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I would like to make a simple Li-Po protection circuit that will prevent a single-cell from going below 3.7V. My circuit needs to have a low current input detector (pref < 5uA.)

My attempt at this was to use a Zener diode backwards in order to test if the battery is above the breakdown voltage.

Schematic below.

enter image description here

This schematic works in the simulator, but I'm not sure if this is actually a viable approach. What improvements can I make to this approach, and what better (preferably simple) approach is better than this?

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  • \$\begingroup\$ Seiko makes a lot of IC's specifically dedicated to the purpose of protecting single cell Lithium Ion battery packs. Start there. \$\endgroup\$
    – user57037
    Commented Jan 11, 2017 at 18:21
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    \$\begingroup\$ Zener is not precise enough and has a temperature coeficient and probably cannot meet your target quiescent current. Also, 3.7 is a pretty high voltage to cut off a Lithium Ion or Lithium polymer cell. The battery will still have a lot of capacity when the cell reaches 3.7V. \$\endgroup\$
    – user57037
    Commented Jan 11, 2017 at 18:23
  • \$\begingroup\$ @mkeith "The battery will still have a lot of capacity when the cell reaches 3.7V. " - depends on the load. At 3.7V resting voltage it is ~90% discharged. At rated current draw the voltage might drop to 3.0V. \$\endgroup\$
    – Bruce Abbott
    Commented Jan 12, 2017 at 0:30
  • \$\begingroup\$ @BruceAbbott, well, I do have practical experience in this area. I would never cut out at 3.7. Even if the load is C/5 or C/10, there will be quite a bit of life left at 3.7V. Even if it is 10%, it will be significant when measured in minutes. And it is also hard to see why it would be necessary or desirable. Any decent LDO can maintain 3.3V output down to 3.5 or so (especially with a light load which we seem to be talking about). \$\endgroup\$
    – user57037
    Commented Jan 12, 2017 at 3:55

2 Answers 2

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What improvements can I make to this approach, and what better (preferably simple) approach is better than this?

Your circuit is basically wrong. You can't use an N channel MOSFET like that because the parasitic diode will always be driven into forward conduction with the positive voltage on the battery. I don't know how your sim managed to convince you it would work but that's another story: -

enter image description here

Get a chip that does this for you.

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  • \$\begingroup\$ can I wire the FET the other way? EG: such that the diode does not pass current through? \$\endgroup\$
    – tuskiomi
    Commented Jan 11, 2017 at 20:11
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    \$\begingroup\$ It would work with a p channel mosfet and a different drive arrangement. An n channel won't work as you suggest because there will be no gate voltage higher than battery voltage to properly turn it on. \$\endgroup\$
    – Andy aka
    Commented Jan 11, 2017 at 20:49
  • \$\begingroup\$ Ah, I see I misinterpreted your statement! thank you! \$\endgroup\$
    – tuskiomi
    Commented Jan 11, 2017 at 20:52
  • \$\begingroup\$ hello, this is an old question, and i'm not intending to make this circuit, but would I need a pull-down between U2 and T1, or is this not an issue due to the zener? Thanks! \$\endgroup\$
    – tuskiomi
    Commented Jun 19, 2017 at 16:31
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    \$\begingroup\$ The circuit doesn't work. I've shown that the mosfet will conduct current thru its body diode so it's a moot point. \$\endgroup\$
    – Andy aka
    Commented Jun 19, 2017 at 16:51
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Apart from using the wrong polarity MOSFET, the basic flaw in your circuit is very soft cutoff because T1 has no voltage gain. However even after fixing those problems it still won't be satisfactory because the Zener will need a lot more than 5uA to regulate properly (and low voltage Zeners also have poor temperature stability).

Here's the simplest circuit I can think of that meets your requirements. It uses a TLV431 shunt regulator which is 'programmed' by R1 and R2 to cut off at 3.7V (when the voltage on its adjust terminal drops below 1.25V).

enter image description here

The TLV431 only draws about 0.15uA bias current, so R1 and R2 can be made large enough to draw less than 5uA. When turned on it drops 1.25V between Cathode and Anode, so the FET must be able to turn on with 2.45V Gate drive (the AO6407 is specified down to 1.8V).

If 'simple' means requiring the fewest parts then you could use a microcontroller such as the PIC10F322. The MCU would be programmed to sleep most of the time (drawing less than 1uA) waking up every few seconds to measure the battery voltage and turn the FET on/off. If your load only draws a few mA then you might be able to power it directly from an MCU I/O pin, and then you don't even need a FET! The MCU could also provide other functions such as auto power off and status LEDs.

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    \$\begingroup\$ Hello, Bruce. Thank you for your answer! I have recently found a certain type of IC called a voltage monitor. They use an average of 5 ua, and only have 3 pins to worry about. I think that i have settled on using one of those, as i only need under voltage protection. \$\endgroup\$
    – tuskiomi
    Commented Jan 12, 2017 at 2:07
  • \$\begingroup\$ It is probably too late to ask this question. Shouldn't we use a N channel MOSFET in the above circuit ? As far as my understanding goes AO6407 will turn ON when there is a zero or negative voltage to it's gate however here the TLV431 exhibits the positive voltage at its cathode when battery voltage is greater than 3.7v and drops to zero only when battery voltage drops below 3.7. I am bit confused, can you please clarify, ? \$\endgroup\$
    – Frank Donald
    Commented Aug 12, 2020 at 7:56
  • \$\begingroup\$ The TLV431 is intended to be used as a shunt regulator (like a Zener diode) so it 'turns on' (pulls down) when voltage on the adjust terminal rises above 1.25V. \$\endgroup\$
    – Bruce Abbott
    Commented Aug 12, 2020 at 20:38

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