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I've a CR2032 battery connected to DS3231 module as backup battery, when module is connected to 5V the battery voltage is at 4.3V and without power 3.7V.

That means the module is trying to charge the nonchargeable battery, i couldn't find a chargeable (LIR23032) battery to buy and after some research someone in forums suggested to remove the diode from the module...

  1. If i keep using the CR2032 battery for long run 24/7, how long will battery last? is there a chance of explosion?

  2. is there any other way than modifying the module it self?

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    Most modules have a bad charging circuit design. I'd advice to modify the board and use non-recharchable batteries, even if you had LIR2032 batteries available. What is the problem with removing the diode? Are you unsure how to do this? Or are you reluctant to permanently change your module? – Gerben Jun 28 '19 at 16:16
  • @Gerben i removed the resistor. – ElectronSurf Jun 28 '19 at 16:53
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If i keep using the CR2032 battery for long run 24/7, how long will battery last? is there a chance of explosion?

Yes. Recharging a non-rechargeable battery is dangerous. It should be avoided at all costs.

is there any other way than modifying the module it self?

You have to prevent the power getting to the battery. That power is fed through a diode (hence 4.3V, which is 5V minus 0.7v for a silicon diode forward voltage). Without knowing what the module is, I can't tell you if there is any other way that modifying it - however I would guess that no there isn't. You need to stop that power getting to the battery, and that can only be done at some point between where the power comes in to the module and the battery itself - and the simplest method is to remove the diode.

The purpose of that diode is to prevent power from the battery from trying to power the rest of your circuit when you remove the main power. Removing the diode will have no adverse affect. You just won't be able to charge a rechargeable battery in future.

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  • i removed the 200 ohm resistor and kept the diode, the voltage is now dropped to 3.8V which is 0.05V more than battery voltage. is that 0.05V gonna be a problem? – ElectronSurf Jun 28 '19 at 15:53
  • I don't know what resistor you are on about. If you are going by the voltage that is printed on the battery: that is only a "nominal" value and is less accurate than the 0.05v difference you are measuring anyway. – Majenko Jun 28 '19 at 16:25
  • i'm using a multimeter to measure the voltage, and i removed THIS resistor. – ElectronSurf Jun 28 '19 at 16:50
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    That looks like it's in series with the diode to limit the charge current. It has the exact same effect as removing the diode - it breaks the circuit. – Majenko Jun 28 '19 at 17:31
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    So the 0.05V is the difference between "no power connected to the Vcc pin" and "5V connected to the Vcc pin"? In that case the 0.05V drop when no power is applied, is most likely due to the battery voltage sagging a bit because a bit power is being drawn from it. I wouldn't worry about that. – Gerben Jun 29 '19 at 10:35

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