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3 NiMH batteries in series with BQ2002 charger IC

Jeffrey Shih
Prodigy 50 points
Other Parts Discussed in Thread: LM317, BQ2002

Hi,

For the  DV2002L2, The pin5(TS) is connected to the R5(RT1) 3.57k with a 10k resistor in series.So I need to get the NTC 3.57k and the regular 10k resistor? Then I calculate  the value of VTS is 5V * (10k/13.57k) =  3.68V, am I right? Also, what's the function for R17(100k) ? If I take out the R17, what will be happened? Also, if the pin5(TS) is going to ground, what will be happened too?

Another, how to do the pin8(CC)? I know the pin is to control the current flow. In the DV2002L2 schematic, it goes to the npn transistor's emitter. Why need to do that? Can I float the pin8(CC)?

 The below link is my currently design. I am not sure the circuit should be worked. Please let me know the issues for the circuit.   My design is only to charge 3 NiMH (2000mAh) batteries (from VBAT+ to VBAT-) in series with 5V adapter input.  

https://docs.google.com/leaf?id=0ByYtrhrwtaGbMTc5MzJkMjgtOWUzMC00MjI3LWFmMDktNDk5YWU2MDE2NGJk&hl=en_US

  • 0
    TI__Guru**** 150790 points

    Please read the UG and datasheet before using these parts. 

    The UG (http://focus.ti.com/lit/ug/sluu007b/sluu007b.pdf) givesrecommended thermsitors.  Use one of them.  This post discusses this more: http://e2e.ti.com/support/power_management/battery_management/f/179/p/74822/272710.aspx#272710

    R17 and C5 create a low pass filter on the TS pin.

    If TS is tied to ground, the TS pin will be 0V.  What the IC does in the case is explained in the datasheet.

    The CC pin's operation is explained in the datasheet.  It is the IC's output pin.

    I recommend that you follow the EVM design or contact a design services company to do your design.  Potential issues that I see with your design are: you need something to limit the charging current--limit the current in Q5; and your gate drive circuit of Q5 and Q6 may not support the desired fast charge current or power lost in Q5.

  • 0 in reply to Chris Glaser
    Prodigy 50 points

    Hi,

    The UG gives recommended a thermistor R5 is 3.57k, but not specified for 10k resistor. I just confused for 10k resistor. Is it using 10k thermistor or just a regulor resistor?

    The input 5V adapter provide 1A, and the Q5 Ic supports to 0.8A. The 1N4148 forward current is only 0.2A.  I think I should change the diode to have bigger forward current up to 1A. Am I right? I don't understand the power lost in Q5. Can you explained that? Thanks very much.

  • 0 in reply to Jeffrey Shih
    TI__Guru**** 150790 points

    Did you look at the user's guide yet?  http://focus.ti.com/lit/ug/sluu007b/sluu007b.pdf

    The picture of the board on page 1 shows the current source used on the EVM--the LM317.  It is bolted to a heat sink that occupies about 1/4 of the PCB area.  This is what I mean by power lost in Q5.

    The EVM is designed to accept a 10k thermistor.  If you don't want to use a thermistor, which is less safe, then you can use a fixed 10k resistor instead.

    You need to get a design services company involved that understands these things and who can make sure your battery charger is safe.  The adaptor may be rated for 1 A, but when Q5 is turned on, then is nothing to limit the charging current.  The adaptor might be able to supply 5A into a short circuit which is very unsafe for the batteries.

  • 0 in reply to Chris Glaser
    Prodigy 50 points

    One more question. The battery I may use is about 2000mAh. For my design, I set up the  pin1 (TM) to be low. That means it will provide 1C( around 2000mAh) fast charge. But if I use the LM317 and set the control current about 500mA, the charge time will be more than 240 minutes instead of 80 minutes. Does it conflict with pin1 (TM) setting?  

  • 0 in reply to Jeffrey Shih
    TI__Guru**** 150790 points

    The bq2002 does not control the amount of charging current.  You must set this externally.  It just turns this current source on or off.

    If you charge at 500 mA and set TM low, then the IC will time out before your battery is full.  You need to charge at at least C/2, per table 1.