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BQ2970: Using BQ2970 as low voltage indicator to put voltage regulator into sleep mode.

Marcell Veszpremi
Prodigy 30 points
Part Number: BQ2970

I would like to use the BQ2970 battery protector to act like a low battery indicator. I am using a voltage regulator which can be put into sleep mode (when a pin is pulled logic LOW - the over discharge mosfet pin would control my regulator perfectly) using a protection IC. I would like to use another device which better suits my application or how I can use this device. The problem I am having is that on the V- pin there will be unexpected behaviour as no MOSFETs are present. Thank you for any help. 

  • 0
    TI__Expert 7035 points

    Marcell,

    Could you explain in more detail what problems you are having on the V- pin? It should not be affected by the absence of MOSFETs.

  • 0 in reply to Shawn Hinkle
    Prodigy 30 points

    Hello,

    I have not actually tested this but strongly suspected that it wouldn't work, so sought support. Considering the data sheet of the device, it says minimum 100mv hysteresis is expected for the device to go back into regular operation after over-discharge condition. A diagram in the data sheet illustrates this with the expected voltages on the V- pin. As I would be connecting DOUT to a standby pin of the regulator I would expect unusual behaviour. I am wrong in my understanding, where could I connect the V- pin to make sure the device function correctly if this device can function without MOSFETs operation.

    Thanks,

    Marcell

  • 0 in reply to Marcell Veszpremi
    TI__Expert 7035 points

    Marcell,

    The V- could be connected to VSS using a 2.2k Ohm resistor. With this connection, the device should never go into shutdown mode but it will also should not detect overcurrent. Since faults cannot be disabled, if any fault occurs the V- pin would have to be manipulated to recover from that particular fault. Refer to section 9.4.3 of the data sheet for the behavior of the device in an overdischarge condition. For the overdishcharge fault, the V- pin would need to be higher than -0.7V to recover.

  • 0 in reply to Shawn Hinkle
    Prodigy 30 points

    Shawn,

    I understand. I would just like to clarify what the hysteresis voltage means then? Also, the V- pin will be higher than -0.7V at all stages as over discharge kicks in at 2.8V, meaning that if V- is connected to VSS with no MOSFETs on the the line, it would be -2.8V. So the V- pin would meet the conditions. I still don't understand what the 100mV hysteresis is required for or if it will be satisfied. Do you believe that the way I would like to use this device, the device will function correctly?

    Thanks,

    Marcell

  • 0 in reply to Marcell Veszpremi
    TI__Expert 7035 points

    Marcell,

    The 100mV hysteresis must happen in conjunction with the -0.7V for the device to recover. This means that if the voltage on the V- pin drops to -0.75V the device will trigger UV condition. If the V- pin rises to -0.68V, the device will not recover because it did not increase by at least 100mV. In this example, the V- pin would have to rise to at least -0.65V to recover.

    The exact requirement for the device to recover is described in Table 7.6. In order for the device to recover, the difference between the BAT pin and the V- pin must be greater than 1 V. 

    If you make sure to correctly connect the device, it should function according to your needs.