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BQ24195: Standalone Vsys behaviour

Victor Claver
Prodigy 175 points
Part Number: BQ24195

Hi, 

A bit of basic information about the set-up:

  - Standalone application. 

  - 18650 battery (discharging @1.3 constant current).

  - No Input Source.

The discharge of the battery looks all good until you reach the 2.5ish volts. I was expecting the FET to switch OFF completely and cut off the current to Vsys. However, I find that the Vsys and Isys are switching at 20Hz approx. 

    - Vsys between 2.45V and 0V (pulse width 1ms)

    - Isys between +40mA and -40mA

Vsys progressively gets lower so does the Isys. However, at the beginning of the switching I saw spikes up to 6A. Does this mean I need a protection circuitry? Am I missing something?

Thanks in advance. 

  • 0
    TI__Expert 8230 points

    Hello Victor,

    At the moment, are you using only a cell without protection or a battery pack based on a 18650 cell? Usually if it's a pack, it includes some sort of protection built into it.

    If it's the latter, I suspect what's happening is the protection FETs in the pack are opening when they reach the battery cutoff voltage (typically around 2.5V).Once this happens, the discharge current will be cut off from the system and the battery voltage will begin rising as the cell relaxes. Protection FETs typically have some hysteresis built into them; as the battery voltage increases pass the hysteresis threshold, the FETs will close and allow discharge again. At this point, since the battery is almost completely empty, will proceed to repeat the cycle immediately until the voltage of the cell doesn't relax above the hysteresis threshold.

  • 0 in reply to Fernando Lopez
    Prodigy 175 points

    Hi Fernando, 

    Thanks for your quick response.

    At the moment I am using a single cell 18650 without protection. Probably wrong, but I assumed that if the BAT FET cuts off at about 2,5V I wouldn't need that protection on the battery. Is that right or I still need a protected battery?

    Thanks again.

    BR

    Victor

  • 0 in reply to Victor Claver
    TI__Expert 8230 points
    Hello Victor,

    You will still need a protector for Li-ion cells.

    The BATFET of the charger and FETs used in protection typically have different hysteresis thresholds as well as different deglitch times to handle such events differently, for example, protection IC for Li-ion cells typically have much faster response for over-current and short-circuit events, etc.
  • 0 in reply to Fernando Lopez
    Prodigy 175 points

    Hi Fernando, 

    Thanks for your comment. 

    The chemistry I am using is: LiNiCoAlO2 - NCA and LiNiMnCoO2 - INR, which are meant to be intrinsically safer than others and hoping to get away without extra protection. 

    Anyway, back to the initial question. Does the BATFET switch off when the battery goes below 2.5V (I believe this is the value, but I might be wrong)? Why am I seeing those glitches? FET hysteresis? That would be poor performance. They happen even when the system is sourcing low current (battery voltage drop is minimum) and they run for a very long time as there is not much battery discharge. 

    If it is recommended extra protection IC, why would I spend extra money in the BMS with all the belts and whistles? In case of using a host, I think I could monitor that voltage (Vsys) and send a command to the BMS to switch the BATFET off,  but unfortunately that is not an option at the moment and the BQ24195 can be used as a standalone device. Am I missing something? Sorry to load you with so many questions.

    Thanks in advance.

    Best Regards,

    Victor

  • 0 in reply to Victor Claver
    TI__Expert 8230 points

    Hello Victor,

    After some testing on the EVM, here's what I've found: during supplement mode without VBUS, the system is getting it's power from the battery. As the battery discharges to ~2.5V, the BATFET will turn off since you have reached the voltage level necessary to operate the charge pump to keep the BATFET on. On a real battery, once the load is remove, the battery voltage will begin increasing. The following is similar to what happens on protection FETs: as the voltage increases the charger thinks it has enough voltage to operate the BATFET and turns on again. Since the battery is almost fully discharged, as soon as the system current is applied, the battery will quickly discharge below the 2.5V threshold and the cycle will repeat.

    This will go on until the battery voltage doesn't relax above the hysteresis (this is the behavior you observed where Vsys and Isys gets progressively lower and lower). The hysteresis on the BATFET when the voltage is rising seems to be from testing ~100mV. It's not unreasonable to have the battery voltage relax back up 100mV when removing a load when taking into consideration the cell's internal impedance + any impedance along the discharge path on the PCB, battery holder, etc. Most Li-ion cells have a cut-off voltage of 2.5V-2.6V so it's not unreasonable to have the cutoff threshold for the BATFET set to it.

    We have couple options here to prevent this behavior:

    • Use the host to monitor the battery voltage and stop discharge when the battery voltage is close to 2.5V to prevent this behavior. This can also be accomplished with a gas gauge, where you can program a reserve capacity to avoid fully discharging the battery and reporting 0% charge at this point.
    • Use a secondary protection IC with programmable hysteresis to have a bigger hysteresis than the BATFET so the protection FET is the one that opens and not the internal BATFET. 

    While the charger does provide primary protection such as over-current, over-voltage, deeply discharge prevention,etc.; when dealing with Li-ion cells safety is always crucial. Secondary protection ICs provide another fail-safe in the event the host isn't responsive and cannot disable operation, or there is catastrophic damage to the primary protection (charger). 

    Hope this helps.

  • 0 in reply to Fernando Lopez
    Prodigy 175 points
    Hi Fernando,

    Thank you very much for your very detailed expanation, I couldn't ask for more and it corresponds exactly with the issue I had. I've taken good notes of your suggestions and I will definitely help me to improve my solution.
    Thanks again.

    Best Regards,
    Victor Claver