THVD1512: termination resistor RT

Termination resistors must be (only) at the two ends of the bus.

(If you removed the transceiver at the right, you would still need the resistors there.)

Hi DDdoor,

Clemens is correct.

For RS-485 systems there are 2 termination resistors suggested (typically 120 Ohms as the termination resistors in parallel equals to 60 Ohms). If there are more than two transceivers on the line only the transceivers on each end of the transmission line are terminated. 

If you have any other questions please let me know!

Best,

Parker Dodson

The end literally is the end of the cable. It is the point where current cannot flow any further.

Additional resistors do not terminate anything, and would just increase the load, i.e., they would reduce the noise margin or even make communication impossible.

Electrically, the THVD1512 and the ADM3491E (and all other transceivers) behave the same.

A termination resistor is not needed at a device that is a transmitter. However, if one of the slaves at the bottom is transmitting, then the two transceivers at the top are not transmitting, and you will get reflections at their unterminated Y/Z outputs. That example configuration would be correct only if the two slaves at the bottom never transmit anything.

Hi DDoor,

Clemens is correct by saying the devices are electrically similar. This configuration is possible but it does come with risks.

In an ideal situation RT = Z0 of the transmission line so that reflections = 0 since the reflection coefficient = (ZL-Z0)/(ZL+Z0). However due to tolerance of the passives and cable characteristic impedance there will be some reflection. Since the RX is not terminated it will reflect any incoming reflection back onto the line without much, if any,  attenuation. This typically isn't enough interference to cause major issues as this reflection will decay pretty quickly, but it could degrade the max distance of the device.

With respect to the two bottom devices in the diagram - if they ever transmit their will be a lot of reflections - as Clemens mentioned as the bottom two devices send the signal to the entire line (essentially both ends of the cable are seen as the "load" - the other bottom device can be ignored - this is explained a little lower in the post) and since one side is unterminated there will be a large risk of data integrity issues due to the reflections from the unterminated side.

See below for a diagram showing reflection pathways:

L1 = L1_bar ; L2=L2_bar. These are the stub lengths of the two bottom devices. If L < (tr/10 * v * c) - where tr is the rise/fall time of the transceiver, v is the signal velocity of the transmission line as a fraction of c, and c is the speed of light - then these devices can be ignored as reflection sources - however if that isn't true they can't.

In the above figure D2 is transmitting; I_D1 is the incident signal into D1, I_D4 is the incident signal into D4, and I_D3 is the incident signal into D3. R_D1, R_D4, and R_D3 are the resulting reflections from each of those signals.

If the length condition is met for both L1/L1_bar and L2/L2_bar we can assume R_D3 ~ 0. Since D1 is terminated R_D1 ~ 0. However, since D4 is unterminated R_D4 ~ -1 * I_D4 which can cause integrity issues.

If the length condition is not met there could be even more reflections possible.

So in conclusion in this setup only D1 and D4 can transmit without many issues as long as L1/L1_bar/L2/L2_bar < (tr/10)*v*c; there will be some reflection but it shouldn't cause problems until you are near maxing out the distance of the transceiver. However D2 and D3 will not be able to transmit cleanly due to the lack of termination of D4; if the length condition is not met there could be even more reflections.

To avoid the above complications terminate both ends of the cable and keep stub length of the bottom devices (in this case D2/D3) less than 1/10th of the fastest signal transition on the bus.

Please let me know if you have any other questions.

Best,

Parker Dodson