LiFePO4 battery Monitor

Discussion in 'The Projects Forum' started by SkyBill, Feb 12, 2010.

  1. SkyBill

    Thread Starter New Member

    Feb 11, 2010
    4
    0
    Hello,
    I built an electric motorcycle using 24 - 90 A-hr LiFePO4 batteries. I would like to build an individual battery monitor that would operate between 3.5V and 2.5V. I have already constructed a basic circuit using the LM3914 chip. I just followed the diagram in the spec sheet and built a 0-3.5V circuit with a 10 element LED. I understand how to calculate the Vref out but I need help with figuring out how to use the R lo (pin 4) to get a 2.5V - 3.5V monitor.
    Could you help me understand how the R Low works and how to determine what resistance I need?
    Right now I am using 2.18 Kohm for R2 and 1.21 k ohm for R1. 0 - 3.5 V works great. Need 2.5V - 3.5 V and would like to understand why.
    Thanks
    PS If you'd like to check out the bike it's at electrovoyageur.com
     
    Last edited: Feb 12, 2010
  2. bertus

    Administrator

    Apr 5, 2008
    15,649
    2,348
    Hello,

    Can you post a schematic?
    With the schematic we can take a look how to adjust it.

    Greetings,
    Bertus
     
  3. SkyBill

    Thread Starter New Member

    Feb 11, 2010
    4
    0
    I should have a diagram uploaded.
    I am using a 9V battery for the power supply and another 9V with a potentiometer to give the signal voltage.
     
  4. bertus

    Administrator

    Apr 5, 2008
    15,649
    2,348
    Hello,

    When you take a look at the internals of the LM3914, you can see there are a string of resistors between Rhi and Rlo.

    [​IMG]


    By putting a resistor between Rlo and ground you can lift the lower detection voltage.
    The value of the string of resistors is about 10K.
    You want to have a range of about 1 Volt.
    The top value is 3.5 Volts.
    The 1 volt must be accross the 10K, that means a cuttent of 1 Volt / 10K = 0.1 mA.
    You want a lower value of 2.5 Volts, The needed resistor will be about, 2.5 Volts / 0.1 mA = 25K.
    You can take a resistor of 22K with a pot of 5K in series to adjust the lower value.

    Greetings,
    Bertus
     
  5. SkyBill

    Thread Starter New Member

    Feb 11, 2010
    4
    0
    I will try it tonight.

    ....
    it worked great. Thanks for the excellent explanation and the help.
    PS
    When the voltage drops below 2.5V the bottom LED goes out. Any way you can think of that would keep it lit when below 2.5V?
    Thanks again
     
    Last edited: Feb 12, 2010
  6. bertus

    Administrator

    Apr 5, 2008
    15,649
    2,348
    Hello,

    If you want an undervoltage detection, you will need an extra comparator.
    Put the - input of the comparator to Rlo, put the + input of the comparator to the input signal at pin 5.
    The output of the comparator is probably an open collector, there you can put a led with a current limiting resistor to the + of the powersupply.

    Greetings,
    Bertus
     
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