How much loading can an auto-polarity RS-485 bus support?

Other Parts Discussed in Post: SN65HVD888

The RS-485 interface is popular because of its robustness; the use of differential signaling effectively rejects common-mode noise on the bus. To make the communication work, the transceiver and receiver have to match the pins: A to A and B to B. In some RS-485 applications, however, the bus polarity information might not be available on the receiver ends or the polarity could be mistaken.

Auto-polarity RS-485 transceivers like TI’s SN65HVD888 can detect the bus polarity automatically so that the two bus wires can be connected during installation either way. For example, A pin of the device can be connected to any one of the two bus wires and B pin can be connected to the other wire. This saves cost and makes the system more reliable. This feature could be beneficial for the systems with high installation rate, such as IP camera, E-meter, air conditioning and lighting. In the event that the RS485 connection is installed backwards, the device can self-correct the polarity, instead of requiring an installation team to re-visit the site and correct the error.

The mechanism of auto-polarity detection relies on generating enough voltage difference on the idle bus in the system using bias resistors, RFS and RT2, shown in Figure 1.

Figure 1: Auto-polarity RS-485 bus network

Because designing this bias (fail-safe) network is sometimes not as straightforward as it looks, in this post I’ll apply two equations to make the calculations. After the resistor values are fixed, I will show the network’s impact on the loading condition of the system.

To design a proper fail-safe network, you need to meet two fundamental conditions. First, the equivalent resistance of the network, RFS and RT2, needs to be equal to the characteristic impedance of the cabling. In this setup, this value is presented by the termination resistor value of the bus, RT1. Secondly, the voltage generated on the idle bus should be bigger than the input threshold voltage, Vit, of the receiver. Provided with the first condition, the voltage drop on RT2 from the network itself should be twice the voltage required. In other words, the calculation is only based on the resistors in the dotted blue box of Figure 1, excluding RT1.

From the first condition, you can redraw the network by connecting the virtual ground, Vcc and GND, of RFS together (Figure 2).

Figure 2: Equivalent circuit of a fail-safe network

By doing this, you can easily calculate the equivalent resistance using Equation 1:

From the second condition, suppose that the minimum Vcc is 4.5V and Vit is 100mV. You can also include 50mV of noise margin on top of Vit (Equation 2):

Now that you have two equations with two variables, you can put them together using Equation 3:

From here, it’s easy to deduce the value of RT2. In this example, it is 128Ω.

After you have obtained all resistor values, you can evaluate their impact on the system. In the RS-485 standard, 375Ω represents the maximum loading of the bus (RINEQ in Figure 3), or (1/375) S in conductance. Assume the bus voltage is 1V. This means that the most leakage current the bus can take is about 2.67mA single-ended.

RS-485 specifies a term of unit load (UL) to represent a load impedance of approximately 12kΩ. If each node has a 96kΩ input resistance (1/8 UL), it would take 256 nodes to generate equivalent leakage. 900Ω RFS takes 1.11mA away from the total current budget (the blue arrow in Figure 3). Therefore, the nodes on the bus can have a maximum leakage of 1.56mA (the red arrow in Figure 3). 1.56mA divided by 104µA (1V/96kΩ) is equal to 149 nodes.

Figure 3: Autopolarity RS-485 bus network showing loading conditions

A proper fail-safe network for SN65HVD888 shows that the bus can support up to 149 1/8UL nodes with the network.

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