I am powering an embedded system using one 3V Lithium battery.

Two similar devices were acting very differently: when they both were on, the voltage read across the batteries were different (2.9 and 2.78--and of course both completely the same when not under load), even though both batteries had the same voltage when not under load. I thought the two devices are different because of the tolerance of the parts (LEDs, photodiodes, and such).

But when I exchanged the batteries, the behaviour of the two devices (and the voltage across the batteries) also changed, So I concluded this has something to do with the battery.

Now, my question is, how I can single out such batteries? And why do they act like this? And what are the techniques to make this problem go away or less significant?

The battery has a storage life of 10 years and has been produced six months ago, both of the batteries have also been used the same amount


2 Answers 2


Open-load measurements of power sources are of only limited use, especially in the case of electrochemical cells. Always measure both the unloaded and the loaded voltage (and preferably multiple measurements with different loads) in order to determine suitability of the power source for the application.

  • \$\begingroup\$ Thanks, where can I read more about it, or better decide on the type of battery I need? (for a device that is supposed to run for two years with 30 micro amps current consumption in sleep mode) because this issue is really hurting my design \$\endgroup\$
    – user135819
    Jun 21, 2017 at 15:29
  • \$\begingroup\$ What kind of regulation do you have going? These sorts of voltage differences are to be expected. \$\endgroup\$ Jun 21, 2017 at 15:31
  • \$\begingroup\$ @JarrodChristman Not much as the circuit should be very simple (I hope it is not too simple): two caps inparallel: 1u and 100n \$\endgroup\$
    – user135819
    Jun 22, 2017 at 8:43
  • \$\begingroup\$ Ah, well there are some pretty simple boost converter chips. Voltage will lower under load as any cell ages and charge dissipates. Using a boost converter will allow you to more efficiently use as much of the energy in the battery as possible. \$\endgroup\$ Jun 22, 2017 at 13:21
  • \$\begingroup\$ @Sina, from there you can simply look at the energy requirement of your application in Watts and compare to a battery with a capacity in Watt-hours. \$\endgroup\$ Jun 22, 2017 at 13:27

A lithium cell can have different open circuit (no load) voltages due to the age of the cell or the difference in chemistry of the cell. Lithium cells are available in a wide range of chemistries to fit various applications and budgets.

When a lithium cell is placed under load, its voltage will drop due to the internal resistance of the cell. This is the case for all cells but the degree of voltage drop will vary from one cell type to another.

In order to determine what type of battery can supply your 30 uA of current for two years, you should consult the datasheets for the batteries. The datasheet will contains curves that show for a certain current draw, how long the battery will last. Some chemistries are better suited to lower current draws over long periods while others optimize short term, higher current capacities.

Here is an example of a chart from a Varta 6216 (IEC CR 1216), lithium manganese dioxide, primary cell

enter image description here

You will notice that the current draw is 95 uA and the life of the cell depends on the temperature. So this cell would last for a little over 200 hours (less than 9 days) at room temperature before the voltage of the cell would drop below 2.75 volts. Clearly for your application, you will require a cell with a much higher capacity. Web sites like Digikey and Mouser have the ability to search based on parameters so you may find them helpful in narrowing your search.

  • \$\begingroup\$ Thanks. regarding the first two paragraphs and how similar batteries (with no load) differ when under load: Is measuring the internal resistance of the battery a good measure to identify how they will react with load? If so what is the best and more accurate way to do so? I simply use a DMM but it is always overloaded (with max 20 Mohms on DMM)@glenn-w9iq \$\endgroup\$
    – user135819
    Jun 22, 2017 at 8:57
  • \$\begingroup\$ @Sina, the internal resistance is just a simplification. In reality, it's more of a measure of the voltage drop under load due to the battery chemistry. This will change as the cell ages and as the cell dissipates energy. \$\endgroup\$ Jun 22, 2017 at 13:24

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