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I am using ADM2482, Isolated RS-485 Transceivers. I need to calculate total max power needed to Drive this circuit.

Specs are 1. Logic side voltage is 5V 2. bus side voltage is 3.3V 3. Speed 1 Mbps. 4. RS 485

In datasheet it it given enter image description here

SO what will be my total power needed to drive this circuits.

= 5V*6mA+3.3V*50mA=195mW ( total power) only ???

what about the short circuit current shown in attached pics, do I also need to consider this current , if yes that which side Logic side or bus side.

if logic side

=5V*6mA+3.3V*50mA + 5v*250mA=1.4W ( total power)

if bus side =5V*6mA+3.3V*50mA + 3.3v*250mA=1.02W ( total power)

Does RS485 need this much power (1W/1.4W) to operate ??

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5V*6mA+3.3V*50mA=195mW (total power) only ???

Yes, this is the correct result, in the nominal case.

The short-circuit current is the maximum amount of current that will flow if the driver outputs are shorted (Y/Z outputs). This is not the nominal case. And it will actually go "in place" of the 50mA Vdd2 supply current, so, in case of short circuit, you'll have:

5V*6mA+3.3V*250mA = 855mW

of power dissipated.

Whether you want to account from this in your design depends on the probability to have a short circuit (basically, on who will do the cabling), or whether you absolutely need your design to survice such a case. But since it is not the nominal case, it is up to you to decide (and document whether this is safe or not).

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195mW is NOT the right answer. The datasheet is telling you how much current is used by THIS CHIP at this data rate. It does NOT include the amount of current that is flowing through this chip and out the output pins. Very few datasheets for digital devices actually include the power drawn from their output pins in the "supply current" section of the datasheet.

You need to also determine the power to drive your data lines with the speed and duty cycle that you expect to "talk" on the bus and add that to your 195mW in order to properly size your power supply.

Of course, if you design for the short circuit current, then that will obviously be higher the normal operating current.

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