I am planning to use this in my project. The AVR microcontroller is asleep most of the time (less than 500 µA), and the current draw raises every minute up to 100 mA, but it lasts 500 ms maximum. So I think connecting the charger directly is not a problem.

In the original LDO circuit there are two Schottky diodes connected. I was told that the first one is to prevent reverse-voltage and the second to protect the LDO when Vout > Vin. I was also told that the first diode's voltage drop will prevent me from using the full capacity of the battery...

So I guess the best way is to remove the first diode. But is this circuit going to work at all?

The AVR battery voltage measurment is a voltage divider controlled by a MOSFET. The voltage divider ratio is 1:2, so the input voltage is always lower than the AVR's VCC.

Enter image description here

  • \$\begingroup\$ What is the battery capacity and normal charge current? \$\endgroup\$ – mkeith Oct 12 '15 at 0:31
  • \$\begingroup\$ Battery is 320 mAh and charge current (in my case) set to about 200 mA. The battery is being charged when power source (5V) connected. This is only happening when recharge is needed. \$\endgroup\$ – Herr Krum Oct 12 '15 at 0:40
  • \$\begingroup\$ 100 mA for a half second is likely significant enough to confuse a LiPo charger which is configured to charge a merely 320 mAh cell. \$\endgroup\$ – Chris Stratton Oct 12 '15 at 5:38

I think you are right. I think this setup will work OK (without the series diode). I am not totally sure what the effect will be of the occasional <500 ms 100 mA burst. It shouldn't prevent the charge from terminating, but it could cause a voltage dip which may initiate a new charge cycle. I think it would be worthwhile to test and see what happens if possible before you commit.

I would like to point out that even without the diode, this setup will not give you the best battery life. Once the battery voltage drops to 3.5 or 3.4, your LDO will not be able to supply 3.3V any more. If it is possible to power your circuitry at 3.1 or 3.0V, that would allow you to use more of the available battery power. Good luck!

  • \$\begingroup\$ I was just thinking - my AVR and other things connected to LDO CAN run up to 5 V so is it possible to remove LDO completely? The battery operating voltage is 3-4.2 V so when at e.g. 3.2 V I should sleep the AVR until recharged. This is better than with LDO but not sure if the unstable voltage source would affect something. I think i need power source switcher to cut off the battery when being charged and power AVR from the external source. \$\endgroup\$ – Herr Krum Oct 12 '15 at 10:27
  • \$\begingroup\$ Some battery charger IC's have a separate connection for system load and battery. They automatically manage all the switching. If you google "power path charger" I think you will find some examples. But it is also possible to manage it yourself with a FET switch to cut off battery from load when 5V is present. The 5V can just go through a diode to the system load. \$\endgroup\$ – mkeith Oct 12 '15 at 17:40
  • \$\begingroup\$ I already found suitable schematic. But I don't know what values the diodes are: electronics.stackexchange.com/questions/194889/… \$\endgroup\$ – Herr Krum Oct 12 '15 at 18:09

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.