Help understanding this circuit

[caphair]

Can't find a Datasheet for this charging ic from an iPhone 6. Hopefully someone can explain its function to me.

Things I'm confused about; Does it take USB power and convert it to both vcc_main and vbatt? Or does it output vcc_main and vbatt is also an input to the chip?

What's q1403's role?

Sorry if the schematic is too small/blurry it's a screenshot from my phone and gets compressed when uploaded

[gabiz_ro]

Re: Help understanding this circuit

I'm not 100 but 99.9% sure that is something like buck converter combined with charger circuit.
When VBUS_DET and OVP are OK then VBUS is used by buck converter to obtain PP_VCC_Main and by charger circuit to charge battery through PP_BATT_VCC in this case Q1403 is off.

Q1403 act like a switch,connect battery to "load" in this case PMU,battery is power source for phone or disconnect battery (and let it be charged) in that case buck controller is power source for phone.

When VBUS is absent or VBUS_DET or OVP are not OK then buck converter is disabled,charging also disabled,and Q1403 is on.
PP_BATT_VCC is connected by Q1403 to PP_VCC_MAIN

This is some basis concept.
More things are required for normal function.
Data communication between U1401 and AP,and PMU,

[caphair]

Re: Help understanding this circuit

So when battery is charging, q1403 is off and main power source to system comes from vcc_main provided by u1401?

With q1403 off won't some vbatt still get through to vcc_main by the body diode inside q1403?

[gabiz_ro]

Re: Help understanding this circuit

I don't think.
PP_VCC_MAIN is higher than PP_BATT_VCC when battery full charged so no current flow through diode.

It's something like in laptops,when charger is connected mains power is 19V,when charger disconnected power is from battery,12.6V (that is just one example for charger 19V and battery with three Li-Ion series group)
In that case there is big difference of voltages but in case of phone is much lower,but even 0.1V make differencies.
If PP_VCC_Main is 4.3V then no way to draw current from battery that reach max 4.2V

[caphair]

Re: Help understanding this circuit

Makes sense thanks.

So vbatt line from u1401 is an input (battery feed to q1403) AND output (u1401 charging vbatt) line connected directly to battery?

[gabiz_ro]

Re: Help understanding this circuit

Pins A1 B1 C1 D1 of U1401 are output of charging circuit.
Pin E2 control state of Q1403
When charger connected,phone is powered by U1401 and battery is charged until full.
When battery full or during charge time U1401 provide power to phone.

When charger disconnected or not present U1401 switch on Q1403 and current flow from PP_BATT_VCC via Q1403 to PP_VCC_MAIN

[caphair]

Re: Help understanding this circuit

Thanks for that.

I'm curious what are the "desense capacitors" in the schematic? Are they different than normal filter caps? Only a few are named that

Is the VDD_Main like after it's been bucked down or before? There's a VDD line and Buck sw line that tie into each other

[caphair]

Re: Help understanding this circuit

Question:

Is VDD a power input to turn the chip on or is that an output that becomes vcc_main?

I’m wondering because if it is an input to turn it on, how would it turn on to start charging and to create vcc_main if the battery was completely dead?

[Agent24]

Re: Help understanding this circuit

Can you post the whole manual or a better screenshot?

[WildPuppy]

Re: Help understanding this circuit

Question:
...
I'm wondering because if it is an input to turn it on, how would it turn on to start charging and to create vcc_main if the battery was completely dead?

It probably wouldn't turn on in case the battery is completely dry. In that case, the only potential solution would be to try to "jump-start" the battery by means of another current source connected directly on the battery poles, hoping it will still pick up some charge. With that little charge in the batt, the normal charger would continue the work.
With this new "smart" batteries, letting them deplete to 0V means throwing them away for the common consumer for just that reason. Some times this "jump-start" works, but not always. I suspect some physical/chemical damage inside the battery happens at 0V but I am not sure as I don't know how they work internally.
But hey, I'm no specialist. There may be others with much more insight to this than me.

[lnux]

Re: Help understanding this circuit

Thanks for that.

I'm curious what are the "desense capacitors" in the schematic? Are they different than normal filter caps? Only a few are named that

Is the VDD_Main like after it's been bucked down or before? There's a VDD line and Buck sw line that tie into each other

VDD MAIN connected internally to BAT line to charge
It's not used to power on the IC

[dj_ricoh]

Re: Help understanding this circuit

Why you didn't put the whole diagram?