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I'm trying to implement a board that will use USB-C Power Delivery. We want to operate with 20V 5A input. If we know the source will always be at 20V, is it necessary to put a PD controller on the board? From what I've read, the PD controller is meant to protect the board from sourcing too much current when different sources are being used - since this is not the case for my application can I get away with not using a PD controller? Is there something that I would be sacrificing?

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  • \$\begingroup\$ You need a PD conttroller to "ask" for 20V 5A, otherwise the bus will always be 5V (or not even that, depending on the cable that you have and if you havent set the CC resistors) \$\endgroup\$ – Wesley Lee May 27 at 4:14
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If VBUS is always >5V, this device cannot meet USB specifications. You are sacrificing USB compatibility and endanger users who might accidentally connect a standard USB compliant device (with PD or without) and fry it. If you want to implement "a board that will use USB Power Delivery", then implement the Power Delivery. In PD there is only one default and safe voltage, +5V. Anything else would be irresponsible hackery. If your source is permanent 20 V and up to 5 A, then it cannot be "power delivery", you should use a proprietary power jack for this.

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  • \$\begingroup\$ st.com/en/interfaces-and-transceivers/… says that USB-C can deliver up to 20V/5A though ?? \$\endgroup\$ – VanGo May 27 at 7:29
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    \$\begingroup\$ No, it says "from 5V/0.5A up to 20V/5.0A". It's up to the devices to negotiate the actual voltage supplied. \$\endgroup\$ – Finbarr May 27 at 7:38
  • \$\begingroup\$ I don't quite understand. If I plug in 20V/5.0A, am I not supplying 20V/5,0A? Doesn't the PD controller just act as a switch, and the CC lines are like the enable/disable? \$\endgroup\$ – VanGo May 27 at 8:05
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    \$\begingroup\$ @VanGo, if you are supplying 20V, you are supplying 20V. If your source doesn't start with 5V, it is not a PD. For PD, you need a complex serial packet-based two-way communication over CC wire, see this short explanation, electronics.stackexchange.com/a/439312/117785 and this electronics.stackexchange.com/q/361391/117785 \$\endgroup\$ – Ale..chenski May 27 at 15:48

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