-1
\$\begingroup\$

A bit of background: my girlfriend's smartphone lately has picked up the habit of turning off while the battery is still at about 35%-40%, or so the system says. For anyone around here who is not knee deep in the Android world there's a thing called battery calibration, the system may sometimes report a wrong reading if it is not properly calibrated. There are 'fixes' for this but I am quite sure this whole story is nonsense.

I think the battery is faulty because I once had the device in my hands when it turned off: some strange artifacts appeared on the screen and shortly after it died. I think such artifacts are due to an extremely low voltage, infact I popped the battery out and measured some 3V without load, and that seems quite low to me since that's a regular Lithium-Ion battery.

Before buying a new one, since this is some six months old, I want to test this one to actually prove it's faulty. I was thinking of fully charging it in the smartphone, then connecting it to a load drawing some constant current (a resistor), while monitoring the voltage. Even if this setup is correct or auspicable there still are some questions that come to mind:

  • how much current can I safely continuosly draw from it? I was thinking of 100mA but maybe I can go a little higher to make the whole process faster
  • how often should I check the voltage? should I expect it to drop suddendly when the battery is nearly empty?
  • how low should the discharged voltage be? I know I can search this but a (very) quick search did not answer this so I'm just putting it here because it makes perfect sense.

What I expect is to see a somewhat normal discharge cycle but for a sudden voltage drop when the battery should not be fully depleted.

Bonus: my girlfriend used to fully discharge her phone, as in "damn the battery's dead flat again". This did not happen often enough to bother me (apart unreacheableness) but I'm guessing it played a role in all this.

\$\endgroup\$
  • \$\begingroup\$ Six months should still be within warranty, no? If so dump it back on the supplier. Once you open it you may void the warranty. I saw a very similar thing happen with a friend who had an iPod battery replaced (probably with an off-market junk battery) by a repair outfit. \$\endgroup\$ – Spehro Pefhany Jun 3 '15 at 18:08
  • \$\begingroup\$ I am not sure that warranty covers batteries, the battery can be removed without invalidating the warranty though. \$\endgroup\$ – Vladimir Cravero Jun 3 '15 at 18:56
  • \$\begingroup\$ 3V is absolute minimum and should be avoided if possible. | 6 months of very heavy use may kill a bad quality LiPo battery | max mA is usually same as mAh figure eg 1000 mA max for a 1000 mAh battery. | Vmax should be 4.3V. Vni say 3V. Using an LM317 in a constant current cct gives linear discharge but a resistor is OK. Vstart = 4.2V/R. Vend >= 3V/R = 71% of I start. | LM317 constant R - connect R = 1250/I_mA R from adj to out and take current from adj. eg for 100 mA R = 1250/100 = 12.5R. Or 1.25R for 1000 mA. | Voltage should fall in S-on-side curve with rapid dropoff at near 3V. \$\endgroup\$ – Russell McMahon Jun 14 '15 at 2:46
  • 2
    \$\begingroup\$ Question is VERY on topic. It is about a cellphone battery but DIRECTLY applies to LiPo & LiIon batteries in any thing. \$\endgroup\$ – Russell McMahon Jun 14 '15 at 2:47
1
\$\begingroup\$

3V is absolute minimum and should be avoided if possible.

6 months of very heavy use may kill a bad quality LiPo battery

max mA is usually same as mAh figure = "C".
eg 1000 mA max for a 1000 mAh battery. Lower does no harm. C/2 is almost always safe.

Vmax should be 4.3V.
Vend say >= 3V.

Using an LM317 in a constant current cct gives linear discharge but a resistor is OK.

Resistor:
Vstart = 4.2V/R. Vend >= 3V/R = 71% of I start.

LM317 constant load:

Connect R = 1250/I_mA Ohms from adj to out and
Take current from adj.
eg for 100 mA R = 1250/100 = 12.5R.
Or 1.25R for 1000 mA.

Voltage should fall in S-on-side curve with rapid dropoff at near 3V.

\$\endgroup\$
0
\$\begingroup\$

how much current can I safely continuosly draw from it? I was thinking of 100mA but maybe I can go a little higher to make the whole process faster

I have tested LiIon batteries, and the slower you discharge them, the better for the battery. Also, discharging them faster would give you a slightly lower capacity reading. 100mA is a very good number for a cell-phone battery; you can go faster on higher-end (say >1500mAh) batteries.

how often should I check the voltage? should I expect it to drop suddendly when the battery is nearly empty?

Yes, the voltage will drop suddenly when the battery is nearly empty. There is an inflection point for every battery (typically around 3.5V, may be around 3.6V for cell phone batteries), after which the voltage drops suddenly. Typically, about 10-15% of the capacity is after the cut-off. So if your battery is rated 900mAh, would would want to check it no less often than (900mAh / 100mA discharge rate, *10%) = .9 of an hour. If you see voltage around 3.5V, then check it every 10 minutes or so until it goes down to the discharge voltage.

how low should the discharged voltage be? I know I can search this but a (very) quick search did not answer this so I'm just putting it here because it makes perfect sense.

If you are discharging at a rate of at least 100mA, 3V is a good stopping point. If you are discharging it slower, you would want, say, 3.1V or so. However, be careful not to go lower than 3.0V. As the other poster pointed out, 3V is an absolute minimum you should go with Lithium Ion batteries.

However, in this case I think your cell phone is really the culprit for your battery failure:

infact I popped the battery out and measured some 3V without loa

If your cell phone regularly discharged the battery to this level without shutting off, then it is to be expected that the battery would not last for very long. If you get a new battery, it will fail just as fast with that kind of usage. Definitely try to calibrate the cell phone, and if it fails then it is the cell phone that you'd need to replace under warranty, not just the battery.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.