SN65HVD3080E: Signal attenuation fail-safe circuits for RS-485

Hi,

SN65HVD3080E has integrated the fail-safe feature. Please refer to FEATURES on the first page of the datasheet. 

Failsafe Receiver (Bus Open, Short, Idle)

Therefore you don't need extra series resistor R3 to attenuate signal. Please check out this app note for the detail of how internal fail-safe works (SN65HVD3080E works similarly as SN65HVD72 in this sense).  

Please let me know if you have any further questions. 

Regards,

Hi, Hao.

Thank you very much.
I understand that the fail-safe circuit is not necessary in this IC.

But I would like to know more about the effect of signal attenuation by R3.
The past model of the product I am in charge of had this fail-safe resistor R3, but I can't get in touch with the designer at that time, so I don't know how much signal attenuation is caused by R3.
This past model is not a TI RS-485 IC.

I will consider changing the IC to SN65HVD3080E, but before I do that, I would like to know how much signal attenuation is caused by R3.

Regards.

It would be helpful if you could share your schematic with 32 nodes. You don't need to draw all nodes. I want to see how resistors are applied on the bus (pull-up, pull-down, series, termination). 

Back to your question, if one node is set up like this,

Shotaro Adachi said:

R1=33kΩ, R2=120Ω, R3=3k,

the attenuation might not change much if you swap ICs. The reason is that the input impedance of bus pins (A/B) is usually high (around 96kOhm for SN65HVD3080E), which is negligible comparing to 120Ohm. 

Regards, 

Hi, Hao.

I have compiled the schematic and questions in Excel.
Please check it and let me know if there is any missing information.

Regards.

20220112_RS-485_Attenuation.xlsx

Thanks for your detailed information. Before we go into the questions, could you confirm if you are able to change the design of each node? My understanding is that you cannot make any change except swapping SN65HVD3080E. Is it correct?  

Back to your questions, 

1) 375Ohm is the maximum load with 2x 120Ohm termination. In your case (all nodes are terminated), you need to make sure the total load (including R1, R2, R3, Rin of all nodes) is not smaller than 54Ohm. BTW, Rin of SN65HVD3080E is around 96kOhm. 

2) Yes, R3 affects the total load.

3) what's Rc? Basically the signal is divided by R1 and R3. 

4) I don't think the attenuation of each node would change with multiple nodes. However the multiple nodes do load the bus more heavily. 

Regards, 

Hi Hao.

Thank you for your reply.

I'm sorry for the confusion. There was a mistake in the schematic I sent you last time. The terminating resistors are not connected to all the nodes, but to one master and one slave.
The fail-safe resistors (R1 and R3) are connected to all the nodes.
I will send you the schematic again just to be sure.

20220118_RS-485_Attenuation.xlsx

Could you please check this as I think it may affect 1) and 2)?

Hao L said:

Before we go into the questions, could you confirm if you are able to change the design of each node?

It is still in the design stage and could be a transceiver IC other than the SN65HVD3080E. Therefore, let me assume Rin=12kΩ in this thread.
If this does not answer your question, or if you have other concerns, please let me know again.

Back to the question.

1) ,2)
In the modified schematic, is the maximum load still 54Ω? Also, how did you come up with the 54Ω value?
Can you also tell me how to calculate nUL from this result?

3) Rc is the DC resistance of the cable. I believe it is not dominant, so I am ignoring it.

4) I understand. Thank you very much.

For 1) and 2), please refer to Figure 7-2 in the datasheet of SN65HVD3082E. 

Basically 2 termination (120Ohm each) become 60Ohm. 375Ohm is the maximum load. In total the impedance is about 54Ohm.

If there is no other load, with 12kOhm input impedance, the maximum nodes are 32 (12000/375). 

3) I agree it's OK to ignore Rc. 

I think the calculation in your xls file looks correct. If you assume 12kOhm input impedance, it's dominant comparing R1 (33kOhm). Basically input impedance is parallel with R1. Please let me know if you have any further questions. 

Regards,

Hi Hao.

Sorry for the late reply.

I could understand about 3). Thanks!

Can you tell me more about 1) and 2)?
I was able to understand about 54Ω.
I remembered that in the RS-485 standard, the maximum load impedance of the driver is 54Ω.

Hao L said:

In your case (all nodes are terminated), you need to make sure the total load (including R1, R2, R3, Rin of all nodes) is not smaller than 54Ohm.

Please tell me about this.
Can you tell me exactly how much the total load (including R1, R2, R3, and Rin at all nodes) will end up being in our circuit?

Hao L said:

If there is no other load, with 12kOhm input impedance, the maximum nodes are 32 (12000/375). 

Also, I understand that if there is no load, the maximum number of nodes will be 32.
I would like to know how to calculate the maximum number of nodes when a fail-safe resistor is attached, as in this case.

Adachi-san,

Thank you for your patience, we will get back with you tomorrow by the end of business CST.

Regards,

Eric Hackett 

The maximum load is 54Ohm, which includes two termination (120Ohm) and all bus load (375Ohm). This is your total budget. If you add any more load on the bus (like pull-up and pull-down resistor), this extra load cannot make the total load less than 54Ohm. You can check out this blog as an example. Rather than impedance, estimating leak current might be easier. 

(+) How much loading can an auto-polarity RS-485 bus support? - Analog - Technical articles - TI E2E support forums