I2C – Maximum Value of Pull-Up Resistor in I2C

The correct pullup resistance for the I²C bus depends on the total capacitance on the bus and the frequency you want to operate the bus at.

The formula from the ATmega168 datasheet (which I believe comes from the official I²C spec) is --

$$\text{Freq}<100\text{kHz} \implies R_{\text{min}}=\frac{V_{cc}-0.4\text{V}}{3\text{mA}}, R_{\text{max}}=\frac{1000\text{ns}}{C_{\text{bus}}}$$

$$\text{Freq}>100\text{kHz} \implies R_{\text{min}}=\frac{V_{cc}-0.4\text{V}}{3\text{mA}}, R_{\text{max}}=\frac{300\text{ns}}{C_{\text{bus}}}$$

The Microchip 24LC256 specifies a maximum pin capacitance of 10pF (which is fairly typical). Count up the number of devices you have in parallel on the bus and use the formula above to calculate a range of values that will work.

If you are powering off of batteries I would use values that are at the high end of the range. If there are no power limits on the power source or power dissipation issues in the ICs I would use values on the lower end of the range.

I sell some kits with an I²C RTC (DS1337). I include 4K7 resistors in the kit which seems like a reasonable compromise for most users.

Hold time is the amount of time for which the signal must be stable after the clock edge. When the signal is low and starts rising on a clock edge it must not cross VIL for 300ns. You have $$0.8 = e^{-300ns/(200pF R)}$$

However, that still does not give you the correct answer. I had a look at the I2C specification here. Interestingly, 300ns only appears as the rise time requirement when you are in fast mode, which is the mode for 400kbps. The time it takes for the signal to rise to VIH is: $$0.2 = e^{-300ns/(200pF R)}$$ Solving for R $$R = \frac{-1}{\ln 0.2} \times 300ns / 200pF\approx 932 \Omega$$