I've seen applications (Nintendo game cartridges, for example) that uses a coin cell battery for backup power to some RAM-chips (for savegames). When the ordinary 5V is gone, the battery takes over and lets the IC retain its contents.

However, won't the CR2032 (3V) take damage when connecting it to a power rail that contain almost its double voltage? Or is there some overvoltage protection within the coin cell that shuts it down when external voltage is applied?

Or how does it work?

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    \$\begingroup\$ "When the ordinary 5V is gone, the battery takes over and lets the IC retain its contents." -- Do you have any concrete evidence that the battery is directly connected to a 5V rail? \$\endgroup\$ – Shamtam Jul 31 at 14:33
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    \$\begingroup\$ Indeed, it would - which is why coin cells are not connected across power rails. Backup batteries typically use diodes or low-loss switching, ie so called "ideal diode" circuits - sometimes that may be internal to the device being given backup power such that the backup power is supplied on a different pin than the operating power. \$\endgroup\$ – Chris Stratton Jul 31 at 14:34
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    \$\begingroup\$ @ChrisStratton Sounds like an answer to me. \$\endgroup\$ – Dampmaskin Jul 31 at 14:43
  • \$\begingroup\$ Agree. @ChrisStratton please make an answer of that comment so I can accept it. \$\endgroup\$ – bos Jul 31 at 21:25

The main supply is not in parallel with the battery. Most often the battery powers RAM chips via a diode, so that main supply can be connected to RAM chips with a transistor. Sometimes both the supplies go through diodes.

  • \$\begingroup\$ The diodes will prevent a current flowing into the CR2032 cell. \$\endgroup\$ – Uwe Jul 31 at 18:35

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