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Where Do These Protocols Best Fit?
 RS232: communication with modems, printers, and other PC peripherals. The typical maximum cable length is 100ft.
 RS422: industrial environments that require only one bus master (driver). Typical applications include process automation (chemicals, brewing, paper mills), factory automation (metal fabrication), HVAC, security, motor control, and motion control.
 RS485: industrial environments for which more than one bus master/driver is needed. Typical applications are similar to those of RS422: process automation (chemicals, brewing, paper mills), factory automation (autos, metal fabrication), HVAC, security, motor control, and motion control.
What Factors Limit the RS485 Data Rate?
The following factors affect how far one can reliably transmit at a given data rate:
 Cable length: At a given frequency, the signal is attenuated by the cable as a function of length.
 Cable construction: Cat5,Cat5e, and Cat6 24AWG twisted pair are very common cable types used for RS485 systems. Adding shielding to the cable enhances noise immunity, and thereby increases the data rate for a given distance.
 Cable characteristic impedance: Distributed capacitance and inductance slows edges, reducing noise margin and compromising the ‘eye pattern’. Distributed resistance attenuates the signal level directly.
 Driver output impedance: If too high, this limits drive capability.
 Receiver input impedance: If too low, this limits the number of receivers that the driver can handle.
 Termination: A long cable can act like a transmission line. Terminating the cable with its characteristic impedance reduces reflections and increases the achievable data rate.
 Noise margin: Bigger is better.
 Slew rate of driver: Slower edges (lower slew rates) enable transmission over longer cable lengths, but reduce maximum achievable data rate.
 PointtoPoint vs. Multidrop: stubs created by the third or more devices on the bus limit achievable data rate, sometime severely.
Termination and Biasing
Termination and Biasing are concepts that only apply to differential wiring. As such, they only apply to RS485 and not RS232.
The Termination Concept
Cable termination is a way of absorbing transmitted energy at the end of a network. This prevents signal reflections from bouncing back towards the transmitter and potentially upsetting signal quality and communications.
The termination resistor should match the characteristic impedance of the cable being terminated. The effective impedance of the RMC7xS’s termination resistor and biasing resistors is 114W. Therefore, cabling with impedance of 100W to 120W is recommended.
Termination should be placed at the end of the network for each wire pair. For RS485 (2wire, pointtopoint or multidrop), terminate the wire pair at each end of the network. The diagram in Serial Network Topologies shows the correct location of the termination.
Termination and Cable Length
Termination is not required on all differential networks, but it does typically extend the maximum cable length. The following chart shows the maximum cable lengths at various baud rates with and without termination:
NOTE:
The maximum cable length is the length of the entire network and not just the distance between nodes on the network.
Termination vs. Cable Length:
Baud Rate 
Max Unterminated Cable Length (ft) 
Termination Requirements 
Max Terminated Cable Length (ft) 
115,200 
475 
Required beyond 475 ft 
3250 
57,600 
950 
Required beyond 950 ft 
4000 
38,400 
1900 
Required beyond 1900 ft 
4000 
19,200 
3750 
Required beyond 3750 ft 
4000 
9,600 
4000 
Not Required 
4000 
4,800 
4000 
Not Required 
4000 
2,400 
4000 
Not Required 
4000 
Cable Length Derivation
The values presented in the chart above are based on 24AWG cable with capacitance of 16 pF/ft and the following reasoning. Signals travel through a cable at approximately 66% of c or 0.66 ft/ns. It is assumed that a signal transition will dampen out after three round trips in the cable. This damping must occur before the bit is sampled or within half a bit time. One bit time is equal to the reciprocal of the baud rate.
Example:
Compute the cable length for 115,200 baud RS422.
First, we compute a half bit time at this baud rate.
Half Bit Time 
=  0.5 * 1 / 115200 

=  4,340 ns 
Next, we convert this time to the distance the signal would travel in this time, assuming a speed of 0.66 ft/ns as described above:
Distance 
=  4,340 ns * 0.66 ft / ns 

=  2890 ft 
Since three round trips are required for the signal transition to dampen and each round trip is twice the length of the cable, the total distance in feet is divided by six to get the final unterminated cable length:
Length 
=  2890 ft / 6 

=  482 ft 
This value is then rounded down to allow for inexact cable velocities and damping rates, giving us 475 ft.
The Biasing Concept
RS485 indicates a binary 1 when the A line is at least 200 mV negative with respect to B, and a binary 0 when A is at least 200 mV positive with respect to B. It is important that the lines always be in a known state, not only when being driven. Biasing forces the network into a known state when the lines are idle and therefore otherwise not driven.
A known state is forced by allowing current to flow across the termination resistor. Therefore, biasing is usually selected on the RMC that also has a termination. However, some masters only have termination, in which case the user may want to only select biasing on an RMC close to the master. The current will then flow across the master’s termination resistor.
The RMC7xS requires biasing in order to be in a known state when the lines are idle. The biasing forces a binary 1.
Example:
This example assumes that there is a single master and two RMCs on the network. Compute the voltage across a 120 ohm termination resistor when using 1150 ohm biasing resistors.
First, we calculate how much DC resistance will be between the biasing resistors. Calculating the parallel resistance of all DC terminations and node input impedances does this. For a single master and two RMCs we have the following components:

Master load: 1 unit load, which is defined as 12 kohm.

RMC loads: unit load each, which is 48 kohm.

Termination Resister in the RMC: 120 ohm
Therefore, putting all of the resistances in parallel yields the following:
Termination Resistance 
=  120 ohm  12k ohm  48k ohm  48k ohm 

=  118 ohm 
Then, we calculate how much DC resistance the network has between power rails:
Total Resistance 
=  1150 ohm + 118 ohm + 1150 ohm 

=  2418 ohm 
Next, we calculate how much current is flowing through this DC resistance:
Current 
=  5VDC / 2418 ohm 

=  2.068mA 
Finally, we calculate the voltage drop across the termination resistor:
Voltage 
=  2.068mA * 118 ohm 

=  244mV 
This value is greater than the 200mV difference required by the TIA/EIA standards and constitutes a valid binary 0 state.