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AMC1100: Question about measuring the voltage of a battery.

Gabriel Cassio Moreira Freitas
Prodigy 85 points
Part Number: AMC1100
Other Parts Discussed in Thread: AMC1301, AMC1200

Hi everyone.

In the last days, I've been searching a component (the IC AMC1100) to use it to isolate the measuring the voltage of a battery. The project is to use the ic as an isolation from the voltage divider to measure the battery voltage to the inputs of the microcontroller (Arduino UNO).

And so, my doubts: as datasheet explains, it has a +- 250mV of input, so, for my application, will be just from the 0 to 250mV? Other part is, that the IC has a fixed gain of 8 (and about 300mV the output has a saturation), so for my inputs in Arduino, do i get the VOUTP to an ADC, and the VOUTN to the GND, or something different?

Any help, suggestion will be very good. Thank you all.

In the last days, I've been searching some components for a project 

  • 0
    TI__Guru**** 181521 points
    Hi Gabriel!

    You have the basics of the AMC1100 understood correctly. The input is bi-polar at +/-250 mV, and the gain is fixed at 8V/V. What is the voltage on the battery you want to sense? You could supply the input of the AMC1100 with a small common mode voltage so that you get a wider input dynamic range. Other parts in this family are the AMC1200 and the AMC1301.

    The outputs of the AMC1100, AMC1200 and AMC1301 are differential, so you would not really want to tie VOUTN to GND, but you could add an amplifier stage to take the differential output and present it to the internal ADC of your micro controller as a single ended signal. This app note explains that process: www.ti.com/.../sbaa229.pdf
  • 0 in reply to Tom Hendrick
    Prodigy 85 points
    Hi Tom, thank you so much about your answer. It was amazing to me. The voltage of the battery is 55V at the maximum. By now, I figured out that it's necessary a voltage divider with a value of 220, to an input of 250mV (max.).

    About the common mode voltage I understood that with a smaller one (Vdd2 = 3.3V) I would get a great output voltage to the ADC of the Arduino. And with this app note, I understood how to take the differential signal to an sigle one. By now, I'm thinking how to take the final signal to the Arduino with the best way.

    Thank you so much about the answer.