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SN65LBC175: min of VOH value

Masayoshi Takahashi
Intellectual 1530 points
Part Number: SN65LBC175

Hi team,

My customer is considering using SN65LBC175. Then I got some question.

1. I can see the min of VOH is 3.5V when IOH = -8mA. The customer would like to know the min value of VOH especially IOH = -10uA. We can find typical value from figure 5.

2. Could you tell me how to connect the unused channel input and output? Should they tied to GND?

Regards,

Yoshi

  

  • 0
    TI__Guru 62920 points

    Hello Yoshi-san,

    Unused channel inputs can be left floating or could be tied off to provide a constant differential input level (for example, "A" pulled to VCC and "B" pulled to GND to force a high-state output).  Unused outputs could be left floating.

    Regarding the VOH level, we can see from the typical curves in the datasheet that the primary factor causing voltage loss from VCC to the output is the resistance of the high-side driver.  (From these curves, I compute a typical driver output resistance of about 31 Ohms.)  We can use the minimum VOH specification in order to compute the worst-case output resistance as follows:

    [4.75 V (minimum operating supply voltage) - 3.5 V (minimum VOH spec)] / 8 mA (test current for minimum VOH spec) = 156.25 Ohms.

    We can then use this value to compute the expected worst-case VOH at other test currents:

    VCC - IOH*(156.25 Ohms) = VOH.

    For such a small test current as 10 uA, the voltage drop will not be very significant.

    Please let me know if this is unclear or if you have further questions.

    Regards,
    Max

  • 0 in reply to Miguel Robertson
    TI__Intellectual 1530 points

    Hi Max,

    Thanks for your quick answer. I understood.

    Regarding unused channel, to decrease the power dissipation the customer should disable the EN pin and floating the input and output. Is this correct?

    Or is there any difference of power dissipation between floating or tied off?

    Regards,

    Yoshi

  • 0 in reply to Masayoshi Takahashi
    TI__Guru 62920 points

    Yoshi-san,

    You are correct.  Disabling the unused channels will decrease quiescent current regardless of how the A/B inputs are handled.

    Regards,
    Max