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Dear all,
There are some "VCONN FET Circuit" in the following application diagrams.
What is the purpose of adding the external driving circuit?
I think driving circuits for supplying eMarker are integrated in the CC1/2 functional block. Correct?
Regards,
Nagata.
Solved! Go to Solution.
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USB EZ-PD Type-C
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Hi,
Yes your understanding is correct.
However in the second part regarding the Vconn control, in 60W the Vconn control is active low and in 100W the Vconn control is active high.
Hence to turn ON the FET in 100W, high signal should be sent
#define VCONN_SWITCH_ON_VALUE (1)
#define VCONN_SWITCH_OFF_VALUE (0)
and in 60W, low signal should be sent,
#define VCONN_SWITCH_ON_VALUE (0)
#define VCONN_SWITCH_OFF_VALUE (1)
Regards,
Abhilash P
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Hi,
CPD3196 (CCG3PA) is targeted for Automotive rear seat charger application.
Vconn supplies power to EMCA chip and EMCA is required if the current is greater than 3A.
Vconn control FETs are not integrated in 3196. This FET is enabled only when there is presence of EMCA chip and the requested current is greater than 3A.
The driving circuit, that is the Rp source is present to detect the sink (Rd). This power is not sufficient to power up that EMCA chip inside the cable. DFP communicates with EMCA before PD contract whether the cable supports high current (5A) or not. Hence this FET is enabled based on the desired bus current.
CCx_VCONN_CTRL pin controls FET to provide power to CC line.
In 2*60W ref board, this pin is active low and in 2*100W ref board, this pin is active high. Hence FW must be changed accordingly.
Please let me know if this explanation has helped you understand the need for external FETs.
Regards,
Abhilash P
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Hi Abhilash,
Thank you for quick reply.
Is the following my understanding correct?
When starting up eMarker, VCONN FETs are turn on and supplying high-current to boot-up.
*CC1/2 is not source to eMarker at this sequence.
In the power of 60W case, VCONN FETs are turn off after the PD contract is completed.
In the power of 100W case, VCONN FETs are turn on after the PD contract is completed .
Regards,
Nagata.
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Hi,
Yes your understanding is correct.
However in the second part regarding the Vconn control, in 60W the Vconn control is active low and in 100W the Vconn control is active high.
Hence to turn ON the FET in 100W, high signal should be sent
#define VCONN_SWITCH_ON_VALUE (1)
#define VCONN_SWITCH_OFF_VALUE (0)
and in 60W, low signal should be sent,
#define VCONN_SWITCH_ON_VALUE (0)
#define VCONN_SWITCH_OFF_VALUE (1)
Regards,
Abhilash P
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Hi Abhilash,
Thank you for your support.
Code sample is helpful.
Regards,
Nagata.