I2C – connecting 5V device to 3.3V host

According to version 4 of the \$\mathrm{I^2C}\$ spec,

"Due to the variety of different technology devices (CMOS, NMOS, bipolar) that can be connected to the I2C-bus, the levels of the logical ‘0’ (LOW) and ‘1’ (HIGH) are not fixed and depend on the associated level of VDD. Input reference levels are set as 30 % and 70 % of VDD; VIL is 0.3VDD and VIH is 0.7VDD. See Figure 38, timing diagram. Some legacy device input levels were fixed at VIL = 1.5 V and VIH = 3.0 V, but all new devices require this 30 %/70 % specification. See Section 6 for electrical specifications." (page 9)

Deeper in the spec, you'll see that this \$ 0.7 \times V_{DD}\$ is the minimum logic high voltage:

For your 5V system:

\$ 0.7 \times 5 V = 3.5 V\$

\$ 0.3 \times 5 V = 1.5 V\$

To me, the 3.3 V pull-up looks marginal, especially if any of your 5V devices use the 'new' standard of \$ 0.7 \times V_{DD}\$ for logic HIGH.

Your mileage may vary, but it's always best to be within the spec wherever possible...

Since your I2C works at 100Khz, but not 400 Khz, it is a good idea to look at the various factors that have an effect on timing.

1: Check that your slave board supports 400Khz.

2: Resistor values are too big.

When the timing is increased from 100k to 400k, the period of the clock drops from 10 us to 2.5 us.

This means that the rising edge of your data/clock signals has a significantly less amount of time to settle. the time taken is calculated as follows:

t = rc

the capacitance on the bus is usually constant and a property of each device. It sounds like you have these. Add them up.

The resistor values are the next variable. Since you have three in parallel, you need to add them using 1/Rt = 1/R1 + 1/R2 + 1/R3 and so on. You only need one resistor on the bus, so having three in parallel is going to lower the total resistance.

You can now calculate t using the above formula. If it is more than 300ns (just over 10% of your clock period at 400k), then the rise time is out of I2C spec. Here, table 5, page 32.

If you'd like to calculate the correct resistor value, you can re-arrange the above formula to get R=t/c and work from there, where T is 300ns or less.