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I have a 40W USB PD compliant charger from my ultrabook. I recently bought a pair of headphones (Sony WH-1000XM4) which can be charged using a USB C Plug. Sadly I can not find in the specifications of these headphones whether they support the USB PD standard. It only says they are supposed to operate at 5V and draw 1.5A, so they are potentially not compatible.

So I'm wondering, what happens when a non USB PD compliant device is charged with a USB PD source? The source can supply 5-20V. I assume the device would be fried if it were to be exposed to the full 20V, but I wonder whether the source recognize that the device is not USB PD compatible and doesn't supply any voltage at all. This would make sense to me, as it seems like USB PD compliant devices "tell" the source which voltage they want.

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    \$\begingroup\$ it makes no sense for non-compliant devices to get anything other than 5 V \$\endgroup\$
    – jsotola
    Commented Aug 16, 2020 at 17:41
  • \$\begingroup\$ Side note, USB PD has fix voltage of 5 9 15 and 20. A 40 W charger should only provide up to 15 V, you shouldn't even get 20 according to the standard. Other fast charge like QC 4.0 allows arbitrary voltage from 3.2 to 20 at 200 mV steps. \$\endgroup\$
    – Passerby
    Commented Aug 18, 2020 at 1:52

3 Answers 3

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By default, all USB chargers and USB hosts provide 5V on VBUS pin.

Once connected, a USB PD charger will then try to communicate on the CC1/CC2 pins using the USB PD protocol. If the attached device does not answer, it stays at 5V.

Your headphones seem to draw 1.5A. This is above the default USB current of 500mA for USB 2 and 900mA for USB 3 applying for USB hosts. For USB chargers, 1.5A is ok.

Your device can distinguish a charger from a host by measuring the voltage on CC1 and CC2. It should (and probably will) draw 1.5A only if it detects a charger.

The voltage is determined by certain pull-up (charger, host) and pull-down (device) resistor values on CC1 and CC2.

You should have no problem connecting your device to the charger.

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  • \$\begingroup\$ Tahnks for the answer. Out of curiosity, you distinguish between "host" and "charger", can you define what the difference between the two is? I'm guessing a host would e.g. be the port of my laptop, when it is connected to the headphones. \$\endgroup\$
    – user257988
    Commented Aug 16, 2020 at 18:29
  • \$\begingroup\$ Yes, your laptop is a host. It can also provide power but likely only up to 900mA. \$\endgroup\$
    – Codo
    Commented Aug 16, 2020 at 19:20
  • \$\begingroup\$ The host would be negotiated. A laptop can be both sink and source for data and power. And some laptops like Mac book pros can do multiple 27W sources. Like 2x 27W and 2x 12W at a time. \$\endgroup\$
    – Passerby
    Commented Aug 18, 2020 at 1:56
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    \$\begingroup\$ This is not true with USB, USB C to C is designed to be VBUS cold until a valid connection is detected via the CC wire. (You don't want any source to provide power if it is plugged into another source) If you have a sink, the bare minimum you need to implement to be PD compliant is 2 5.1k resistors, so the source can turn on \$\endgroup\$
    – Ferrybig
    Commented Jun 7, 2023 at 8:42
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The question is about USB Type-C connectivity.

The Type-C connectivity provides two methods of determining source capability.

The primary method is the value of pull-up on HOST side on CC pins. Type-C specifications define three levels of capability: 500/900 mA (56k pull-up to 5V), 1.5 A (22k pull-up), and 3A (10k pull-up). The connecting device pulls down this with 5.1k to ground, and the resulting voltage level tells the device how much current it can take over the particular connection. When the host sees the pull-down, it will turn on "+5Vsafe" VBUS. This is per Type-C protocol. So the Sony device is perfectly compliant to the basic +5V safe supply.

The secondary method is provided by nearly independent Power Delivery specification. If the consumer implements PD, it still need to follow Type-C specifications for CC pull-up-down protocol, and will receive "+5Vsafe" VBUS.

Only then the consumer will send serial PD-defined messages over CC pin to discover source capabilities. If provider responds, then negotiations for power contract will proceed.

If the consumer is not PD-agnostic, no messages will be generated and no responses will be returned, and no contract will be negotiated. The link power will stay at the default "Safe+5VBUS" power schema, per DC levels on CC pins.

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  • \$\begingroup\$ I read this document regarding USB-C PD cables - cuidevices.com/blog/… and it suggests they are using CC pins in one cable (pins A5 for CC1 and B5 for CC2) but no CC pins in the second (higher wattage) cable. Does this make any sense to you? Their 100 W cable uses A1, A4, A9, A12, B1, B4, B9, B12 - but NOT A5/B5! \$\endgroup\$
    – Steerpike
    Commented Jul 3, 2022 at 1:31
  • \$\begingroup\$ @Steerpike, this is a horrible hack on CUI side. Their "8-pin 100W" cable grossly violates the entire TYPE-C specification. A normal USB port will ignore any voltage if CC pin is not active high, it will not work. Their 8-pin receptacle also won 't work with certified Type-C source, because the source won't engage any voltage if CC pin is not sensed active \$\endgroup\$
    – Ale..chenski
    Commented Jul 3, 2022 at 21:49
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If the device doesn't support USB-PD, then it doesn't have a controller to negotiate the USB-PD voltages and the plug will act as a normal USB.

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