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I'm making a battery voltage meter with a Wemos D1 Mini. The Analog input has a Vref of 3.2v. The battery that i will measure is a Lithio one of 72v, minimal 68V to max 84V, the Analog Input of the Wemos give to me (3.2v / 1024) of precision and i need more precision because with a resistor divisor i have a range from 0-90v but i will not use 0-60v, so my question is if with a OpAmp (i have a LM324) and some resistors i can map 60v-90v to 0v-30v and then use resistor divisor to down to 0-3.2v and get more precision with the ADC. The resolution that i want get is aprox 3mV-5mV

Update

I was used the document "Designing Gain and Offset in Thirty Seconds" with this values: Excel with values

And used proteus to simulate the circuit of positive m and negative b

enter image description here

Like you see Vref' not is same (even closed) to calc and Vout is outside 3.1v range of calc... what I'm doing wrong here?

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The voltage conversion could be implemented like this:

enter image description here

To obtain:

enter image description here

Vref can be generated using precision adjustable reference (e. g. tl431) and R2 can be a trimpot. If you have specific requirements regarding the opamp power rails, reaching 0V could require some changes.

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  • \$\begingroup\$ Thanks for the response... \$\endgroup\$
    – Luis Felipe Dominguez Vega
    Commented May 17, 2021 at 22:14
  • \$\begingroup\$ the only problem is high current drawn from the battery. I have same problem and I do not to draw more than 70nA from the battery. Any idea? \$\endgroup\$
    – 0___________
    Commented May 18, 2021 at 0:45
  • \$\begingroup\$ @0___________ how much mA draw from battery?? \$\endgroup\$
    – Luis Felipe Dominguez Vega
    Commented May 18, 2021 at 4:16
  • \$\begingroup\$ @vangelo Why 13.6v for Vref?? can be 12v?? \$\endgroup\$
    – Luis Felipe Dominguez Vega
    Commented May 18, 2021 at 4:17
  • \$\begingroup\$ @LuisFelipeDominguezVega To change the original range to the (initially) desired one we have: (84 - 68) / 3.2 = 5. So the minimum voltage will be 68 / 5 = 13.6, which needs to be subtracted. The maximum current required from the battery is: 84/(47000+5222) = 1,609 mA. \$\endgroup\$
    – devnull
    Commented May 18, 2021 at 14:37

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