Jason Garner

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9 years, 10 months ago.

LPC11U6X USB pins connections

Am I overthinking this?

I emailed embedded artists, but I also want to get feedback here in case anyone needs this information too.

I'm looking at http://www.embeddedartists.com/sites/default/files/support/xpr/LPC11U68_Xpresso_v2_Schematic_Rev_B.pdf (page 3) and I'm wondering how necessary PRTR5V0U2X is? In my application, the only one touching my circuit is the system installer.

I also am wondering about how USB_CONNECT and USB_VBUS are connected as opposed to the mbed LPC11U24 design.

I had used a 10k resistor for USB_VBUS, but they use a 470 ohm. I don't really see why we would need the extra current.

This is how I had the LPC11U35/LPC11U24 connected for USB_CONNECT /media/uploads/elevatorguy/capture_1.png /media/uploads/elevatorguy/capture2.png

but PIO0_6 is not connected as USB_CONNECT in the embedded artists design. Does this matter?

Also, the 33 ohm resistors and 18p caps are not connected. I wonder if the PRTR5V0U2X removes the need for that, or if it is simply in the chip now? (page 282 of UM10732 shown below)

/media/uploads/elevatorguy/capture3.png

I originally thought I could use the same circuit from the LPC11U24 design, but I'm getting confused now:/

1 Answer

Tom Russell
9 years, 6 months ago.

The PRTR5V0U2X protects the LPC11U68 from transients so I would keep it. The resistors affect the rise and fall time of the USB signals. The LPC11U68 has USB_CONNECT built-in so does not need the external transistor (see page 280 of user manual). Did you get an answer from EA? If you have a working schematic, can you post it?

Tom