I've got a question regarding the internal resistance of a CR2032 coin cell. Here's some background info on the project I'm working on:

I'm working on an electronics project in which I hope to have battery powered, more specifically I would like it to be powered with a CR2032 coin cell battery (https://na.industrial.panasonic.com/sites/default/pidsa/files/crseries_datasheets_merged.pdf). However, I've also decided to be open to other coin cells that are more appropriate for my application.

After a few early failed attempts to power my design, I decided to do a little more research into the characteristics of coin cells. From my research, I discovered that the rated coin cell capacity is only valid for the continuous drain rating which is typically low current draw ( < 0.25mA). The reason that this rating is so low is due to the fact that coin cells are not ideal voltage sources (duh), but instead they have a relatively high internal resistance. So when high current is drawn from the coin cell, this leads to a high voltage drop across the internal resistance thus reducing the available voltage across the coin cell terminals.

Furthermore, I've come across design recommendations that suggest putting a capacitor in parallel with the coin cell if your load can be periodically (pulsed) powered, this seemed to work very well. This setup is intended to allow for slow current draw into the capacitor in between power pulses, and then have the capacitor be the primary power supply during high current demands.

However, my design is still not perfect. Here are my problems:

  • Initially when I was powering my design without a parallel capacitor, the board would stay powered for about 3 minutes before it died. My initial thought was that I was drawing too much power from the battery and had completely drained it. However, I realized that I could wait 10 minutes, and then the coin cell would be able to provide enough power until the board died again after 3 minutes of operation. From this I concluded that I had not used all of the battery's capacity, rather I was just drawing too much power which temporarily reduced the coin cell terminal voltage.

Question: Why is this happening? I theorize that it is due to the battery's internal resistance, however, I had the assumption that the battery's internal resistance was constant. The effect I described above seems to suggest that the internal resistance increases as the battery is discharged. And as soon as you stop drawing power and let the coin cell rest, it appears that the internal resistance decreases such that the coin cell is functional again.

  • After adding the capacitor in parallel with the coin cell and converting my load such that is draws power periodically (pulsated), I was able to improve the lifetime of the design to about 20 minutes. However, the same phenomenon as described previously seemed to happen again. The board died, but as I let the coin cell rest, after about 10 minutes it was able to power the board again for about 20 minutes. This again suggests that the design isn't dying due to expended capacity, but rather the internal resistance is increasing.

So my question: What is this effect? I'm theorizing that the battery's internal resistance must be a function of integrated current such that it increases over time. Also I'm curious as to what is causing this? Is it just from me drawing too much continuous power from the coin cell? Even though I'm using a capacitor and pulsating power, my capacitor must not be taking the full brunt of the power?

Thanks for the help!

  • \$\begingroup\$ Are you describing your experiences with or without a parallel capacitor (and if with, which capacity)? And how much current does your circuit draw? \$\endgroup\$ Commented Oct 22, 2016 at 15:18
  • \$\begingroup\$ @WoutervanOoijen Both. When I added the capacitor, it seemed to help but it only delayed the original problem. Essentially the useful time duration was extended from 3 minutes to 20 minutes after adding the capacitor. My circuit draws approximately 8mA at pulse bursts. I'm assuming my capacitor isn't spec appropriately, period between pulses are too short, or the pulse durations are too long. There still a lot of possibilities. I still need to do more analysis. \$\endgroup\$
    – Izzo
    Commented Oct 22, 2016 at 15:23
  • \$\begingroup\$ Please give us the full info! How much current does your circuit use? 8mA pulse is only part of the info: how long is this pulse, how often, what does it draw between the pulses, and what voltage drop is tolerable during a pulse? \$\endgroup\$ Commented Oct 22, 2016 at 17:51
  • \$\begingroup\$ @WoutervanOoijen Like I said, I need to do more analysis and this includes measuring the pulse duration. This question was primarily focused on getting an answer regarding the general cell internal resistance properties. I can keep you posted as I collect more data. \$\endgroup\$
    – Izzo
    Commented Oct 22, 2016 at 17:59

1 Answer 1


A cell is a bunch of chemicals between two electrodes. A chemical reaction produces an excess of electrons on one terminal, and a lack of electrons on the other. Once the potential difference reaches a certain level (the voltage of the cell), the reaction slows to a halt.

If you connect a load to the battery, the electrons can travel between the two terminals. The reaction re-starts and generates more electrons.

But there's a limit to how fast the reaction can run. Take too much current, and the voltage drops. To a reasonable approximation, this looks like the cell is a constant voltage source in series with a resistor. But there isn't actually any resistor inside the cell.

As the cell ages, the chemicals run out, and this limits the rate at which it can generate current. Effectively, the internal resistance has gone up.

If you overload a primary cell, the chemicals in contact with the electrodes are used up faster than the diffusion within the cell can keep up. The internal resistance increases, and the cell looks flat. But give it a few minutes (particularly in a warm place), and the chemicals can re-mix again, and the cell recovers.

  • \$\begingroup\$ Thanks for this, so it sounds like I was thinking of it someone correctly. Is there any comprehensive documentation on this topic targeted for non-chemists? For example, if you operate a cell at anything above its rated continuous drain, will you eventually run into this problem? If for example I'm using the a CR2032 with a 0.2mA continuous drain, if I were to operate at 0.5mA continuous, would I still experience this problem? I'm primarily curious as to 'how much I can abuse the battery' while not completely disabling it. \$\endgroup\$
    – Izzo
    Commented Oct 22, 2016 at 15:34
  • \$\begingroup\$ @Teague The datasheet you linked to has handy graphs of capacity vs. load resistance. If you go from 0.2mA to 0.5mA, you'll see a slight drop in overall capacity. The cell isn't specified above about 2mA. Be aware that they are assuming that the cell is flat once it drops to 2V - well below the initial 3V. \$\endgroup\$
    – Simon B
    Commented Oct 22, 2016 at 15:42
  • \$\begingroup\$ So that's one thing that still confuses me. Is that rating assuming 2mA continuous? Or is 2mA pulsed such that the average current draw is still at 0.2mA (the rated continuous drain)? \$\endgroup\$
    – Izzo
    Commented Oct 22, 2016 at 18:02
  • 1
    \$\begingroup\$ That graph would be for a continuous 2mA, until the voltage drops below 2V. \$\endgroup\$
    – Simon B
    Commented Oct 23, 2016 at 12:53

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