2
\$\begingroup\$

I am using a PHY that has internal pull-ups and pull-downs on certain pins.

On this particular pin, there is a pullup that the datasheet says can range from 33 kΩ to 93 kΩ.

For the entire range, I would like to calculate the actual voltage that will be on this pin. The pin is pulled up through this internal resistor to +2.5 V. I am assuming that this pin will be at something less than +2.5 V due to some leakage current through the resistor.

The datasheet mentions that there is 85 μA of leakage current (including the pullup resistor), but I am not sure how to use this value since 85 μA * 33 kΩ = 2.8 V which doesn't make sense.

Can someone help me figure out the voltage on this pin for the 33 kΩ to 93 Ω range?

The datasheet can be found here.

enter image description here

\$\endgroup\$
3
  • 2
    \$\begingroup\$ It's an input pin with a pullup resistor that's designed to ensure it gets read as high if you don't connect it. Why would you care what the exact voltage is? \$\endgroup\$
    – Finbarr
    Commented Jul 26, 2022 at 12:44
  • \$\begingroup\$ I am just curious to know how you would calculate this voltage \$\endgroup\$
    – Matty
    Commented Jul 26, 2022 at 12:54
  • 2
    \$\begingroup\$ If it's not connected there's nowhere for (leakage) current to flow. No current means no voltage drop so it'll pretty much be 2.5V .If something is connected to the pin it'll likely be much smaller impedance and thus the voltage will be dictated by that external circuit. \$\endgroup\$ Commented Jul 26, 2022 at 13:01

2 Answers 2

3
\$\begingroup\$

The +/- 85 μA figures in table 8-3 on page 65 of the data sheet are max and min limit values hence, they are meaningless for trying to estimate what the typical voltage is on the input pin. Forget about trying to calculate that this way.

However, you can make something from these numbers; you can assume that when the input pin is taken to ground, that there might be up to 85 μA flowing and that implies a minimum input pull-up resistance of 2.5 volts / 85 μA = 29.4 kΩ. This value of resistance is the physical internal pull-up in parallel with the input gate's natural leakage resistance (equivalent).

And, when you have the internal resistor configured as a pull-down, it might be as low in value as 29.4 kΩ.

But, other than that, there is nothing you can do to estimate the voltage on the input pin.

\$\endgroup\$
2
\$\begingroup\$

The listed leakage is a maximum number and it needs to be something that is testable. Since it is listed as "+/-", they are going to test for leakage to ground and Vcc. With some type of current measuring device, the tester will ground the pin and measure the current in the pin, then tie the pin to Vcc (2.5V) and measure the current.

If they are testing the resistor value, that would be in another test. There is no way to know what the open-circuit voltage will be from this specification.

For critical applications (military, space), you shouldn't depend on weak internal pullups. For consumer devices, not so important.

\$\endgroup\$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.