Safe to charge a Lithium Battery with a Voltage Limit?

  • Thread starter NTL2009
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All the references I find refer to safely charging lithium cells by a method like this:

https://www.powerstream.com/li.htm

The lithium ion battery is easy to charge. Charging safely is a more difficult. The basic algorithm is to charge at constant current (0.2 C to 0.7 C depending on manufacturer) until the battery reaches 4.2 Vpc (volts per cell), and hold the voltage at 4.2 volts until the charge current has dropped to 10% of the initial charge rate. The termination condition is the drop in charge current to 10%.

The next page shows the effects on capacity of charging to less than the 4.2 V terminal cell voltage. For example, charging to 4.0 V still provides 73% of the capacity achieved at 4.2V, with a 4x improvement in life cycle. That might be a good trade-off if getting that last 27% of capacity isn't a priority.

https://www.powerstream.com/lithium-ion-charge-voltage.htm

I think the above means that it would be safe to design a very simple charger, just a 4.0 V source, with enough source resistance to limit current to ~ .2 C at the lowest voltage ( ~ 2.8 V) . No specific current changes at the top, since you would never reach the 4.2 V terminal voltage. With a 4.0 V ~ 4.1 V source, current would drop to zero before hitting that terminal 4.2 V.

I'm thinking about circuits to provide a few hours of back up to a low draw (~40 mA?) 5 V device, just to protect against short term power outs, or unplugging the device to move it. The charge time could be long, these would be rare events.So for a 150mA cell, a divider off a 5 V source, with a source R of ( (5 - 2.8) / (.2 * .15) ) ~ 73 Ohms. So 5 V feeding a 75/300 Ohm divider would do it. Ahhh, that 300 Ohm would draw ~ 13 mA when power is out, reducing battery up time by ~ 8%. If that's an issue, feed a BJT emitter follower with a voltage divider @ 4.7 V to get 4.0 out, and the b-e junction will isolate the battery when power is out. The emitter follower would be lower impedance, so you can use higher values in the divider, and put the 75 Ohm between the emitter and the battery.

Maybe increase the low voltage C rate closer to 1, it will be dropping as the battery charges, and approaching zero as you reach the terminal 4.0 V level.

These small, cheap 150 mAH have high/low voltage cut-offs (BMS) built in ( < $2 each at aliexpress << link to search). But I kinda hate to rely on their high voltage cutoff for something that will be on 24*7*365, plus, it goes open circuit for high voltage cut-off, so you might need to regulate your source anyhow. I've got a half dozen of these little batteries for my micro-quad copters, and have tested the cut-offs. For me the advantage is simple, common parts, versus buying a specific charger circuit plus BMS, waiting for shipping, etc. And cheaper, but that's minor for one-offs

Thoughts?
 

Answers and Replies

  • #2
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I'm thinking about circuits to provide a few hours of back up to a low draw (~40 mA?) 5 V device, just to protect against short term power outs, or unplugging the device to move it. The charge time could be long, these would be rare events.So for a 150mA cell, a divider off a 5 V source, with a source R of ( (5 - 2.8) / (.2 * .15) ) ~ 73 Ohms. So 5 V feeding a 75/300 Ohm divider would do it. Ahhh, that 300 Ohm would draw ~ 13 mA when power is out, reducing battery up time by ~ 8%. If that's an issue, feed a BJT emitter follower with a voltage divider @ 4.7 V to get 4.0 out, and the b-e junction will isolate the battery when power is out. The emitter follower would be lower impedance, so you can use higher values in the divider, and put the 75 Ohm between the emitter and the battery.

Maybe increase the low voltage C rate closer to 1, it will be dropping as the battery charges, and approaching zero as you reach the terminal 4.0 V level.

These small, cheap 150 mAH have high/low voltage cut-offs (BMS) built in ( < $2 each at aliexpress << link to search). But I kinda hate to rely on their high voltage cutoff for something that will be on 24*7*365, plus, it goes open circuit for high voltage cut-off, so you might need to regulate your source anyhow. I've got a half dozen of these little batteries for my micro-quad copters, and have tested the cut-offs. For me the advantage is simple, common parts, versus buying a specific charger circuit plus BMS, waiting for shipping, etc. And cheaper, but that's minor for one-offs

Thoughts?
Divider configuration works, but i do not recommend it. High heat production (increase cycle waste heat by ~70%), and prone to under-performance in case of substandard cable and/or connectors. Also, built-in even modest self-discharge is very inconvenient in operation.
The emitter follower is not good for charger at all, because it may suffer a thermal runaway. The hotter the charger, the larger voltage it will supply.

Simple yet relatively safe & good performance solution for compact charger is parallel voltage regulator. A zener diode for 4.0V. And a ballast resistor from power supply to prevent overheating of zener diode.
 
  • #3
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Thanks for the reply.

Yes, dividers waste power, but we are talking about ~ 13 mA at 5V, ~ 65 mW, and this is powered by the line, not the battery, so I don't see it as a practical concern. And if I use an emitter follower, the values will be at least 10x higher.

I'm not familiar with thermal runaway in an emitter follower, as it has a gain of less than 1.

There is this effect:

The base-emitter voltage required for a given collector current will decrease. This decrease is about -2.5 mV/°C.

At these low currents, there won't be much heating of the junction. But even a 20 C rise would be a voltage increase of .0025 * 20 = .050 V. So a 4.0 V supply might rise to 4.05, seems like plenty of margin from 4.2V. And the current in the transistor decreases as it reaches 4.0 V, just hard to imagine much heating at all.
 

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