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BQ24650 troubleshooting

Alexander MacDonald
Prodigy 180 points
Other Parts Discussed in Thread: BQ24640, BQ24650

I'm having trouble getting my circuit to work. Both LEDs are off, which means that it must be in either charge suspend, over-voltage, sleep or battery absent. I'll address these cases individually;

Charge suspend activates when the voltage at the TS pin is not within a certain threshold, about 1.6 V through 2.5 V. The voltage I'm reading at the TS pin is 2.1 V.

My input voltage is 25 V, and my output voltage is hovering around 0.5 V. I'll mention right now, that I'm using a large, 1 mF electrolytic capacitor in lieu of a battery. This should not make a difference.

The IC goes into sleep mode when the voltage at the SRN pin comes within 100 mV of VCC. VCC = 25 V, SRN = 0.25 V.

Battery absent is a bit more complicated. As I understand it, the chip injects a current out of the SRN pin, and measures the voltage at VFB. If that is below a certain threshold (1.55 V), then it increases the current and measures the voltage at VFB again, if the voltage went up beyond a certain threshold (2.1 V) then the battery is absent, otherwise the battery is present. This can be overcome by shorting VFB to ground. With VFB connected, the voltage is around 50 mV.

I'll go through the enable / disable criteria.

-"Charge is allowed (MPPSET > 175 mV)" -- my MPPSET pin is at 1.2 V.

-"Device is not in UVLO mode and VCC > VCClowv" -- my VCC pin is at 25 V.

-"Device is not in sleep mode" -- see above.

-"Device is lower than AC over-voltage threshold" -- see above.

-"30 ms delay is complete after initial power-up" -- NA

-"REGN and VREF are at the correct levels" -- my Vref pin is at 3.3 V, and my REGN pin is at 5.95 V.

"-Thermal Shut is not valid" -- this occurs at 145 oC. I'm pretty sure the ambient temperature is not that hot, and the IC isn't warm to touch.

-"TS fault is not detected" -- I'm not using a thermistor, rather I have my TS pin pinned to 2.1 V.

Disable conditions:

-"Charge is disabled (MPPSET < 75 mV) -- nope

-"Adapter is removed, causing the device to enter Vcclowv or sleep mode" -- nope

-"Adapter voltage is less than 100 mV above battery" -- nope

-"REGN or VREF" are not valid" -- nope

-"TSHUT IC temperature threshold is reached" -- not to the best of my knowledge

-"TS voltage goes out of range" -- nope

I'm getting rather fed up with debugging this circuit, and if anyone could help, I would greatly appreciate it. Thanks

  • 0
    Prodigy 180 points
    I'd like to update this question. Nothing has changed, per se, but I have some more information.

    1) The STAT1 LED is on. My partner changed my circuit, and had it connected wrong. In fact, the STAT1 is on even if I have nothing connected.
    2) The BTST and HIDRV pins are lower than one would expect. They are rising in time, but very slowly (1 mV every 5 mins) and the current passing through the inductor is tiny.

    This leads me to my next question; what sets the voltage at BTST? Thanks,

    -Alexander
  • 0 in reply to Alexander MacDonald
    TI__Expert 5565 points
    Alexander,

    Are you using the EVM? If not, could you share the schematic?

    How are you forcing 1.2V on MPPSET? Or is it being set by the resistor divider? Is the PH node showing any signs of switching (in other words, is there a switching signal on the HIDRV and LODRV pins?)?
  • 0 in reply to Alexander MacDonald
    TI__Expert 5565 points
    Alexander,

    The BTST voltage is determined by the PH node voltage. The purpose of the BTST node is to be the power supply for the high side (HIDRV) FET driver. Since we are driving NMOS FETs, we need a gate voltage higher than the source voltage of the FET. By using an external diode and capacitor tied to the REGN (which is the output of a 6V LDO - see page 11 of the functional block diagram), we are able to put 6V on top of the PH voltage in order to get enough gate voltage (with respect to source voltage of the high side FET) to turn on the high side FET.

    So to approximate - the BTST voltage will be ~6V higher than your input voltage, minus the input blocking diode voltage drop. Therefore, if your input voltage is 25V, BTST should be ~ (25V - 0.6V + 6V) = 30.4V.
  • 0
    TI__Expert 5565 points
    Alexander,

    One additional comment - if you are forcing 1.2V on MPPSET, try increasing the voltage. The 1.2V on this pin should correspond to the input voltage point where you want charging current to fold back in order to maintain the solar panel's output voltage (ensuring that the solar panel voltage doesn't crash).
  • 0 in reply to Smriti B
    Prodigy 180 points

    BQ.pdfI'm forcing the MPPSET pin with a resistor divider, and I'm not seeing any sign of switching on either the HIDRV or LODRV pins.

  • 0 in reply to Smriti B
    Prodigy 180 points
    Thanks for the clarification. The next question would by, why the PH node is not being drawn up to 24 V.
  • 0 in reply to Smriti B
    Prodigy 180 points
    I've tried increasing the voltage, to no avail. In a separate note I tried using the BQ24640 with an almost identical circuit (I had to make a couple changes), and was getting the same behavior. I have two final thoughs;

    1) Could it be my FETs? I'm using NVD5890NT4G made by OnSemi.
    2) Could it be because I'm using a breadboard and this; www.proto-advantage.com/.../product_info.php
  • 0 in reply to Alexander MacDonald
    Prodigy 180 points
    I would just like to revise; I'm using a 100 mO sense resistor, not a 24 mO one. Also, I don't have the 0 O resistor installed. The FETs I'm using are NVD5890NT4G.
  • 0 in reply to Alexander MacDonald
    TI__Expert 5565 points

    Alex,

    Can you check to see what the REGN voltage is? If there is no voltage on REGN, then the HIDRV pin will not put out a voltage, which means no switching.

    Also, STAT1/2 are open drain pins. Therefore, to know the status of the charger, the STAT1/2 should be pulled up to a voltage. See the application circuit on page 2 of the datasheet for an example.

    Additionally, I would recommend a 10uH inductor instead of the 3.3uH you are using to ensure that the LC resonant frequency is within 12-17kHz.

  • 0 in reply to Smriti B
    Prodigy 180 points
    The voltage at REGN is 6 V. I actually had STAT1 and STAT2 connected to VCC, and not GND as shown in the schematic. I tried using a 35 uH inductor and got the same behaviour. Below you can find the voltages at each pin.

    VCC: 24.8 V
    MPPSET: 1.20 V
    STAT1: 0.24 V
    TS: 1.65 V
    STAT2: 23.6 V
    VREF: 3.3 V
    TERM_EN: 3.3 V
    VFB: 6 mV
    SRN: 61 mV
    SRP: 61.5 mV
    REGN: 6 V
    LODRV: 6 mV
    PH: 1.5 V
    HIDRV: 1.5 V
    BTST: 5.8 V
    V_OUT: 61.5 mV

    The STAT1 LED is on, and STAT2 is off, however, if I remove the supercap from the output, the STAT1 LED remains lit, when it should turn off. Thanks for your help!
  • 0 in reply to Alexander MacDonald
    TI__Expert 5565 points
    Alex,

    The voltage you measured on the TS pin (1.65V) does not match the calculated voltage from your supplied schematic (where I calculate 1.95V). I would recommend ordering the bq24650EVM to debug. Once you have changed the external component values on the bq24650 EVM to get the voltage and current you need, compare it to your prototype at the relevant voltage nodes to isolate the problem.
  • 0 in reply to Smriti B
    Prodigy 180 points
    I rebuilt the circuit and used 2 x 5 kO resistors as a voltage divider across TS instead of a 5 kO and a 7.5 kO. Either way the voltage at TS is within the range. Unfortunately my team is working on an extremely limited budget, and we can't afford the EVM. Regardless, pretty much all the nodes are connected to pins, and the only one that is suspect is HIDRV. Since BTST is 5.8 V, shouldn't HIDRV also be around 5.8 V?

    I'm going to remove Q1 from the circuit, and observe the behavior of HIDRV and LODRV.

    It is also very strange that STAT1 would remain lit, even when the battery has been removed.
  • 0 in reply to Alexander MacDonald
    Prodigy 180 points
    I know how they interact, but I still don't fully understand how PH, BTST and HIDRV are set. I understand that BTST and PH are the high and low drivers for HIDRV, which means that BTST > HIDRV > PH, but HIDRV and PH also form Vgs for Q1. Let's begin with a chip coming out of shut down (assuming normal operation);

    BTST is set to 6 V by REGN, which then sets HIDRV to 6 V, which then sets PH to 6 V. What happens next? You said that PH sets BTST, but BTST sets HIDRV, which sets PH, which sets BTST, which sets HIDRV... You see where I'm going with this.
  • 0 in reply to Alexander MacDonald
    TI__Expert 5565 points
    Voltages for PH, BTST, and HIDRV for the chip coming out of shutdown:
    PH = 0V
    BTST = REGN - (BTST diode drop)
    HIDRV = 0V

    With the above voltages, the high side FET remains off. For the high side FET to turn on, HIDRV will go from 0V to BTST. As the high side FET turns on, the PH node will start to rise. When the high side FET is fully on, PH = VIN (which is the drain of the high side FET). Since PH went from 0V to VIN, the BTST cap reference point (the point tied to PH), but the voltage at BTST is PH+~6V since the BTST cap was charged to REGN minus diode drop before the high side FET was turned on. As long as a ~6V delta is maintained across the BTST cap, the high side driver has the required voltage to give the high side FET the Vgs it needs to turn on.

    Therefore, when the high side FET has gone from fully off to fully conducting, here are the voltages:
    PH = drain of high side FET minus I*R losses of the high side FET
    BTST = PH + REGN - (BTST diode drop)
    HIDRV = BTST - (losses in the IC)

    Figure 6 in the datasheet is a good example of PH and HIDRV waveforms.
  • 0 in reply to Alexander MacDonald
    TI__Expert 5565 points
    Alex,

    It seems odd that the PH node is 1.5V, but the SRP node is only 61.5V. Are you sure the inductor is connected? If the inductor is connected corrently, then PH = SRP.
  • 0 in reply to Smriti B
    Prodigy 180 points
    I accidentally had the inductor between SRP and SRN, not SRP and PH. That would explain that discrepancy. However...

    I think I cracked the problem. When V_in was at 25 V, PH, SRP and SRN were stuck at 75 mV, however when I turned the input voltage up to 26 V or higher, they jumped to 2.08 V, but stayed there. This would indicate to me that the problem was with the MPPT biasing. Fixing that is not going to be hard.

    Another "problem" did come up. Once PH, SRP and SRN got to 2.08 V, they stayed there, however, if I turned the supply off and on (or if I partially charged the caps prior to turning the supply on), they would charge to 24 V. Bearing in mind that for the time being I'm using caps, and not a battery, I believe this is just a matter of the precharge, which should fix itself once I get the batteries in. Please confirm that this is indeed a precharging issue.

    There was another issue, but I believe it to be a capacitor issue. I'm getting the batteries this weekend, and will begin re-testing next week. I'll update you when I know more.
  • 0 in reply to Smriti B
    Prodigy 180 points
    While it wasn't the only problem, using a 35 uH inductor did help.
  • 0 in reply to Alexander MacDonald
    TI__Expert 5565 points
    No, this is not pre-charge mode. Assuming that you are still using a 1mF cap as your load, the bq24650 is stuck in battery detection mode. See pages 18 and 19 in the datasheet for detailed flow and graphical explanation.