Working on a portable boombox powered by four 18650 Panasonic NCR 3400 mAh cells in series. I bought a PCB and I have a small 4s balance charger. The PCB will be secondary if anything fails and cuts voltage at 3 V.

That's to low for me during daily use. I want a primary cut off at 3.5 V (14v in series). I was thinking maybe this can be done with a Zener diode and a relay?

Don't know if it is a good idea? Or is there any other simple way of doing this. I'm not very good at electronics but understand some basics and can solder.

  • \$\begingroup\$ See my added example of a self disabling low voltage cutout at end of my answer. \$\endgroup\$ – Russell McMahon May 4 '16 at 17:08

3V is the generally accepted end of discharge point for LiIon.
3.5V is definitely too high - you have a very significant amount of charge remaining at 3.5V.
To be conservative 3.1V or 3.2V is safe.
Note that Vmin depends somewhat on load.
For high to very high loads eg C/1 up somewhat lower voltages are acceptable due to internal IR drop giving lower than true cell voltage at the terminals.

A zener is a VERY rough voltage reference - OK if V is non critical or well away from limits. A reference diode such as a TL431 is far safer. A TL431 can operate a small relay or drive a transistor or op amp when it switches and cam be programmed from 2.5V on up.



  • With careful design a Zetex TLV431 needs well under 1 uA when in the fully off state.

  • Check data sheet for other brands - specs may vary.

  • Note that TLV431 often has an 18V Vcathode max whereas TL431 is often 36V. No problem here. Again - this spec can vary by brand.

For lower power a TLV431 (1.24xV ) draws substantially less than a TL431.
Reference current is under 1 uA across temperature.
I'll give stats for the Zetex part - do check as Zetex usually manage to be on the good side of average.
I_cathode_off is 0.001 uA typical and 0.1 uA max.
Device comes into guaranteed regulation at Icathode 50 uA typical 300 uA max over 0-70C
BUT the "saving grace" is that for currents below guaranteed Ireg_min Vcathode_ground is less than Vreg. ie if the device is fully off Icathode is under 0.1 uA and probably closer to 0.001 uA and I ref is under 1 uA and probably closer to 0.15 uA.
When the device is "just" turned on the cathode current rises until it is fully on at 50-300 uA range. Above that Icathode rises as required to maintain requlation and Vcathode = Vreg.

All the above means that if you use the TLV431 to monitor battery state, once vbat drops slight;ly below trip point the TLV431 draws under 1 uA all up. (Bias resistor draw is up to you).
At 1 uA you get 1000 hours per mAh =~ 40 days and in a year it takes 8.8 mAh or (probably) less to much less.

Example - Low voltage cutout with self disable:

This is a portion of a "real world" circuit.

Here is an example of a TLV431 being used to implement both low voltage battery cutoff and self disable - so no current is drawn by the voltage monitor once the low voltage level is reached. Relevant circuit portions have been marked in darker/thicker black. Thin lines are irrelevant to this example.

Q2 supplies battery to the target circuit.
Q1 enables reference VR1 (= TLV431 = SPX432)
VR1 when operated turns on Q2
Q1 is operated by line /GO being pulled low.

/GO low turns Q1 on.
Q1 on applies battery tp R7 + R16 divider.
If R7 R16 junction is >= 1.25V VR1 turns on.
If battery voltage is too low VR1 will not turn on and no further action occurs.

If VR1 is on R8 applies drive to Q2 base, Q2 on supplies battery to main circuit.

/GO is initially taken low by a user pushbutton action and then held low by a processor powered via Q2.(OR via thin line from top of R7.)

When battery falls below trigger level of VR1 driven by battery-Q1-R7-R16, VR1 turns off, Q2 off, power removed from processor, Y'All sleep.

enter image description here

  • 1
    \$\begingroup\$ Mostly agree except The TL431 will consume a fair amount of power. I would be hesitant to design it into a circuit like this (battery operated). Also, I have designed a number of products that connected a 3.3V LDO to the lithium battery. If you run until the LDO drops out (3.5 or 3.4) you actually do get most of the energy out of the cell. Especially if you are discharging at C/8 or C/10 (8 or 10 hour battery life). \$\endgroup\$ – mkeith May 1 '16 at 3:47
  • \$\begingroup\$ Thanks for quick feedback! Maybe aiming dor 3.3v is better? Class d amp is 40w but difficult to say how much it will discharge. Im lookih at this circuit at the moment. homemade-circuits.com/2013/10/… \$\endgroup\$ – Robert May 1 '16 at 7:04
  • 1
    \$\begingroup\$ Well, if it really is 40 Watts, that is more like C/1, and in that case you probably want to go fairly low, exactly as Russel says. Also, if you use 200 uA to keep the TL431 biased, who cares? You are going to be drawing several Amps from the battery. So, again, Russell's answer is looking pretty good. It would be nice if you could design the circuit so it "latches" off when it cuts out, and de-powers the TL431 at that time. \$\endgroup\$ – mkeith May 1 '16 at 22:55
  • 2
    \$\begingroup\$ @Robert see my TLV431 addition to my answer. \$\endgroup\$ – Russell McMahon May 2 '16 at 7:05
  • 1
    \$\begingroup\$ @mkeith See added example \$\endgroup\$ – Russell McMahon May 4 '16 at 17:09

you can use a low voltage detector with a voltage divider (bringing the 14V down to the detector low voltage) driving a mosfet.


┌┴┐                             │
│R│                             │  
└┬┘     ┌───────────┐        ┌──┴───┐
 ├──────┤ detector  ├────────┤mosfet│
┌┴┐     └─────┬─────┘        └──┬───┘
│R│           │                 │
└┬┘           │               ┌─┴─────┐
 │            │               │load   │
 │            │               └─┬─────┘
 ┴            ┴                 ┴

may be this will work for you, try it.this is low voltage battery cut off circuit


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