5
\$\begingroup\$

It seems to be connected to D+ and D- pins of mini usb connector.

enter image description here

\$\endgroup\$
3
  • 3
    \$\begingroup\$ An ESD protection chip? \$\endgroup\$
    – ErikR
    Jun 9 at 10:26
  • 3
    \$\begingroup\$ Is it connected to VBUS and GND as well? A TVS or similar would make sense there. \$\endgroup\$
    – Colin
    Jun 9 at 10:29
  • 1
    \$\begingroup\$ Same thing on the usb a female connector. \$\endgroup\$
    – Passerby
    Jun 9 at 10:32
3
\$\begingroup\$

That is a common-mode choke, as the USB data pins go into one end and comes out of the other end.

It is not an ESD protection device, as the TVS diodes are the two components right next to the CM choke, connected to the connector side of the USB data pins.

The exact same arrangement is seen on the other USB connector too; four-pin CM choke and two ESD protection devices.

\$\endgroup\$
4
  • \$\begingroup\$ Are you sure those aren't the capacitors for d+ and d- ? \$\endgroup\$
    – Passerby
    Jun 9 at 17:12
  • 1
    \$\begingroup\$ I first thought of a common mode choke, but usually the differential input and output traces would connect on opposite sides of the CMC. Here they appear to route through the CMC at 90 degrees to the connector traces. It is difficult to tell exactly what the pin out of the device is from the photo, but if it is a device with more than 4 pins, then it probably is an ESD protection device. I expect there is less concern about common mode immunity in a consumer device as there is ESD when plugging in the cable. \$\endgroup\$
    – TimB
    Jun 9 at 18:39
  • \$\begingroup\$ @TimB I interpret from the picture that they do connect to opposite sides, but the traces go in from the side to pins in order to swap the pins without vias. The CM chokes may be required to pass official USB certification tests if it fails to pass the tests without CM chokes and it is enough to add them. And ESD would be taken care of the small two-terminal devices connected from data pins to ground, they are most likely low-capacitance TVS diodes that are first line of protection at the connector. \$\endgroup\$
    – Justme
    Jun 9 at 18:47
  • \$\begingroup\$ @Justme, excellent point about swapping the differential pair. I can certainly see that being a possibility here. The component certainly looks like a typical CMC and with your input I agree that this is most likely a CMC. Thanks. \$\endgroup\$
    – TimB
    Jun 11 at 3:00
6
\$\begingroup\$

No way of knowing 100% for sure, but it's likely it's a ESD ic, essentially just some diodes (TVS or otherwise). Part of the standard for USB, or best practices. You won't always see it on cheaper products. It's on both the mini and full size usb port. Per the other answer, it may be common mode chokes in an IC package. Only the designers know (or requires testing and or removing parts to see).

One variation of possible connection. There are multiple types of these ICs: enter image description here

More information on USB ESD protection at https://blog.semtech.com/esd-protection-of-usb-2.0-interfaces

Image source: https://www.digikey.com/en/product-highlight/w/wurth-electronics/usb-solutions

\$\endgroup\$
3
  • \$\begingroup\$ As you said, no way of being 100% sure, but if we compare the diagram you posted to the circuit it is likely to be CM choke. If there are capacitors, they would be at the chip end, and the ESD parts would be at the connector end, between connector and CM choke. The mystery parts at both USB connectors looks like to be of ferrite material instead of plastic, and shaped like a typical CM choke as there is the narrow void part in the middle. The terminals look like they are at the four corners only, thus no pads under it, and tracks can be routed under the part for pair swap without vias. \$\endgroup\$
    – Justme
    Jun 9 at 18:13
  • \$\begingroup\$ Reasonable assumptions. Updated to reflect that possibility. The diagram posted is only one possible variation. \$\endgroup\$
    – Passerby
    Jun 9 at 19:30
  • \$\begingroup\$ I also removed my original comment about your answer. I think your answer is also a very reasonable assumption about the circuit. \$\endgroup\$
    – Justme
    Jun 9 at 19:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.