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For an upcoming ESP8266 project, I need to use a relatively large (or several smaller) batteries and am hoping to get at least 10Ah. To achieve long life, I will be utilizing one of the ESP8266's available sleep modes. In the deep sleep mode, it utilizes somewhere between 10 and 70 microamps.

I know that most lithium ion batteries are recommended to discontinue discharge at a voltage of around 3.0-3.2V, for the sake of battery health. However, do they also have any kind of limitation such that the battery will not supply current if the load goes below, say, 1mA? I apologize if this is a nonsensical question, but I know very little about batteries, and I want to ensure I can put it into deep sleep while using a battery like this.

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    \$\begingroup\$ Batteries themselves have no cutoff values, managing circuitry around them has. \$\endgroup\$ – PlasmaHH Aug 4 '16 at 17:32
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    \$\begingroup\$ Please edit your question its a little confusing, you can draw a battery to near zero volts if you continue drawing current out of it. Which will kill the battery \$\endgroup\$ – Voltage Spike Aug 4 '16 at 17:58
  • \$\begingroup\$ Lithium, lithium ion (Li+) and lithium polymer (LiPo) batteries all have different characteristics. It looks like you mean lithium ion. \$\endgroup\$ – jbarlow Aug 4 '16 at 21:19
  • \$\begingroup\$ @laptop2d, edited to be more clear \$\endgroup\$ – Josh Aug 4 '16 at 22:20
  • \$\begingroup\$ @jbarlow, edited to be more clear \$\endgroup\$ – Josh Aug 4 '16 at 22:22
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It sounds like the question is whether there is some sort of hysteresis in a lithium (ion) battery such that a load this too resistive will not draw current.

In general the answer is no, there is no minimum supply current needed to stabilize the output of a battery. (Switching power supplies do have a minimum current.) Experimentally, a voltmeter across the battery terminals will have a typical input impedance of 10 MΩ, so it would draw only 0.3 µA of current at 4 V.

In terms of the physics, the minimum current you can have is the resistance of the air between the two external terminals in parallel with any internal leakage. The resistivity of an air gap is ~\$10^{16}\$ Ω·m which will also works to out a resistance of ~\$10^{16}\$ Ω between two terminals on a battery, or about 0.4 fA (femtoamperes) at 4 V.

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  • \$\begingroup\$ Thank you for your answer -- upvoted but not enough reputation to publicly display it. I am curious though--if a switching power supply does have a minimum current, is a regulated supply the best way to go, generally, for a circuit that will have a small current draw? \$\endgroup\$ – Josh Aug 4 '16 at 22:24
  • \$\begingroup\$ Normally you use a linear LDO regulator for small current draws such as a standby power device, and turn on the switching supply when it's needed. You almost certainly want some kind of regulator, unless your MCU and any accessories are designed to work over the entire operating range of the battery (i.e. has its own regulator). \$\endgroup\$ – jbarlow Aug 4 '16 at 23:02
  • \$\begingroup\$ "not enough reputation to publicly display it"?? - I'm pretty sure you can accept answers at reputation 1 if you choose to. \$\endgroup\$ – jbarlow Aug 4 '16 at 23:03
  • \$\begingroup\$ I upvoted, and must have a 15 reputation for that to be displayed publicly, as I said; I mentioned nothing about marking as accepted. Your answer was thorough, but one hour after posting and receiving an answer is not a sufficiently long period of time for me to deem an answer "Accepted," as others may come along and wish to post, but may be deterred in doing so if there is already an accepted answer. I'm marking as Accepted now. \$\endgroup\$ – Josh Aug 5 '16 at 15:55

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