Modbus – Kabel, Termination, Biasing

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Bron: maximintegrated – Website: maximintegrated.com

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Kabel

Where Do These Protocols Best Fit?

• RS-232: communication with modems, printers, and other PC peripherals. The typical maximum cable length is 100ft.
• RS-422: 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.
• RS-485: industrial environments for which more than one bus master/driver is needed. Typical applications are similar to those of RS-422: process automation (chemicals, brewing, paper mills), factory automation (autos, metal fabrication), HVAC, security, motor control, and motion control.

What Factors Limit the RS-485 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 RS-485 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.
• Point-to-Point 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 RS-485 and not RS-232.

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 RS-485 (2-wire, point-to-point or multi-drop), 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

RS-485 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.