About input resistance of DE terminal of ADM3064E

Hi,

I am thinking of connecting the ADM3064E to the DE terminal from VIO (3.3V) via a resistor.

In the data sheet, INPUT CURRENT is -2uA ~ 2uA,
Is there a recommended value for the pull-up resistor between the DE terminal and VIO terminal?

Also, please tell me the input resistance value of the DE terminal with respect to GND.

Thank you for your reply

Best Regards,

Knj.

  • 0
    •  Analog Employees 
    on Sep 30, 2020 12:05 PM 2 months ago

    Hi Knj, 

    10kΩ is a good value for a pullup resistor, which adds 330uA to the VIO supply current. Higher values can be used to reduce the power consumption, but too high a resistance may cause issues in noisy environments. 

    The input resistance of the DE terminal to GND can be calculated from the input current. The input current is a maximum of 2uA when VIO = DE = 3.3V. This equates to a minimum of 1.65MΩ input resistance to GND.

    Regards,

    Neil 

  • Hi Neil,

    Thank you for your answer.
    It worked correctly by setting the pull-up resistor to 10kΩ.

    I have another question.


    When the pull-up resistor is set to 47kΩ, it seems that the input current is sufficient, but when the power is turned on / off, the DE terminal voltage becomes 0.6V.

    As you answered, I think that the reason is that if you increase the resistance, the current consumption will decrease, but it will be vulnerable to noise.


    Can you tell me what kind of function is working internally when the DE terminal becomes 0.6V?

    Best Regards,

    Knj

  • 0
    •  Analog Employees 
    on Oct 6, 2020 11:25 AM 1 month ago in reply to knj

    Hi Knj, 

    When the VIO pin of the ADM3064E is unpowered (VIO = 0V) the voltage on the DE pin should remain within the maximum ratings of the device, -0.3V to +0.3V. In this case the DE pin is high impedance. Internally it is connected to GND and VIO through a reversed biased diode. 

    When the ADM3064E is powered (VIO = 3.3V), with either a pull-up resistor to VIO of either 10kΩ or 47kΩ, I would expect the DE pin voltage is 3.3V

    During the transition from the unpowered to powered state, there is an internal function that activates a pulldown bias current to hold the DE pin in a logic low state, for the duration of the device power up. This is to ensure that parasitic current which can flow to the DE pin during power up does not enable the transmitter unintentionally. The interaction of this feature with the 47kΩ pullup resistor could cause the DE pin voltage to equal 0.6V, but this should only persist during the power up sequence. 

    Under what specific power on/off conditions do you observe 0.6V on the DE pin? 

    Regards,

    Neil 

  • Hi Neil,

    Thank you for your answer.

    I saw about 5 units with the phenomenon of 0.6V mounted on the board, but there were variations that occurred or did not occur.

    As a condition

    1. Power on

    2. Power off after 5 seconds

    3. Power on after 5 seconds

    4. Symptom confirmation

    5. Repeat steps 1 to 4 How many times does it occur 10 times?

    The symptom disappeared just by reducing the resistance between the DE terminal and the 3.3V power supply line to 39kΩ or less.
    It may have been just near the threshold where the phenomenon is likely to occur.
    We have not tested resistance values over 47 kΩ.

    As a power input condition

    Vcc = 12V => DC-DC 5V => LDO 3.3V => ADM3064E

    The ripple noise at 3.3V was about 40mVp-p.

    Best Regards,

    Knj

  • 0
    •  Analog Employees 
    on Oct 13, 2020 7:45 PM 1 month ago in reply to knj

    Hi Knj, 

    Thank you for the detailed description. I will investigate this in the lab and see if I can recreate what you are seeing. 

    40mV is low ripple and shouldn't cause any functionality issues. 

    In the meantime I'd recommend using a pullup resistor of less than 47kohms if the power consumption is tolerable in your design. It has the advantage of improved noise immunity when the DE pin is undriven. 

    Regards,

    Neil