AM26LS31: Estimate Power Consumption

user5163865 said:

Pd_o = {(Vcc-Voh) x Io + Vol x Io} x 4ch
= {(5V-3.13V) x 29mA + 0.3V x 29mA} x 4
= 252mW

This basically gives the power of the RS422 resistors when the driver is idle (like a DC power).

user5163865 said:

Pd_static = 160pF ​​x (5V^2) x 1MHz x 4ch = 16mW

This not a static value. This is an AC power which tells the power loss due to the load capacitor's charging and discharging. I wouldn't add this power to the DC power (the one above). I would pick whichever one is the higher one. In this case it's the driver's DC power. But if the load capacitance or the switching frequency were higher, the AC power could become worse/larger.

I would use do:

total power = Device's supply current power + RS485 system power

Device's supply current power (worst case) =  Vcc x Icc = 5V x 80mA = 400mW (this is probably an over estimate)

Device's system power (includes driver and the resistor)= I_RS422 x Vcc x 4ch= 20mA x 5V x 4 = 400mW (I used 20mA as the current value but you can use larger to over estimate)

The above is the DC idle power when it's driving low/high but not toggling. We will use this as the worst case but if we calculate the switching power, 

AC switching power loss: Cpd x VCC^ 2 x f x 4ch , as a rule of thumb I usually say 1 meter of cable is 50pF. If you have 100 meters, you have 5000pF 

If we use 1MHz (data rate is 2Mbps) and a 100 meter cable on all 4 channels then

Cpd x VoD^ 2 x f x 4ch = 5000pF x 3^2 x 1Mhz x 4 = 180mW (VoD is the voltage across the resistor or Y-Z, I chose 3V this is the voltage the load capacitor will charge up to it won't charge up to Vcc)

180mW > 400mW (DC we calculated earlier)

So I would choose the DC power since it's larger (pick AC power loss or DC power loss, which ever is bigger for the final calculation)

Power = 180mW (AC power) + 400mW (from device's supply current power) = 580mW

-Bobby