1
\$\begingroup\$

This question already has an answer here:

I am working on a 3V battery powered arduino project. I need to monitor the battery so that I could send a signal when battery is dying. At first it seemed tricky because the same voltage is being used by the micro-controller but later on I came to know that there is some sort of internal 1.1V reference in Arduino. However I need to use a voltage divider to reduce the voltage. This is the circuit described on the webpage:

enter image description here

PIN_A1 is where the battery voltage is being read. High value resistors have been used to avoid excessive current draw.

My question is - What is the aim of using a voltage divider here?

EDIT:

It has been figured out that voltage divider is bringing the battery voltage within internal voltage reference range i.e. 1.1 V.

(Credit goes to Vladimir)

\$\endgroup\$

marked as duplicate by Ignacio Vazquez-Abrams, Ricardo, Daniel Grillo, Adam Haun, Community May 19 '15 at 22:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    \$\begingroup\$ 3V*470kohm/1470kohm=.959V < 1.1V that's why \$\endgroup\$ – Vladimir Cravero May 19 '15 at 8:23
  • 4
    \$\begingroup\$ Not answering your question, but... there's a way of measuring Vcc without using any external circuit at all \$\endgroup\$ – tomnexus May 19 '15 at 9:05
  • \$\begingroup\$ @Vladimir - Now I see the blunder I made. :( \$\endgroup\$ – Whiskeyjack May 19 '15 at 9:39
  • 1
    \$\begingroup\$ Do read the link above - you can just configure the ADC so it uses Vcc as reference, and measures the 1.1 V internal reference as the signal. This achieves the same thing but without any resistors, leakage, etc. \$\endgroup\$ – tomnexus May 19 '15 at 9:48
  • 2
    \$\begingroup\$ As a side note, don't use ground but instead tie that to an output. Drive it low, take the measurement, then drive it high again. And I'd use much smaller value resistors, to actually place a load on the battery for the split second it is measured. \$\endgroup\$ – rdtsc May 19 '15 at 16:44