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I'm designing a peripheral circuit between a sensor (U1) and a MCU (U2). U1 has an open-drain output pin to indicate its status. This pin will be staying at either HIGH or LOW constantly (not frequently toggling). U2 requires 3.3V as input.

I only have 5V supply and therefore a voltage divider is used to produce 3.3V. Since the voltage divider already has high enough resistors (100k and 200k), the current flows through R1 will be really low.

In this case, do I still need a pull up (R3) for the 3.3V? In other word, can I replace R3 with a wire/jumper?

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ You have 5V supply then how do you power U2? \$\endgroup\$ – Rohat Kılıç Oct 31 '18 at 4:39
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No, you don't need R3.

Be aware of the current draw of your input. Input circuitry impedance isn't infinite, and some logic circuits might struggle to "understand" the signal if the resistance is too high.

I don't know what is U2, but your device should not suffer if you were to use a single pullup resistor. It's common for ICs to have over-voltage protection circuitry inside, and some 3v3 powered ICs even have 5V tolerant inputs

schematic

simulate this circuit – Schematic created using CircuitLab

For input current draw and input voltage tolerance, you should read the datasheet.

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You don't need the additional resistor. The 100k will make the rise time slow-ish but since the line isn't toggling often that should be ok. The additional 10k would make it even slower.

How are you powering U2? It seems odd it isn't powered with 3.3V yet that's what its inputs require.

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