Electronic – 2C lines use open drain driver instead of tri-state drivers

i2copen-collectoropen-draintri-state

My understanding is that I2C lines use pull-up resistors to passively pull up the bus to logic high because the drivers used on the bus are active drivers, namely open collector/open-drain. Since open collector/open-drain drivers can drive the line low but not high, the issue of bus contention is mitigated.

My question is however, why does the I2C protocol use these drivers as opposed to tri-state drivers? If you have multiple tri-state output drivers connected to the same bus, as long as the enable signals for the tri-states are mutually exclusive, shouldn't we be able to take care of bus contention and also achieve faster rise times in comparison to open-collector/open-drain topologies?

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...as long as the enable signals for the tri-states are mutually exclusive...

The trick is how to do this without adding another wire, or multiple wires to tell each peripheral when it is allowed to drive the bus.

The main advantage of I2C is that it only uses two wires, and two pins on each chip connected to the bus.

If you're willing to trade pins for speed, consider using SPI, which can generally achieve higher speed than I2C, but needs 3 or 4 pins per device.

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Electronic – PIC18F2550 I2C Open Drain

You can do open drain on any digital pin on the PIC yourself, it's easy! I'll show through a quick chunk of example code using RB0.

_TRISB0 = 1; // Set the pin to high impedance
_LATB0 = 0;  // Set the output low (this would be _PORTB0 on some pics)
// As long as you're using pin RB0 as a open-drain, don't touch _LATB0

// To output a "low" (drain) do this:
_TRISB0 = 0; // Set the pin to output, it's already low, so it will "drain"

// To output a "high" (open) do this:
_TRISB0 = 1; // Set the pin to high impedance, the pullup resistor will pull high

The I2C module will do something similar to this internally, without you doing anything. In other words: yes! You'll need a pullup resistor!

EDIT: To answer your second question which I did not directly address, the pin does not need a pullup resistor during normal digital output operation. My example was to show you how a normal, fully functioning digital I/O pin could operate as a open-drain without any additional behavior or behaving unexpectedly. I should also note that I have done this exactly what I showed here to do open-drain in real programs, so this isn't a toy example, it's really how you do it!

Voltage level in an open drain bus (I2C)

The maximum voltage that can be interpreted as a low per the I2C spec is 30% of VCC. For your circuit to operate correctly (assuming that the Raspberry Pi follows spec) the voltage will need to be less than 30% of 3.3V or 0.99V.

You'll need to do some circuit analysis to determine the voltage that will be seen.

The ATTiny13 data sheet says that for the GPIO lines, a low can be as high as 0.7V depending on the pin and the amount of current sunk.

Add in the forward drop of the diode you've placed (which will be conducting when the ATTiny13 is conducting) and you may end up in trouble.

You may need to reverse the logic on the ATTiny13 and have it drive an external logic level MOSFET on, instead of relying on the sinking capability of the pin. You should also go with a schottky diode for D1 to lessen the forward drop.

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Electronic – PIC18F2550 I2C Open Drain

Voltage level in an open drain bus (I2C)

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