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How do i calculate the discharge time for an lithium-ion battery at a specific load?

Let's say i have a lithium-ion battery with a nominal voltage of 3.7 V, a cut off voltage of 3.0 V and a nominal capacity of 450 mAH. The battery is discharged with a load of let's say 20µA.

Now i want to know how long it theoretically takes until the battery is discharged. Can I use the formula discharge time = capacity(mAh)/load(mA) as for (NiMH-) batteries?

Thanks in advance.

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    \$\begingroup\$ Yes you can.... \$\endgroup\$ – Brian Drummond Sep 30 at 12:11
  • \$\begingroup\$ use Puekert's Law en.wikipedia.org/wiki/Peukert%27s_law#Limitations \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Sep 30 at 12:16
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    \$\begingroup\$ Peukert's law is for lead acid batteries. \$\endgroup\$ – JRE Sep 30 at 12:40
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    \$\begingroup\$ And it doesn't work so well at low currents like 20uA \$\endgroup\$ – Finbarr Sep 30 at 12:42
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    \$\begingroup\$ 20uA is pretty low. If the battery has a protection circuit built in, the protection circuit will also draw a few uA. So don't expect your calculation to be perfect, but it is a good first approximation. I recommend against discharging all the way to 3V at low current. This is a severe deep discharge. If possible, stop at 3.4 V or 3.3V when you are discharging at 20 uA. \$\endgroup\$ – mkeith Oct 1 at 0:44
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The formula you quoted is a good approximation. It ignores a lot of things that will reduce the effective capacity, but it will give you an upper limit on the run time.

Given a steady discharge rate of 20uA, you are getting into an area where you need to consider the self discharge of the battery.

Over the theoretical 90 days your example battery would run with a load of 20uA, it would lose maybe 10% of its charge due to self discharge. That reduces your run time by maybe 9 or 10 days.

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  • \$\begingroup\$ 1. Ignoring self discharge, the formula comes closer to providing a LOWER limit on run time, as rated mAh will be at >> 20 uA and mAh at 20 uA will be higher. 2. Self discharge of 10% in 90 days for most LiIon is 'rather higher' than I'd expect. NimH may be higher. Low discharge nimH much lower. YMMV. \$\endgroup\$ – Russell McMahon Sep 30 at 21:58
  • \$\begingroup\$ Thanks. That helps me to get a rough estimate of how long it takes for the battery to be empty for a given load (in the current use case the mentioned 20µA). \$\endgroup\$ – mcoc7n Oct 1 at 9:19
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Run time = mAh/mA
is a reasonable approximation for LiIon cells at around 1C discharge rate and 20 degrees C. At very low discharge rates if the effects of self discharge are ignored then the effective mAh capacity will be higher. Increaases in capacity with decreasing discharge rate is less pronounced that with head acid chemistry.
With LiIon an increase in capacity in the 10% - 20% range is liable to be experiences compared to rated capacity.

Note that the discharge range is 4.2V to 3V - 3.6V or 3.7V is the mean voltage during discharge.

For lead acid chemistry Peukert's law may be used to estimate battery capacity.
It is generally considered that Peukert's law is far less applicable to LiIon chemistry.

This paper claims that Peukert's law is 'differently applicable' to LiIon with alpha close to unity and a polynomial law said to be more appropriate. Observation indicates that LiIon has a much lower change in capacity with load than LA.

This paper discusses applicability to LiFePO4 cells.
Paper re applicability to LiFePO4

This paper argues the law cannot be used even for LA except at constant current and temperature.

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