• State Resolved
  • Locked Locked
  • Replies 9 replies
  • Subscribers 64 subscribers
  • Views 590 views
  • Users 0 members are here
Support feedback
Options
Options
Related

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

BQ24075T inrush current value.

Franco Zampicinini
Prodigy 110 points
Other Parts Discussed in Thread: BQ24075T

Hi to all,

I'm testing the BQ24075T EVM.

The only problem I found it the input inrush current.

Here the configuration:

- battery disconnected;

- Vin = 5V

- load: variable resistor with 100uF capacitor in parallel.

If I configure EN1 = 0 and EN2 = 0 input current is limited to 100mA in normal condition and at power-up; with EN1 = 1 and EN2 = 0 the limit changes to 500mA in both conditions; if I set EN1 = 0 and EN2 = 1, with Iset to about 1.5A (with 1.1K resistor) in running condition the limit is correct (1.45A), but at power-up the maximum I can have is 500mA (or the output became 1V - short circuit mode).

 Reading the datasheet  (table 1) and  the detailed functional description seems (or I understood..) that with the last configuration the maxim input current should be setted by resistor, not limited to 500mA. This is a big issue for my circuit (I could have 1A of current at the power up...).

Where is the mistake?

Thanks in advance,

Franco.

  • 0
    TI__Mastermind 32350 points

    Hi Franco,

     

    ISET: is used for fast charge current programming.

    ILIM: Is used for current limit programming.

     

    What is the resistor values used in both of these pins? in your post, you are talking about current limit but you set your Iset pin.

     

    Thanks!

    Tahar

  • 0 in reply to Tahar Allag
    Prodigy 110 points

    Hi Tahar,

    both resistors are 1.1Kohm.

     

    In my post there is a mistake, my problem is Ilim, not Iset (sorry...):

     

    Thanks.

  • 0 in reply to Franco Zampicinini
    TI__Mastermind 31120 points

    there is an initial start up algorithm that does shorts checks with 100mA supplied...see data sheet for more information

  • 0 in reply to Charles Mauney
    Prodigy 110 points

    My question is: why with EN2 =1 and EN1= 0 the inrush limit is 500mA as when EN2= 0 and EN1=1? Reading the datasheet the maximum input current, with EN2 =1 and EN1= 0, is determined by resistor on Ilim on all conditions, why at startup is only 500mA? Where is this on datasheet?

     

    Thanks.

     

     

  • 0 in reply to Franco Zampicinini
    TI__Mastermind 31120 points

    Not sure I understand your question fully, but if you refer to the funciton block diagram you will see a few control loops.  The basic design is that maximum current/voltage will be applied until some control loop indicates it is too high and starts to reduce the current/voltage.  There is another charge loop programmed by ISET and this will try and charge the battery at the programmed setting if the availabe input current will allow.

    The intial start up is 100mA for a far less than 1 second.

    The 100mA start up is explained under Power On on page 17 and in flow diagram on page 18.

    EN2 EN1 Maximum input current into IN pin

    0 0 100 mA. USB100 mode

    0 1 500 mA. USB500 mode

    1 0 Set by an external resistor from ILIM to VSS

    1 1 Standby (USB suspend mode)

  • 0 in reply to Charles Mauney
    Prodigy 110 points

    Now I understood...on page 18 of datasheet the startup flow diagram shows me the light!

    Before to choose the current limit settings the IC tests if the output voltage is less than 1V on output with 100mA of input current.

    Why this? I think 300mV should be better as voltage limit..example (worst case):

    Vin = 4.5V

    My load: 4.5 ohm (equals to1A)

    Input current limit: 1.5A (permitted)

    The IC never starts because at startup, with 100mA, measures 450mV.

    Another example (medium case...):

    Vin = 5V My load: 8 ohm (equals to 0.625A)

    Input current limit: 1.5A (permitted)

    The IC never starts because at startup, with 100mA, measures 800mV.

    In real application I have to powerup a DC/DC converter, 500mA could be insufficient as starting current...what can I do?

     

    Thanks in advance.

  • 0 in reply to Franco Zampicinini
    TI__Mastermind 31120 points

    There are two basic solutions:

     

    Disconnect the load prior to the charger start-up or supplement the current with the input via a FET.   Have a RC time constant or some delay on turning off the FET from the input to the system (probably need back to back FETS).

     

  • 0 in reply to Charles Mauney
    Prodigy 110 points

    Have TI a Charger with power-path, SYSOFF, 1.5A max input, but without the 100mA short circuit test?

     

    Thanks.