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I'm currently doing a schematic that has an IC that operates at 3.3V (HDC1000 for those curious), and the rest of my digital operates at 1.8V. I have a level shifter for the I2C lines, but there is an active low data ready pin that is an open drain output that's used as an interrupt on the MCU.

My question is, would I need a level shifter on this line as well, or since it's an output, can I just pull this line up directly to 1.8V and have it sink the current to interrupt the MCU?

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    \$\begingroup\$ so long as it goes low enough, fast enough, that would be fine. \$\endgroup\$ – Jasen Jan 14 '16 at 9:32
  • \$\begingroup\$ What does the data sheet say about this? \$\endgroup\$ – Andy aka Jan 14 '16 at 9:35
  • \$\begingroup\$ @Andyaka, the data sheet does not address this directly. It's written under the assumption that you're using one voltage domain for your design. My first iteration for this design is going to be dev board style so we can validate all the components, so I could feasibly try both implementations with a combination of 0 ohm and DNP resistors, but the form factor implementation is very small so if I can save myself an IC and a couple passives completely then I'm all for that. \$\endgroup\$ – maksschmidt Jan 14 '16 at 9:44
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Not directly to 1.8V, but through a pull-up resistor of about 10K.

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  • \$\begingroup\$ Yeah, by directly I meant have a pull-up resistor directly to 1.8V as opposed to having a level shifter with a pull-up resistor to 3.3 on the IC side, and a pull-up resistor to 1.8V on the MCU side. \$\endgroup\$ – maksschmidt Jan 14 '16 at 9:42
  • \$\begingroup\$ OK. That will work perfectly well, I do that all of the time. \$\endgroup\$ – Steve G Jan 14 '16 at 9:51

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