4 ms·

I've shocked myself with the end of a USB C cable too many times to believe this (power negotiation) is implemented well.

• duskwuff 3y ago
  The metal "end" of the cable is connected to ground, not power. If you're getting shocked by that, you've got bigger problems.
  • 05 3y ago
    It's the EMI shield coupling inside the power supply, see [0]. Shouldn't happen with properly grounded chargers (but the charger in question is probably two-pronged without a ground pin, like the majority of chargers). [0] https://electronics.stackexchange.com/questions/476895/ac-to-dc-power-supply-earth-ground-capacitance-on-dc-output https://electronics.stackexchange.com/questions/476895/ac-to...
• NavinF 3y ago
  That has nothing to do with the voltage/current inside the cable. You can't feel 48V on dry skin.
  • lxgr 3y ago
    This is assuming that only 48 V (relative to ground) were present on the cable. There are many ways in which low quality power supplies can expose up to full main power voltage on the secondary side. For example, with 230 V on Vbus and 225 V on "ground", an USB device might not even be able to tell the difference.
• lxgr 3y ago
  A compliant USB-C device or power supply is not allowed to expose anything other than 3-5.5V of Vconn on one of the configuration pins. Only if it senses an upstream-facing port of another device will it expose 5V of Vbus, let alone 20 V (or more recently 48 V) via PD, which would require a complex handshake to prevent exactly that situation (i.e. electric shocks or arcing) from arising. As others have already said, what you're experiencing is likely an improperly designed power supply (that can happen even with plain old USB-A battery charging and has caused deadly accidents in the past!) or static electricity.