What small IC could give switchable I2C pullups?

Question

My goal is to replace a DPST on pullup resistors with something simple, cheap and very small that can be activated with the I2C bus voltage. I thought that this might be a common enough requirement that I would find a simple IC for this task, but after hours searching through datasheets I am not finding a solution as ideal as my gut suggests should exist.

The question on how to Use GPIO for switchable Pullups of I2C has the basics of the thing I am trying to achieve on a very small PCB. I don't necessarily want to use a GPIO to turn the pullups on and off, my goal is flexibility of the design so the pullups could be turned on by a single jumper or a GPIO.

^{simulate this circuit – Schematic created using CircuitLab}

Inspired by the Bus Pirate which does something similar using a CD4066BC, quad bilateral switch I found the TC7W66F, dual bilateral switch which vaguely fits my price and space goals but I imagined something similar to a SOT-23 package.

The question mentioned above has an answer suggesting using FETs and I did find some dual FET packages with source already tied together that seemed like they might work but I'm primarily a software guy and have to admit not feeling 100% comfortable using something like this without further advice, or if it would require further components killing my space desires?

Answer 1

I would suggest "prebiased" BJTs

Dual MOSFETs tend to have a lot of Drain-Source capacitance which will affect the rise time of your I2C bus. BJT switching transistors tend to have less. For example, these ones have Cob of only 3pF typically at 10V (it will be 2-3 times higher at low voltage, which they don't tell you and you're supposed to know) but that's still pretty modest. Add the pullup resistors to the collectors, the emitters to Vdd, and connect the bases to your /enable line and you're done (one part plus the resistors, and the package is only 2.0 x 2.1mm). Very cheap in volume, and not much worry about ESD.

^{simulate this circuit – Schematic created using CircuitLab}

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So why not MOSFETs?

They're lower resistance, right? Well the saturated BJTs will drop 50~100mV most likely at the currents you'll be using them (compares well with 4066 switches), and compare the output capacitance of an FDS6312P MOSFET (Coss)- typically several hundred pF near 0V, which is nearly as high as the 400pF maximum for all devices on the bus itself.

Answer 2

One option is two P-Channel MOSFETS. These can be connected as follows:

^{simulate this circuit – Schematic created using CircuitLab}

Basically, if you use logic level MOSFETS you can feed the EN_n (not-enable) signal with +V volts to disable the pull-up resistors, and 0V to enable them. When disabled the resistors will essentially not be there. You will have body diodes from the I2C lines to V+, but those shouldn't cause any issue.

As @bitsmack points out, you could also add an optional pull-up resistor (R3) on the EN_n pin which would keep the MOSFETS disabled if the pin is left floating. This would allow the input to be open drain - simply short to ground to enable, or leave floating to disable.

It is possible to get very small (SOT23-6 or SOT23-5) packages which contain two P-Ch MOSFETS (e.g. this) which can then be wired up as shown above - usually they are arranged in ways which make routing quite simple. You can basically join the gates together and sources together making a very small essentially 4-pin package. Even if you throw in the pull-up resistor as say an 0603, the whole circuit would probably be smaller than a 2x2pin 0.1" jumper.

Answer 3

Practical approach

All the breakout boards I'm plugging together are under my control [...]

Keep it simple. Rip out the I2C pull-up resistors from every breakout board that you have. Install pull-up resistors with reasonable values on the microcontroller board.

my basic desire is to be able to turn-on the PU where I want it with a single jumper rather than 2

Trying to reduce the number of jumpers from 2 to one is lots of do beyond diminishing returns, if I may say so myself.

If a practical approach does not appeal

You can do something along the lines of active constant-current pull-up circuit.

\$I=\cfrac{V_Z-0.65}{R_{307}}\$

\$R_{307}=R_{308}\$

Any general-purpose small-signal PNP transistor would do.

Answer 4

How about just using an I2C bus accelerator with an enable, like the LTC4300A-1

This can isolate parts of the bus, and is meant to do so. The downside is that you need your pullups on both sides. The upside is you wouldn't have to wonder!