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BQ2970 seems not suitable for applications with removable batteries, any suggestions?

Valentin Goverdovsky
Prodigy 30 points

Other Parts Discussed in Thread: BQ2970

We would like to have under voltage protection for a removable lithium battery which doesn't have any protection circuitry by itself, e.g. coin cell in CR2450 form factor. I've tried using the BQ2970 chip and it works as expected, provided you start it up by following the suggestion in the Caution: 

When the battery is connected for the first time, the discharging circuit may not be enabled. In this case, short the V– pin to the VSS pin.

Alternatively, connect the charger between the Pack+ and Pack– terminals in the system.

The problem is that it's really cumbersome to be shorting the pins every time the battery is inserted, since we'd need to have a button on the board to do this, which will take up valuable space. We also charge the batteries on a separate charging block, thus a charger is never connected to the actual device.

Does anyone have any suggestions for a workaround or an alternative chip which would provide under voltage protection and would work as soon as the battery is inserted?

  • 0
    TI__Expert 4640 points
    Hi Valentin,

    The bq2970 is a protection device that is typically inside a battery pack before it is shipped or sold by the pack maker. While a protector can be placed on the system side there are certain considerations to keep in mind for such applications. Now with a removable battery is a new challenge. Can you let me know what battery pack you are using? As I mentioned before typical battery packs come with protection inside already. 

    Miguel
  • 0 in reply to Miguel Angel Rios
    Prodigy 30 points
    Hi Miguel,

    Thanks for your reply. Yes, now I understand that BQ2970 is for battery packs and is to be installed during manufacturing. We are using these guys: www.farnell.com/.../1475807.pdf They don't have any protection built in, therefore at the moment our device goes to sleep when it detects that the voltage has dropped to below 3.1V, but even during sleep it consumes somewhere around 200-500 uA, which implies that the battery will eventually and pretty soon be discharged to below the recommended 2.8V. Ideally what we want is for the device to detect 3.1V and go into low battery mode whereby it would flash an LED at certain rate notifying the user that battery needs replacing, but eventually the battery should be completely cut off if the user ignores or doesn't notice the indication. Any suggestions would be great!

    Valentin
  • 0
    TI__Expert 4640 points
    Hi Valentin
     
    This seems like something funky is going on here, given that if the bq2970 was truly in sleep mode then it would be consuming in the nA instead of uA (let alone of hundreds of uA).
     
    You should be able to do this easily. Which spin version are you using? bq297XX?
     
    Have you isolated the system and just focused on the protector to make sure that it is working properly? It should only need 1.5V to stay powered on. If you can confirm that it is reading 1.5V at the BAT pin and the V-pin that would help to understand if this device is really in power down mode.
     
    Best Regards,
    Miguel
     
  • 0 in reply to Miguel Angel Rios
    TI__Genius 10785 points
    Valentin,

    Power consumption for normal mode is 5.5uA(max) for bq2970. Shutdown is 100nA(max).
    Something else is drawing the 200ua-500uA current not bq2970. Yes, I agree that it will drain your battery pretty fast.
    I think you should focus on other components.

    Bq2970 will not cause 200uA of current.
  • 0 in reply to Miguel Angel Rios
    Prodigy 30 points

    I appreciate your replies, but it seems like you misunderstood my problem here, probably because I've given too many details and used the word 'device' ambiguously', for which I apologise. When I said 'device' I didn't mean just the BQ2970, but rather the whole device that I'm building, which has MCU, ADC, SD card etc. BQ2970 is just one component of that device whose only intended purpose is to detect low battery voltage (2.8 V) and shut off the battery from the rest of the circuits to prevent deep discharge.

    I KNOW that BQ2970 is not drawing 200-500 uA and I KNOW which other components of my device are responsible for this current draw!

    MY PROBLEM:

    BQ2970 DOES NOT activate discharge circuit when I insert the battery (the battery is the LIR2450 rechargeable coin cell with no protection) and this behaviour is exactly as per the datasheet. I'm asking for suggestions of alternative chips which can provide under voltage protection or under voltage protection circuits that I can build, which would ACTIVATE AUTOMATICALLY upon battery insertion.


    THE PROBLEM YOUR SUGGESTIONS ARE TRYING TO SOLVE:

    Relatively high power consumption in sleep mode. To reiterate, I know which components draw this current in sleep and there isn't much that I can do about it since this current draw is as per the data sheets of those components. Therefore, thank you, but I'm not looking for suggestions related to the current draw 'problem'.

    Hope this clears things up.

    Valentin

  • 0
    TI__Expert 4640 points
    Hi Valentin
     
    I see what you are wanting to do. By design most (if not all protectors) behave in a similar fashion. What you are seeing is that the protector is not designed to power on the moment a cell is connected immediately, as you point out, per the DS you must short V- and Vss together to jump start the protector. This is more of a safety feature as during assembly a manufacturer would apply this voltage across the V- and Vss pins for the customer. A way around this is to require the end user to apply a charger instead of always applying this differential voltage. We don’t have any other devices that can support a hot swappable feature.
     
    BR,
     
    Miguel
     
  • 0 in reply to Miguel Angel Rios
    Intellectual 375 points

    Hi,

    this short between V- and VSS is a little confuse to me. Is it a permanent short  or just a temporary and quick one? In case of being temporary, how is the process envolving it? It would be of great value if you could answer that.

    Regards,

    Vítor Pereira

  • 0 in reply to Vítor Pereira
    TI__Expert 6085 points

    Hi Vitor,

    the resistor value of ~2.2K provides isolation  for  the V ( -) pin from the PACK- transients.

  • 0 in reply to Vish Nadarajah
    Intellectual 375 points
    So the circuit is going to work if it's exactaly like it is shown in the datasheet, or there is some other issue I should concern about?

    Vítor Pereira
  • 0 in reply to Vítor Pereira
    TI__Expert 6085 points
    Yes; the EVM available @ TI website should give all insights.
  • 0 in reply to Vish Nadarajah
    Intellectual 375 points
    Ok, thanks!

    Vítor Pereira