USB0_VBUS current draw higher than expected

Discussion created by lpcware Employee on Jun 15, 2016
Latest reply on Jun 15, 2016 by lpcware

Content originally posted in LPCWare by jsdmichaud on Wed Feb 25 13:39:56 MST 2015


We are working on a product that can be powered by USB. The uC used is the LPC4357. The VDDIO of the LPC4357 is seperate from the USB power. It can happen that the USB power is present on the board while the VDDIO is not present (regulator shutdowned). Looking to the LPC43xx user manual and the AN11392 application note, we learned that the USBx_VBUS (USB0_VBUS in our case) pin are 5V tolerant only when VDDIO power of the LPC is present.


So we needed a way to reduce the USB voltage at the USB0_VBUS. In the AN11392 application note there's a recommended solution by using a voltage divider with a 24kohm and a 39kohm resistor. This solution doesn't work for us because the voltage measured at the USB0_VBUS pin is a lot lower than expected.


So we tried another solution by using a series resistor and a simple diode between the USB0_VBUS pin and our 3.3V power bus, which is the VDDIO voltage of the LPC. Attached is the little circuit we used toreduce the voltage at USB0_VBUS. Here are results of what we measured:


- When VUSB is at 5V and the 3.3V bus (VDDIO) is not active, voltage measured at USB0_VBUS is 1.7V. We get this result because the 3.3V is around 1V (not 0V) because of the voltage going accross the protection diode in the LPC4357 because VDDIO is not present. So 1V + Vf of the diode = 1.7V


- When VUSB is at 5V and the 3.3V bus (VDDIO) is active and we are in the bootloader of the application (no initialization of the USB interface), voltage measured at USB0_VBUS is 4V. **We were expecting (3.3V + Vf of the diode) = 4V. So this is correct at this point.


- When VUSB is at 5V and the 3.3V bus (VDDIO) is active and the initialization of the USB interface is done, voltage measured at USB0_VBUS is 2.9V. **We were expecting (3.3V + Vf of the diode) = 4V. At this point USB0_VBUS pin seems to draw too much current. (5V - 2.8V)/ 1000ohm = 2.2mA which seems a lot for this kind of pin.


This current draw could also be the cause of the problem with the voltage divider solution recommended in the AN11392 application note.


Is there any other solution ? And, is it normal that this pin is acting like that when the USB is initialized in the application code.


Thnaks a lot