We have made a board using an LPC4078FBD100 with the following circuit:
Using Keil MDK 5.24a I have copied the example for an Embedded Artists LPC4088-32 Developers Kit and modified it for the LPC4078. Since the FBD100 packages only has U1 on the pins (I believe) I've changed the port functionality to U1=device, U2=host.
After setting up the USB pins and trying this I cannot get any life from the USB port. Plugging it into Windows 10 and using usbview.exe there is nothing detected on the port.
Should the LPC4078FBD100 U1 port be able to work as a USB device? It is not clear from the Product Data Sheet Section 13.1 (https://www.nxp.com/docs/en/data-sheet/LPC408X_7X.pdf).
My Project files:
Any help would be appreciated,
Thanks
Hello Ian,
In your scheme, is missing connected pin "USB_Connect" to connect 1k5 resistor on DP signal, I think.
Example can you find in datasheet (fig. 25, fig. 26) https://www.nxp.com/docs/en/data-sheet/LPC408X_7X.pdf
Some MCU (ex. LPC1549) has internal 1k5 resistor. But LPC408x not.
The 1k5 resistor is used to allow the PC to detect that a new full speed USB device has been connected to the USB port. This resistor must be used for the device to enumerate over USB. This is defined in section 7.1.5.1 of the USB specification.
Peter.
Ah thank you, I was puzzled by that as there previous MCU (LPC43XX) did not have such a thing in its USB device circuit. Your explanation cleared that up. So we need the self-powered circuit, but I don't know what the "SoftConnect switch" is specifically (Sorry I'm a software engineer).
Any advice on what component we need for this?
Thanks for the help.
Hello Ian,
there are two ways.
A. With SoftConnect feature:
B. Without SoftConnect:
Use 1k5 resistor directly between USB-DP and VCC
Have a nice day.
Regards.
Peter