Battery protection circuit
I am using two 18650 cells in series to power a RC car. Is the voltage goes below 6.2 V, the battery should be disconnected to prevent over discharging the battery. At the same time I want to implement reverse battery protection into the circuit. Since n-channel MOSFETs will be used for the H-bridge, therefore I want to use one for the reverse battery protection as well:
What kind of IC can be used to drive the MOSFET, so that when the voltage drops below 6 V the battery disconnects from the load? Is it possible to use a low voltage detector, for example R311x or TPS3840 or other, with a simple voltage divider? Or is there a better solution to this problem?
EDIT:
Would this circuit work? EDIT 2.0: The load and battery cells are supposed to be switched in this schematics.
Another question, what will be the difference if instead of 10 kOhm and 20 kOhm, 10-times bigger values would be used?
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What kind of IC can be used to drive the MOSFET, so that when the voltage drops below 6.2 V the battery disconnects from the load?
That reverse protection MOSFET works fine for reverse protection but, it cannot be used to turn-off the load when the battery is correctly connected. Look closely at your diagram; can you see the little diode symbol associated with the MOSFET? That diode symbol tells you that there is an internal diode in every MOSFET that will conduct even if the MOSFET is supposedly deactivated.
In other words it won't prevent over-discharge of the battery.
Is it possible to use a low voltage detector, for example R311x or TPS3840 or other, with a simple voltage divider? Or is there a better solution to this problem?
You'll need a separate MOSFET to your reverse protection MOSFET and, it will need to be in series with the load (like the reverse protection MOSFET) but, with drain and source positions swapped. You can then use a low-power comparator to deactivate the added MOSFET should your battery voltage drop below a prescribed threshold. There are some very low power comparators that barely take a microamp that can be used here.
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This is the circuit I have decided to use:
The cut-off voltage will be just above 6 V. MOSFETs used are SiRA14BDP and the voltage supervisor is TS831. Maximum current consumption of TS831 is just 12 uA.
For the voltage on one cell to hit dangerous level it needs to drop from 3 V to 2.5 V. From discharge rate characteristics of any 18650 cell we can get its capacity in voltage range 3 - 2.5 V. It will be anywhere from 0.1 Ah to 0.5 Ah. If we assume the worst case, the approximation for how long it will take for the cell to hit dangerous level is:
$$Time = \frac{Capacity}{I} = \frac{0.1 ~ Ah}{1\cdot 10^{-6} ~ A+12\cdot 10^{-6}~ A+\frac{6 ~ V}{270000~ \Omega}} ~ \approx ~ 2839~ hours$$
Where: $$I ~= I_{DDS} + I_{CC} + I_{voltage~divider}$$ Idds is zero gate voltage drain current. Icc is current consumption of the voltage supervisor.
Using this approximation we get that it would take over 100 days for the cells to hit 2.5 V. Using better 18650 cells that have 0.5 Ah capacity left at 3 V the time would be extended to almost 600 days.
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