bypassing the battery and powering the device directly from the wall socket

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kelogs

Member
Jun 9, 2010
24
0
Hi,

any way to accomplish this ? To be more clear, batteries will get a shorter life from the current work regime that I put them to, Unfortunately, the USB data cable of most phones also acts as a charger. I am using the phone for development, so this USB data cable is always attached to the phone and to the dev machine, thus forcing the battery to always charge, even at the slightest 1% discharge. It would be really good if I could take out the battery and still be able to run the phone.

Thank you!
 

adamo3957

Senior Member
Aug 17, 2011
112
27
Hi,

any way to accomplish this ? To be more clear, batteries will get a shorter life from the current work regime that I put them to, Unfortunately, the USB data cable of most phones also acts as a charger. I am using the phone for development, so this USB data cable is always attached to the phone and to the dev machine, thus forcing the battery to always charge, even at the slightest 1% discharge. It would be really good if I could take out the battery and still be able to run the phone.

Thank you!

Depends on the phone, I know on the moto defy there was a cable mod that would bypass the battery

Sent from my GT-I9300T using xda app-developers app
 

Renate

Inactive Recognized Developer / Recognized Contrib
Feb 3, 2012
2,817
1,267
Boston
Nexus 7 (2013)
Moto E5
Some (most?) battery circuits are designed to deal with a dead or shorted battery.
The circuit is not arranged in a direct line between charger, battery, load.
Disconnecting a battery connector also disconnects the temperature-measuring thermistor.
With a NTC thermistor, it would think that the battery is ice cold.
A resistor of the correct value would fool it into believing it's a reasonable temperature.

I tried disconnecting the battery on my Nook and it wouldn't power up.
 

djbxp

Member
Apr 19, 2011
26
9
36
Limassol
3,7V supply circuit as battery

You can supply the device with 3,7V (like the battery) from an external source. The only thing bad is that you have to attach wires to the gold plated battery slots on the device, or you can do it with small crocodile clips to avoid soldering. (better)

If you are ok with this, here is how you do it.
Take off your battery and measure voltage DC with a multimeter or voltmeter between battery leads. Now you know what your battery gives to the device. Example for 3 leads battery: you have 2 positive leads with the ground as reference (one slightly lower than the other) and the actual ground. So you have to supply with 3,7V the same leads that the battery was supplying. You can check while inserting the battery back to the device.

Hot to have the 3,7 Volts supply:
You will need 2 resistors, some capacitors, LM317 regulator, heat sink for that and a higher voltage DC power supply 6-12V.
Get and android device and go to play store. Install "Electrodroid" application. This will help you on sizing the LM317 regulator. Have in mind that this is programmable regulator, so you need 2 resistors to set the output voltage as 3,7V. LM317 is a linear voltage regulator, so it will act as uninterrupted 3,7 Volt battery. Be carefull to get a big heat sink, depending on the current you will be supplying and input voltage, also you can read this device datasheet online.
You can build this circuit on a breadboard if you are familiar with electronics or you can solder point to point the parts, or make a pcb if you can.
 
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Rebellos

Senior Recognized Developer
May 13, 2009
1,353
3,427
Gdańsk
You're lucky it's N7000
full schemas are there - http://forum.xda-developers.com/showthread.php?t=1813315

PMIC's pin "_DETBAT" is connected to "VF" of battery connector. Perhaps if you pull it high it'll bootup.

And btw - usually HW is fully capable of starting off USB power. The thing is that bootloader does check if battery is present and, if not, turns off the phone. Actually this is because phone, especially during bootup, can peak to much more than 500mA current, and battery is there to compensate "missing" power.


//edit:
However, in case you don't provide any power into battery pins, it might try to charge it and U607 - Switching Charger might not really like working without load. This can generate alot of noise around AFAIK so modding kernel somehow to disable charging would be good choice.
 
Last edited:

Entropy512

Senior Recognized Developer
Aug 31, 2007
14,095
25,088
Owego, NY
You're lucky it's N7000
full schemas are there - http://forum.xda-developers.com/showthread.php?t=1813315

PMIC's pin "_DETBAT" is connected to "VF" of battery connector. Perhaps if you pull it high it'll bootup.

And btw - usually HW is fully capable of starting off USB power. The thing is that bootloader does check if battery is present and, if not, turns off the phone. Actually this is because phone, especially during bootup, can peak to much more than 500mA current, and battery is there to compensate "missing" power.


//edit:
However, in case you don't provide any power into battery pins, it might try to charge it and U607 - Switching Charger might not really like working without load. This can generate alot of noise around AFAIK so modding kernel somehow to disable charging would be good choice.
The system is powered off of Vbat - As a result, the charger MUST be active. Also, the N7000 is EASILY capable of drawing more than the maximum input current limit from Vbus, mandating extraction of power from the battery in some operating regimes.

The only way to achieve what the OP wants (total battery removal) would be with a dummy battery that had an external 4.0 volt power supply. Bad Things could happen if the device is connected to USB in this state.

However, to satisfy the OP's stated reasons for removing the battery (lots of time on USB), the likely best solution would be to disable the charging circuitry in the kernel at high states of charge. For example, one could set it up so that the charger would only be enabled when Vbat was below 4.0 volts, or when the fuel gauge SoC is below X per cent. See Ezekeel's "BLX" implementation for the Galaxy Nexus as one example of this.
 
Last edited:
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prietpraat1972

Senior Member
Jul 15, 2012
50
21
nijmegen
thanks!

you're lucky it's n7000
full schemas are there - http://forum.xda-developers.com/showthread.php?t=1813315

pmic's pin "_detbat" is connected to "vf" of battery connector. Perhaps if you pull it high it'll bootup.

And btw - usually hw is fully capable of starting off usb power. The thing is that bootloader does check if battery is present and, if not, turns off the phone. Actually this is because phone, especially during bootup, can peak to much more than 500ma current, and battery is there to compensate "missing" power.


//edit:
However, in case you don't provide any power into battery pins, it might try to charge it and u607 - switching charger might not really like working without load. This can generate alot of noise around afaik so modding kernel somehow to disable charging would be good choice.

thanks for the link1
 

kelogs

Member
Jun 9, 2010
24
0
Thank you all for sharing knowledge and experiences. I have made the decision to just go for some cheap ebay replacement batteries due to some advice I got from a friend, which I am sharing below:

Do not fiddle with such a fine piece of hardware (i.e. smartphone) by attaching some exposed wirings to it. The gadget could easily slip from your hands and cause the loosely hanging wirings to short-circuit upon landing on the floor. Definitely not a good perspective.
 

dany_mid

New member
Jun 22, 2012
2
1
a

Thank you all for sharing knowledge and experiences. I have made the decision to just go for some cheap ebay replacement batteries due to some advice I got from a friend, which I am sharing below:

Do not fiddle with such a fine piece of hardware (i.e. smartphone) by attaching some exposed wirings to it. The gadget could easily slip from your hands and cause the loosely hanging wirings to short-circuit upon landing on the floor. Definitely not a good perspective.


Oh give a f***ing break! It's a phone not an egg shell. And short-circuiting the wires would at worst damage the power supply not the phone.
 
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joederp

Senior Member
Sep 8, 2012
190
57
The system is powered off of Vbat - As a result, the charger MUST be active. .

This just isn't true. Modern PMIC not only have the option to stop charging the battery, but to charge the phone only on AC and also do things like send power the the USB OTG. So it is a matter of the PMIC knowing what to do. What I wish was possible was a nice app that told the PMIC to stop charging at 80% and then go to trickle mode. This would save your battery life by a lot, instead it appears to charge to 100% then back off the voltage a little.
 

jrs03

New member
Mar 16, 2013
3
0
bypass battery on unibody phone and run directly from charger

The system is powered off of Vbat - As a result, the charger MUST be active. Also, the N7000 is EASILY capable of drawing more than the maximum input current limit from Vbus, mandating extraction of power from the battery in some operating regimes.

The only way to achieve what the OP wants (total battery removal) would be with a dummy battery that had an external 4.0 volt power supply. Bad Things could happen if the device is connected to USB in this state.

However, to satisfy the OP's stated reasons for removing the battery (lots of time on USB), the likely best solution would be to disable the charging circuitry in the kernel at high states of charge. For example, one could set it up so that the charger would only be enabled when Vbat was below 4.0 volts, or when the fuel gauge SoC is below X per cent. See Ezekeel's "BLX" implementation for the Galaxy Nexus as one example of this.

This would just disable the charging until the SoC dropped to the level you set though, correct? I.e. the phone is still running from the battery..? If so you still end up with more or less the same issue (although with some potential benefits depending from cycling at lower SoC).

I have a HTC One X so removing the battery and adding some circuit trickory isn't an option. But bc of this damn unibody even more motivation to run the phone directly from the charger bc I can't feed it with replacement batteries. ( which he is right thats the best options for the OP)

Anyone know if this is even possible with software mods based on the design of the phone charging system? Or any sources for literature on this. I really wanna save this battery if its the only one I got!
 

EwokIng

Member
Jan 19, 2015
24
4
I am currently experimenting with this. I suppose I can only get 3 volts from the 5V input of a usb charger. Going to need to hook up to a 12v power source I suppose. Built this power supply, linear variable voltage regulator. Going to still want that data transfer. I am using diodes to make sure no power goes the wrong way into my electrolytic capacitors. I will try to post a thread if it works because I have not seen one yet.
 

creatip

Member
Nov 24, 2017
24
1
Hello, sorry to post in an old thread, but this is the closest problem to mine that I could find.

So I'm using a 4G mi-fi modem on my PC. It's plugged in constantly 24/7 through USB cable to my PC. You can imagine the effect on the battery. I threw out 2 batteries coz they've gone bad. Sadly the modem won't turn on if it's not detecting a battery.

So I'm considering the capacitor route, just to fool the modem into thinking that a battery is installed, the real power comes from USB data cable anyway. It's a Huawei E5577, the battery got 4 terminals on it. The outermost terminals are (+) and (-), while I'm guessing the middle two are used to read the battery status (voltage, etc). So what's the simplest schematic to achieve this, using simplest capacitor circuit to fool the modem into thinking the battery is installed and working well.

Thank you
 

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    3,7V supply circuit as battery

    You can supply the device with 3,7V (like the battery) from an external source. The only thing bad is that you have to attach wires to the gold plated battery slots on the device, or you can do it with small crocodile clips to avoid soldering. (better)

    If you are ok with this, here is how you do it.
    Take off your battery and measure voltage DC with a multimeter or voltmeter between battery leads. Now you know what your battery gives to the device. Example for 3 leads battery: you have 2 positive leads with the ground as reference (one slightly lower than the other) and the actual ground. So you have to supply with 3,7V the same leads that the battery was supplying. You can check while inserting the battery back to the device.

    Hot to have the 3,7 Volts supply:
    You will need 2 resistors, some capacitors, LM317 regulator, heat sink for that and a higher voltage DC power supply 6-12V.
    Get and android device and go to play store. Install "Electrodroid" application. This will help you on sizing the LM317 regulator. Have in mind that this is programmable regulator, so you need 2 resistors to set the output voltage as 3,7V. LM317 is a linear voltage regulator, so it will act as uninterrupted 3,7 Volt battery. Be carefull to get a big heat sink, depending on the current you will be supplying and input voltage, also you can read this device datasheet online.
    You can build this circuit on a breadboard if you are familiar with electronics or you can solder point to point the parts, or make a pcb if you can.
    1
    You're lucky it's N7000
    full schemas are there - http://forum.xda-developers.com/showthread.php?t=1813315

    PMIC's pin "_DETBAT" is connected to "VF" of battery connector. Perhaps if you pull it high it'll bootup.

    And btw - usually HW is fully capable of starting off USB power. The thing is that bootloader does check if battery is present and, if not, turns off the phone. Actually this is because phone, especially during bootup, can peak to much more than 500mA current, and battery is there to compensate "missing" power.


    //edit:
    However, in case you don't provide any power into battery pins, it might try to charge it and U607 - Switching Charger might not really like working without load. This can generate alot of noise around AFAIK so modding kernel somehow to disable charging would be good choice.
    The system is powered off of Vbat - As a result, the charger MUST be active. Also, the N7000 is EASILY capable of drawing more than the maximum input current limit from Vbus, mandating extraction of power from the battery in some operating regimes.

    The only way to achieve what the OP wants (total battery removal) would be with a dummy battery that had an external 4.0 volt power supply. Bad Things could happen if the device is connected to USB in this state.

    However, to satisfy the OP's stated reasons for removing the battery (lots of time on USB), the likely best solution would be to disable the charging circuitry in the kernel at high states of charge. For example, one could set it up so that the charger would only be enabled when Vbat was below 4.0 volts, or when the fuel gauge SoC is below X per cent. See Ezekeel's "BLX" implementation for the Galaxy Nexus as one example of this.
    1
    a

    Thank you all for sharing knowledge and experiences. I have made the decision to just go for some cheap ebay replacement batteries due to some advice I got from a friend, which I am sharing below:

    Do not fiddle with such a fine piece of hardware (i.e. smartphone) by attaching some exposed wirings to it. The gadget could easily slip from your hands and cause the loosely hanging wirings to short-circuit upon landing on the floor. Definitely not a good perspective.


    Oh give a f***ing break! It's a phone not an egg shell. And short-circuiting the wires would at worst damage the power supply not the phone.
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