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PCA9306: VREF2 with no pull-up resistor

Part Number: PCA9306

Hello there,

In the datasheet the "Figure 2. Typical Application Circuit (Switch Always Enabled)" is given:

The datasheet states the following:

Do I understand correctly that if we have open-drain pins connected to the SCL1/ SDA1/ SCL2 /SDA2, then VREF2 will not need to control the EN pin? Will this circuit then work correctly?

  • Łukasz,

    Thanks for reaching out! First let me ensure we are looking at the same datasheet. I'll be referring to the one located on the PCA9306 product page. This means that your "Typical Application Circuit (Switch Always Enabled)" will be Figure 16.

    Please note that the EN pin and the VREF2 pin must be connected to VDD (shown in Figure 16 as VDPU) through a 200 kΩ resistor, as shown in Figure 16. These two should also be tied together. Figure 17 shows how to use an enable signal to control the device.

    Your schematic shows VREF2 tied directly to VDD. This will not allow the device to be operational. You’ll need to tie EN to VREF2 and connect these to a 200 kΩ pull-up resistor. Please also note that VREF2 > VREF1 (i.e. VEXT > VDD … it looks like this is fine on your current schematic).

    Regarding the sentence you highlighted – that paragraph is simply describing the need for directional control on the lines. Bus contention occurs when one device is attempting to force/drive high while another is attempting to force/drive low. That paragraph is stating that you’d need to prevent that from happening on your bus using either unidirectional communication or other direction control. If your I2C bus is driven only by open-drain sources, then there will not be a bus contention situation, since no device is able to drive high.

    I hope this clears things up.

    Best,

    Danny

  • Hi Danny,

    Thank you for the answer. I must have had some old datasheet at hand. The one you have shared consist of a lot better explanation about the EN/ VREF2 connection case. Thank you.