BQ24610: Clarity ON Power up Sequence

Karthik B

Part Number: BQ24610

Hi,

Please can anyone clarify how the IC comes to know that ADAPTER is connected ?

Because in datasheet it says that the when VCC is greater than SRN voltage then it will turn ON the ACFET by driving the ACDRV pin .But How the IC will come to know that VCC is (greater than SRN) increased because the adapter voltage is not appeared at VCC pin since FET is OFF. Now if FET is OFF how the VCC will be greater than SRN if that is not happened how adapter connection will be detected , its really confusion and am not able to understand how this will work?

Similarly what is he purpose of BTST and PH pins? And How battery charging will occur?Does PH pin charges the battery?

Please suggest.

Attached reference from datasheet

I am lagging with design of it please provide your valuable feedback ASAP.

Regards,

Karthik

over 7 years ago

0 Kedar Manishankar over 7 years ago

TI__Mastermind 24075 points

Hello Karthik,
Even if the FET is off, it can still conduct through its body diode. As you can see the PVCC pin at the source of the 2 PFET is connected straight to the VCC pin. When battery only mode you can see PVCC = Vsys - Body diode voltage drop. As ACPWR increases, PVCC (meaning VCC as well) will eventually be higher than SRN, trip the internal SLEEP comparator and exit SLEEP Mode. When exiting sleep mode PFET between BAT and SYS will be turned off and the ACFETs will be turned on.

NFET are used as the FET for the buck converter. These require gate voltage larger than source voltage (PH is source of high side FET). Gate voltage is obtained by charging capacitor to value greater than PH (source) to provide voltage for driving HSFET, which is what the BTST pin is used for.

For PH pin this needs basic understanding of buck converters. Please read this document for further understanding of what the switching node(PH) does and how current can be supplied to the load. www.ti.com/.../slva057.pdf

0 Karthik B over 7 years ago in reply to Kedar Manishankar

Expert 1310 points

Hi Manishankar,

Thanks for your reply.

I have attached my design could you please review it and provide your valuable feedback.

System load requires 1.5A to 2A current and battery need to completely charged within 4hrs , Battery capacity is 10AH and maximum charge Voltage and current is 8.4V(max) and 5A(max).

Input adapter supply is 9V DC.

**Basically when I tried to search this part in webench I was not able to find it .Could you please suggest why?

Regards,

Karthik

0 Kedar Manishankar over 7 years ago in reply to Karthik B

TI__Mastermind 24075 points

Hello Karthik,
This is one of our older parts and does not have WEBENCH support.

0 Karthik B over 7 years ago in reply to Kedar Manishankar

Expert 1310 points

Thanks for your reply.

Any concern regarding the circuit(schematics)?

Please suggest.

Regards,

Karthik

0 Kedar Manishankar over 7 years ago in reply to Karthik B

TI__Mastermind 24075 points

Hi Karthik,

You need to ensure your output capacitance is high enough that the LC output filter has resonant frequency between 12kHz - 17kHz.

10uF - 20uF input capacitance is suggested for 3A - 4A charging current. It is a good idea to use the schematic in the user guide as a reference for optimum values.

0 Karthik B over 7 years ago in reply to Kedar Manishankar

Expert 1310 points

bqstroller calculation tools_V1.5.xlsHi,

I have used a excel sheet "bqstroller calculatin tools_V1.5" shared by TI.

same values I have used which is suggested by that calculator.

I have also attached the excel sheet for your reference

Regards,

Karthik

0 Karthik B over 7 years ago in reply to Kedar Manishankar

Expert 1310 points

Ok can I use it in my new design?

0 Kedar Manishankar over 7 years ago in reply to Karthik B

TI__Mastermind 24075 points

Hello Karthik,

Yes if your output capacitance is 20uF you will be fine. I cannot see the values clearly but it looks like you have followed the EVM schematic and it looks good.

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BQ24610: Clarity ON Power up Sequence