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BQ34Z100-G1: Current peak on sampling

Gabriel Lucas
Prodigy 230 points
Part Number: BQ34Z100-G1

Hi TI,

Our product is embedding a BQ34Z100-G1. I've noticed a current peak and therefore a voltage drop when the IC is sampling. I know it is during sampling as while the IC is in FULLSLEEP (current used by design < 1mA), the current peak appears every 20s which is the FULLSLEEP sampling rate. The current peak doesn't appear on the RShunt resistor so it sure happens on the bq34z part of our design.

Here is the electronic scheme designed

Do you see something that could cause this issue ?

Best regards,

Gabriel Lucas

  • 0 in reply to Gabriel Lucas
    TI__Mastermind 24870 points

    Hello Gabriel,

    Can you attach another image? The one didn't come through.

  • 0
    TI__Guru 61206 points

    It's likely the inrush current when the gauge re-charges the 1uF capacitor on REG25 when it exits FULL_SLEEP. You can try limiting the current into the gauge (and subsequently the capacitor on REG25) with a series resistor on REGIN.

  • 0 in reply to Kang Kang
    Prodigy 230 points

    Hello Kang,

    It didn't when I first updloaded it but now it seems displayed on the first post.

    Let me know if it is still not the case.

    Best regards,

    Gabriel Lucas

  • 0 in reply to Dominik Hartl11
    Prodigy 230 points

    Hello Dominik,

    I am not sure I understood well. What is the state of the 2.5V regulator during FULL_SLEEP ? I am measuring a stable 2.5V on the 1uF capacitor during the 20s of FULL_SLEEP. Then, when the gauge wakes up to sample the battery voltage and the current, the regulator should not have much to recharge on the capacitor. Are you telling me that the REG25 pin is sinking before the regulators enables and therefore the capacitor is discharged then recharged in a small amount of time ?

    What calls me to mind is that the application circuits in the datasheet doesn't suggest any series resistor and I can't believe that this behaviour wouldn't be noticed at TI and pointed out on the datasheet.

    I am going to try what you mentionned and let you know the result.

    NOTE: The current peak is around 30mA for 125ms.

    Best regards,

    Gabriel Lucas

  • 0 in reply to Gabriel Lucas
    Prodigy 230 points

    The board consumption has been measured with and without the 1uF capacitor:

    As you said Dominik, the issue is with this capacitor. Thank you for this correct guess.

    I have few questions adding to those of my last reply:

    - I need to reduce this behaviour at short-term without any PCB modification, can a 100nF capacitor take the place of the 1uF ?

    - For the long term, what resistor value can I choose that would minimize the consumption but doesn't prevent the gauge from working properly ?

    - Finally, what is the impedance on the REG25 pin ? I can't find anything about this in the datasheet.

    Best regards,

    Gabriel Lucas

  • 0 in reply to Gabriel Lucas
    TI__Guru 61206 points

    A 100nF capacitor will not going to reliably provide the power for the gauge during full sleep. The gauge relies on the capacitor during full sleep (it will not keep the regulator constantly on during full sleep) so if you use a 100nF capacitor, you'll likely cause the voltage on this capacitor drop too much during full sleep and the gauge will go through a power on reset.

    The series resistor depends on the internal resistance of your power source (battery). Right now, the internal battery resistance seems too high for the inrush current and the voltage outside the battery drops as a result. The gauge itself won't add a significant resistance to the path (a few mOhm) so you can calculate the inrush current and duration with your battery's internal resistance and a 1uF capacitor. Then select a series resistor that guarantees that the voltage at the battery connector doesn't drop below your required minimum voltage. Make sure it doesn't get too high otherwise the voltage measurement from the gauge will become a function of the gauge's own current draw (keep this below 1mV) which will interfere with gauge performance.

  • 0 in reply to Dominik Hartl11
    Prodigy 230 points

    Hello Dominik,

    Thanks for your reply, that makes things clearer. Still, it seems that the inrush current (30mA peak spreading on 125ms and drawn from a 3.6V supply) is quite high if only related to the capacitor charge and discharge cycles. As I am measuring 2.5V during the 20s of FULLSLEEP, the capacitor hasn't discharged much, supposing the regulator is off. So it looks like it is somehow discharged heavily when waking up which also causes a heavy charge afterwards. A serie resistor should reduce that but I want to make sure I don't have an additionnal issue.

    I need to confirm if the gauge is behaving as expected. The datasheet indicates a nominal FULLSLEEP current of 30uA, is this the current drawn on REGIN plus the one drawn from the 1u capacitor ? Are the capacitor charge/discharge cycles taken into account in this value ?

    Best regards

  • 0 in reply to Gabriel Lucas
    TI__Guru 61206 points

    The gauge will charge up this capacitor with just a few uOhms of resistance inside the gauge so if your external circuit doesn't limit the current, it can be very high.

    The nominal FULLSLEEP current is the average current during full sleep over a long time (several minutes) that the gauge draws from the power source. This includes any current that is used to charge up that capacitor. It will not be a constant 30uA current but fluctuate.