CCG4-Dead Battery Case : How is Chip powered ?

[Anonymous]

Hello Everybody,

       

I have a system using the CCG4.

   

In dead-battery mode, Rd is detected via the CC-Line and VBUS gives 5V on the line.

   

But How does the chip get powered ? The chip can be powered only through VDD and VDDIO. Not over CC (If I believe the bloc diagramm)

   

Can the CC-Line (powered over VCONN) activate the GPIO and switch the FET to enable VBUS ?

       

This sounds to me unclear.

   

Please give me posted and thanks in advance.

   

Regards

[Anonymous]

In case of a CCG4 The Rd is always advertised. the consumer path is always on even in dead battery condition.

   

For your reference I have attached the CCG4 daughter card schematic. In the schematic if you look at the right-top most part in page 3, the circuit is for dead battery. As the Rd is always advertised and the VBUS is supplied by the external supply USB2_VBUS and  VDDD gets powered to 5V. Here the V5P0 becomes 5V and powers the VDDD, you can see in page 4 how they are connected. The VDDIO is also powered in a similar way. Hence the Chip works.

[ShifangZ_26 · Accepted solution]

If your question is based on CCG4, the answer for your question is The dead battery support shall be supported by external circuits, the example circuits can be referred Type-C_VBUS - D18-U13 - V3P3 - V5P0 on CY4541 based board.

The main thought about deadbattery is, CCG4 will have Rd inside of CC1 and CC2, when CCG4 powered off, the RD will also present on CC1/CC2, so that the 5V will be present on VBUS from power source. So that we could get 3.3V and 5V for CCG4 and powered on.

The other way is: Make sure VBUS_C_CTRL be enabled when powered off, which is means VBUS_C_CTRL = LOW, the power consumer path is ON.

Best regards,

Lisa