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I would like to wire a device I'm making with USB-C input for power but I don't need its SuperSpeed capability. The port will only act as a power input (to this device) and USB 2.0 in. Do I need a chip that communicates (over CC) what should be sent or will it automatically receive 5V at 1.5 amps-ish. I have investigated many chips but they always include CC and support higher power standards that I can't use.

All help accepted, please ask me to clarify if I was vague.

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In a standard (500mA) situation you don't need any ICs to make your device with Type-C connector. All you need is to connect D+/D- from Side-A and Side-B together, and put two pull-down resistors (5.1k) on BOTH CC1 and CC2 pins.

However, if you plan to take 1.5 A (or 3 A) out of the cable, you will need to check the voltage level on CC1 or CC2 pins. This level will depend on host port advertised capability. A Type-C host with 3A capability will have 10 k pull-up to 5 V, a 1.5 A capability will be advertised with 22k pull-up, and the standard port capability (500 mA) will have 56 k pull-up. If you don't want to overload a legacy host, you should watch for CC levels, and limit your power consumption accordingly.

Limiting power intake might be challenging without some intelligent IC, but there are several manufacturers (Texas Instruments, NXP, Microchip, Rohm, etc.) who can offer a solution.

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  • \$\begingroup\$ thank you very much. Are all of those amperages at 5v ? \$\endgroup\$ – Vesperk38 Apr 30 '17 at 18:27
  • \$\begingroup\$ Yes, the default Type-C CC pull-up signature scheme is solely 5V. For higher voltages the PD specifications should be deployed. \$\endgroup\$ – Ale..chenski May 1 '17 at 1:32

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