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BQ24278

leo14
Intellectual 350 points
Other Parts Discussed in Thread: BQ24278

I completed BQ24278 test circuit with the following configuration:

Firstly, I drove bq24278 without battery connected. When both CD & CE/ was connected to Vcc, LEDs were on and Vsys was about 1.7V, where system drew less than 1mA. When CD was pulled down to GND, LEDs became OFF, Vsys=0V, where system drew 41mA current from power supply. Then, I disconnected TS pin from the design, and pulled TS pin to DRV. In this case, the current bq24278 draws increased rapidly to the limit of power supply(1A for example), then operated with constant voltage which was less than Vcc. 

There is a great problem with the design as you understand. It is most likely due to grounding of the chip. I was told by Mr. Jeff that power, system, and battery ground should have been connected to PGND, whereas PMID, DRV, and VDPM should also be connected to PGND. Which pins should I connect to PGND? Due to some manufacturing problems, I couldn't implement design into PCB. Does it create this much operation problem? What do you propose?

Thanks.

  • 0
    TI__Guru*** 143910 points
    Resistors R53 and R52 are not sized correctly for a typical 10k thermistor. Please use the datasheet equations to size those resistors for your thermistor. Also, with input current limit set to 1A, the charge current will never much larger than 1A, even though you have set the charge current to the maximum.
  • 0 in reply to Jeff F
    Intellectual 350 points

    Vdrv=5.2V, Vcold=3.12V, Vhot=1.56V, Rcold=32.65k (at 0C), Rhot=2.48k (@ 60C). I re-calculate two values and results are the same. I don't think there is something wrong with the resistances (as I said previously, I also connected TS to DRV as you suggested for another person). I think problem is due to grounding. What you are saying about which pin to connect either PGND or AGND does not match with the datasheet. Which one should I take into account? 

    Have a nice day.

  • 0 in reply to leo14
    TI__Guru*** 143910 points
    Power ground (PGND) is for the buck converter so only the IN, PMID, SYS and BAT capacitor grounds should ideally connect to it. All other grounds should connect to analag GND (AGND). AGND and PGND should connect to each other at the IC. I have seen other customers not follow this advice and see no issues with charger operation. Is the input voltage too low and therefore triggering the VINDPM circuit to reduce input current limit?
  • 0 in reply to Jeff F
    Intellectual 350 points

    I don't think so. I supply 5V Vin, and 1.20V appears at VDPM pin; however, there exists no current drawn by the circuit. When CD and CE/ are high, approximately 1.3V is read from SYS pin, which is far from being normal operation. In addition, \PG and \CHG LEDs are flickering instead of normally shining. One last problem is when a battery is connected. It pulls SYS pin to its voltage and the circuit still draws no current. I couldn't have found any solution yet. 

    Note: TS pin is directly connected to DRV.

  • 0 in reply to leo14
    TI__Guru*** 143910 points
    With CD high and no battery attached, SYS does not regulate to VBATREG but float to slightly above 1.0V due to leakage current. With /CE high, charging is disabled. Take CD and /CE low for normal operation.
  • 0 in reply to Jeff F
    Intellectual 350 points

    Hi Jeff,

    As you said earlier, operation failure is due to input voltage at 5V which results in low DPM threshold at the VDPM pin. However, it quite confused me. Supplying 8V that creates 1.9V at VDPM pin seems violation for the VDPM threshold which provided between 1.18 and 1.22V at the datasheet. Is V_in_dpm voltage equal to Vin, or is it the value that should be arranged with resistor divider between Vin and VDPM pin?

    Thanks.

  • 0 in reply to leo14
    TI__Guru*** 143910 points

    The resistor divider should be set to give 1.2V typical when V(IN) is has fallen to the voltage at which you want input current limit to be zero (i.e., the maximum rated voltage of the source).