a 10v zener on the -ve input? Any suggestions on how to do that? My analog skills are very limited! \$\endgroup\$ – mprowe Feb 12 at 8:38
  • \$\begingroup\$ Oh... And the op-amp will have a single 3v supply. \$\endgroup\$ – mprowe Feb 12 at 8:57
  • \$\begingroup\$ My analog skills are very limited! Yeah, that's clear. Why do you want to use this ADC? Would it not be easier to use some microcontroller's ADC input? Then the uC can be connected to something else as well and control that as well. You're making life hard on yourself by using that specific ADC. Example: this ADC needs a 5 V supply, where is that coming from? There will be plenty of examples on the internet showing you how to do this using an Arduino for example. With an Arduino you only need 2 resistors to make a voltage divider and you're done. \$\endgroup\$ – Bimpelrekkie Feb 12 at 14:55
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It sounds like you need a voltage divider. Any reason you can't use a pair of properly valued precision resistors? Let the ADC do the work of dealing with the offset so you don't have to calibrate it or buy too many precision resistors to go with your opamps.

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    \$\begingroup\$ If you don't remove the offset before the ADC you will sacrifice the resolution of the measurements...you're only using a small part of the ADC's input range. \$\endgroup\$ – Elliot Alderson Feb 10 at 23:16
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    \$\begingroup\$ @ElliotAlderson - Yes, you lose half your resolution. Now it's only 15 bits instead of 16. For a 15-volt full scale divider, resolution will be 0.45 mV. I find it difficult to believe that this is too coarse for lead-acid charge monitoring. \$\endgroup\$ – WhatRoughBeast Feb 11 at 0:28
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    \$\begingroup\$ @WhatRoughBeast I don't know what resolution the OP needs, that information was not provided, and I didn't suggest that this answer was incorrect. Nor did I down vote. I was only trying to answer Andrew Macrae's question "Any reason you can't..." \$\endgroup\$ – Elliot Alderson Feb 11 at 0:40
  • \$\begingroup\$ Thank you Andrew, As I understand it, yes, I can use a voltage divider to reduce the 0v to 15volt range to 0v to 2volt range. But I am only interested in the top 30% of the input - 10v to 15v. So with a simple divider, I only get working range of 0.6v (30% of 2v) from the 0v to 2v output of the divider pair. \$\endgroup\$ – mprowe Feb 11 at 8:14
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Allright, How about using a 10V voltage reference to set up Battery voltage-10V and then a voltage divider to create a 0-2V ADC input? You'll need to compensate for temperature in a way the voltage divider wouldn't require but it would give you full scale of readings over your target range.

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