In recharging a battery-powered headlamp, I exceeded recommended voltage by .3 volt, and that fried the circuitry. Why? The headlamp is a Mifine, and it takes two 18650 batteries, which function at 3.7-4.7 volts, and can be recharged through a 3.4mm port on the battery pack. I finally found a 3.4mm connector coming off a 5-volt ac-dc power supply, so I tried that. Result: tiny curls of smoke coming out of the battery box so I quickly disconnected everything. I assumed that the cells had been destroyed, but checked everything up to the front-end LED. Finding nothing amiss, even with the cells, I disassembled the battery box and found the problem: inside the box, the 3.4mm plug and socket were wired to what I assume is a tiny recharging circuit (see photo) which also connects the battery to the cable delivering power to the front LED. The wires connecting everything to everything else were fried. The circuit board looks OK, but there is now nothing connecting the 18650 cells to anything else. I did some research at that point and discovered that maybe I was better off burning out the wires rather than handing the cells 5 volts rather than 4.7 or less, owing to the nature of 18650 batteries. My question is: was burning out the wiring a (uncredited) safety measure designed into the recharging system? Is it normal for such a minute over-voltage (.3 dc volt) to wipe out a circuit?
At the moment, my plan is, since I was responsible for stupidly using the wrong voltage, to get another Mifine headlamp like the first one, and reverse-engineer the recharging wiring. If it turns out that that is more difficult than I am ready for, I can drop the idea of recharging through the battery box, connect the cells' positive and negative plates directly to the front-end LED switch, and leave recharging to a separate 18650 recharger, which I have since bought.
I have also recharged the original cells, and they hold a charge, so they apparently weren't damaged in the mishap. So does this sound like a good idea, and should I keep using the original cells?

 

18650 batteries, which function at 3.7-4.7 volts

Are you sure about the 4.7V figure? The fully charged voltage I see published is usually 4.2V. So 5V would be seriously too high.

 

It appears that you are right. I checked the specs on the headlamp, and the Working Voltage is 3.7-4.2 VDC. So the over voltage was just shy of one volt. It sounds as if the range of acceptable voltages is proportional to the stated voltage? OK, that clarifies my first question, although, as I mentioned, the batteries seem serviceable, but I guess the circuit wasn't built for that much voltage. Thank you for the clarification.

 

Over-voltage shouldn't cause that level of damage - its current that melts wires and a 5v USB charger would be current limited. Looking at the wiring it would appear that the charger connection is directly onto the battery with no charge controller - presumably thats in the proper charger. However, there is one possibilty and thats if the polarity was reversed.. looking at that wiring there's a strong possibility that was wired as centre negative (the B+ was clearly connected to the barrel outer contact). If that was the case and the other charger was wired conventially as centre positive (as most chargers are these days) then its likely your battery was shorted through the reverse protection diode likely to be in the other charger - that would easily pass enough current to melt those cables. That's why you must always use the correct charger...

 

I tried connecting the correct charger versus the wrong charger I had used that ruined the circuit, and I wish I could tell you the polarity was reversed, but it wasn't (see below). Both are positive center, negative barrel, in spite of that mysterious "B+" on the circuit board. That error would have explained the results, but it doesn't, unfortunately. Which leaves the excessive over-voltage as the culprit. 5 V is about 120% of 4.2V-which doesn't seem like too much to me, but that's why I'm here asking questions. Clearly, I did use the wrong charger, in any case. Thanks for stepping up with a possible explanation. I've ordered some repair parts, and my first step will be to attempt to reconstruct the original wiring between the board and the power socket. I traced it out, and it's pretty straightforward.
EDIT: The results I gave above came from my [inexpert] use of a good voltmeter. Then I took another look at the photo I included (above) and tried re-tracing the connections between the damaged power socket outputs and the identifiable parts of the battery case. Result: the center contact on the socket DOES go to the negative battery pole. So regardless of my earlier conclusion, the socket does require a center-negative-barrel-positive 3.5mm plug. You had it right. The "B+" note does indicate barrel-positive. Thank you.

 

Volts doesn't melt wires, amps does. What is the rated output of the charger you plugged in?

LiPo charging is initially constant current, the CC phase. The voltage is irrelevant as long as its higher than the battery voltage (otherwise there's no 'push' to get the current flowing). The charger limits the current to an acceptable charge rate. A LiPo battery can by charged at several amps if the charger permits (I often charge my big packs at 50A). Once battery volts have climbed back to 4.1/4.2v (about 90% charged) the charger will switch to constant voltage for the final (CV) phase of charging, during which the charge current drops to a few milliamps and the charger switches off. The last phase can take anything from a few minutes to several hours to go from 90% to 100% depending on the size of the battery.

So, unless the charger you plugged in was capable of several amps the higher voltage didn't melt those wires - from your description of the event that needed a current of probably more than 10A to do that damage that quickly. That suggests a short circuit of your 18650 cells through the charging plug...

 

The charger is rated at 3 Amps.

 

So, unlike the usual rule, in battery charging the amperage must be tightly controlled? The battery being charged doesn't draw as much as it needs, and no more?

 

So, unlike the usual rule, in battery charging the amperage must be tightly controlled? The battery being charged doesn't draw as much as it needs, and no more?

There are two possibilities here. The headlamp expects a constant voltage and has the charging circuit built in. Or it expects the charging circuit to be external. A “cell phone charger” is not a battery charger, it is a constant 5V supply, and the cell phone has the charger built in. You cannot assume that is the case with your headlamp, it might require an actual charger.

In answer to your question, though, it is true that a battery will not limit the current to a safe level if you connect it to a voltage source with no current limiting. It will draw enough current to destroy itself, assuming the voltage is higher than the terminal voltage of the battery .

Bob

 

It sounds as if I will be better off buying rather than trying to design a battery charger. No surprises there. I also discovered an unexpected re-design in a newer version of the same headlamp, one that will change how I recharge the batteries: I bought the new one to try to reverse-engineer the old circuit. Luckily, the old circuit is still traceable in its battery box, because the Mifine company has changed the polarity of its recharging circuit. I looked carefully, and the center connector from the socket now goes to the positive pole of the batteries and, of course, the positive side of the whole circuit. The board itself is now smaller and only sports one IC. I can't tell you more because the new board annotations are even smaller and less clear than the annotations on the older board.
So, in order to recharge batteries for the older lamp I will need to recharge the batteries separately from the lamp itself. I hope I can remember to do that. And, if I had made the mistake of trying to recharge my older headlamp with the new recharger, the results would have been quite familiar-smoke and fried wiring.

 

Addendum-I managed to rewire a charger so that the center output is now negative, as in the older design.