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To keep RF as low as possible in a music synthesizer, I power down the SoC (XIAO SAMD21) which performs autotune once the autotuning procedure is completed.

On one side of the TCA9548A there is the SoC (XIAO SAMD21), with its I2C pulled up to 3.3 V, and on the other side there are many MCP4728s with their I2C pulled up to 5 V.

When I power down the SoC, I'd like to power down the TCA9548A too (to avoid a second 3.3 V LDO just for the TCA), while the MCP4728 are still powered and with the I2C pulled-up to 5 V.

Is it OK for the TCA to have pull-up resistors still supplied when its power is off?

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  • \$\begingroup\$ What does the data sheet say about the SDA and SCL pins? Do you need to communicate on the same bus with unpowered chip? \$\endgroup\$
    – Justme
    Commented Dec 28, 2022 at 14:37
  • \$\begingroup\$ No, once the procedure is completed there is no more communication on I2C busses (the DACs stay at the same value until a new procedure is initiated). And there's nothing about this in the datasheet... \$\endgroup\$
    – Synthlink
    Commented Dec 28, 2022 at 15:34
  • \$\begingroup\$ If I understand your setup correctly, the powered up pull up resistors are going to forward bias the input ESD protection diodes of the TCA (assuming it has them on its inputs). If this is the case, you just have to know what DC current the ESD diodes can sustain, and what the current supplied through the pull up resistors is. \$\endgroup\$
    – SteveSh
    Commented Dec 28, 2022 at 21:15

1 Answer 1

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Here is a Texas Instruments engineer's answer : "Having a pull-up voltage above VCC is OK, with any VCC, even 0 V."

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  • \$\begingroup\$ Synthlink - Hi, Thanks for coming back with an answer to your question. In order to effectively mark the topic as solved, please consider "áccepting" your choice of the best answer (i.e. click the "tick mark" next to an answer - your answer or another one, if one is written - to turn the relevant tick mark green). This shows that you don't need more help and future readers can quickly that there is a confirmed solution. Thanks. \$\endgroup\$
    – SamGibson ♦
    Commented Jan 6, 2023 at 23:51

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