I recently got an inexpensive (~$4 USD) Li+ 18650-powered flashlight from banggood.com, one of my favorite sources for inexpensive Chinese electronics to tear down and hack for fun 🙂
I wanted to validate its safety circuit before considering putting it into active service. I'm not liking what I found, but am wondering if I'm being too conservative. I'm having a hard time shaking visions of a Li+ battery charging on my counter bursting into flames suddenly 🙂
The flashlight takes a single 18650 non-protected Li+ cell. The flashlight won't turn on (draw any more than 10uA) when the voltage is below 2.37V. However, once on, it continues to draw current down to about 1.95V. Also, until it reaches that shut-down threshold, the turn-on "hysteresis" is defeated; that's basically how I've defined shut-down.
Here's my test setup. Only the LED holder and switch assembly are shown; everything else is mechanical (and ground conductor):
Here's a table of the various voltages (on the way down, lower brightness setting):
V+ I (mA) ---- ------ 4.2 65.5 3.8 61.4 3.4 45.2 3.0 22.2 2.7 6.3 2.6 2.5 2.5 165uA 2.4 83uA 2.3 60uA 2.2 46uA 2.1 36uA 2.0 28uA 1.95 0.2uA
And on the way up:
V+ I (mA) ---- ------ 1.5 0.0uA 1.7 0.1uA 1.9 0.1uA 2.0 0.3uA 2.1 0.6uA 2.2 1.1uA 2.3 1.9uA 2.4 70.3uA 2.5 165uA 2.6 2.5mA
After this on the way up the V-I characteristics match those on the way down.
So I do note that the current consumption falls precipitously at about 2.5V on the way down (2500 -> 165 uA), and that nothing much happens at all until 2.4V on the way up. What I'm wondering is whether this is enough to constitute adequate protection.
The scenario I'm imagining is that someone places the flashlight vertically, lens down, on a desk, and then forgets it's turned on since the light isn't visible. If left for a couple weeks like this, I'm thinking the battery would discharge to levels that would make it prone to violent misbehavior on re-charge.
What I rather expected to see was the current drop a little sooner, say 2.7V, since the discharge rate wouldn't be great enough to allow much recovery. Then not take more than protection IC current up to about 3.0V on the way up.
Am I being over-conservative?