Hello:
I have below system:


may I know in system A, what supply should I used for MAX13431 VC? can I connect it to 3.3V or I must connect to 5V? and what is the maximum current for this power supply?

 

This question should be answered by the datasheet. Do you have a copy of the datasheet?
There you can find out if the receiver inputs are 5V tolerant if Vc is 3.3Volts.
There you can find out if the receiver inputs have enough sensitivity to handle a degraded 3.3Volt signal if Vc is 5V.
I can't think of anyplace on the planet where you will find more information than the datasheet, unless you are invited to dinner at the chip designers's house.

 

Right at the top of the datasheet in big bold letters.


Wide +3V to +5V Input Supply Range

Low-Voltage Logic Interface +1.62V (min

Even someone like me that struggles with datasheets can figure that one out.

Or are you asking if you can have mismatched supply voltages on each end? If so you should be clear about what you are asking.

 

Hi, Papabravo:
thank you for reply.
I did check the datasheet, but I got confused, in the datasheet it just give logic interface for DI/DE/RE, but not mentioned A/B which is the receiver input, so I am not clear, if I connect VCC to 3.3V, whether the input 5V signal will damage the input or not...



thanks

 

Then you need to dig deeper. You might have to dig up a copy of the RS-485 specification to see just what the requirements are. I alluded to the two questions with respect to the bus interface. In more detail you should worry about the available signal level from a 3.3 V bus interface after it has traveled down a length of cable, become distorted and reduced in amplitude. Will the receiver on the other end be OK with that? The opposite of that question is if the part with the 3.3V bus interface can tolerate a nearby part which is putting out 5V signals.

In the datasheet under absolute maximum ratings
It says the 13430 and 13431 parts have a maximum receiver input voltage (A, B) of -8V to +13V
It says the 13432 and 13433 parts have a maximum receiver input voltage (A, B) of -25V to +25V

This latter requirement is consistent with my recollection of the original requirements that drove the RS-485 standard. So you're good on that score.

All we have left is the input threshold.
Under DC Electrical Characteristics, look at Differential Driver Output and make sure you understand what Figure 1 is telling you.
Under DC Electrical Characteristics, look at Receiver Differential Threshold Voltage, and make sure you understand what that number is telling you.

Now you can begin to consider the effect of cable length on data rate and signal integrity. How much distance do you have to cover anyway?

Got it?

 

Hi,
Thank you very much for advise !
from the datasheet:

1. system B use Vcc=5, so its output will be 2.25-->5V, it will not damage System A input if Vcc=3.3V in system A ( 13431 parts have a maximum receiver input voltage (A, B) of -8V to +13V)
2. system A output will be 2--3.3V (VCC=3.3V in system A)
and system B should be ok with this level due to :


3. in our application, the cable will be less than 0.5m, it should be ok , right?

so the conclusion is, I can use VC=3.3V , I am right?

Thanks

 

Hello:
I have below system:
View attachment 99602

may I know in system A, what supply should I used for MAX13431 VC? can I connect it to 3.3V or I must connect to 5V? and what is the maximum current for this power supply?

Try it and see if it works. Why all the theories?

 

Try it and see if it works. Why all the theories?

WTF theory are you talking about? Do you even know what you're talking about? No theories here, his initial concern, a valid one, was to avoid destruction by connecting parts with different Vcc voltages. We have put that concern to rest by careful examination of the datasheet. Now with half a meter of cable there should be no issues with signal degradation and excess capacitance -- so go ahead and run the experiment to see what happens. It should work just fine for a half duplex point to point connection.

Remember to put your one termination resistor on the cable. Never, ever put it in the product. You only need one termination resistor because the cable is so short.

 

Remember to put your one termination resistor on the cable. Never, ever put it in the product. You only need one termination resistor because the cable is so short.

Interesting. I've been using RS485 chips for years now. And it is my understanding that the termination resistors are there to dampen "signal bounce" when using long lines. But this is the first time I've heard that only one resistor is recommended for short cable lengths (makes sense, btw)

My question here is, what do you mean the resistor should be placed on the cable and never in the product?
Do you mean that the resistor should not be mounted too close to the RS485 chip? What could happen if it were mounted in the product?
My naive assumption is that maybe a little distance is required between the resistor and the product (the RS485 chip) to make sure that no reflection reaches the product... am I right?

 

WTF theory are you talking about? Do you even know what you're talking about? No theories here, his initial concern, a valid one, was to avoid destruction by connecting parts with different Vcc voltages. We have put that concern to rest by careful examination of the datasheet. Now with half a meter of cable there should be no issues with signal degradation and excess capacitance -- so go ahead and run the experiment to see what happens. It should work just fine for a half duplex point to point connection.

Remember to put your one termination resistor on the cable. Never, ever put it in the product. You only need one termination resistor because the cable is so short.

"Read the data sheet"? BS what will the device do outside what is on the data sheet? No math about VCC required. If it works, it works, even if the parameters are not in the data sheet.
The data sheet for TTL says VCC should be between 4.5 and 5.5 V. BS. I have ran them at 9 V before. It just won't perform within the parameters of the data sheet.
If you want to know if the RS485 transceiver will work on 3.3 V and/or 5 V plug it into a breadboard and try it. You have your answer in minutes.

 

Hi, Papabravo:
in our system both system A and B are terminated like:


what do you mean not put terminate resistor on product?

 

It should work just fine for a half duplex point to point connection.

 

Interesting. I've been using RS485 chips for years now. And it is my understanding that the termination resistors are there to dampen "signal bounce" when using long lines. But this is the first time I've heard that only one resistor is recommended for short cable lengths (makes sense, btw)

My question here is, what do you mean the resistor should be placed on the cable and never in the product?
Do you mean that the resistor should not be mounted too close to the RS485 chip? What could happen if it were mounted in the product?
My naive assumption is that maybe a little distance is required between the resistor and the product (the RS485 chip) to make sure that no reflection reaches the product... am I right?

The termination should be on the cable system and not in the product because, in the product there will be a jumper, and it could be in or out. In a multi-drop application, if too many of those jumpers are in what happens? That's right you get a dozen or so 'termination resistors' in parallel and your transmit levels get seriously degraded. Next question, when things go wrong and you suspect there are too many termination resistors, how quickly can you make the determination? How many boxes do you have to open? You get the idea. Terminations are an integral part of the cable system and there is nothing that says the last device has to be at the end of the cable system.

If you have a cable system that is 100 meters long of twisted pair and you put 120Ω at each end you have a 60Ω impedance for each driver. Shorten the cable to 0.5 meters and it looks less and less like a transmission line to signals with ordinary data rates and you can get away with a single 60Ω terminator. Works like a champ.

Good questions BTW

 

Here's an interesting short article about how to calculate RS485 termination resistors:
http://ltxfaq.custhelp.com/app/answ...rs485-&-rs422---when-is-termination-required?

I found this fact very interesting, btw:
"... for cables 2000 ft or less termination is not needed at 9600bps."

 

Here's an interesting short article about how to calculate RS485 termination resistors:
http://ltxfaq.custhelp.com/app/answers/detail/a_id/1382/~/rs485-&-rs422---when-is-termination-required?

I found this fact very interesting, btw:
"... for cables 2000 ft or less termination is not needed at 9600bps."

Q: Using the standard calculation for frequency and wavelength, what is the wavelength of 9.6 kHz?
A: 31,250 meters

Now assume a square wave that is rich in harmonics up to the 9th and we have 31,250 / 9 = 3472 meters, and you begin to see why cable length makes a difference.

Now one other point to consider is that you might want to use terminations on a shorter cable to allow the line in a multi-drop configuration to return to a DC level when no transmitters are active. Receiver thresholds can do strange things when there is a high common mode voltage on them.

 

Receiver thresholds can do strange things when there is a high common mode voltage on them.

Would you believe me if I told you that I have not yet fully understood what common mode voltage is, and why it is an undesirable thing? I mean I've read technical articles about it, but I have yet to find one that explains things to me in simple, plain english.

 

CAUTION: there may be complex situations where the following is not entirely accurate.

It happens in differential signaling. In very simple terms, you can think of it as a DC offset with an AC information signal riding on top of it. The differential receiver ignores the DC component of the input as long as it is within the common mode range, and only pays attention to the difference between the two inputs. This is exactly what you want over a long cable where the difference between two grounds can upset a single ended receiver with a fixed threshold.

Since long cables look like distributed capacitors they will hold a charge for a long time if there is no low impedance discharge path. It is the termination resistors that keep both long and short cables from maintaining a DC offset. With no differential transmitters driving the cable, both differential signal lines will go to the same potential.

You can formulate a more rigorous definition, but let's stick with simplified for the time being.

 

CAUTION: there may be complex situations where the following is not entirely accurate.

It happens in differential signaling. In very simple terms, you can think of it as a DC offset with an AC information signal riding on top of it. The differential receiver ignores the DC component of the input as long as it is within the common mode range, and only pays attention to the difference between the two inputs. This is exactly what you want over a long cable where the difference between two grounds can upset a single ended receiver with a fixed threshold.

Since long cables look like distributed capacitors they will hold a charge for a long time if there is no low impedance discharge path. It is the termination resistors that keep both long and short cables from maintaining a DC offset. With no differential transmitters driving the cable, both differential signal lines will go to the same potential.

You can formulate a more rigorous definition, but let's stick with simplified for the time being.

Thanks!... DC offset with AC information signal riding on it... that already says a lot.
What are the situations in which no differential signals are being used and yet common mode voltage is something to be concerned about?

 

If you have an untwisted signal and ground. Any noise will be capacitivly coupled to both lines and may or may not cause problems at the other end.