Timeline for Measure Lithium ion battery voltage (thus remaining capacity)
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Jun 17, 2018 at 2:33 | history | tweeted | twitter.com/StackElectronix/status/1008175722995777536 | ||
| Sep 4, 2012 at 9:09 | vote | accept | boardbite | ||
| Sep 4, 2012 at 9:06 | history | edited | boardbite | CC BY-SA 3.0 |
schematic updated based on @stevenvh and @Nick's suggestions
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| Sep 3, 2012 at 18:57 | answer | added | stevenvh | timeline score: 16 | |
| Sep 3, 2012 at 18:42 | answer | added | Nick Alexeev | timeline score: 9 | |
| Sep 3, 2012 at 18:00 | history | edited | boardbite | CC BY-SA 3.0 |
Updated schematic with P-MOSFET
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| Sep 3, 2012 at 17:54 | comment | added | boardbite | Finalized schematic with the P-MOSFET; thanks a bunch! I'd still appreciate a pointer regarding the N-MOSFET alternative, just for future knowledge. | |
| Sep 3, 2012 at 17:51 | history | edited | boardbite | CC BY-SA 3.0 |
Changed to P-MOSFET high-side switch
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| Sep 3, 2012 at 17:43 | comment | added | boardbite | The high-side switching makes sense. Also, I used an N-MOSFET because I happened to have one; can't I make it work with one? | |
| Sep 3, 2012 at 17:37 | history | edited | boardbite | CC BY-SA 3.0 |
deleted 9 characters in body
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| Sep 3, 2012 at 17:29 | comment | added | stevenvh | You've got the wrong FET on the wrong side. If you switch this one off the full voltage will be on the input pin, since R1 won't draw current. You need a P-MOSFET on the high side, so that switching it off pulls the I/O pin to ground. | |
| Sep 3, 2012 at 17:26 | comment | added | boardbite | @stevenvh: Understood. I have refined the Question with the updated details and a schematic. | |
| Sep 3, 2012 at 17:25 | history | edited | boardbite | CC BY-SA 3.0 |
updated based on @Jonny and @stevenvh's suggestions, and added schematic
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| Sep 3, 2012 at 17:11 | comment | added | stevenvh | I guess so. The FET will have a leakage current when off, but that will be a few orders of magnitude smaller than the 5 \$\mu\$A the divider takes. (I think that charger is very sensitive). | |
| Sep 3, 2012 at 16:21 | comment | added | boardbite | With the FET in place, in the off "state" of the divider, do you think this might also solve the problem of the charger IC's mis-reading of battery presence? (BTW, what a coincidence that that person posted the question today as well!) | |
| Sep 3, 2012 at 16:14 | comment | added | stevenvh | Or use a FET to switch on and off the divider, like I suggested in this answer to a similar question. | |
| Sep 3, 2012 at 16:11 | answer | added | Christoph B | timeline score: 6 | |
| Sep 3, 2012 at 16:09 | comment | added | boardbite | So I want low current through the divider but high enough that it's at least an order of magnitude above the maximum leakage current. | |
| Sep 3, 2012 at 16:02 | comment | added | stevenvh | "I used large resistor values". The input pin may have a leakage current, a typical worst case value is 1 uA. With low current through the divider this may distort the reading. | |
| Sep 3, 2012 at 15:57 | history | asked | boardbite | CC BY-SA 3.0 |