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Firefox with Leste (10)
to Maemo 7 / Leste by teroyk - 4 days, 2 hrs ago - more...
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2019-03-21
, 10:29
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Posts: 3 |
Thanked: 16 times |
Joined on Jun 2010
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#2
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We have succeed with battery-less setup in Maemo Leste. You only need to have 5V connected to MicroUSB port to have device running.
| The Following 14 Users Say Thank You to Spinal For This Useful Post: | ||
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2019-03-21
, 12:24
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Posts: 158 |
Thanked: 340 times |
Joined on Jul 2010
@ middle of nowhere
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#3
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Originally Posted by Nokia 5700
Can you disassebly the battery?
Hey guys,
I haven't posted here in a long while, but I still have a functioning Nokia N900 at hand. The problem is that it is becoming increasingly difficult to maintain batteries for it. The first battery started to expand, and so I had to toss it and get a new one (not Nokia-branded), only to find that it lied about its capacity.
This made me want to approach using a DC power supply to drive the device in place of the battery. I'll skip the reasoning and etc. and get to the case:
- If you connect a stable DC power supply set at 3.7V, uncapped amperage, the device *does not* turn on.
I started to research the so-called BSI-pin on the battery, which is (to my understanding) a circuit between GND and BSI with a 100kOhm resistance, meant to identify the battery characteristics.
I found the following table of resistances online:
http://pinoutguide.com/CellularPhone...y_pinout.shtml
I interpreted this in two ways:
- Does it need a 100kOhm resistor between BSI and GND?
- Is the internal resistance on the device terminals (BSI and GND) supposed to be 100kOhm?
I tried both of these, using a 100kOhm for the first interpretation, and a 430kOhm resistor to achieve 100kOhm resistance on the battery terminals (of course, capacitance made this a pain to measure). In both cases, the device responds in the following way:
- I hold down the power button
- The LED "charges up" in a blue color
- The screen turns on, black, as if NOLO was loading
- The device proceeds to turn off as soon as I release the power button
The power supply displayed an amperage of 250mA for a certain period of time while "booting", but flats out to 100mA afterwards.
After doing this, the device would still turn on when I inserted a battery.
I tested the same setup on another Nokia device (a Nokia 5700) only to get the same response (screen turns on, turns off after letting go of power button), including the same amperage readings. (And of course I adjusted the resistor for the battery type).
Most of my understanding of the battery circuit is based on the BSI reporting *battery model*. A single article online said that BSI was implemented with a termistor, not a resistor, but it also referred to NiMH batteries.
I didn't try ramping up the voltage. I know that some Li-Ion batteries have different voltages at different charges, but I didn't want to get risky.
Has anyone had success with this?
Inside the battery you will find a small board connected to the battery cell, just disconnect the cell and connect your power supply to the + and - of that board instead.
| The Following 7 Users Say Thank You to ryu1 For This Useful Post: | ||
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2019-03-22
, 16:38
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Posts: 298 |
Thanked: 197 times |
Joined on Dec 2009
@ Norway
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#4
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Originally Posted by Spinal
This sounds very interesting! Is the change purely software-based, and if so, where do I start? I assume you mean a regular 5V power source for the microUSB port?
We have succeed with battery-less setup in Maemo Leste. You only need to have 5V connected to MicroUSB port to have device running.
Originally Posted by ryu1
I'm not very comfortable with disassembling a battery, and I'd like to avoid doing this. Do you know any posts that detail doing this successfully?
Can you disassebly the battery?
Inside the battery you will find a small board connected to the battery cell, just disconnect the cell and connect your power supply to the + and - of that board instead.
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2019-03-22
, 21:06
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Posts: 16 |
Thanked: 12 times |
Joined on Jan 2013
@ Poland
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#5
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Wow, it's been a while.
Recently I dug out my N900 and started messing with it, I soon found out that the battery has lost almost a half of it's capacity.
With not much to loose I gently pried off the top plastic from the battery where the connector is. Then I carefully cut off the metal tabs holding the plastic (and the protection board embedded) to the actual cell. Be careful not to short any of the tabs to the casing when cutting, or be ready to scram
After that I connected a USB cable through a regular rectifier diode. USB is 5V, while a fully charged battery is around 4.2V, so a 0.7V drop across the diode puts me in the ballpark.
With that done the N900 boots right up.
Recently I dug out my N900 and started messing with it, I soon found out that the battery has lost almost a half of it's capacity.
With not much to loose I gently pried off the top plastic from the battery where the connector is. Then I carefully cut off the metal tabs holding the plastic (and the protection board embedded) to the actual cell. Be careful not to short any of the tabs to the casing when cutting, or be ready to scram
After that I connected a USB cable through a regular rectifier diode. USB is 5V, while a fully charged battery is around 4.2V, so a 0.7V drop across the diode puts me in the ballpark.
With that done the N900 boots right up.
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| hacking, nokia n900 |
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I haven't posted here in a long while, but I still have a functioning Nokia N900 at hand. The problem is that it is becoming increasingly difficult to maintain batteries for it. The first battery started to expand, and so I had to toss it and get a new one (not Nokia-branded), only to find that it lied about its capacity.
This made me want to approach using a DC power supply to drive the device in place of the battery. I'll skip the reasoning and etc. and get to the case:
- If you connect a stable DC power supply set at 3.7V, uncapped amperage, the device *does not* turn on.
I started to research the so-called BSI-pin on the battery, which is (to my understanding) a circuit between GND and BSI with a 100kOhm resistance, meant to identify the battery characteristics.
I found the following table of resistances online:
http://pinoutguide.com/CellularPhone...y_pinout.shtml
I interpreted this in two ways:
- Does it need a 100kOhm resistor between BSI and GND?
- Is the internal resistance on the device terminals (BSI and GND) supposed to be 100kOhm?
I tried both of these, using a 100kOhm for the first interpretation, and a 430kOhm resistor to achieve 100kOhm resistance on the battery terminals (of course, capacitance made this a pain to measure). In both cases, the device responds in the following way:
- I hold down the power button
- The LED "charges up" in a blue color
- The screen turns on, black, as if NOLO was loading
- The device proceeds to turn off as soon as I release the power button
The power supply displayed an amperage of 250mA for a certain period of time while "booting", but flats out to 100mA afterwards.
After doing this, the device would still turn on when I inserted a battery.
I tested the same setup on another Nokia device (a Nokia 5700) only to get the same response (screen turns on, turns off after letting go of power button), including the same amperage readings. (And of course I adjusted the resistor for the battery type).
Most of my understanding of the battery circuit is based on the BSI reporting *battery model*. A single article online said that BSI was implemented with a termistor, not a resistor, but it also referred to NiMH batteries.
I didn't try ramping up the voltage. I know that some Li-Ion batteries have different voltages at different charges, but I didn't want to get risky.
Has anyone had success with this?