Sunday, August 28, 2011

Using RS-485 transceivers

The beauty of using RS-485 (more formally known as the TIA/EIA-485 standard) for long distance serial communication is that because it uses differential lines it's largely immune to common mode voltages and noise. It's also utterly simple electrically--at least in a half duplex setup--in that it uses only one twisted pair cable and a common ground.

Here's a setup for perhaps the simplest possible configuration: there's only one transmitter and one or more receivers. In other words communications is a simplex type--information goes only one way. The transmitter sends out data while the receivers intercept the data stream but do not transmit anything.


In the above circuit the non-inverting input of the unity gain buffer is connected to a sensor circuit with an analog output. The MCU's ADC converts the sensor reading to a digital value and sends it out serially. U1 is a RS-485 transceiver. Pin DE and /RE are controlled by the MCU, switching them high only when transmitting.

At the heart of the RS-485 is the transceiver. In fact RS-485 is merely an electrical standard. And the particular electrical characteristics of the transceivers determine whether one has a RS-485 circuit or not. There are a lot of transceivers out there and I would rather use something like Maxim's MAX48x series, but the price is simply eye popping! They're as expensive as instrumentation amplifiers. Instead I've been using the jurassic SN75176B which is over five times cheaper. Texas Instruments says it's rated for 10Mbps throughput. But hey my requirements are 2 to 3 orders of magnitude lower.

The problem with the 757176 is that it's a gas guzzler. An informal test shows that it draws some 16mA @Vcc = 5V when it's set up for reception only (pins DE and /RE are low). This shoots up to around 32mA when it's in transmission mode (pins DE and /RE are high). As Dave Jones would say, you can fly to moon on 32mA!

To limit the amount of juice it gulps, we can--at least in a simplex arrangement--turn off the 75176 when no transmission is being performed. As can be seen in the schematic above, a PNP transistor is used to power it up/down. When transmission of a data packet is about to commence the transceiver is switched on and right after the last stop bit is sent, the 75176 is powered down. Average power consumption of the transmitter circuit will depend on how frequent transmissions are made. It will also depend on the baud rate--the higher, the less time every transmission takes and therefore the shorter the time the transceiver will be on.

In half duplex systems where drivers (transmitters) are also receivers powering down the transceiver will probably be not an option.

A very important issue with these transceivers is to ensure that they will be fail-safe. Fail safe in this context means that even if the transmitter's DE pin is grounded--ie., the driver is disabled and lines B and A become tri-stated to high impedance--or if power to the transmitting transceiver is cut off as it is in the circuit above--any and all the receivers will still see a differential voltage between B and A that's > 200mV. As per the RS-485 standard, if -200mV < B - A < +200mV, then output of pin R will be indeterminate. To guard against this bias resistors are added to lines A and B. In the schematic above these are pull-up resistor Rb1 and pull-down resistor Rb2.

Very important note: The RS-485 standard states that when the difference between the voltages between B and A is > +200mV (i.e, B - A > 200mV) then a logic 1 is output on the receiver, and when B - A <. -200mV then a logic 0 is output. But chip manufacturers like Texas Instruments and Maxim have reversed this. And so B is A and A is B on their transceivers. No problem as long you don't mix transceivers that use the RS-485 standard labeling. From hereon I'll follow the TI / Maxim convention. 

The values of the bias resistors determine the differential voltage A - B. How much bias resistance should be installed depends on how many receivers are connected to the network. For the transceiver used above, each receiver has a minimum input resistance of 12Kohms. The more receivers there are the less the total resistance there is since the input resistances are in parallel. This translates to lower value bias resistors. Here's how to compute the values:

Let:
Ritotal = total input resistance
I = current through Ritotal
VCC = power supply voltage to which pull-up resistor will be connected
Rb = bias resistor value
RT = total resistance which includes the bias resistors and Ritotal

We require that the voltage across the total input resistance be greater than 200mV:

I Ritotal > 200mV

Solving for current:

I > 200mV / Ritotal

The bias resistors and Ritotal are in series and form a voltage divider and we define their sum as the total resistance:

RT = 2Rb + Ritotal

It's also equal to:

RT = VCC / I

Since we're looking for the value of the bias resistors:

Rb = (RT - Ritotal)/2

Substituting and simplifying we obtain:

Rb < (5VCC - 1)Ritotal / 2

Example: Let's say we have two SN75176B transceivers on the network configured as receivers. VCC = 5V. Each transceiver has an input resistance of 12Kohms. Because they're in parallel, Ritotal = 6K. Using the formula above, we find that bias resistors should be less than 72Kohms each in order for A - B to be > 200mV when transmitter is disabled (lines A and B are in a high impedance state).

If termination resistors are present then total resistance would be = Ritotal|| termination resistors. So if two 120 ohm termination resistors were present (termination resistors are usually between 120 and 150 ohms when using UTP cables) and the Ritotal= 6K then the total resistance = 6K || 120 || 120 = 1/(1/6000 + 1/120 + 1/120) = 59.4 ohms. Computing for the bias resistors we obtain 720 ohms. Because the very low resistance values of termination resistors, as much as possible we don't want to install them since they significantly increase power consumption.

Termination resistors are used across lines A and B for impedance matching to prevent/minimize signal reflections leading to ringing. So when do we need termination? Well, it depends if we have a short or long line/cable. One heuristic for determining whether a line is long or short is as follows.

Let:
tt = transition time also known as rise/fall time of transceiver's transmitter signal in seconds
RP = propagation rate of electrical signals in copper = 0.2m/ns = 2 x 108m/s
L = cable length in meters

A line is considered long if:

tt < 2(L/RP)

(see Maxim's equation for a more conservative figure).

According to the SN75176B datasheet its tt = 20ns. Let's solve for L to see how long the cable can be before it's deemed long.

L = ttRP/2 = (20 x 10-9)(2 x 108)/2 = 2 meters.

So if we're using the SN75176B and our cable is longer than 2 meters termination resistors are necessary. That's not the end of the story, however. We may still be able to get away without having any termination even at cable lengths much longer than this:
If the rise time is unknown, another way of deciding whether a line is long or short is to compare the shortest expected bit width and the 1-way cable delay. This method considers two factors: the reflections may bounce back and forth several times before settling, and the bit rates at the transmitter and receiver may vary slightly from each other. As a general guideline, if the bit width is 40 or more times greater than the delay, any reflections will have settled by the time the receiver reads the bits. (Axelson, p.111)

The key to obviating the need for termination is to have a sufficiently slow baud rate so that reflections in the transmission line (a "long line") have already died out by the time the microcontroller reads the bit. The 1-way cable delay is merely L/RP. Thus for a 100-meter wire, the 1-way cable delay = 500ns or 0.5┬Ás. Using the above guideline we can determine the maximum cable length given a particular transmission baud rate--the bit width is the reciprocal of baud rate.

Let:
B = transmission baud rate in bits/sec
RP = propagation rate of electrical signals in copper = 0.2m/ns = 2 x 108m/s
L = cable length in meters

The guideline states that:

1/B >= 40(L/RP)

Solving for L we have:

L <= [(1/B)/40]RP

L <= RP/(40B)

For a baud rate of 19,200, L = 260 meters. So theoretically no termination is necessary if cable length is less than this. Just out of curiosity let's see the maximum baud rate at which we can transmit for the 2-meter cable above.

B = RP/(40L) = 2.5Mbps


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References:

Janet Axelson, Serial Port Complete: COM Ports, USB Virtual COM Ports, and Ports for Embedded Systems, 2ed, Lakeview Research LLC, 2007, Chapters.6 & 7.

Polarities for Differential Pair Signals (RS-422 and RS-485)


RS-422 and RS-485 Application Note (computation of bias resistor values)

AN1090 Methods for Trimming the Power Required in RS-485 Systems

1 comment:

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