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SN65HVDA1040A-Q1 - question about circuit

Other Parts Discussed in Thread: SN65HVDA1040A-Q1

Hi all,

Would you mind if we ask SN65HVDA1040A-Q1?
 
The detail of question is as following;
6862.3108.20141125_SN65HVDA1040.pdf
We can not judge this circuit has problem or not.

So, we need your help.

Kind regards,

Hirotaka Matsumoto

  • Matsumoto-san,

     

    I have moved your post to the interface forum for a better response. Since this week is a holiday in the US, it may take a little longer than normal for your question to be answered. Thanks for your patience.

  • Hi Hirotaka,

    Vod can suffer in amplitude in a setup like this when the bus is long and heavily loaded.By changing the termination resistors to 10K the corner frequency of the 3dB point is being moved from 1MHz to ~333kHz and extra loading is being placed on the system. This causes a smaller common mode filtering effect as well as forces the Vod to take an extra hit on amplitude due to the additional loading. Please refer to this wiki on the topic for more information: http://e2e.ti.com/support/interface/industrial_interface/w/industrial_interface/2513.can-termination-techniques.aspx

    In situations where there are not to many nodes and the distance of the bus is not that great your customers termination scheme should work. Please let me know if you have any more questions?

  • Michael san,

    Thank you for your reply and your cooperation.

    We would like to confirm what you mentioned. 
    You have two apprehensive points as followings; 

    1. By changing the termination resistors to 10K the corner frequency of the 3dB 
       point is being moved from 1MHz to ~333kHz.
       This causes a smaller common mode filtering effect as well as forces the Vod
        to take an extra hit on amplitude due to the additional loading.

    2. There are not to many nodes and the distance of the bus is not that great
        your customers termination scheme should work

    About2, it means that it cannot connct many nodes and long distance, right?

    Kind regards,

    Hirotaka Matsumoto

  • Michael san,

    We has two additional questions.
    We guess that your days so busy.
    However, could you let know us followings?

    <Question1>
    The combined resistance is 12kohm because of the differential-input resistance 30kohm of CAN tranceiver
    and 20kΩ(=10kΩ×2).
    Our customer told us that this circuit satisfies ISO11898-2. Is it correct?

    <Question2>
    There are 20kohms on the branch node, so why does the cut-off frequency change from 1Mhz to 333kHz? 
    So, could let know us the calculation method of the cut-off frequency?

    We need your help.

    Kind regards,

    Hirotaka Matsumoto

  • Micheal san,

    Could you let know us above reply?
    Especially, we need your reply of <Question2>.
     
    Kind regards,

    Hirotaka Matsumoto

  • Hi Hirotaka san,

    The ISO11898 CAN standard specifies that the bus lines be terminated at each end with a characteristic impedance of 120 ohms. There are two common bus termination techniques that are used on CAN buses:

    The first method call for a single termination resistor of 120 ohms while the second method calls for a split termination of two 60 ohm resistors with a capacitor to ground to filter the common mode noise. The resistors that your customer is looking to use is 100 times larger than then recommended value of 120 ohms. This will cause the Vod to take a larger hit in amplitude to due the voltage drop across the termination resistors.

    Now the 3dB roll off point can be calculated using:

    So you can see that the larger termination resistance is going to move the corner frequency to a lower value. I hope this helps answer your questions.

  • Michael san

    Thank you for your prompt reply.
    About calculation Fc= 1 / 2πRCL, R means the resistance between CAN-H and CNA-L.
    Is it correct?

    Kind regards,

    HIrotaka Matsumoto

  • Hi Hirotaka,


    For split terminations the R used when calculating the corner frequency is simply 1/2 of the differential pair. In your case R would be 10k yielding an Fc of ~3.4kHz where as a 60 ohm termination resistor would yield an Fc of 564kHz.

  • Michael san,

    Thank you for your reply!
    We'd like to confirm that you mentioned before as following;

    "Vod can suffer in amplitude in a setup like this when the bus is long and heavily loaded.
    By changing the termination resistors to 10K the corner frequency of the 3dB point
    is being moved from 1MHz to ~333kHz and extra loading is being placed on the system."

    About frequency, it is being moved from 564kHz to 3.4kHz. Is it correct?

    Kind regards,

    Hirotaka Matsumoto 

  • Hi Hirotaka,


    I am confident in my calculations. You can confirm my findings by reviewing the following wiki: http://e2e.ti.com/support/interface/industrial_interface/w/industrial_interface/2513.can-termination-techniques.aspx

    Can you please give a description of the bus topology if you would like me to comment further? I would like to know number of nodes, distance between nodes and total distance of the bus.

  • Michael san,

    Of cource, we trust your calculation, but there is the difference of each value of corner frequency which you caluculated.
    (you shows us on Nov 25 2014 14:51 PM and Dec 03 2014 08:30 AM).

    That's why we asked it.

    We need your help.

    Kind regards,

    Hirotaka Matsumoto

  • Michael san,

    We apologize that our request makes you confuse.
    And we appreciate all of your help.

    Now we confirm followings;
    -number of nodes
    -distance between nodes and total distance of the bus

    And then, would you mind if we ask split termination?
    If our customer only uses it as divergence node(there is two 120ohm terminals on bus line),
    is it required to use split termination?

    We need your help.

    Kind regards,

    Hirotaka Matsumoto

  • Michael san,

    We confirmed followings.
    However, our customer's product uses as node divergenve(brunch), so they don't know them.
    It means that customer's product is included their end user's system.
    -number of nodes
    -distance between nodes and total distance of the bus
    (Then, we found that there are two Rt=120ohm on the bus terminal.)

    About ISO11898-2,  
    -Cable: pallarel 120ohm
    -Max data rate: within 1Mbps
    -Max cable range: within 40m(in case of 1Mbps)
    -Terminal on the bus: It is only recommended  two Rt=120ohm on the bus terminal

    As the result, if the system need to satisfy ISO11898-2 exactly,
    our customer has to remove 10kohm x 2.
    Is our recognition correct?

    Could you let us know some advice?

    Kind regards,

    Hirotaka Matsumoto

  • Hi Hirotaka san,

    You are correct, if you exactly follow what ISO11898 says than the bus is simply teminated with 120 ohms at both the two end points and all the nodes in the middle do not use partial termination networks like the 2 x 10k ohms that we have been discussing. Partial terminations will not prevent the system from working (i.e. 2x 10kohms), I just recommend that you do not that this.

    Partial terminations are implemented by designers to try to reduce the amount of reflections on the bus that are caused by excessive stub lengths. If the ISO11898 standard is followed correctly than there are no excessive stubs and these partial terminations are not needed.

  • Michael san,

    Thank you for your reply!
    We explained TI's recommended design for Vsplit of CAN bus as following document.
    http://www.ti.com/lit/an/sloa101a/sloa101a.pdf

    Our cusotmer almost agreed to our explanation.
    However, our customer showed us other reference circuit about Split termination as following document.
    Please refer to P72.
    http://www.nxp.com/documents/report/AH1014_Application_Hints_TJA1042_43_48_51.pdf

    There is "Split termination for stub node (optional)" on the datasheet.
    We judged depending on the stub lenght and number of stub whether we use "Split termination for stub node" or not.
    Is our recognition correct?

    Kind regards,

    Hirotaka Matsumoto

  • Hi Hirotaka san,

    You are correct.

    Split termination is implemented in situations where there are long stub lengths and a high density of nodes on the bus. These factors weigh into the overall bus impedance and may cause reflections. In a system with a large amount of nodes and long stubs, impedance mismatches are common so implementing partial termination or what NXP calls "split termination for stub nodes" will help to dampen out the reflections providing an overall better frequency response.
  • Michael san,Thank you for your reply.OK, we got it.Kind regards,
    Hirotaka Matsumoto