PU pins on the MSP430f5359

Roy Stegers said:

Hi All,

For one of our products we want to use the MSP430F5359 as MCU. After looking at the datasheet we saw that the MCU has two PU pins. The data sheet says that these pins are high current: 

The Port U Pins (PU.0 and PU.1) function as general-purpose high-current I/O pins. These pins can only be
configured together as either both inputs or both outputs. Port U is supplied by the LDOO rail. If the 3.3-V LDO is
not being used in the system (disabled), the LDOO pin can be supplied externally.

Does this mean that the current is 3.3v only, there is no way there could be 1.8v on these pins when setting them "high"? 

When not using usb, these pins function as normal I/O pins but with a higher voltage?

Is it possible to set these pins around 1.8 volt when the pins are high? The reason that i post this question is that we want to use one of the PU pins as a output pin. On the other end want to connect a Input pin of the PAN1326 Bluetooth chip. We want to connect the nshutdown pin to PU1.

But in the PAN1326 data sheet in the recommended table we see a lower recommeded  I/O voltage (max 1.92).

Does anybody have a sugesstion? Is using a normal I/o pin a good suggestion?

Below is the table of the PAN1326 datasheet:

Does anybody have an idea if it is possible to connect a ~1.8 - 1.9 volt input pin of the PAN1326 to one of the PU pins? Because of the higher voltage, in the MSP datasheet says that the minimum output voltage is 2.4 volt.

Jens-Michael Gross said:

You can always drive a lower-voltage input by a higher voltage port pin. Just connect the input with a pull-up to its supply (1.8V then) and put a diode between the output and the input. This way, you turn the output into an open-collector output that will either drive the input low or release it to its pull-up.
However, it takes two external parts to do so. Of course, this way you can only sink a current, not source it.
If you want to source a current, you can limit the output voltage by putting a zener diode in series. For a low-impedance load, this will cause a voltage drop so the ‘output’ voltage is low enough and still a significant current flows (other than with a series resistor).

Don’t think just digitally. After all, output voltages and currents are analog things.

Hi Jens-Michael,

Thank you for your reply! We have a quick follow up question about the PU pins. When we use the MSP430F5359 where the USB is disabled and we feed the MCU with a external current of 1,85v. One of our hardware people says that by giving a 1,85v external  current that the I/O voltage is then also 1,85.

But we are not sure of this because the datasheet has the following table:

In this table we see that the minimal output voltage is 2,4 volt, this will cause a problem when connecting to our other device. So when we supply the MCU with 1,85v and using the PU as GPIO pins will this have a output voltage of 1,85 or a minimal 2.4 volt?

Also in the datasheet we found the following passage:

The Port U Pins (PU.0 and PU.1) function as general-purpose high-current I/O pins. These pins can only be
configured together as either both inputs or both outputs. Port U is supplied by the LDOO rail. If the 3.3-V LDO is
not being used in the system (disabled), the LDOO pin can be supplied externally.

This means that the PU pins are special high current pins? And have not the same current as the normal GPIO pins?

Hi Jens-Michael,

Thank you for your answer, it is very clear!