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This question is based on the Philips I2C specification UM10204.They gave (section 7.2.1, page 56) for calculating the maximum frequency of the bus:

$$F_{max}=\frac{1}{T_{low(min)}+T_{high(min)}+T_{risetime}+T_{falltime}}$$

\$T_{risetime}\$ depends upon the RC time constant of the bus and \$T_{falltime}\$ depends on lowest output drive on the bus. How are \$T_{falltime}\$ and the lowest output drive related?

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It is the weakest driver on the bus that determines the (longest) Tfalltime: if you consider that driver as a switched resistor (good approximation for a FET output stage) it is just another RC time, but now with a different resistor (and probably a different threshold voltage too).

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  • \$\begingroup\$ You might want to clarify the "different threshold voltage" comment. If pull-ups are stiff and an output stage is feeble, the voltage when driven might not approach 0 volts, but instead exponentially approach some point which is significantly above zero. If one is using a 5-volt supply, the lowest V(IL) of any part is 1.2 volts, one has a 1K pull-up and a 176-ohm pull-down, then the pull-down would try to pull 5 volts to 0.75 volts. Thus, the capacitor would start 4.25 volts above its final voltage and have to discharge to 0.25 volts above--a factor of (4.25/0.25)=17. \$\endgroup\$
    – supercat
    Mar 27, 2013 at 16:51
  • \$\begingroup\$ Thus, the fall-time would be ln(17)=2.83 RC time constants. Not humongous, but significantly longer than one RC time constant. \$\endgroup\$
    – supercat
    Mar 27, 2013 at 16:52

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