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I want to make a simple battery powered development board for my AVR's but the problem is that the regular 3.7v batteries or very big in size, so I decided to use a smartphone battery from one of my old phones, but can't seem to understand the logic how it works and I can't find datasheet anywhere for this product.

Here are the specs:

OEM HTC Li-Ion Standard Battery (35H00128-00M)
Rating: 1230mAh / 3.7V
Charging: 4.2V

enter image description here

I can't seem to find the charging current, but my power source gives out 1A, I guess it should be good enough, since it's a standard USB (from wall socket).

The battery has 3 pins, positive and negative terminals and additional negative terminal. By connecting multi-meter to the positive and negative terminal, I got 3.7V, everything was good for a while until it started to give out 0V, so I re-connected multi-meter to the second negative terminal (middle pin) and got 3.53v.

I'm assuming second negative terminal is used for battery level detection, but when I connected 4.21v/1A to the main battery terminals, after 20 minutes the voltage didn't increase and still displayed 0v at the main terminals and 3.53v at the 'negative status terminal'.

Does that mean I have to connect the charger to the 'other' negative terminal (the one which reports battery level) and drain the battery only through the main terminals? If this is the case, how would I be able to 'read' the battery status, if it would always report 4.21v while the charger is plugged in?

How can I charge this battery? Thanks!

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  • \$\begingroup\$ Battery packs will normally have a thermistor. I think the middle pin is one of the terminals of the thermistor. The other terminal of the thermistor will be connected to the negative terminal of the battery. \$\endgroup\$ – R. Hirur Oct 20 '16 at 11:53
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    \$\begingroup\$ I'm assuming second negative terminal is used for battery level detection Why would you assume that ? If you don't understand things in electronics, don't just assume but research it. It is better to say "I don't know" then to just assume. Ask the technician who touched the 1000 V line "assuming" it was 0 V. \$\endgroup\$ – Bimpelrekkie Oct 20 '16 at 11:53
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    \$\begingroup\$ and 3.53v at the 'negative status terminal'. That terminal does not exist, the 2 terminals between which you measured the 3.53 V are the main battery terminals. By connecting that 4.21 V to that other terminal you could have blown something up inside the battery. Next time you fiddle with something, inform yourself first about how it works instead of just trying "something". \$\endgroup\$ – Bimpelrekkie Oct 20 '16 at 12:02
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    \$\begingroup\$ I would open the battery, you can close it back with normal tape. If the outermost contacts show 0V, maybe the internal protection circuit detected an overdischarge and cut (sometime irreversibly) the connection. This would be my first guess, since after providing 4.21V to the outermost contact the voltage hasn't changed. You can try to provide the 4.2V to the middle- and to the + contacts (in case the middle one is the power supply and not thermistor), but you must limit the current to very low values (10 mA?) to avoid destroying the thermistor if the assumption is wrong. \$\endgroup\$ – FarO Oct 20 '16 at 12:24
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    \$\begingroup\$ how can I detect the charge level You cannot detect the charge level !! It is not a "charge level" pin. The pin is connected to a thermistor (temperature dependent resistor) to get an indication of the temperature of the battery. Not the charge level. The voltage of the battery indicates the charge level. When charging and full, the battery becomes warm this is detected by the charger through the thermistor. Again you made an assumption and went up the wrong tree. Only assume if you can verify if that assumption is correct. \$\endgroup\$ – Bimpelrekkie Oct 20 '16 at 12:44
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In Why are there 3 pins on some batteries? is clearly stated that the middle pin is the negative to use for charging.

If you powered the other negative one, you blew the chip, thermistor or whatever it's inside.

Now you have a dumb battery, no more smart. At your own risk...

To detect end of charge for LiIon you check the current: when the charging current drops below a threshold (50 mA for example), it's done.

LiIon charger boards (check Chinese suppliers like DX) have a LED with selectable threshold for that purpose. Limit the current to safe values (0.5C) and voltage to 4.1V (little is gained from 4.2V). Stop discharge at 2.9V (you can read the supply voltage using the AVR and no additional components).

You can also open the battery and check yourself, it's just glued. Close it back with normal tape.

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    \$\begingroup\$ Now you have a dumb battery, no more smart. At your own risk... Explain to me how that battery was ever smart. It wasn't. The charging's intelligence is in the charger circuit. That extra pin is only for temperature feedback. \$\endgroup\$ – Bimpelrekkie Oct 20 '16 at 12:53
  • \$\begingroup\$ You are right, but now it even lost the possibility of being smart. That's what I meant. \$\endgroup\$ – FarO Oct 20 '16 at 14:43
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Since this phone battery is a Li-Ion battery, the only safe way to charge such a battery is by using a LiIon battery charger circuit.

You can buy a ready made module on ebay for this, they're cheap only $ 1.50 or so.

enter image description here

If you try to charge the battery in any other way you risk overheating, damage, smoke and fire.

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    \$\begingroup\$ I would agree if the circuit you provided were able to read the thermistor, but it only provides two output contacts, therefore it's just a fixed voltage power supply. Any power supply capable of providing a regulated and accurate voltage of 4.1V would be fine and as safe as the one you showed. If you want real safety, you should use a circuit capable of reading back the thermistor, but it's not the one of the photo. \$\endgroup\$ – FarO Oct 20 '16 at 12:20
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    \$\begingroup\$ @OlafM You only need the thermistor (or temperature feedback) if you want to charge the battery quickly. For slow charging at 0.2 C for example, temperature feedback is not needed. This module is intended for slow charging. \$\endgroup\$ – Bimpelrekkie Oct 20 '16 at 12:34
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    \$\begingroup\$ @0x29a You want to learn, excellent ! Then get reading, learn how a LiIon cell must be treated at Battery University: batteryuniversity.com Then see how charger chips work, like the TP4056 in the module above. Basically you need voltage and current limiting. You will need some extra components besides the microcontroller to do this. That in total will not be smaller than that module. \$\endgroup\$ – Bimpelrekkie Oct 20 '16 at 12:39
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    \$\begingroup\$ That module is not just a fixed voltage source. Li-Ion batteries are only safe if you treat them correctly. That's why this phone battery will have a protection circuit. But it does not provide 100% safety (not idiot proof). If you have a lab supply with adjustable current limit then you can safely charge this battery, use 4.1 V and 100 mA and you're good to go. Ask Samsung about their Note 7 and if they think Li-Ion batteries are always safe. \$\endgroup\$ – Bimpelrekkie Oct 20 '16 at 12:49
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    \$\begingroup\$ @0x29a re " I asked a very simple question and I'm sure the answer doesn't involve digging as deep as 'battery university" -> Apparently you ar not ready to learn afterall.| Why do you thing the question was "simple? How hard can it be to read a site aimed at addressing your problem? You have already (probably) destroyed part of your battery. Keep going as at present and the battery proper will die as well. |Note that Boeing an Samsung both underestimated LiIon battery issues. \$\endgroup\$ – Russell McMahon Oct 20 '16 at 14:49

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