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I have very low background in electronics.

I have a Li-Po battery connected to an electric motor. The question is:

  • Why does battery voltage drop under load?

  • Why is each cell of a Li-Po battery limited to 3.0 V (discharging)?

  • Why is each cell of a Li-Po battery limited to 4.2 V (charging)?

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closed as too broad by Eugene Sh., Wesley Lee, uint128_t, ThreePhaseEel, Voltage Spike Feb 22 '17 at 5:17

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    \$\begingroup\$ 1 - Internal resistance. 2 & 3 - Chemistry. When you overdraw the battery you reduce its lifetime. When you overcharge the battery you can risk explosions depending on high you go. \$\endgroup\$ – Wesley Lee Feb 21 '17 at 22:16
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Batteries can be practically modeled as a voltage source and its internal resistance, i.e. a resistor. In an ideal world it wouldn't have any resistance, but practically it does. As more current is drawn from a battery, increased voltage is dropped across this internal resistance, lowering the voltage available from the battery.

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  1. There's no such thing as a perfect voltage source, everything (that doesn't superconduct) has resistance and resistance means you lose a bit of your no-load voltage when current starts to flow.

  2. The lower cut-off voltage is not an exact figure, there's a bit of a trade off between the lower cut-off voltage and the battery lifespan, but below 2.5V there are some irreversible chemical reactions that occur that wreck the battery permanently. Lead-acid car batteries have similar issues (but for different reasons). I don't fully understand what reactions take place when you fully flatten a Li-Ion battery although I think there are a few kinds of lithium intermetallic compounds formed that once formed, don't then easily un-form. Much like how with Lead-Acid batteries you get sulphation on the plates (not all chemical reactions are easily reversible, it's very hard to un-rust a lump of iron).

  3. As for the maximum charging voltage, you run the risk of forming metallic Lithium "whiskers" (among other unpleasant things) which can puncture the electrode separator and lead to a short circuit. Seeing as Li-Ion short circuits tend to involve copious amounts of current, lots of heat is usually generated too, and heat + Li-Ion electrolyte = boiling and fire, and once air gets in, the Lithium catches fire too. 4.2V just happens to be the voltage required to fully charge a Li-Ion battery and any more voltage will start to cause unwanted side reactions. There are a variety of electro-chemical reactions that don't really get started unless there's a certain voltage gradient present. There's a similar limit in Lead acid batteries, apply too much voltage and you start to electrolyse the water into Hydrogen and Oxygen gas which (seeing as it'll be in the perfect 2:1 ratio) is highly explosive and will eventually result in all the battery's electrolyte drying out and more of those unwanted one-way chemical reactions.

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  • \$\begingroup\$ Thank you all so much for the response. I really appreciate it : ) \$\endgroup\$ – Karn Feb 22 '17 at 7:24
  • \$\begingroup\$ So when you run a electric motor with Li-Po battery, you have to monitor both current capacity and voltage, right? Also, the electric motor should stop running when either the battery capacity ran out or battery voltage in each cell is getting close to 3V? \$\endgroup\$ – Karn Feb 22 '17 at 7:30
  • \$\begingroup\$ @Karn The two quantities are interlinked, the voltage will drop as you use up the battery's stored energy. While you can get a more accurate measurement of the battery's state of charge by monitoring both the voltage and the used charge (load current x time), for most applications, it's not necessary to keep track of the stored energy to that level of precision so the 3V per cell cut-off is usually good enough. In fact, there are charts where you can work out (roughly) the remaining Li-Ion battery capacity (in %) from it's open circuit voltage, just search for "Li-ion voltage capacity chart" \$\endgroup\$ – Sam Feb 22 '17 at 10:21

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