What are the 4 contacts used for on a typical consumer product (eg. Smartphone) LiPo?

[salukikev]

I would like to run a smartphone using a primary battery instead of rechargeable for a project I'm experimenting with.
Most phone batteries will have a connector of some type with 4 contacts. Of course two of those I'd expect to be +- Voltage, but the other two compose some sort of mystery input to the phone (thermistor? signal? I don't know). I was hoping to substitute a primary battery instead of the factory one for my project (I have a 3.6v primary battery), but my device always seems to have a problem with it, presumably due to the battery protection circuit. I thought I might just leave the protection circuit in place and wire power to both input and output- that way whatever input the phone needs to see is still there, but even that approach seems not to work. Please help me bypass this system so that I can use a custom battery in this device. Here is a diagram with some measured resistance/voltage values. If a value between points isn't shown, you can consider it an open circuit (although I didn't measure battery resistance)

I've actually tried fooling the device by installing resistors of that value(s) appropriately but still it won't power up. Also, I tried just powering it through the usual USB charging port (5v) but that's so far been very inconsistent. Is there a way to find the main power terminal in the PCB and go directly to it? For reference- the donor phone here is an older Motorola Droid Razr

 

[3DDave]

https://electronics.stackexchange.c...es-this-mobile-phone-battery-have-4-terminals

https://electronics.stackexchange.c...re-3-pins-on-some-batteries?noredirect=1&lq=1

https://sayyedhamid.blogspot.com/2011/05/all-about-battery-connector-pinout-of.html

Google.

 

[Comcokid]

Since you have a lot of posts and replies on Eng-Tips, I'll assume your not trying to make an IED controller.

In modern electronics with Lithium Ion batteries, the batteries cells have to be monitored carefully during charge and discharge to get the maximum performance and avoid issues (undercharge, overcharge, battery failure).

At a minimum a battery will have internally a charge control circuit which will coulomb-count energy input/output during charge/discharge, and monitor each cell for operation and balance charge to the cells. At a maximum, the charge control circuits will also include a small microprocessor with the charge control.

In either case, the phone will digitally talk to the battery, and the battery will respond. The structure of this communication is going to be almost certainty unique to each phone design.

I've never tried this myself, but you might be able to trick the phone into thinking it is talking to the battery pack when it's actually being powered from another source. Maybe by connecting the cell phone battery two extra pins and the negative to the phone while simultaneously supplying external power to the negative and positive terminals. You can experiment with this but no guarantee. Internal phone software may be told that the battery pack has one voltage while the phone measure a different voltage and decide there is a battery issue and refuse to operate.

 

[salukikev]

Thanks for the suggestions! I was pressed for time and had to go to different tech, but now I have another (longer) opportunity so I'll try to work this out a bit more. I think I mentioned that the resistance values at least that I read in that diagram, I put back into the simulated circuit via SMT resistors, but even with that this system wasn't fooled. At one point though I got the phone to boot up using an alternative scrap cell phone battery that wasn't from the same model or manufacturer and the battery icon turned into a question mark. This droid settings menu shows that it keeps specs on the particular battery with history of total hours run, etc. so I assume that means at least one of these additional contacts relays a digital serial # of some type.
Thanks!

 

[MacGyverS2000]

Yes, there is digital data, handshaking, etc. You won't defeat it with a simple resistor.

Dan - Owner

http://www.Hi-TecDesigns.com