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ISOW1412: Can performance being tested by simulation?

Part Number: ISOW1412

Hi Team, 

Customer wants to perform test when connecting multiple nodes(more than 64 nodes), can customer do it by simulation such as using equivalent circuit?  

Thanks.

Regards,

Jo

  • Hi Jo,

    Thanks for reaching out.

    Yes, multiple node connection can be simulated with an equivalent circuit. Each ISOW1412 node represents a load of 12kΩ, hence, connecting a 12kΩ on the bus should simulate one node connection.

    Let me know if you have any further questions, thanks.


    Regards,
    Koteshwar Rao

  • Hi Rao, 

    Thank you. 

    Datasheet says following, what does it mean? I thought it means input impedance is 12k*8=96kOhm for ISOW1412. So should customer connect multiple 96kOhm resistors parallel on the Y and Z pin of ISOW1412(used as driver) to test? 

    "The ISOW14x2 devices have 1/8 UL impedance transceiver and can connect up to 256 nodes to the bus. "

    Thanks.

    Regards,

    Jo

  • Hi Jo,

    Sorry, I accidentally referred 12kΩ as the equivalent impedance of ISOW1412. 12kΩ is the Unit Load (UL) defined by the RS-485 standard while ISOW1412 is 1/8th of a UL. Hence, as you stated, the equivalent resistance of ISOW1412 is 8 times 12kΩ which is 96kΩ. Thanks.


    Regards,
    Koteshwar Rao

  • Hi Rao, 

    Thank you. 

    1. Is connection in attached file correct? 

    2. Is there any limitation for cable type to use with ISOW1412? For example, it seems twist-pair cable is required for RS485, is it correct? And is shield necessary? The longest distance in customer application can be 1.2km.

    ISOW1412 connection.pptx

    Thanks.

    Regards,

    Jo

  • Hi Jo,

    Please see my inputs below, thanks.

    1. Yes, the resistor connection is correct in the diagram attached.
    2. There is no limitation on the cable type that can be used but different tables will allow different communication distances. Twisted pair minimizes common-mode noise and shielding cable also protects the bus from any noise pick-up.


    Regards,
    Koteshwar Rao

  • Hi Rao, 

    Thank you. Customer wants to know if they can achieve 2km distance with following RS485 and cable spec. 

    Is it able to infer it from following information based on theory? Since customer don't have time and resource to prepare cable and test it. 

    ※RS-485 spec

    Transmission signal half-duplex transmission

    Electrical specifications RS-485 compliant

    communication protocol MODBUS-RTU

    Transmission system Asynchronous

    connection type Daisy chain

    Baud rate 38400bps

    Nodes 128 nodes

    Maximum communication distance 2000m 

    Terminating resistor 120Ω, 1/2W

    ※Cable spec

    Type Shielded cable

    Number of pairs 2 or 3

    Conductor resistance(20) 88Ω/km or less

    Insulation resistance(20) 10000MΩ-km or above

    Capacitance(1Hz) 60nF/km or less

    Characteristic impedance(100kHz) 110±10Ω

    Thanks.

    Regards,

    Jo

  • Hi Jo,

    Thanks for following up.

    The cable length that can be achieved for a given datarate depends on many external factors and the overall system design. There are many factors that affect the timing specifications and some due to parasitic elements in customers system, customer will have to consider all those into effect to determine overall impact on timing specifications.

    At the same time, I would like to confirm that 38.4kbps datarate on 2km cable with 60nF/km capacitance is achievable. Customer can go ahead with this cable for their application but we do recommend them to test it sufficiently to make sure it meets their requirements in all their test conditions. Thanks.


    Regards,
    Koteshwar Rao

  • Hi Rao,

    Thank you. I understood customer should eventually test on their own system. 

    1. May I know what exactly do you mean "timing specifications"? 

    2. What do you mean by "I would like to confirm" below? And is there any material explaining the relationship between datarate, cable length and capacitance? I would like to explain to customer. 

    At the same time, I would like to confirm that 38.4kbps datarate on 2km cable with 60nF/km capacitance is achievable.

    Thanks.

    Regards,

    Jo 

  • Hi Jo,

    Please see my inputs below,

    1. By timing specifications, I am referring to rise / fall times, jitter and other parameters that are critical to determine if a signal is good for a given distance.
      1. 38.4kbps datarate has a bit width of about 26µs. A signal can be considered good when the rise / fall times are <30% of bit time although many customers would be much lower rise / fall times.
      2. When the cable becomes too long, the rise / fall times could become >50% of bit time. In such a case the data is not readable.
      3. 60nF / km doesn't seem to make the rise / fall time go beyond 30% for 38.4kbps and hence, I said this is achievable.
      4. Please note that the typical distance that can be achieved by RS-485 is 1.2km at 100kbps. Anything beyond this needs to be carefully designed and tested. For more details, please refer to below application note.
        1. https://www.ti.com/lit/pdf/slla272
    2. The previous explains the relationship between various terms in determining datarate for given cable length and vice-versa. You could also refer to the below application note that present test data for various timing parameters at multiple test conditions.
      1. https://www.ti.com/lit/pdf/slla431

    Let me know if you have any further questions, thanks.


    Regards,
    Koteshwar Rao

  • Hi Rao, 

    Thank you for the explanation! 

    Sorry but I couldn't get the relationship between 60nF/km and rise/fall time, could you explain little more or any material I can refer to? 

    60nF / km doesn't seem to make the rise / fall time go beyond 30% for 38.4kbps and hence, I said this is achievable.

    Thanks.

    Regards,

    Jo

  • Hi Jo,

    The 2km cable has a total capacitance of 120nF with a characteristic impedance of 120Ω. The rise and fall times are slowed down by R and C of the cable which can be approximately expressed as a product of R and C, i.e., delay caused by the cable = R * C = 14.4µs.

    Although rise / fall times are marginal, I believe it will still be acceptable to support 38.4kbps datarate. Please let me know if this answers your question, thanks.


    Regards,
    Koteshwar Rao

  • Hi Rao, 

    So do you mean both rise and fall time of signal is 14.4us? As you mentioned below to determine a signal is good or not, 30% of 26us will be 7.8us, meaning rise and fall time is larger than it, so it should not considered good signal?

    38.4kbps datarate has a bit width of about 26µs. A signal can be considered good when the rise / fall times are <30% of bit time although many customers would be much lower rise / fall times.

    Based on both, I still couldn't get why you say this cable is acceptable to support 38.4kbs datarate.

    Although rise / fall times are marginal, I believe it will still be acceptable to support 38.4kbps datarate. Please let me know if this answers your question, thanks.

    Thanks.

    Regards,

    Jo

  • Hi Jo,

    Your understanding is correct, 14.4µs is the rise and fall times. Due to this, you will not see a full swing on the bus but will be a shorter signal but communication can still happen. It is customer who needs to decide how much of a swing is acceptable for them and how much of rise/fall times are acceptable. Due these dependencies, it is the designer of customer product who needs to define what performance acceptable to them and accordingly determine if 38.4kbps at 2km cable length is okay for them.

    Like I mentioned earlier, my expectation is that 38.4kbps is possible for 2km but on a customer system, they would have to evaluate the performance and see if the achieved performance is okay for them.

    Let me know if you have any further questions. Thanks.


    Regards,
    Koteshwar Rao

  • Hi Rao, 

    Want to double confirm, so "rising time" + "falling time" = 14.4us, so we can assume "rising time" and "falling time" is 7.2us respectively, which is smaller than 30% of 26us(7.8us) so you're saying it should be okay, correct? 

    BTW, is following mentioned in any material? I couldn't find it on the Internet. 

    A signal can be considered good when the rise / fall times are <30% of bit time

    Thanks.

    Regards,

    Jo

  • Hi Jo,

    BTW, is following mentioned in any material? I couldn't find it on the Internet. 

    No, this is not a requirement but is a general unsaid guideline. Ideally, fastest rise/fall times are preferred for clear data bit / square wave but if the rise / fall times are a little slower, it isn't really a problem in most cases. Rise / fall times with >30% of bit time can start to very distorted and not many would like them. But this is not a defined limit by anyone, 30% is what I and many other designers consider to be fair limit beyond which waveform start to distort.

    But when customer really wants to still achieve communication even with >30% rise/fall times, it is still possible. The waveform may look distorted but the signals are still read by devices. Hence, a customer needs to take a call on whether they are okay with >30% rise/fall times. The RS-485 standard supports 1.2km of range but if customer needs to meet 2km which is outside the standard, then they would be required to go outside of typical rise/fall times.

    Want to double confirm, so "rising time" + "falling time" = 14.4us, so we can assume "rising time" and "falling time" is 7.2us respectively, which is smaller than 30% of 26us(7.8us) so you're saying it should be okay, correct? 

    Rise time = fall time = 14.4µs. This is 55% of bit time which is much more than the 30% recommendation I made earlier. Hence, the waveform will not look square wave and will look more like a triangular wave but still can be read by devices. The RS-485 standard only requires the VOD signal (differential output) to be > +200mV to be read as HIGH and < -200mV to be read as LOW. Thanks.


    Regards,
    Koteshwar Rao