I purchased a Xiaomi laptop PRO computer, which uses a USB-C plug for its power input. I recently measured the plug's shell with a AC voltmeter, and found to my surprise a voltage between 90 to 100 volts (AC)! May be that's why I always felt the sting of a slight electrical shock every time I used the computer. I understand the shell is used for signal shielding and therefore should not be carrying any voltage. Why then is there a voltage? How can I get rid of it? Was this a design or manufacturing error? (The other end of the USB-C cable is connected to a 2 prong non-grounded 60W transformer plugged into the wall.)
Electronic – Why does the USB-C plug shell carry a voltage
powerusb-cvoltage
Related Solutions
Will USB-C plug work without chassis of the connector?
Yes, it will most likely work with short distances, on your desk, as long as at least one side still has chassis connected.
Will it still comply to USB?
No, USB requires a chassis connector for numerous reasons.
- For signal integrity, the chassis of the connector shields all signals of the cable from the outside world inside the cable. The cable has shielded foil around all conductors. Breaking this shield is not recommended.
USB 3 is very high speed with relative weak signals. - For physical integrity, the chassis provides all the latching features and strength the connector has. It also prevents misalignment of the connector.
If you want your socket to be flush you will have to use a metal enclosure with your custom socket. Or use any of the existing options that will extend trough the enclosure.
Shield is at approximately the same potential as GND, but it is not used as return path for power.
The Power Delivery protocol uses slow, 300 kbps signaling rate, with edges limited to 300 ns. And the driver uses 50 Ohm impedance. With 400 pF load the time constant of this network is just 20 ns, so the cap doesn't have any effect on PD digital communication. Yet the CC pins are multi-functional, so smoothing the signal a bit doesn't hurt. In fact, the BMC PD receiver does require the 200-600 pF cap per PD specifications, see Table 5-19. Note that the old (faster) BFSK signaling mode is depreciated in PD 3.0 Specifications.
Related Topic
- Electrical – What voltage does a laptop provide over USB C
- Electrical – Is the voltage supplied by an LCD backlight inverter in a laptop dangerous in any way
- Electronic – Does USB-A connector have the pins required for USB power delivery (PD)
- Electronic – How do usb-c connectors carry 100W at 20V without melting
Best Answer
It's from an interference suppression capacitor between the shell and the mains AC. The capacitor provides a path for high frequency noise to be shunted to the power supply, but it also allows a tiny amount of current to pass through it. Because your multimeter is very high impedance, it registers as a high voltage, but if you connect it to something the voltage will drop a lot. Your observation of a slight electric shock fits with this as the capacitor won't allow enough current to provide more than a slight tingle.