From Serial Port Complete, Second Edition:

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The total current used by an RS-485 network varies with the impedances of the

components including drivers, cable, receivers, and termination components. A

low output impedance at the driver and a low-impedance cable enable fast

switching and ensure that the receiver sees the largest signal possible. A high

impedance at the receiver decreases the current on the line and thus reduces

power consumption.

When used, termination components can greatly increase the current a line

consumes. Many RS-485 lines have a 120Ω resistor across the differential lines

at each end of the line. The parallel combination of the resistors is 60Ω. The terminations

create a low-resistance path from the driver with a logic-high output,

through the terminations, and into the driver with a logic-low output. On short

cables at slow bit rates, you may be able to eliminate the termination entirely

and greatly reduce power consumption.

With no termination components, the receivers’ input impedance has the greatest

effect on the series resistance of the line. The total input impedance varies

with the number of enabled receivers and their input impedances. Chapter 7

explains how to decide whether to use a termination and if so, what components

to use.

In addition to a 60Ω parallel termination, an RS-485 driver can drive receivers

that draw up to a total of 32 unit loads. TIA-485-A defines a unit load in terms

of current consumption. A receiver equal to one unit load draws no more than a

specified amount of current at input-voltage extremes specified by the standard.

When the received voltage is as much as +12V greater than the receiver’s signal

ground, a receiver equal to 1 unit load consumes no more than 1 mA. When

the received voltage is as much as 7V less than the receiver’s ground, the same

receiver consumes no more than -0.8 mA. To meet this requirement, a receiver

must have an input resistance of at least 12k between each differential input

and the supply voltage or ground, depending on the direction of current flow.

The resistance at each of the two differential inputs of a 1-unit-load receiver is

thus 12k. (This is the resistance from an input to ground or the supply voltage,

not the resistance between the two inputs.) Add a second receiver, and the parallel

resistance of the combination drops to 6k. With receivers equivalent to 32

unit loads, the parallel resistance of the combined inputs is just 375Ω, or slightly

less due to leakage currents.

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http://janaxelson.com/spc.htmFor current loop:

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Because you add the extra current loops in parallel, the power supply’s voltage doesn’t need to change; you only need to make sure it can source the required current. You can mix and match different types of transducers as long as each transducer and power supply’s operating voltage match.

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http://www.ni.com/white-paper/6940/en/#toc1