I have a "host" device that does not have a 5V rail for VBUS but has a lithium battery for power.The battery voltage meets alternate mode VCONN specifications (2.7 V to 5.5 V on VCONN). Is it possible to power a "slave" device over VCONN without implementing a PD control scheme and using VBUS?


2 Answers 2


As I understand, you mean that your host has a 5V internal battery supply that can afford only 200 mA to source (VCONN is 5V @ 1W), right?

The USB PD protocol indeed is the correct thing to advertise "Battery Supply Power Data Object" (Table 6-9 of PD specs) with B[9..0] = 0x004. The use of VCONN is not defined/allowed to power any slave USB devices, only e-markers and PD chips, and VCONN is not connected to VBUS anywhere.

However, based on declared backward compatibility with USB 2.0, you can declare your host as "low-power embedded host", see section 7.2.1 of USB 2.0 specifications (or Section 11.4.1 in USB 3.0), "Classes of Devices", which says, in "Root port hubs" sub-section,

"Battery-powered systems may supply either one or five unit loads."

So, one unit is legal. If you declare your port as "low-power" (in a formal USB-IF declaration for certification), you can legally supply VBUS with only one unit load capability, 100mA for USB 2.0 protocol, and 150mA for USB 3.0 protocol.

However, there is a gap in USB Type-C specifications. While the Type-C specifications explicitly refer to USB 2.0 standard with regard to "default power", the Type-C connector does not define a special pull-up value to advertise the "low-power" port capability. So you can use only the standard CC pull-up of 56k, for "default power". Here is the rare place where the USB device restrictions come to play, to draw only one unit until enumerated.

To make it work without blowing up your host, your host software driver must deny any device that specifies its power consumption above your port capacity (200mA in your case), and hardware must include an overcurrent protection of your port at 200mA as well, to prevent badly-designed USB devices from drawing too much current without being enumerated/configured.

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    \$\begingroup\$ This is not quite my problem. My host has no 5V supply. I only have a lithium battery (3-4.3V) and lower logic voltages (1.8V ect). Is there a way I can establish power over VCONN to an accessory without using a 5V rail and PD controller? \$\endgroup\$
    – EasyOhm
    Commented Feb 15, 2017 at 19:30
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    \$\begingroup\$ @Gonzik007, if your accessory is non-standard and works from battery levels, you can do whatever you want. But if you mean to connect standard USB devices (mice, keyboard, RF dongles, pen driives), you need to boost your supply to +5V. Also, saying that you have a VCONN-compliant power is misleading - VCONN must be 5V. \$\endgroup\$ Commented Feb 15, 2017 at 19:40

Normal USB devices are not allowed to be powered by VCONN. The only VCONN-powered accessories that are allowed are direct-attach Alternate-Mode UFPs. If the device doesn't negotiate an Alternate Mode, the DFP is allowed to turn off VCONN.

Downstream facing ports (DFPs) with a Type C connector are not required to supply VCONN unless they support SuperSpeed data, so you can't actually count on VCONN being present at all.

Even though VCONN is normally 5V (when provided), a VCONN-powered accessory is required to work with a VCONN voltage range of 2.7V to 5.5V.


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