Why I2C designed to work with pull-up resistors and not pull-down ones?

Question

I understand that in I2C, SCL and SDA lines use pull-up resistors and the pin drivers are open collector NPN devices which can drive pins to ground. This gives I2C an advantage that the same bus now can be shared with multiple slaves, and even if two or more slaves accidentally try to drive the bus at the same time it won't cause any damage to the system.

But this can also be done using PNP open drain drivers and pull-down resistors on SDA and SCL lines. Things like clock-stretching and multi-master arbitration can be achieved with this too.

Does the current implementation of I2C protocol gives any benefits over the above suggested alternative implementation?

Answer 1

Electrically it makes sense because ground is the one common connection to all devices on a IIC bus. That's a lot less of a restriction than forcing power to be the common connection to all IIC devices, as would be required if the lines were driven high and floated low via pulldowns.

Note that IIC devices don't all need to be powered from the same net or the same voltage. This would not be true if both bus lines had to be driven to the single common power voltage.

Answer 2

In the good old days, TTL drivers were much better at pulling a signal down than pulling it up. Therefore, protocols like I2C, but also interrupt lines, reset, and others, were all implemented using a pull-up with distributed pull-down.

Answer 3

It's easier to use ground as a common reference among subsystems that might have varying supply voltages. If you use PNP transistors to pull up to a supply voltage, all subsystems would have to be connected to the same supply.

Answer 4

Good answers abound here, but there is also another reason.

If the quiescent state of the bus is at ground, there is no way to tell if the bus is connected or just hanging in space.

It is normal for the pull-up to be located at the master device. Slaves usually do not have a pull-up. This is because the pull-down current that would be required to assert a low level would increase with the number of devices connected to the bus.

A slave, when plugged into the bus, can then detect that the line is pulled-high (Assuming it is not being used) and know that the bus is actually there and quiet. That would not be the case with a ground biased bus.

Answer 5

If I understand the question correctly one aspect is:

• Why do you use pull-up resistors and NPN transistors instead of pull-down resistors and PNP transistors?

First of all you should note that you don't use bipolar transistors (NPN, PNP) but MOSFETs (which exist in four different variants).

Devices using the "pull-up and NPN" variant use a n-channel enhancement MOSFET. Because the source of this MOSFET is connected to ground the gate-source voltage (controlling the current flow) is equal to the voltage between gate and ground. So the MOSFET can be controlled using a voltage between 0 and Vdd.

There would be three possibilities to implement the "pull-down and PNP" variant:

• Using a p-channel enhancement MOSFET

  On an NMOS or CMOS IC p-channel MOSFETs with comparable characteristics (resistance etc.) require more space than n-channel MOSFETS.

  In microelectronics space is money so p-channel MOSFETs avoided if possible.

• Using an n-channel enhancement MOSFET

  This would require the output of the logic circuit driving the transistor to have a "LOW" voltage of the supply voltage (e.g. +5V) and a "HIGH" voltage above the supply voltage (e.g. +10V when the rest of the circuit is supplied with +5V).

  The reason: The source-ground voltage will be Vdd when the MOSFET is conducting. The gate-source voltage must be positive so the voltage between gate and ground must even be higher.

  You would need two voltage supplies - and a circuit shifting the output of the logic circuit from 0...+5V to +5V...+10V ...

• Using an n-channel depletion MOSFET

  Unfortunately I can't tell you much about this solution. However I found some page using Google saying that depletion MOSFETs are more difficult to produce than enhancement MOSFETs and they are avoided for this reason.

  I know from power electronics (not microelectronics) that the "two power-supply" variant described above is even preferred over depletion MOSFETs. (But I cannot tell you why.)

  EDIT Using n-channel depletion MOSFETs you would probably need a negative voltage (e.g. -5V) so you would also need two supply voltages...

Answer 6

There is also one more added benefit for having common ground and pull-up data lines (over having common VCC and pull-down):

Even if the original intention was to connect devices on the same PCB at span of few inches only, it was successful enought so now is not uncommon to have the lines long couple of feets and connecting "devices" which could be computers or something of equal complexity, with some devices having its own power sources (of different quality, say you connect something wall-plug powered with something battery powered). It is better, if connection works "at least good" even in not ideal and out-of-spec conditions.

And lot of such connected devices may be somehow connected also by other means, then only I2C communication. Usually when connecting devices together you connect it with common ground - sometimes as part of other functions, sometimes just because it is mounted on metal case and the devices are ground-connected with the case too (or with common cooler or something like that) or there may be shielded cable with grounded shield inside - which also connects the grounds.

If you also directly connect power lines (VCC) of such devices, you will get problems when those lines would be on different voltage naturally (sure, it may say 5V here and there, but depending on the construction and part tolerancies of power sources it could be also 4.9V or 5.2V or even changing, if it is battery powered and sometimes running some motors, making the power drop and rise over time).

In such case there is effectively short circuit betwenn those power sources of part a Volt and depending on the sources (and resistance of the ways) there could flow relatively high currents resulting not only in energy waste and heat rising, but maybe even in damaging (or shortening life) of some of those sources. Which is not good.

Having common ground and pull-ups avoids such problems - ground is ground and pullup resistors allow for only really small cross current even if the VCC differs a lot over the devices.

Answer 7

You don't have to send so much power through the chip, if it's pull up.

Because the chip doesn't drive anything, it's just creating a short to bring the bus to 0, and doing an open to bring it to 1.

If it was pull down, you'd have to send power through the chip to drive the bus to 1. If the bus happens to be an accidental short, that might be a lot of power you drive through that chip to try to push it up to 1.

Disclaimer: I'm a pretty lousy EE at this point.