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I would like to use a CR2032 battery to power the internal RTC of the SIM900 gsm module. The datasheet (given in the link) says that the battery should be rated at 3V. They also give the following reference circuit for non-rechargable batteries (like CR2032). enter image description here

I am confused about the diode in this circuit. As far as I know the forward voltage drop of a diode is 0.7V at average, which leaves only 2.3V for the RTC. Will that cause any problems, knowing that the datasheet says the battery should be rated at 3V?

Marko Scekic
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  • Only silicon diodes have a 0.7V drop. If it's a concern, you would use other types. Schottky and Germanium have a 0.2~0.4V at full draw. See http://conceptselectronics.com/diodes/comparison-si-ge-gaas-diodes/ – Passerby Jan 11 '16 at 22:46

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Well, that's the recommended circuit. Keep in mind that at the 2uA current draw of the RTC backup, even a 1N4148 will have only about 0.3V drop and it will work down to Vrtc = 2V.

A CR2032 lithium primary cell depleted to 2.3V at 2uA is practically dead, so you're not leaving many (micro)joules on the table.

Edit: In explanation of the final paragraph above, let's refer to the Panasonic datasheet for their CR2032 battery:

enter image description here

This does not directly answer the question of what the discharge looks like at 2uA, but at 190uA you can see that the voltage drops fairly abruptly and a 'cutoff' voltage of 2V vs. 2.3V buys you minimal additional hours. Of course you would expect the times to be about 100x longer for 2uA vs. 190uA but the principle is similar. You're extracting almost all the energy (as measured in joules) from the battery if you consider end-of-life to be 2.3V and setting it at 2V or 1.5V buys you very little additional time. So the diode drop of 300mV nominal is really not very important to the total battery life on standby.

Spehro Pefhany
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  • Oops, I was writing my answer as you were posting yours. I wasn't trying to steal your thunder... Shall I delete it? – bitsmack Jan 11 '16 at 21:55
  • @bitsmack No probs. It happens! That's what I get for actually finding the Vf of a 1N4148. ;-) – Spehro Pefhany Jan 11 '16 at 21:56
  • Thanks :) (_this is just typing to exceed the character threshold_) – bitsmack Jan 11 '16 at 22:00
  • @bitsmack Up to you. If you do you'll lose that valuable point. – Spehro Pefhany Jan 11 '16 at 22:09
  • Can you please explain what do you mean by "A CR2032 lithium primary cell depleted to 2.3V at 2uA is practically dead, so you're not leaving many (micro)joules on the table. ". Sorry if it's a noobish question, but I'm relatively new to the world of electronics. Thank you in advance :) – Marko Scekic Jan 11 '16 at 23:07
  • @MarkoScekic Please see my edit above. This one is a bit more wooly because I don't have the characteristics of the CR2032 at 2uA discharge current, but I think you can understand the principle at work here. It's not like the battery drops linearly in voltage- it holds fairly steady during discharge then drops rapidly as the internal resistance skyrockets at end-of-life. – Spehro Pefhany Jan 12 '16 at 00:03
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    I think I will delete it, losing those three valuable points :) But at least I get a cool internet badge! – bitsmack Jan 12 '16 at 06:53