0
\$\begingroup\$

I am trying to design a protection circuit for a single cell lithium battery, and I was hoping for some advise on what things I need to consider.

So far, my thoughts are to provide low voltage cutoff using a comparator. A fixed voltage would be provided to the comparator using a diode, and a variable voltage from a potential divider between the battery terminals. When the voltage from the potential divider goes below the voltage accross the diode, then the comparator would turn off a MOSFET that allows power to the load. So, while the battery is above ~3V, the MOSFET stays on and powers the load, otherwise turns off and then the only current draw on the battery is from this circuit, which should be very low, and hopefully take a long time to deplete the battery to a problematic level.

I plan on having a battery charging module connected directly to the battery terminals. Is it possible that while not charging, this would allow the battery to discharge?

Can anyone tell me if the design I described is a suitable way to implement battery protection, and if there is a better way. Are there any ICs that would be recommended specifically for this purpose, and what are their advantages?

Thanks in advance.

\$\endgroup\$
1
\$\begingroup\$

Most battery protection circuits use a chip (eg. Seiko) with a pair of external MOSFETs for charge and discharge voltage protection. Also overcurrent protection with PTCs.

Specific product recommendations are off-topic, but you can look at Seiko's line-up and compare.

Advantages are that most of the design work has been done for you, the parts are guaranteed with reasonable voltage tolerances, appropriately low operating current (uA) and they are inexpensive in large volume. Parts count and size is also reasonable enough that they can be included in the cell housing.

\$\endgroup\$
  • \$\begingroup\$ Thanks Spehro. I'm now looking at the DW01 protection IC, that seems to be regularly paired with the 8205A. ic-fortune.com/upload/Download/FS8205A-DS-12_EN.pdf I am a little unsure about something regarding the dual MOSFET chip. It seems that when the battery is either charging or discharging, the body diode of one of the MOSFETs is being used to conduct. And my understanding is that for the 8025A, the maximum current the body diode can allow is 830mA. Does this mean that's the highest current I can supply to my load using this setup? \$\endgroup\$ – Steve Apr 11 '16 at 17:02
  • \$\begingroup\$ When 'on', the body diode is in parallel with the conducting channel, which has much lower voltage drop in non-fault conditions, so it does not have to conduct, except perhaps momentarily. \$\endgroup\$ – Spehro Pefhany Apr 11 '16 at 18:14
  • \$\begingroup\$ @Steve I have seen this in the datasheets and they are just trying to make it clear which is the drain and which is the source, since the two FETs are in opposing directions. A casual glance at the typical application circuit, most users might miss this without those body diodes being drawn. Not used for current conduction. \$\endgroup\$ – Joel Wigton Jun 10 '16 at 14:07

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.