I wanted to convert an old 1990s boombox to be self-powered using two 18650 cells. It was designed to be powered by 6 C-cells, but no one ever did that because it didn't run very long, and it was a terrible waste throwing out all those batteries. Six primary cells provides 9 volts, and two 18650s provide between 6.2 and 8.4 volts. A quick test with my bench supply revealed that there was no change in level or distortion from 9 volts all the way down to 6.0 volts. At that voltage, the box simply shut down, although it did continue to consume enough power that I couldn't count on that to save the batteries from over-discharging.

So two 18650s will work perfectly.

Since I'm using free 18650s from old laptop battery packs, and since these are not protected, I wanted a circuit that would shut off the power as soon as the batteries got below 3.1 volts per cell. The circuit also had to consume virtually no power (less than 100 uA) once the circuit was shut down, and I wanted it to add less than 10% to the load when the circuit was active. Finally, it had to have a few tenths of a volt hysteresis so once shut down, it would stay shut down.

The circuit below fulfills all these design objectives. I used a high hFE transistor biased with really large resistors. This minimizes the current used when the circuit is active. The 220 ohm emitter resistor provides feedback, like what is done in amplifier design, and this provides the hysteresis. I added the capacitor and resistor across the relay contacts after I found that the relay got sticky when the amplifier was turned on with the volume up. The capacitor eliminates any arcing. Finally, the Schottky diode provides battery reversal protection. I wasn't planning on using a Schottky because it isn't in series with the load, but as I was doing measurements during the breadboarding, I found that the zener I used still conducted a little bit for several tenths of a volt below its threshold. By using the Schottky with its lower forward voltage, I changed the point on the curve at which the circuit shut down, and it only required a few tenths of a volt below the shutdown before the leakage current got down to almost 10 uA, a very acceptable standby drain.

Since I know that a lot of people design using two 18650s in series, I thought this circuit might be useful.

 

Welcome to AAC!
Providing the two cells have identical characteristics that circuit should prevent their over-discharge. However, particularly with salvaged cells, one cell is likely to have (or develop) less capacity than the other so would go below its recommended minimum discharge voltage before the other one did. Ideally you want individual protection for each cell.
What is the DC current break rating of the reed relay?

 

Providing the two cells have identical characteristics that circuit should prevent their over-discharge. However, particularly with salvaged cells, one cell is likely to have (or develop) less capacity than the other so would go below its recommended minimum discharge voltage before the other one did. Ideally you want individual protection for each cell.

That is an excellent thought. While the whole point of this over-discharge protection is designed to avoid ruining the cells by discharging them too far, I agree that it would be a shame to still have one get ruined. However, since they are free cells that I salvaged from laptop battery packs that were thrown out, I am playing with "house money" and if a few die an early death it doesn't matter.

What is the DC current break rating of the reed relay?

Since these are ancient parts from when I worked at Hewlett-Packard in the early 1970s and they are stamped with the HP part number, it took awhile to get the specs. Fortunately, one of the dozen that I still have had an Elec-Trol part number and I was able to get the specs. The coil is 500 ohms (measures at 520) and the contacts are rated to 500 mA. The max draw of the boom box at full volume is 325 mA, so I have a pretty good margin. However, when I first used the circuit attached to the boom box, the relay stuck in the closed position. I figured it had to be inrush current, so I added the capacitor/resistor combination across the relay switch and that seemed to fix that problem.

BTW, the one thing you didn't mention, and which I did not test is what happens to my arc supression capacitor/resistor combination if I do indeed put the batteries in backwards. That 10 uF cap is obviously a polarized electrolytic. I think it will survive a short reverse voltage event, but I don't know how much current leakage might happen and whether it would screw up my diode protection scheme. I guess I could wire the load to go through the Schottky diode and that would solve that problem, at the cost of 0.3 volts less voltage delivered to the load.

 

I guess I could wire the load to go through the Schottky diode and that would solve that problem, at the cost of 0.3 volts less voltage delivered to the load.

You can also use a P-MOSFET (below) to block reverse voltage with no significant forward drop (only from it's on-resistance).
Since a MOSFET conducts equally well in both directions, M1 will conduct in the reverse direction when the battery polarity is positive, but blocks a negative voltage.

 

You can also use a P-MOSFET (below) to block reverse voltage with no significant forward drop (only from it's on-resistance).

That was my original design, but since retail electronic stores have totally disappeared, even from Silicon Vallley, and since I didn't want to place a $20 order for a $2 part at Digi-Key or Mouser, I looked at my parts bin and cooked this up.

While I don't have any P- or N- MOSFETS in my parts bin, if you want some 7400 TTL NAND gates, I'm loaded.

 

I didn't want to place a $20 order for a $2 part at Digi-Key or Mouser,

You could buy some (possibly questionable) cheap MOSFETs from the likes of Amazon or ebay.