• State Resolved
  • Locked Locked
  • Replies 3 replies
  • Subscribers 21 subscribers
  • Views 567 views
  • Users 0 members are here
Support feedback
Options
Options
Related

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.

CDCE913's S0 pin

Tim Baker
Intellectual 540 points
Other Parts Discussed in Thread: CDCE913

Hi,

There is very little description on the function of S0 pin in the datasheet, and the only I found is that S0 controls whether output is three state in Fig.6 and Table-4.

 

However, Table-4 only listed Y1, not Y2 and Y3. So does S0 control Y1,Y2 and Y3's Enable/3-state at the same time?

And if I want Y1, Y2 and Y3 to be enabled at all time, should I tie S0 to [pin 3, VDD] directly, or via a resistor? If a resistor is needed, what should its resistance be?

Additionally, controlling whether output is 3-state is a very simple operation; and since CDCE913 already has I2C ports equipped, it can easily be designed to let an I2C register bit control Y1, Y2, Y3's output Enable/3-state. So why there still bothers to a separate S0 pin at the cost of both increased footprint and added connections on the external circuit?

 

Tim

  • 0
    TI__Mastermind 20900 points

    Hi Tim,

     

    if you look to the right side of table 4 you will see the configuration of Y2/Y3. This table shows the factory default setting. The CDCE9xx features the possibility to have a pin-programmable device. (for example you can program a "power down mode" that you can enable/disable with S0)

    If you just want to have the outputs running and you do not change the programming of S0, you can pull S0 to VDD through a 1kOhm resistor.

    Best regards,

    Julian

  • 0 in reply to Julian Hagedorn
    Intellectual 540 points

    Julian,

    The datasheet says S0 has internal pullup of 500K. Then why it cannot be left disconnected? And if you do connect it to VDD or GND, why is a resistor still needed? I think it would be redundant?

    And if you look at the screenshot in the first post: S0 is directly connected to VDD or GND, there is also no intervening resistors.

     

    Tim

  • 0 in reply to Tim Baker
    TI__Mastermind 20900 points

    Hi Tim,

    you are right. you can leave it open as well. To be better isolated against noise, you should use 1kOhm resistor.

    Best regards,

    Julian