As far as i know rs485 specifies a bus common mode voltage of: -7V to +12V

the terminating resistor is 120ohm , using VxV/R , yields high power rating to the resistor , certainly more than 1w . (12x12/120 =1.2w)

However looking at many products and evaluation boards , i see that they use 0805 resistors which are about 0.25w only. They might be pulse withstanding type though.

What am i missing here ? the only explanation i could come up with is that the continuous power rating of this resistor is much lower (so my calculation is wrong), since there is some break between transmitted bytes.

But in this case what happens if transmission is continuous?

  • 2
    \$\begingroup\$ RS485 has +/-2.5V (5vpp_max), not 12V. Common mode doesn't matter, only differential as there is the resistor, \$\endgroup\$ Feb 10, 2018 at 20:00
  • \$\begingroup\$ @MarkoBuršič , can you provide reference that the differential voltage is only +/-2.5v ? \$\endgroup\$
    – ElectronS
    Feb 13, 2018 at 9:59
  • 1
    \$\begingroup\$ All these calculations are for normal operation. What about resistor ratings to accommodate for short circuits to power lines in for example 24V systems? \$\endgroup\$
    – Asrar
    Feb 7, 2022 at 8:53

2 Answers 2


I suspect the answer is in bold in the first line of your question. The resistor is between the data lines but the differential voltage is only a few volts - the common mode voltage can be anything but won't cause an increase in potential across the termination resistors.

enter image description here

Figure 1. Extract from Linear's TIA/EIA-485-A Standard.

The table above says that the limit is ±5 V. Rerunning your calculation gives \$ P = \frac {V^2}{R} = \frac {5^2}{120} = 0.208\ \mathrm W \$ max.

The 0.25 W resistor should be fine.

  • \$\begingroup\$ thank u that makes sense now , it seems i goofed up the definition about the common mode voltage . which is the voltage from one of the lines ( A or B ) with respect to GND. \$\endgroup\$
    – ElectronS
    Feb 10, 2018 at 20:34
  • \$\begingroup\$ actually looking at TI's SLLA070D, section 2.1.1, RS-422 allows up to 10V transmitter output differential voltage. On the other hand, the receiver should operate on a max differential input voltage of 12V (table 1 - note 1). I also checked an online scanned copy of the TIA/EIA-422-B standard (MAY 1994) and it confirms TI's statements. Linear Tech. on the other hand didn't even bother to reference the table you mentioned. \$\endgroup\$
    – fhlb
    Feb 12, 2018 at 15:39
  • \$\begingroup\$ Ah, but the question is about RS485! (I don't know much more about it.) \$\endgroup\$
    – Transistor
    Feb 12, 2018 at 17:40

@Transistor's table says +/-1.5V min loaded and +/-5V max unloaded so with 120 Ohms the actual power will be slightly less depending on driver ESR which tends to be around 25 Ohms for 3.3V technology and a bit more up to ~50 ohms for 5V technology , so the \$V_{OD} \$ differential will be less than 5V or 0.2W, which "may" still ok.

But how hot can the resistor get?

Abs max ratings ~ 150'C depends on p/n. with power derating after 70'C is common for SMD. or a 80'C temp rise above 70'C to result in 0W rating at 150'C

So how do you estimate the temperature rise?

Using the temperature slope from R spec, [-'C/W] the max power rating drops with rising ambient temp. This is the same as a fixed room temp and rising chip temp with Used/Max power rating. Then you need to know the inside ambient in the packaging design and ambient specs.

So if an SMD chip rated for 0.25 Wmax and 0.2W is used this results in 80% of 80'C rise or > 25+64'C = 89'C at room temp.

Generally system design criteria limits a component hot spot to 85'C @ 25'C which can still burn fingers so marginally ok.

However when 3.3V technology is used the power dissipation in the terminator is Vod_max=Vcc so Pd=9/120 W, so 3.3V technology is cooler than 5V technology with a small SMD with a lower source impedance and adequate margin.


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