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Can anyone help identify the component that is missing from this circuit board? The board is the main controller board for a music keyboard. The keyboard is meant to be able to be powered over USB but due to the missing component (presumably) this doesn't work. The keyboard can also be powered via a separate DC jack, which takes power from a 9V 200mA DC adapter (this does work).

Below is an image of part of the mainboard, the USB connector is centre-bottom. Just above that is the missing component, marked L19. The bottom pad of this is connected to the +5V pin on the USB connector. To the right of the USB connector is the DC jack.

Next to the missing L19 are two components labelled L20 and L21 - I presume these are the same as the missing component, but I don't know what they are. A larger image of these is below.

I have contacted the manufacturer but they won't tell me what the component is, so I'm hoping someone here might be able to help me.

View of the circuit board with missing component circled

The missing component and two existing components that are probably the same

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2 Answers 2

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The missing component is a ferrite bead. Without a schematic, service manual or bill of materials, we can't possibly know the exact type of component is, and thus the current rating and impedance are unknown, and we can't suggest a replacement.

However, given that we know it only used to filter USB 5V power, feel free to put almost any ferrite bead there that is rated for 1A-2A and does not drop too much voltage. Or just put a 0R resistor, blob of solder, or a piece wire there to bypass it.

Just check first with a multimeter that there is no short circuit if the ferrite bead acted as a fuse and blew itself to pieces. However there is no sign of that, the damage looks purely mechanical, no blown component residue anywhere.

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    \$\begingroup\$ I understand that lukily this worked for the op. But, please for the future don't recommend solutions like "replace with a wire" before checking the board side for a short-circuit. \$\endgroup\$
    – Dorian
    Feb 3, 2020 at 11:00
  • \$\begingroup\$ @Dorian yes good point, and indeed that would be obvious - same as replacing a fuse without checking if the short circuit still exists. I'll add that to my answer. However the damage looks purely mechanical; there seems to be no sign of short circuits or residue of blown up components. \$\endgroup\$
    – Justme
    Feb 3, 2020 at 17:52
  • \$\begingroup\$ Thanks very much for this. I bridged the contacts (with a crocodile clip) like you suggested and my PC picked up the usb device. So I ordered a 30 ohm 1A ferrite bead, and while it went a bit wonky when I soldered it in, the keyboard is now working fine over USB power. \$\endgroup\$
    – Dave Kennard
    Feb 4, 2020 at 21:19
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You can verify your hypothesis by desoldering L20 (and/or L21) and use LC-meter/component-identifier to identify it (there are cheap ones like this LCR-T4 tester, for example) and buy replacement ones at local electronics components store. If you desolder both, you can compare if they are indeed the same (note that trying to test them while connected to the rest of the board usually won't give correct results)

L is usual marking for inductor, so as Justme said, it is probably ferrite bead used to filter EMI noise.

Also take note of how this element was lost? If it was mechanical incident where it was chipped of, then it is OK to just replace; but if it has blown due to electrical problem, there might be underlying issue causing it to blow again.

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    \$\begingroup\$ No, it is not a good idea to remove them. The two other ferrite beads are for USB data wires. They may (or may not) have wildly different ratings compared to the one that filters USB supply. Soldering them back and forth just increases risk of more damage, USB data lines are a sensitive impedance controlled differential pair. Besides, a 10 dollar component identifier will not give out anything useful about the ferrite bead. They are not inductors, ferrite beads are specified to have their rated impedance at 100 MHz, so it will take quite an expensive meter to get a measurement. \$\endgroup\$
    – Justme
    Feb 2, 2020 at 19:17

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