0
\$\begingroup\$

I am working on a project in which a microcontroller (specifically an RFduino) is powered by a LiPo battery and the microcontroller needs to monitor the voltage of the LiPo. Five of the RFduino pins are already being used for sensors so I only have two pins available for power management. If possible I’d like to use one pin to detect when the battery is charging and the other to measure the voltage.

Currently this device is using the MCP73871 chip for LiPo charing/power management and a 3V LDO voltage regulator to regulate the voltage powering the RFduino. I can run the power good signal from the MCP73871 to one of the two free RFduino pins - as seen in the diagram - to detect when the battery is charging. I would like to use the remaining free pin on the RFduino to measure the battery voltage.

Unfortunately I can not measure the battery voltage directly on the RFduino because the LiPo goes to 4.2V when fully charged and the maximum voltage the RFduino can handle is 3.6V. I was hoping to be able to uniformly shift the voltage below 3V using diodes or a voltage divider but from a couple google searches it looks like current fluctuations would make that impossible. I am not an electrical engineer and I’m way out of my element at this point. Is there a way I can modify my existing circuit to monitor the LiPo voltage on the remaining free pin? I’m out of ideas and any help is greatly appreciated! Thanks!

\$\endgroup\$
  • 1
    \$\begingroup\$ The schematic isn't showing for me (could be a network issue on my side) but if the extra pins are analog to digital converters, you can use a resistor divider to lower the 4.2V to a level that is acceptable to the RFduino \$\endgroup\$ – scld Jan 29 '16 at 19:55
  • \$\begingroup\$ @Wes The schematic isn't showing for me either. Could you post the schematic as a picture (jpg, png, etc). \$\endgroup\$ – Nick Alexeev Jan 29 '16 at 19:59
  • \$\begingroup\$ @Wes, your schematic failed to link through. if you email it to my username (minus the HAM callsign) at my username.com, I can get it converted/posted for you. \$\endgroup\$ – Robherc KV5ROB Jan 29 '16 at 20:24
  • \$\begingroup\$ @RobhercKV5ROB thanks for offering! Since the question was able to be answered without the schematic I went ahead and took down the bad link. \$\endgroup\$ – Wes Wilson Jan 31 '16 at 19:32
3
\$\begingroup\$

I was hoping to be able to uniformly shift the voltage below 3V using diodes or a voltage divider but from a couple google searches it looks like current fluctuations would make that impossible.

That sounds like BS, do you have a source on that? A simple resistive voltage divider can be used to linearly scale down the voltage:

schematic

simulate this circuit – Schematic created using CircuitLab

This would drop the 4.2V maximum voltage down to 2.89V, well within the range of the input. The divider output has an impedance of 69KΩ, which should be low enough for the analog to digital converter but I could not find the ADC specs anywhere in order to make sure.

\$\endgroup\$
  • \$\begingroup\$ lol, you beat me by 30s 'cuz I'm on a cell & had to pull up a separate diagramming program. \$\endgroup\$ – Robherc KV5ROB Jan 29 '16 at 20:22
  • \$\begingroup\$ The voltage divider worked great! I must have completely misunderstood a couple posts I read on dropping voltage using diodes or voltage dividers. Thanks! \$\endgroup\$ – Wes Wilson Jan 30 '16 at 22:58
0
\$\begingroup\$

This voltage divider circuit would give ~2.8v at the 'duino pin from a fully charged battery, with a ~1ma current through the voltage divider: Voltage Divider Circuit

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.