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I am designing a circuit and needs to place some pull downs on the address pins of PCA9546 and usually my go to resistor values for these kind of situations is either 10k ohms or 100k ohms. These values work almost all the time.

Having said that, I thought that if I were designing this circuit where power consumption is critical surely the datasheet must have some information that tells me the maximum pull - up / down. But I could not find such information on the datasheet.

What section of the datasheet does information belong? If the IC that I have given does not have this information (considering its not meant for low-power applications) can you give an example datasheet that has this information?

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    \$\begingroup\$ the datasheet states that the current will be \$1 \mu\$A max if address pins are tied to Vcc or to Gnd \$\endgroup\$
    – jsotola
    Apr 6 '20 at 19:22
  • \$\begingroup\$ Does the address need to be changeable? 0R strap resistors would work in this case too. Or omit the resistor and just connect to VCC or GND. Only if the address needs to be dynamically set via switch or IO pin it might need pull-downs or pull-ups. \$\endgroup\$
    – Justme
    Apr 6 '20 at 20:45
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Look at the maximum leakage current into these pins:

enter image description here

You need to choose the pull-up resistor value small enough that the leakage current flowing through it doesn't push the pin input voltage out of the valid range for the logic level you're trying to pull to.

Since \$1\ \mu A \times 10\ k\Omega\$ is only 1 mV, your default choice is fine.

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  • \$\begingroup\$ but what would be the formula for finding the highest value can go? You used ohms law, V= IR, but what would be the target voltage so that R will be the only one left? or looking at it in a different way, the current must be higher than 1uA, R < V/I , R < 3.3v/1uA , R < 3.3M ohms is it correct that that is the highest value? \$\endgroup\$
    – Jake quin
    Apr 7 '20 at 6:04
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    \$\begingroup\$ @Jakequin A voltage drop of 3.3 V over the resistor would be able to change a 0 signal to 1. The voltage must be small enough to be harmless. (Look at V_IL and V_IH.) \$\endgroup\$
    – CL.
    Apr 7 '20 at 11:33
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    \$\begingroup\$ @Jakequin If your actual low-level signal has, for example, 0.1 V, then you can afford at most 0.89 V until you reach V_IL. \$\endgroup\$
    – CL.
    Apr 7 '20 at 12:25
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    \$\begingroup\$ @Jakequin You have to compute how much voltage is dropped over the resistor; this is the error offset voltage that the leakage current can introduce. \$\endgroup\$
    – CL.
    Apr 7 '20 at 13:22
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    \$\begingroup\$ The leakage current is specified as "± µA", so it can flow into or out of the pin. And if the original signal is not exactly at 0 V, you have less headroom. \$\endgroup\$
    – CL.
    Apr 7 '20 at 13:39

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