In this device, I'm using an internal PHY mode and this pin nPHY_INT is an active-low interrupt output (wrongly mentioned in the datasheet as active-high).

I just want to understand how to generate an output (low signal) from this pin.

The default value mentioned in this register "Internal MII Visibility" on page 123 is 0b. So, if the bit is 0b, that means, the nPHY_INT pin will be 0 V (active low). Or will it be HIGH?

If I want to output HIGH on that pin, should I change that bit to 1b?

Please help me understand.

  • \$\begingroup\$ For what purpose would you need the interrupt pin on a standalone USB to Ethernet chip? \$\endgroup\$
    – Justme
    Commented Nov 2, 2022 at 9:12
  • \$\begingroup\$ To get this information to an MCU to know that there was an interrupt from the device. \$\endgroup\$
    – user220456
    Commented Nov 2, 2022 at 9:17
  • \$\begingroup\$ You forgot to mention what would be the purpose of MCU being able to detect interrupt pin of a standalone USB to Ethernet chip. The MCU can't do nothing about it. \$\endgroup\$
    – Justme
    Commented Nov 2, 2022 at 9:20
  • \$\begingroup\$ Just to know the status or what's the device is conveying. Apart from that, there is know significant use \$\endgroup\$
    – user220456
    Commented Nov 2, 2022 at 9:26
  • \$\begingroup\$ @Justme , any idea with my question, pleasea? \$\endgroup\$
    – user220456
    Commented Nov 2, 2022 at 9:40

1 Answer 1


The table for the hardware configuration register (HW_CFG) in chapter 7.3.5 says for bit 7:

Internal MII Visibility Enable (IME)

This register enables a subset of the MII interface to be visible on unused pins when configured for the internal Ethernet PHY mode. The pins controlled by the IME bit are comprised of the pins listed in Table 3-1, “MII Interface Pins,” on page 12 and the nPHY_INT pin.

0 = The MII signals are not visible. The MII pins function as inputs.
1 = The MII signals are visible. The MII pins function as outputs.

Please note that the data sheet is correct, if it says "The internal PHY interrupt signal is active-high." The lower-case "n" marks the pin as negated from the actual signal.

To use the pin as an output, you need to use the internal PHY, according to note 1 in the table entry.

Then you need to set the bit to 1, but this "just" makes the pin an output.

To effectively generate an interrupt (pin becomes low), the condition for this interrupt needs to be true. I leave the research on the condition to you.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.