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BQ24192: How to select IR compensation resistor setting and voltage clamp in BQ24192

Shaquille Chen
Intellectual 1770 points
Part Number: BQ24192

Dear Team,

My customer are working on design of BQ24192.

One of setting in register is about IR compensation.

May I know how to set those register REG[7:2]?

The spec. as below:

Vin: 5V,

Vcharge : 4.2V

I charge: 3A

inductor DCR: 20mohm.

Thanks 

Shaq.

  • 0
    TI__Genius 10240 points
    shaq,

    The IR compensation is to reduce the impact of the resistance from BAT pin to the battery cell allowing the charging to stay in CC mode as long as possible. The resistance could be from PCB routing, connector and sensing resistor etc. The theory behind this is that the charger determines the CC and CV transition by sensing the BAT voltage and comparing to VREG setting. When the BAT Pin voltage reaches the VREG, the battery cell has VREG - Ichg x R. The IR compensation is to add the Ichg x R to the VREG, so the battery cell could get the desired VREG.

    REG06[7:5] is to set the conservative resistance value on the path from BAT pin to the battery cell. REG06[4:2] sets the maximum allowed compensation voltage. The voltage on top of VREG is the LOWER voltage value of the Ichg x the resistance set by REG06[7:5] AND the voltage set by REG06[4:2].

    One example: the minimum resistance from the charger BAT pin to battery cell is 57mohm. The REG06[7:5] should set as 101 (50 mohm). The REG[4:2] is the designer's choice setting and let's use 110 (96mV) as an option. The fast charging current setting is 3A. The BAT pin voltage changes based on the real time charging current. The charger will regulate the BAT pin voltage to VREG (use 4.2V as the setting below) + 96mV when the real time charging current is 3A (3A x 50 mohm > 96 mV). The BAT voltage is about 4.2V + 1A x 50mohm = 4.25V when the real time charging current is reduced to 1A (1A x 50mohm < 96mV).

    Best regards,

    Eric