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I'm designing a compact addressable LED strip driver circuit. Since USB Type-C chargers with PD should be able to provide up to 100W @ 20V, I want to use a USB cable as my main power source to power both the MCU and LED strip.

However, I ran into a roadblock when trying to route the PCB. If we're pulling 100W @ 20V the current will be 5A. So I need to make the VBUS traces pretty wide for such a small PCB design - between 1.5 mm and 3 mm, depending on the copper layer thickness. However, even if I was able to find the space for these traces on my PCB, I wouldn't even be able to connect them to the USB port, because the pads for VBUS pins are less than 1 mm in width.

How are you expected to make use of the full 100W when the pads are so small you can't even connect a correct width trace to it?

enter image description here

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    \$\begingroup\$ put a block of vias, like 6 8x16 vias down to internal layers and use planes to get the current out \$\endgroup\$
    – TQQQ
    Commented Mar 20 at 13:05
  • \$\begingroup\$ Should the vias go directly on the pad or how should they be connected to the pad? \$\endgroup\$
    – Tadej Gašparovič
    Commented Mar 20 at 13:09
  • \$\begingroup\$ USB-C relies on using all four Vbus pins in parallel and all four GND pins in parallel to deliver its maximum power. \$\endgroup\$
    – Finbarr
    Commented Mar 20 at 13:36
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    \$\begingroup\$ Don't do via-in-pad, it's expensive in production. Just put close to the pad, from both sides even. \$\endgroup\$
    – TQQQ
    Commented Mar 20 at 13:48
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    \$\begingroup\$ You can take a track from both sides of each pad to join to a larger track. As long as it's short enough the heat generated will be minimal and it will flow to the larger track. As it stands, you've got all your Vbus power heading down one small track. \$\endgroup\$
    – Finbarr
    Commented Mar 20 at 13:53

2 Answers 2

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A very simple solution is to use a through hole connector:

enter image description here

Source: https://gct.co/connector/usb4085

Then you can fill a solid power/GND plane each on it's own layer and obtain very low resistance. This also has the advantage of more securely attaching the connector than an SMD part with only 2 or 4 mechanical mounts where lifting pads is sometimes possible.

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  • \$\begingroup\$ Great stuff! For some reason it didn't cross my mind that through hole connectors exist. I think this is the simplest solution to get the power onto its own plane. \$\endgroup\$
    – Tadej Gašparovič
    Commented Mar 20 at 14:14
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The current is split between four conductors, and therefore needs to be collected in parallel on your PCB too. You've made a comment that there are only two pins, but this isn't usually true - instead, two pins are brought out to one pad, and two pins to a second pad. The mechanical drawing for USB-CD16HMxTR (rated for 5A) is shown below for clarity - the footprint has two power pads, but the connector has four physical pins.

mechanical drawing of a USB-C connector, showing four pairs of pins

Your routing has these pads connected serially, which means that half the current will be borne by the link between the two. Instead, if you're dealing with this level of power, I'd suggest you immediately use vias to connect to a plane on another layer - possibly on both sides of the pad if you can fit it.

Here's an example layout I've used before which fits within reasonably cheap processes (and only Full Speed USB), though admittedly it doesn't draw nearly so much current. Here, we have a ground plane, and a VBUS plane (highlighted in the second image).

USB-C power layout example

same layout, with the net highlighted

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