I'm planning on powering a 3.3V I2C sensor and its pull up resistors with a linear LDO regulator.


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The output of the LDO aswell as the sensor's power pins require bypass capacitors. Wouldn't said capactiors together with the pull up resistors act as low pass filters and thus mess up the signals? Calculating the cutoff frequency and lowering the bus frequency accordingly would be one way to solve this I guess. But wouldn't this problem occur with every I2C sensor since every supply voltage trace has buffer or bypass caps on them?

  • \$\begingroup\$ No. The signals are isolated from the capacitors by the pullup resistors. Now capacitance from SDA or SCL directly to GND is a problem. This happens when you run the I2C bus over too long cables. \$\endgroup\$ – user_1818839 Jul 4 '20 at 18:54
  • \$\begingroup\$ I believe I made a mistake. The capacitors and the resistors are in parallel. Though I don't know what you mean by "The signals are isolated from the capacitors by the pullup resistors". How are they isolated? \$\endgroup\$ – crispus Jul 4 '20 at 19:05
  • \$\begingroup\$ The signals have whatever R1 and R2 are between them and the capacitors. \$\endgroup\$ – Wesley Lee Jul 4 '20 at 19:20

No, the capacitors are on the power supply pins, and the capacitors will have no effect on the I2C bus pins. That's exactly how it is supposed to be done.

  • \$\begingroup\$ But why exactly do those capacitors not effect the bus if the pull ups share a net with the capacitors? I fail to understand. \$\endgroup\$ – crispus Jul 4 '20 at 19:37
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    \$\begingroup\$ The capacitors really do not share a net with the resistors at all. Imagine there is an ideal voltage supply. As it is ideal, capacitors can be thought as open circuit, and thought as not being there at all. As much as the bus wants to toggle high and low, it has no effect on the supply or ground, there will always be 3.3V at supply and 0V at ground. The bus would be affected if there was capacitance from bus to ground, or from bus to supply, but there isn't, there's just pull-up resistors to ideal 3.3V supply. \$\endgroup\$ – Justme Jul 4 '20 at 19:39
  • \$\begingroup\$ Why can capacitors be seen as open circuits with ideal voltage supplies? \$\endgroup\$ – crispus Jul 4 '20 at 19:57
  • \$\begingroup\$ Capacitors have infinite impedance at 0Hz DC. \$\endgroup\$ – Justme Jul 4 '20 at 19:59

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