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LMH6601: What would happen if the SD_PIN (5PIN) was not at a constant level?

Shinichi Inoue
Guru 20486 points
Part Number: LMH6601

Tool/software:

Dear Specialists,

My customer has a question about SD pin.

I would be grateful if you could advise.

---

According to the LMH6601 datasheet, p. 20, 7.2.1 Shutdown Capability and Turn On/Off Behavior,

---This pin must be tied to a level or otherwise the device state would be indeterminate.---

What would happen if the SD_PIN (5PIN) was not at a constant level?

Is there a possibility of damage?

---

I appreciate your great help in advance.

Best regards,

Shinichi

  • +1
    TI__Mastermind 25335 points

    Hello Shinichi,

       If SD_Pin (5PIN) is not tied to a voltage (left floating or disconnected), then device state would be indeterminate. The reason this sentence is in the datasheet is because a lot of our devices have the capability of leaving PD pin floating, and internally there is a pull-up resistor to a supply to most likely leave device enabled even with pin disconnected. However, this device, it is advised not to leave PD pin disconnected, otherwise, the device will either be enabled or disabled. 

       There is no possibility of damage, rather, if PD pin is not biased, then device can either be in shut-down mode or on. PD pin can be connected to different voltage levels and does not have to be constant, but it should be within the supplies of the amplifier to avoid damage.

    Thank you,
    Sima 

  • 0 in reply to Sima Jalaleddine
    Guru 20486 points

    Hi Sima,

    Thank you for your reply.

    I understand that there is no risk of damage as long as the SD pin does not exceed the power supply voltage.

    Could you please let me know the absolute maximum rating of SD pin.

    Is is possible to think Voltage at Input/Output Pins (V +)+0.5,  (V −)−0.5V as a SD pin specification.

    I appreciate your great help and cooperation.

    Best regards,

    Shinichi

  • +1 in reply to Shinichi Inoue
    TI__Mastermind 25335 points

    Hello Shinichi,

      Yes that is correct! It is (V +)+0.5,  (V −)−0.5V, due to ESD diodes internal to the device. 

    Thank you,

    Sima

  • 0 in reply to Sima Jalaleddine
    Guru 20486 points

    Hi Sima,

    Thank you for your reply.

    I understand.

    There is no question.

    Let's close this thread.

    I appreciate your great help and cooperation.

    Best regards,

    Shinichi