I am understanding the basics of RS422. Now, consider a driver (RS422) connected to a reciever. Now, we have a termination resistor of about 100Ohms or 120Ohms.

Termination resistor

Now, imagine the driver sends out Voc 2V (the differential voltage). Now, if I understand it correctly line A if more positive than line B, then the reciever outputs a logic 1 and vice versa. So, if Vod is 2V, then assuming line A is 2V and Line B is 0V. Please right me if wrong. Vod

In this case there is a 100 Ohm termination resistor. Then shouldnt the current path be from the A line into the termination resistor and back into the driver via the B line.

Is my understanding correct till now.

Now, how does the RS422 receiver work - The symbol of the receiver(and the driver) is that of an amplifier. SO, I am assuming that it takes in the difference and amplifies the difference thereby acting as a difference amplifier. Now, I was wondering, the whole current will be sunk via the line B into the driver (please understand that I am connecting a driver to a receiver). Now, the amplifier(or in basic an op-amp) has infinite impedance which means that no current is sunk into the input terminals of the amplifier or rather receiver.

Is my understanding correct on this topic please.

  • \$\begingroup\$ It would help if you add a schematic/block diagram so we know what you mean b e "Voc", "Line B", etc. \$\endgroup\$ – The Photon Oct 30 '15 at 4:32
  • \$\begingroup\$ Sorry sir. I just put the images. \$\endgroup\$ – Board-Man Oct 30 '15 at 4:45

That's mostly correct except for the part about the receiver having infinite input impedance. It is a differential op-amp so it does mostly just care about the voltage level at the input pins and not the amount of current flowing into the receiver. Wire up a typical op-amp with no feedback and what happens? The output will be pegged to one of the rails. For a communications signal this is exactly what you want to happen. No ambiguity, the output is either full positive voltage or full negative voltage.

  • \$\begingroup\$ So, the points are - 1) the current return path is as I said from the A line of driver,via the termination resistor ,and into the driver via the line B. 2) Since the amplifier (diff op amp) is voltage level based, there is no issue of current playing havoc here. The importance of termination resistor and its value is still unclear. Some websites have 100R, while others have 120R. Some circuits also have resistors on the A and B lines. WHy would that be present. \$\endgroup\$ – Board-Man Oct 30 '15 at 4:39
  • \$\begingroup\$ Also, are RS422 NR. So, when there is no data it is raised to the Vcc of the driver. But, then what if the driver IC itself is not powered. WHat would the lines return to. I believe that if there is no data - both the lines will NR and be pulled up to Vcc. Now, since to register a logic high or low there should be a minimal of +200mV or -200mV they will be not registered as valid datat as both will be Vcc. \$\endgroup\$ – Board-Man Oct 30 '15 at 4:44
  • \$\begingroup\$ Also, my point of the input impedance to the diff amplifier is valid as well,right. \$\endgroup\$ – Board-Man Oct 30 '15 at 4:47
  • \$\begingroup\$ Yes, current flows mostly from one driver pin out through the lines and back. The resistors are for termination so that there is a set current path. Removing that termination causes issues since you're driving a low output impedance driver into a high input impedance receiver and this will generate signal reflections and ringing. In addition, by having the input pins tied together via a resistor when the driver is not powered the pins will develop the same voltage and the receiver output will go to 0. \$\endgroup\$ – Chris Ryding Oct 30 '15 at 4:51
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    \$\begingroup\$ You also seem to be getting hung up on the voltage to the individual lines. It doesn't really matter as long as it's within range of the components. The difference between the lines is what is important. Finally, the value of the resistors is to impedance match the output of the drivers to provide the best signal integrity. \$\endgroup\$ – Chris Ryding Oct 30 '15 at 4:53

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