• State Resolved
  • Locked Locked
  • Replies 7 replies
  • Subscribers 34 subscribers
  • Views 839 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.

SN65LBC180 Fail safe operation RS485

Joseph Hessel
Prodigy 215 points

Other Parts Discussed in Thread: SN65LBC180, SN65LBC184

I need to know what the fail safe operation is for the SN65LBC180 when used with RS485 signals.

Fail safe operation was not listed in the SN65LBC180  data sheet and I am wondering if it is simular to the SN65LBC184.

Thank you.

  • 0
    TI__Mastermind 28265 points

    Joseph, the rise time of the LBC180 is only provided as the rise time for the differential output Vod with 10ns typical.

    The LBC184 however has slo rsie time s beteen 0.25 and 1.2 us.

    Regards, Thomas 

  • 0 in reply to Thomas Kugelstadt
    Prodigy 215 points

    Page 5 of the RS-485 Design Guideline, states the following:

    “Section 1 of the graph presents the area of high data rates over short cable length. Here, the losses of the transmission line can be neglected and the data rate is mainly determined by the driver’s rise time. Although the standard recommends 10 Mbps, today’s fast interface circuits can operate at data rates of up to 40 Mbps.”

     

    How can I determine the data rate for a RS-485 cable using the SN65LBC180 driver? i.e., what is the data rate for an RS-485 cable that is 40 feet (12m) in length?

     

    Also, how do I calculate the electrical length of the stub for an RS-485 using the SN65LBC180 driver? Lstub <= tr/10 *v*c (Page 3 of the RS-485 Design Guideline) Do I just use the value given for the rise time of the differential output Vod instead of the driver rise time tr?

  • 0 in reply to Joseph Hessel
    TI__Mastermind 28265 points

    1) Data Rate

    If you have nthing but the Vod rise time, then that's what we are going to use to determine the data rate. A rule of thumb suggest the maximum rise time to be 30% of  the minimum bit-time (which occurs at the maximum data rate) tumax = t-bitmin x 30. With the data rate being DR = 1 / tu, you can insert the above expression for tu and get:

    DR = 0.3 / tr = 0.3 / 20ns = 15 Mbps.

    2 Stub Length

    for caclulating the stub length you need the min. rise time of 5ns. Then for a vecloity of 78% or 0.78, this length is

    Lstub = 5ns / 10 x 78% x 3 x 108= 0.117m or 12cm.

    Regards, Thomas

     

  • 0 in reply to Thomas Kugelstadt
    Prodigy 215 points

     

    Thomas,

    Thank you for your continued help with this. However, I am still a little confused.  Please see my response in blue.

    1) Data Rate

     

    If you have nthing but the Vod rise time, then that's what we are going to use to determine the data rate. A rule of thumb suggest the maximum rise time to be 30% of  the minimum bit-time (which occurs at the maximum data rate) tumax = t-bitmin x 30. With the data rate being DR = 1 / tu, you can insert the above expression for tu and get:

    Which expression you are referring to when you say “you can insert the above expression for tu”?

    DR = 0.3 / tr = 0.3 / 20ns = 15 Mbps.

                    Where does the value of 20ns come from?

     

    2) Stub Length

     

    for caclulating the stub length you need the min. rise time of 5ns. Then for a vecloity of 78% or 0.78, this length is

    I just need you to break this down a little further for me and explain where each of the values in the equation come from.  

     

    Lstub = 5ns / 10 x 78% x 3 x 108= 0.117m or 12cm.

                    Where does the 3 come from?

                     Why not 9.8x10^8 instead of just  10^8

                    According to the RS-485 Design Guideline, the equation for stub length is:

                    LStub ≤ *v*c

                    Where

                    LStub = maximum stub length (ft)

                    tr = driver (10/90) rise time (ns)

                    v = signal velocity of the cable as factor of c

                    c = speed of light (9.8 ´ 108 ft/s).

    Thanks again I truly appreciate all of the time and effort you are investing into helping me develop a better understanding of these concepts.

  • 0 in reply to Joseph Hessel
    TI__Mastermind 28265 points

     

    Okay, let's do this again and hopefully remove any confusion.

    1) Data Rate

     Because the data sheet does not give any single-ended rise times I use the differential output transition time tt(OD) which is specified on page 4 under switching characteristics with: MIN = 5ns, TYP = 10ns, MAX = 20ns

    The rule of thumb is that the maximum rise time should be maximum 30% of  the minimum bit-time: trmax = t-bitmin x 0.3. Solving this equation for the minimum bit time yields: t-bitmin = trmax / 0.3. Because the maximum data rate DR-max is the resiprocal of the minimum bit time: DR-max = 1 / t-bitmin you can also sove this equation for the minimum bit time: t-bitmin = 1 / DR-max.

    Now you can set both t-bitmin terms equal: trmax / 0.3 = 1 / DR-max.   Then solving for DR-max gives: DR-max = 0.3 / trmax .  Now inserting the above 20ns for trmax  results in a maximum data rate of   DR-max = 0.3 / 20ns = 15 Mbps.

     

    2) Stub Length

     for caclulating the maximum stub length you need the fastest transition time on the bus which is the minimum rise time of 5ns. Then the signal vecloity of a cable is given in percent of the speed of light. So a velocitiy of 78% means the signal travels through the cable at 78% the speed of light. the speed of light in vacuum is approximately 9.8 x 108 ft/s or 3 x 108 m/s.

    Then using the equation from my RS485 Design Guide Lines and inserting the numerical values yields:

    Lstub-max   =   tr/10 x v x c    = 5ns/10 x 0.78 x 9.8 x 108 ft/s = 0.38 ft     or   Lstub-max = 5ns/10 x 0.78 x 3 x 108 m/s = 0.12 m

    what that all means is:

    1) A stub is a piece of unterminated transmission line which will cause reflections.

    2) By limiting the length of this unterminated piece of line one ensures that the signal transition (send by the driver at the end of the stub) moves towards the cable trunk where it is reflected and then moves back to the driver, all within one tenth of the driver's minimum rise time.

    I hopethis explanation hits the nail on the head. Please let me know if it does not.

    I also apologize for the delaid repsonse because my hard drive crashed and I'm currently working from a loaner PC with extreemly weak VPN stability.

    best regards, Thomas

  • 0 in reply to Thomas Kugelstadt
    TI__Mastermind 28265 points

    I also apologize for the spelling mistakes I discovered : resiprocal should be reciprocal, vecloity should be velocity, and extreemly should be extremely.

     

  • 0 in reply to Thomas Kugelstadt
    Prodigy 215 points

    Thank you this asnwers my question and I truly appreciate the clarification. Now I am confident in my understanding of the concepts.