Differential line terminator power

[bremenpl]

Hello there,
I have a question regarding differential line resistor power in quadrature encoders applications.



Assuming RT is 120R and VCC of IC's is 5V, my understanding is that (with steady state, no data transfer) power dissipation on RT is:

P = 5V^2 * 120R = ~0.2 W

The question is- Am I right about this? Do I understand correctly that the outputs of AM26C31 (or any other generic differential line driver like this) are push-pull type? Or is there an additional resistance on its outputs? My assumption is from the AM26C31 datasheet:



Does this mean that terminating resistors, in fact have to be power resistors, at least 1206 (0.25W)? In some applications like this I have seen 0603 resistors are placed and thats what alarmed me, that maybe I dont get something and that approach is ok.
I would appreciate all help regarding this question.

 

[FvM]

The question will be best answered by the output voltage specifications of the driver datasheet. It says typical differential output voltage magnitude with 100 ohm load is 3.1 V.

 

[bremenpl]

Does that mean that the outputs work as current sources? How to understand this?
Edit: Ah, its because output levels are not 5V - 0V, but 3.4V - 0.2V.

 

[KlausST]

Hi,

Yes.
That´s one benefit of differential signalling. The power through the dataline is independent of high or low state.
Thus the driver VCC/GND current is constant. (Only giltches during switching, but these are compensated by the VCC capacitor).
Therefore there is no significant change in GND supply current. All the signal current, DC as well as AC is constant through the two signal lines.

In opposite with single ended signalling: The GND current changes with signal states. So in all impedance calculations you have to take care about GND wiring.
This makes calculations difficult, and the signal quality worse.

Klaus