0
\$\begingroup\$

I'm currently building a Li-Ion battery charging circuit which has a 5V/2A input. It should implement load-sharing in order to keep the output alive during charging.

For charging, I found the BQ25606 IC from Texas Instruments and it seems to fulfill all of my requirements. However, one aspect confuses me when looking at the datasheet. It states "58-μA Low Battery Leakage Current with System Voltage Standby". The leakage current is only mentioned at the front page and I've had a hard time thinking of what it might refer to.

Does that mean that when the charging cycle is terminated a continuous current flow of 58μA slowly discharges the battery? The Li-Ion battery I want to use discharges less than 5% per month. Do these two discharging processes add up? Or am I totally mistaken and the 58μA refer to something completely different?

\$\endgroup\$
  • 1
    \$\begingroup\$ It'll be the chip's own power drain from the battery, in addition to the battery's own 5%/month. Calculate what it adds to the drain in %/month; it's probably insignificant. \$\endgroup\$ – Brian Drummond Mar 12 '18 at 20:19
5
\$\begingroup\$

Yes and no. That is the quiescent current when there is no power supply. Page 7 of the datasheet says 5uA max when Vbus exists (charge terminated, charger still connected to power). 85uA (58uA typical) is when there is no power to the charger.

58uA drain on a fully charged 2000mAh battery would deplete it to 20% in 27586 hours, or 3 years.... to put this in context.

And yes, it is in addition to the battery self drain.

\$\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.