• State TI Thinks Resolved
  • Locked Locked
  • Replies 4 replies
  • Answers 1 answer
  • Subscribers 33 subscribers
  • Views 1106 views
  • Users 0 members are here
Support feedback
Options
Options
Similar topics

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

Original question:

PCA9306: PCA9306 EN and VREF2

PCA9306: PCA9306 EN and VREF2

Dan Markoff
Prodigy 20 points
Part Number: PCA9306

Why is there a FET from VREF1 to VREF2?  It seems like that would always be a problem.  We do not currently have a 200k series resistor on the VREF2 side and things seem to be working fine.  This is in reference to this previous post: https://e2e.ti.com/support/interface/i2c/f/390/t/629454?tisearch=e2e-quicksearch&keymatch=PCA9306%20VREF2

  • 0
    TI__Mastermind 49010 points

    Hey Dan,

    Just to clarify, the FET between Vref1 and Vref2 is internal to the device.

    The FET between Vref1 and Vref2 is meant to help establish a reference voltage for other circuitry in the device when Vref2 pin is shorted to Enable pin of the device. This is done to get lower propagation delays.

    "We do not currently have a 200k series resistor on the VREF2 side and things seem to be working fine. "

    If the FET is not turned on (Vref2<=Vref1+Vth) then the device will still function however if the FET is turned on then you basically have a very small impedance between Vref1 and Vref2 which can be simplified to a short circuit.

    Thanks,

    -Bobby

  • 0
    TI__Genius 17195 points

    Hello Dan,

    Do you have a diagram or schematic of how you connected it?  We can give you better feedback once we know how you connected it in your system.

    -Francis Houde

  • 0 in reply to fhoude
    Prodigy 20 points

    Francis,

    We have EN and VREF2 tied together and directly to 3.3V.  3.3V is always enabled.

    VREF1 is tied to 1.8V which is powered up and down as needed.

    Dan

  • 0 in reply to Dan Markoff
    TI__Genius 17195 points
    Hello Dan,
    The 200k ohm resistor is there for two reasons.
    1) The 200k ohm resistor is there to limit the bias current for generating the reference voltage. You can have smaller values to you end up just burning extra current for little to no reason.
    2) The 200k ohm resistor is there to limit the current to protect from being exposed to the max current rating of the FET. If you have a fast turn on when turning on the supplies then the act of charging and discharging the capacitances on both Vref1 and Vref2 could cause a large current spike to go through the FET. If there is a large dv/dt then the Ic passing through the fet could be large if not limited by a resistor.

    I am guessing that your 1.8V supply has a soft start (slow controlled start up) which might be protecting your part, but it is still a bad idea to rely on that.

    To me, it seems surprising that Vref1 is't being raised up because the internal FET is partially turned on because gate is at 3.3V and the source is 1.8V, that means the Vgs is 1.5V, which means it should be close to full on. Please look into this. Your really shouldn't be doing this.

    -Francis Houde