I saw this weird symbol in a schematic and I couldn't find what it means.

It was connected from one side to the shield and ground of a USB port. From the other side it was connected to the ground of a chip.


2 Answers 2


That would appear to be a solder jumper (aka solder bridge). On the circuit board there will be two pads like that and you can either leave them open or put a blob of solder (or a short piece of wire) across them. This is done to allow for different configurations to be made at the time of populating the circuit board or later.

  • \$\begingroup\$ Then I would wonder why they would put it on the Arduino Mega2560. Thankfully it's not something crucial \$\endgroup\$ Nov 8, 2023 at 19:55
  • 4
    \$\begingroup\$ @TadeášHrudka It appears to be for connecting the USB Ground (UGND) to the Ground (GND) on the ATMEGA16U2-MU chip and defaults to open so the two grounds are isolated. I haven't found anything in the documentation that explains why you would have it one way or the other but I only skimmed the datasheet so I may have missed it. \$\endgroup\$
    – GodJihyo
    Nov 8, 2023 at 20:29
  • 6
    \$\begingroup\$ Another common use is at the output of a power-conversion section of your circuit board. When you first test the board, you can make sure the output of that section is as expected before putting the solder connection in place. If there is anything wrong with the power-conversion section, this can help you find it before applying the output downstream and destroying dependent devices such as microprocessors and FPGAs that tend to be expensive, hard to replace, or both. \$\endgroup\$ Nov 9, 2023 at 0:31
  • 2
    \$\begingroup\$ @GodJihyo If you’re connected to a computer (or other externally-powered USB device) via the USB connector, but are powering the board from the DC jack, the grounds need to be isolated to avoid the possibility of things like ground loops. But if you’re only ever going to power it via the USB port (and have a way to ensure that is the case), you may want a guarantee that the grounds are equipotential depending on what you’re actually doing. \$\endgroup\$ Nov 9, 2023 at 14:08
  • 1
    \$\begingroup\$ @GodJihyo Breaking ground loops is a thing but you need to have your signals remain connected with optical, magnetic or capacitive isolators. You can no longer pass power or ground referenced signals trough interfaces that do not have a valid ground in common. Special cases can exist if you assume your grounds do not exceed the common mode input voltage of a differential signal but that is asking for trouble as often happens with people using ungrounded RS-485 systems and hoping no ground references float too much. \$\endgroup\$
    – KalleMP
    Nov 9, 2023 at 21:46

It's a solder bridge, they look like this on a PCB:

enter image description here
Source: https://diptrace.com/forum/viewtopic.php?t=11687

  • 1
    \$\begingroup\$ In particular, note how the schematic symbol is a representation of the second type shown in this image. \$\endgroup\$
    – Hearth
    Nov 9, 2023 at 14:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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