I'd like to power up a PCB that I'm design with USB-C. The main reasoning is that the connector is low-profile, reversible and can carry 5A at 5V. This PCB will have an ATmega2560 chip onboard, which will be programmed via USB.

What do I need to take into consideration to do the following:

1) Power the PCB via USB-C, providing 5A (where possible)

2) Use the USB-C to program the chip

Does some sort of negotiation need to be done between the sink (my PCB) and source (wall transformer)? Alternatively, can I just read the necessary voltage on one of the CC pins? Also I only need to use the D+ and D- lines of the USB-C as I only need to run at USB2 speeds.

As an aside, this similar question mentions:

Keep in mind that a typical Type-C receptacle needs a thin PCB (0.8 mm), which is fairly inconvenient for DIY projects

I can't see any mention of that in data sheets I've seen. Any further info?

  • \$\begingroup\$ I have no info on the way to handle 5A, but for the point 2 just keep in mind that an ATmega2560 like the one on the arduino mega uses a USB-UART interface to be programmed (no native USB), so you won't be able to program it directly \$\endgroup\$
    – frarugi87
    Dec 7, 2017 at 16:15
  • \$\begingroup\$ Comment because I can't dive into it myself, but have you looked at the USB Power Delivery specification? Can find it in the USB 3.1 specification document: usb.org/developers/docs \$\endgroup\$
    – Shamtam
    Dec 7, 2017 at 16:18
  • \$\begingroup\$ If you doubt the 0.8mm info, just get any on-line source (Digi-Key, Mouser), and get drawings for Type-C connectors and their suggested PCB layout/requirements. You will see. \$\endgroup\$ Dec 7, 2017 at 17:33
  • \$\begingroup\$ Plan is to use an ATmega16u2 (as the Arduino does) as a USB-UART interface @frarugi87 \$\endgroup\$ Dec 7, 2017 at 22:10
  • \$\begingroup\$ @CircularRecursion so it is not an Atmega2560 as stated in the question \$\endgroup\$
    – frarugi87
    Dec 8, 2017 at 21:02

2 Answers 2


With USB Type C the power can be up to 3A @ 5V without any negotiation. To go beyond that (up to 5A @ 20V) requires negotiation over the CC lines. There are USB type C connectors that work with a 0.062 board such as Wurth 632723300011.

Here is a good article on USB Power I found:


  • 2
    \$\begingroup\$ "without any negotiations" is incorrect. Even in baseline Type-C specification there are DC-level "negotiations" regarding power source capability, 500/900mA, or 1.5 A, or 3 A. See this answer, electronics.stackexchange.com/a/327946/117785 \$\endgroup\$ Dec 7, 2017 at 17:21

USB Type-C sources can be of wide variety in terms of capability to supply current. The basic capability (500 mA - 1.5 A - 3.0 A) is determined by value of a pull-up resistor (or proper current source) on host side. So not every Type-C port you find can do 3 A, likely nearly none. Betting on having this level of power will severely limit employment of your device.

To get 5 A out of Type-C port (if it supports that much), the cable must be "electronically marked", and supply to host proper information (over CC pin) that the cable itself can carry 5 A of current. Not every cable can do this, and 5 A host is very rare.

An alternative (mostly theoretical) is to design in so-called "Power Delivery" functionality, which (if supported on host side!!!) can jack-up VBUS voltage (using less current), so you would need a step-down (buck) converter to get back to desired 5V (at high current). But you will be really pushing your luck in finding Type-C hosts that support PD (as of end of 2017).

More practical at this time is to consider Qualcomm QuickCharge standard for negotiation of higher power over USB connector. Their list counts nearly 1000 devices that have this technology, while the number of devices with PD technology is likely less than a dozen, mostly in area of high-end laptop manufacturers.

  • \$\begingroup\$ From my understanding, the USB spec requires Power Delivery mode (USB-PD) to supply up to 5A. All of the other USB3 specs specify a maximum current of 3A. \$\endgroup\$
    – Shamtam
    Dec 7, 2017 at 17:48
  • \$\begingroup\$ Long term, the power delivery will be from a wall charger (or equivalent). Short term, high current isn't required, I just wanted to avoid having a DC power input and mini-USB when I can replace them with just a USB-C. \$\endgroup\$ Dec 7, 2017 at 20:18
  • 1
    \$\begingroup\$ @Shamtam, USB spec doesn't require anything. USB 3.0/3.1 specifications, Power Delivery Specifications, and Type-C Connector Specifications are loosely related INDIVIDUAL specifications. A port can be a combination of anything, and 3 A is not a given if you take an arbitrary Type-C port. \$\endgroup\$ Dec 8, 2017 at 0:53

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