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Regarding bq24072 Li-ion charger with System Load

Pavan Kumar
Prodigy 130 points
Other Parts Discussed in Thread: BQ24072, BQ24070, TPS54240

I am considering this part bq24072 for GSM Module and GPS based tracker application, where GSM Module needs power supply ranging from 3.4V to 4.2V with nominal 3.8V with 2A peak and average current under 600mA or so.

It is to be used for battery charging and system load at the same time, and it should automate as battery backup feature. Vin is a 5V regulated supply as input to this IC.

As per datasheet System Load voltage Vout is specified as Vbat + 225mV, I have a question here as:

1. What will be the Vout if there is no battery connected

2. What will be Vout if battery is faulty or short

3. Can this circuit provide 2A peak current requirement

4. Is this best part suitable or other option is much better for this application?

 

Regards

 

 

 

  • 0
    TI__Mastermind 32350 points

    Hi Pavan,

     

    When the BAT voltage falls below 3.2V, OUT is clamped to 3.4V. If the battery is not connected or shorted, the sys output will be 3.4V.

     

    The IC can support up to 1.5A charge current & input current.

     

    The part has DPPM mode.

     

    Did you look at the bq24070/1?

     

    Thanks

    Tahar

     

     

     

     

  • 0 in reply to Tahar Allag
    Prodigy 130 points

    Thanks Tahar for your reply.

    I checked bq24070 and other similar parts also, but those provide Vout as 4.4V, which is extreme high as per GSM Module, the recommended range is 3.4 to 4.2V, with typical value as 3.8V, which fits well with bq24072 only.

    For bq24072, the package has 2 pins as Vout, that seems it can provide required current for the load, but it has only one pin for Vin and at the same time Input current is stated to be within 1.5A. What will happen if battery is not connected or defective, then whole load current has to come from Vin only, so will that be OK, 1) only one pin for Vin in the package, 2) Max current 1.5A? Or in that case load current will also be limited to 1.5A only.

     

    Regards

    Pavan

     

  • 0 in reply to Pavan Kumar
    TI__Mastermind 32350 points

    Hi Pavan,

     

    The bq24072 has DPPM mode where you can have you system supplied by the battery and the input at the same time as sown in figure 37 & 38 in the DS. However, the max input current is 1.5A. 

     

    I am assuming you are interested only on the linear topology. You can use the bq24070 and use additional buck converter to step sys down to the desired voltage.   

     

    Thanks,

    Tahar

  • 0 in reply to Tahar Allag
    Prodigy 130 points

    Hi Tahar

    OK, I understood that, but think of the case when system load is looking current of the order of 2A or more and battery is not connected, what will happen in that case as the input current is limited to 1.5A, so the output (load) can not be more than 1.5A I believe. And in case if battery is connected, it can supplement the additional current. 

    While using bq24070 with additional buck converter, this will limit power utilization from battery due to the dropout of the additional buck converter and at the same time add one more stage. 

    Before the charger stage, I need to use other part as tps54240, which can provide up to 2.5A, but using bq24072 will not be able to use its full current delivering capacity of 2.5A.

     

    Regards

    Pavan 

     

  • 0 in reply to Pavan Kumar
    TI__Mastermind 32350 points

    Hi Pavan,

     

    Regarding the charge current: The part can charge up to 1.5A charging current. Please take a look at electrical characteristic table on page 5 at the ICHG.  

     

    The input current is limited 1.5A and it is adjustable using ILIM pin by connecting a resistor from ILIM to VSS with EN1=0 & EN2=1. The input load includes the charging current of the battery and the system load. In your case, if the battery is disconnected, all the input current will be at system load. Since your input voltage is 5V and the output voltage is 3.4V, you should be able to get more than 1.5A at the system load.  

     

    Thanks!

    Tahar

  • 0 in reply to Tahar Allag
    Prodigy 130 points

    Hi Tahar

    OK, Yes, I got it. 

    Input power is 1.5 x 5.0 = 7.5W, so Load current can be 7.5*0.85/3.4 = 1.875 and while a capacitor is connected across battery, then load voltage = 4.2V and then load current can be 7.5*0.85/4.2 = 1.52A, considering 85% efficiency.

    At 5.5V input and 3.4V output the current can be 2+ Amp at load.

     

    Regards

    Pavan 

  • 0 in reply to Pavan Kumar
    Intellectual 445 points
    Hello Pavan,
    Were you able to get the GSM + GPS module working with BQ24072 part ? What was the resistor values you have used for ILIM, ISET ? Whats the battery capacity you have used ?

    Thanks
    Hardik Sanghavi