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BQ2970: BQ29700 not turning on pass FETs

Part Number: BQ2970
Other Parts Discussed in Thread: CSD16406Q3

I have the following circuit in a prototype board:

At this point, I have verified that all footprints and layout are correct, all connections are good. However, when I plug in the battery, the two FETs in Q7 do not turn on and no return path is established. I have also gotten to the point that both U10 and U11 are removed to totally isolate just the BQ29700 function. Still will not turn on the FETs. One thing I did find is that the dual FET chosen for Q7 actually has 1k gate resistors built into the package. I am wondering if this is causing sufficient turn-on delay that the chip sees a high voltage between GND and BAT_RTN terminals and thinks there is a fault condition. However it doesn't try to recover.

One other thing - I tied a 100k resistor between GND and BAT_RTN. With this in place, it will turn on after ~1sec delay, but then it is stable. I can see the proper voltages at the gates of both FETs and I can then remove the 100k resistor and it stays on. It just needs some sort of kick-start to get it stabilized. Anyone encounter this before? Could it be the 1k gate resistors? The part - MTM78E2B0LBF - is specifically designed for this application, so I don't understand the behavior.

Thanks,

Jamie

  • Hi Jamie,

    We are currently hosting the BMS Deep Dive training session this week, but we will get back to you by Friday afternoon.

  • Hi Jamie,

    Thank you for your patience. When you plug in the battery, what is the value on the COUT and DOUT pins? That can help us determine if the issue resides in the device or in the FETs. It also may be a good idea to walk through the fault conditions and make sure none of them are occurring.

    Additionally, make sure to walk through the detailed design procedure outlined in section 10.2.2 in the datasheet. Since the Rds_on of the FETs control the current protection thresholds, FET selection is extremely important. I noticed the typical Rds_on value for your FET is 21.5m Ohm, this is the value you can use for your calculations.

  • Hi Shawn,

    G-S voltage for both FETs is 0V (measuring w.r.t their associated GND). At this point, I have totally isolated the device so that there is no load at all, so this can't be an overcurrent issue due to FET Rds_on. Also, I thought maybe the FET package integrated gate resistance (1k Ohm) might be a problem, so I wired in two discrete FETs (IRLML6346) with no gate resistors and the behavior is the same. Any suggestions for tests I can perform to further diagnose this?

    I want to point out again that I can bridge the GND and VBAT_RTN with a 100k resistor and the device will turn on after about a second. I can then remove the 100k resistor and it remains on in a stable state. Further, if I just directly short the two grounds by touching a couple of wires together, it turns on instantly and stays stable. This suggests to me that there is some capacitance somewhere that needs to charge up in order for the device to be stable, but I can't really make sense of that in the context of the circuit. What do you think?

    -Jamie

  • Hi Jamie,

    I still suggest measuring the value of the COUT and DOUT pins directly on the device. This can help us determine how the device itself is behaving regardless of the FETs. 

    We suggest using the CSD16406Q3 FET with this device. We have a note in the datasheet on page 17 that recommends shorting V- to VSS when the battery is first connected if the circuit is not enabled. Alternatively, you can connect a charger for the same effect.

  • Hi Shawn,

    The note about shorting the V- to VSS is something that I missed and dovetails with what I'm seeing in the application. I think this may be the answer, but I am curious as to why this happens - can you elaborate so I understand? Thanks.

    -Jamie

  • Hi Jamie,

    Sometimes a fault is falsely detected when the battery is first connected. The device uses the differential voltage between V- and VSS to determine if some faults are occurring. Shorting these two pins will clear the fault and allow the device to resume normal operation. 

  • Hi Shawn,

    This is exactly the behavior I am seeing and I totally missed that note on page 17. So I think this is resolved. However, for my own curiosity, can you explain why this happens? Thanks.

    -Jamie

  • Hi Jamie,

    See my last post for explanation.