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PCA9306 EN/VREF2 pullup resistor + power and/or leakage calculations

Toshiyuki
Mastermind 7075 points
Other Parts Discussed in Thread: PCA9306

Hi,

Our customer has a question about EN/VREF2 pullup resistor.

In the data sheet of PCA9306 is discribed as follows,

"the EN input must be connected to VREF2 and both pins pulled to high-side VDPU through a pullup resistor(200kΩ)".

However,they would like to use 180kΩ or 220kΩ because of reducing the number of components.

May I apply 180kΩ or 220kΩ resistor?

If possible,please let me know which is better 180kΩ　or 220kΩ.

Best Regards,

Kato / Japan disty

 

  • 0
    Mastermind 7075 points

    Hi,

    I apologize for rushing you.

    Could you please give me your opinion?

    I appreciate if I could have your comment soon.

    Best Regards,

    Kato / Japan disty

  • 0 in reply to Toshiyuki
    Prodigy 100 points

    Hi,

    I would also be interested in knowing about this. I am looking at using this part. I understand that the 200K pull up resistor is placed to bias the EN/VREF2 to be ~0.6V above VREF1, but how is this achieved?

    My concern is implementing this circuit as shown in the example schematic on the datasheet, i.e. between 1.8V to 3.3V conversion, but with the EN as a 3.3V signal.

    If I drive EN/VREF2 with a LVCMOS output to 3.3V that only guarantees 2.4V Voh(min) (although typically it will be close to the 3.3V level.) Would I be at risk of the circuit failing to operate? My input would only be 0.6V above the VREF1 input (of 1.8V) before accounting for any voltage drop across the 200K pull up?

    Thanks and regards,

    Alan.

  • 0 in reply to Alan Williams1
    TI__Mastermind 30260 points

    The size of the resistor chosen (200k) is large in an attempt to limit leakage current from VREF2 to VREF1. They are effectively connected together through a diode-configured FET.

    Therefore:

    ILEAK = (VREF2 - (VREF1+VT))/R; where VT is the threshold voltage of the FET and R is the size of the resistor.

    ILEAK = (VREF2 - (VREF1+0.6))/(200k)

    As a result, as long as VREF2 > VREF1 + 0.6 the part will be 100% stable, and you can use the equation to see that the current through the resistor merely reduces to 0 when: VREF2 = VREF1 + 0.6

  • 0 in reply to Brian Berner
    Prodigy 100 points

    Hi Brian,

    Thank you for the reply and the explanation. 

    Cheers,

    Alan