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SN65HVD3083E: Method of calculating unit load number in external fail-safe circuit

Kaji@PAN
Mastermind 7815 points
Part Number: SN65HVD3083E


Hi,

I previously asked questions about SN65HVD3083 E in the following thread.
I got an answer of the following contents to be informed.
Among them, there is u.l. = 5.6. How do you calculate this "5.6"?
Can you teach me the calculation method?

Best Regards,

  • 0
    TI__Guru 62920 points
    The unit loading has to do with the equivalent input resistance of the circuit. Since a unit load is defined by a current/voltage characteristic that results in a leakage current of 1 mA at a test voltage of 12 V, it is often thought of as a 12-kOhm resistance. Circuitry that results in a lower input resistance would then correspond to higher unit loading. We will put together some more information on the exact calculation and get back to you later this week.

    Max
  • 0
    TI__Intellectual 1925 points

    Hi Kaji,

    a good way to understand this is to look at the topic like that: The fail safe resistor network gives you the additional safety feature, but it R1 (equivalent to R_FS) is constantly causing a current flow which is seen as load for the bus.

    To understand where the 32 UL maximum comes from, please take a look here:

    https://e2e.ti.com/support/interface/industrial_interface/w/industrial_interface/2575.the-unit-load-in-rs485?keyMatch=rs-485%20unit%20load&tisearch=Search-EN-Everything

    Referring to previous posts, the maximum ULs can be viewed/described as a 12 KOhm load. With this insights, please take a look at the last equations of the application note: 

    /cfs-file/__key/communityserver-discussions-components-files/138/5282.2287.slyt324.pdf

    With the value R1 = 2500 Ohm the first equation gives you R_INEQ ≈ 442 Ohm. Put this in the next equation and this will give you the ULs  (n_max ≈  27) besides you can apply to bus with this particular fail safe resistor network.

    That means this fail safe network reduces the systems maximum number of ULs from 32 to 27 (fail safe network represents 5 ULs). Please consider that there are probably some margins added in order to benefit/ensure this solution, which relates in the final value of 5.6 Uls.

    I hope this helps to understand the calculations.

    Kind regards

    Dierk