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I'm a noob. Please correct my mistakes.

I'm making a circuit where I'd like to protect my Li-ion battery (3.7V, 3500mAh) from discharging too much. The batteries have a protection inbuilt which shuts of at 2.5V. I'd like to stop the discharging at 3V.

The battery connects to the PCB with a JST connector. The TPS78001 is a voltage regulator programmed to output 2.8V. I'd have a switch at the battery to turn the device on and off.

circuit

There's a lot of questions regarding this topic but I couldn't really come to any conclusions.

I feel like I have two options:

  1. The EN pin is a shutdown pin for the regulator. I could drive it below 0.4V to shutdown the regulator or above 1.2V to enable it. Currently it is always on as it's connected directly to the battery. With a voltage divider certain setup (according to Dejvid this won't work) could achieve a shutdown of the regulator if the battery voltage drops below a certain level. This reduces the operating current to 18 nA. But this still drains my batteries if the user forgets to turn off the switch.
  2. Find a way to completely cut the power.

Should I go for option 1 because the 18nA draw is negligible (if the user forgets to turn of the device). Or should I go for options 2? If so how?

link full circuit

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  • \$\begingroup\$ I deleted my answer since I realized you can't really get the EN ON/OFF limits working with just a simple voltage divider? \$\endgroup\$
    – Dejvid_no1
    Commented Dec 2, 2017 at 16:44
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    \$\begingroup\$ Um... if the regulator is 3.3V with .25V dropout, why are you waiting till Vin gets down to 3 and not 3.53V? \$\endgroup\$
    – Trevor_G
    Commented Dec 2, 2017 at 17:28
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    \$\begingroup\$ YEs I think its a non issue. But you have the dropout thing backwards. If the voltage falls below 3.53V the regulator is just a resistor. \$\endgroup\$
    – Trevor_G
    Commented Dec 2, 2017 at 18:59
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    \$\begingroup\$ @Trevor I had an error in my question. I wrongly stated that the output should be 3.3V instead of 2.8V. Thank you for the valuable insight. I'm going to go with your solution of not cutting the power since the queiscent current is so low. \$\endgroup\$
    – cjsd
    Commented Dec 3, 2017 at 0:50
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    \$\begingroup\$ The regulator quiescent current (Iq) is negligible, but it could still be advantageous to disable it when VBATT gets low. Because any time the regulator is enabled, the total battery current will be Iq + Iload. But if you do a good job with power management it may not be necessary. \$\endgroup\$
    – user57037
    Commented Dec 3, 2017 at 1:08

2 Answers 2

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The most cost effective solution, since your processor has an ADC, is to monitor battery voltage and pull enable low when the battery voltage gets to a certain point (determined by code).

schematic

simulate this circuit – Schematic created using CircuitLab

When the regulator is enabled, M3 is ON, which turns on M2. When M2 is on, the processor can sense battery voltage through the ADC_IN. When the voltage gets too low, the processor drives GPIO1 high, which turns on M1, pulling enable low, and turning off the power to everything. In this state, the regulator is off, and the battery load is zero (other than the regulator quiescent current).

To turn the processor back on, user pushes button SW1, forcing enable high, and starting up the processor. If you want SW1 to work like a normal power button (on and off) then you need to add more circuitry. I thought it was complicated enough as is, so I didn't do that. See if you can trace everything through as-is.

Obviously this is untested, but I have done similar things in the past in products that shipped in high volume.

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  • \$\begingroup\$ I like this. I'd need an on and off switch. How much more work would this be? To be honest, this schematic is already hard to understand for me. \$\endgroup\$
    – cjsd
    Commented Dec 3, 2017 at 17:55
  • \$\begingroup\$ I'm gonna implement something like this. Accepted your answer. \$\endgroup\$
    – cjsd
    Commented Dec 3, 2017 at 18:49
  • \$\begingroup\$ why is the capacitor placed there? \$\endgroup\$
    – cjsd
    Commented Dec 5, 2017 at 23:36
  • \$\begingroup\$ To make sure M1 stays on long enough to totally shut down the regulator. \$\endgroup\$
    – user57037
    Commented Dec 6, 2017 at 5:15
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You could use a micro power comparator like the LTC1998 to control the EN pin. That particular IC would draw 2.5 uA which is of course more that the standby current of your regulator. But If you do the figures for the battery life it will last a pretty long time.

There are other comparators with even lower current draw, but the packages might not suit you if you're a hobbyist.

LTC1998

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  • \$\begingroup\$ I couldn't find a "LTC1988" on either digikey or google. It would be nice if you linked to the datasheet next time you're talking about a specific component. \$\endgroup\$
    – Harry Svensson
    Commented Dec 2, 2017 at 23:44
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    \$\begingroup\$ Pardon my bad spelling skills Harry. Updated with the correct name and link. \$\endgroup\$
    – Dejvid_no1
    Commented Dec 3, 2017 at 0:24
  • \$\begingroup\$ While I originally intended to make use of a comparator mkeith's solution seems easier. Especially since the low power comparators I found are of sizes 1x2 mm. I also prefer making use of the Atmega that's already there. Still thank you. \$\endgroup\$
    – cjsd
    Commented Dec 3, 2017 at 17:49

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