[BQ24158] SmartPhone charge curve decrease continuously in the Current Regulation Phase.

user4733531

Other Parts Discussed in Thread: BQ24158

Hi,

As the chart shows, the charging current decrease continuously in the Current Regulation Phase that is expected to correspond closely to constant current.

BQ24158 register values are as follows:

register00: 0xd0

register01: 0xf8

register02: 0xa8

register03: 0x51

register04: 0x7a

register05: 0x13

register06:0x77

How to analyze and solve this problem? thanks

over 7 years ago

0 Jeff F over 7 years ago

TI__Guru*** 143950 points

Is the IC getting hot? It might be thermal regulation. To how much copper are the ground, VBUS and PMID bumps connected for heat sinking?

0 user4733531 over 7 years ago in reply to Jeff F

Prodigy 100 points

Hi Jeff F，

I have turned off thermal function of charge , BQ24158 registers keep the same vaue during the whole charging process.
By the way, phone is in power-off state.

0 Jeff F over 7 years ago in reply to user4733531

TI__Guru*** 143950 points

What is the voltage at the charger's CS pin? If =VBATREG then the charger is tapering early due to high impedance from CS pin to battery.

0 user4733531 over 7 years ago in reply to Jeff F

Prodigy 100 points

Hi Jeff F,

CS PIN refers to CSIN or CSOUT or both?

VBATREG is 4.34V , from register02. If =VBATREG， does it enter Voltage Regulation Phase? Charger current decrease from the early time as Vbat is about 3.4V.

We try a test with impedance, DC supply 5V voltage to Vbus PIN Directly, the result is follows. Does this mean high impedance from usb to Vbus?

0 Jeff F over 7 years ago in reply to user4733531

TI__Guru*** 143950 points

CSOUT senses battery voltage. If CSOUT is near 4.34V, the IC enters CV and starts tapering. High impedance from CSOUT pin to battery can cause the IC to enter CV prematurely.

0 user4733531 over 7 years ago in reply to Jeff F

Prodigy 100 points

Hi Jeff F,

How to test impedance from CSOUT pin to battery?

CSOUT is 3.89V, Vbat is 3.87V, charging current is 1.15A.

thanks

0 user4733531 over 7 years ago in reply to Jeff F

Prodigy 100 points

Hi Jeff F,
High impedance from CSOUT pin to battery can cause the IC to enter CV prematurely.

What reason is this? Could you introduce something about theory？thanks

0 Jeff F over 7 years ago in reply to user4733531

TI__Guru*** 143950 points

Resistance = (3.89V-3.87V)/1.15A = 17mohm. The article www.ti.com/.../slyt569.pdf explains a bit. Also, all batteries have an internal impedance that varies with battery voltage.

Power management

Power management forum

[BQ24158] SmartPhone charge curve decrease continuously in the Current Regulation Phase.