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Ask an Expert Forum Replacing a Li-ion battery with wallpower
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Replacing a Li-ion battery with wallpower

Former Member
Former Member over 9 years ago

I've taken apart a bluetooth speaker that ran on a 3.7V  1500 mAh battery. I want to run it off of wall power, but the battery has three terminals.

 

image

I'm assuming the middle terminal from the battery is to check the charge, but how would I bypass that? It doesn't work without the middle terminal.

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  • mcb1
    mcb1 over 9 years ago in reply to Former Member +3 suggested
    how did you decide which size capacitor to use A good rule of thumb for smoothing is 1000uF/Amp. Since you are replacing a battery and the device could have high demand you can easily double or treble…
  • jw0752
    jw0752 over 9 years ago in reply to Former Member +1 verified
    Hi Chas, Great pictures and they make me more confident in my first advice. mcb1 has a good suggestion to add a capacitor across the power input line. I would suggest a 470 uF to 1000 uF. at 16 Volts.…
  • Former Member
    Former Member over 9 years ago in reply to jw0752 +1
    I'm going to order a few parts and test some of your suggestions. Thank you for your help! I'll let you know how it goes
  • clem57
    0 clem57 over 9 years ago

    Chas Irish

    If you read this as an example, you see the middle is a common ground.

    Clem

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  • jw0752
    0 jw0752 over 9 years ago

    Hi Chas,

    There is a red and black wire that come from the battery interface board. I suspect that it will work if you put  plus 3.7 volts on the red wire and the negative terminal on the black wire. Be careful if you are using a wall wart as the board may not be able to tolerate much more than 3.7 volts. Test the power supply for the proper voltage and also make sure it has the current capacity to run the speakers amplifier. A battery can deliver quite a lot of peak current. If you want take a couple more pictures and post them so I can get a better look at the circuit before you go further it wouldn't hurt.

    John

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  • mcb1
    0 mcb1 over 9 years ago in reply to jw0752

    John is quite right.

    I would also add a large capacitor as the battery will be capable of supplying high peak currents.

    The power pack might be capable depending on the rating.

     

    Mark

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  • Former Member
    0 Former Member over 9 years ago in reply to jw0752

    Thanks for looking into this!

     

    I too, though that I could power the board through the red and black terminal. I had some 3.7V batteries from adafruit that have the same connector but when I plug them in, I get nothing. I checked them with a multimeter to make sure they're giving off the correct voltage and they are.

     

    The small circuit board that has the 3 battery terminals on it doesn't seem to have any components, but some how the leads go from 3 to 2.

     

    imageimageimageimageimage

     

    Just to be clear, I want to completely eliminate the battery from this device while still being able to power it. The micro-usb port on one of the boards is only there to charge the battery. The device will not function while it is plugged in.

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  • jw0752
    0 jw0752 over 9 years ago in reply to Former Member

    Hi Chas,

    Great pictures and they make me more confident in my first advice. mcb1 has a good suggestion to add a capacitor across the power input line. I would suggest a 470 uF to 1000 uF. at 16 Volts. Make certain that you connect the negative to the negative and the positive to the positive. I do not see any reason why your other battery didn't work but you could verify that you have the requisite 3 to 4 volts on the BT2 terminals by taking your meter and checking when the battery is attached. Good luck and let us know when you get it working.

    John

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  • mcb1
    0 mcb1 over 9 years ago in reply to Former Member

    Three pins at the battery end are often for charging purposes.

    They detect the voltage at the battery end of the cable and whatever is inside the battery protection circuitry, which overcomes any cable resistance.

     

    I suspect this is where your 'three to two' is coming from.

     

    There are plenty of points to connect a supply to on the boards in the photos.

    Yes sometimes the USB is just a charger connection.

     

    Mark

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  • Former Member
    0 Former Member over 9 years ago in reply to jw0752

    I'm going to order a few parts and test some of your suggestions. Thank you for your help! I'll let you know how it goes

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  • Former Member
    0 Former Member over 9 years ago in reply to jw0752

    This worked! Thank you! If you don't mind me asking, how did you decide which size capacitor to use? I'm sure I'll need to use a few more later on in this project to clear out some noise.

     

    mcb1 Thank you for your help as well

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  • jw0752
    0 jw0752 over 9 years ago

    Hi Chas,

    As with many things it is a choice between two limiting factors. In this case we want enough capacitance so that a sudden demand for current can be met without drawing the input voltage too low. The other factor is that too high a capacitance will look like a short to your power supply when it is first turned on. This can cause some power supplies problems. The real engineers  (I'm just a technician with many years of experience) can calculate the optimum capacitance if they have all the parameters of your power supply and application. Experience suggested that it was a good value to begin with. I am glad that things worked out for you.

    John

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  • mcb1
    0 mcb1 over 9 years ago in reply to Former Member

    how did you decide which size capacitor to use

    A good rule of thumb for smoothing is 1000uF/Amp.

     

    Since you are replacing a battery and the device could have high demand you can easily double or treble that.

     

    Capacitors nowdays are much smaller, able to handle higher ripple currents and cheaper than the older days.

    Because larger value electrolytic (or even Aluminium) capacitors don't handle high frequency spikes, adding a 0.1uF (100nF) across them is also good practice.

     

     

    Mark

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