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https://www.allaboutcircuits.com/technical-articles/the-i2c-bus-hardware-implementation-details/
I read from this article that for the open drain there's nontrivial power consumption when the input is HIGH and the output is LOW. The current flows from power supply, through the resistor and through the nmos to GND.
How is this non trivial power consumption when compared to the push/pull configuration where the inputs are HIGH and Vout is connected to GND through an NMOS?
A CMOS input (for example, VIN in the circuits above) does not source or sink any current (except for a little bit when switching), so when a push/pull output is connected directly to an input, no power is consumed for either low or high signals.
If the output does not drive another CMOS input but a load like a LED, then a push/pull output and an open-drain output would consume the same amount of power.
In push-pull, no current flows in either output state.
In open-drain, current has to be drawn through the pullup resistor continuously to maintain the output low state.
The pullup resistor can be increased to reduce current, but not without limit. At some point, it still needs to source enough current to charge line capacitance quickly enough, and to drive any loads on the line.
Whether the current is trivial or not depends on your context. If you are using properly designed CMOS, you can expect sub-uA currents, which might give you years of operation on a small battery. A 5k resistor to 5v will draw 1mA, which if it reduces operating lifetime 1000 fold is definitely non-trivial.
