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RS485 Modbus RTU Interfaces

DonetteCovey926998352025.04.16 11:33조회 수 1댓글 0

However, if the receiver considers the undriven wires to be a binary 0, when the driver is turned on and set to transmit a start bit, which is also a binary 0, the receiver will not see a transition, and therefore will not see the start bit. The exact voltage level that a logic device considers ON or OFF varies by logic type, but when the voltage is high (usually but not always approaching the IC's supply voltage), the output is on and a binary 1 is on the wire, and when the voltage is approaching 0 the output is off and, a binary 0 is on the wire. You can design the system to accomplish this if you pay careful attention to logic function of the driver being used, as well as the labeling of the terminal on the device. 200mV as undefined, but the IC manufacturer can put the threshold for a 0 and a 1 anywhere they want. Look at the data sheets from each manufacturer to determine what signal is on what pin and if the two devices can be connected either directly or with an adapter.


And since a 120 Ω termination resistor can cause enough signal loss with an extremely long 120 Ω cable to stop the network from functioning, a 500 Ω or even a 1kΩ termination resistor may improve the signal quality enough without causing too much signal loss for the network to function. Often, use of shielded CAT5/6 cable can give an adequate signal ground, although this is not recommended. Adding the biasing resistor to every device on the network can cause termination problems. This allowed other devices on the RS-485 network to transmit. Since data rates on a PC serial port can vary from 300 baud (26mS for 8-bits) up to over 100k baud (0.08mS for 8-bits), the timer will have to either be limited in the baud rates that it can work with, or keep control of the RS-485 network for much longer than it needs. This example of biasing would be a bad thing on an equivalent RS-485 network.


The second thing is that this is effectively an infinite voltage divider. Another thing to note is that biasing may not be required. Other receivers on the network that do not have this shifted receive threshold may require that external bias resistors added to the network. As mentioned above in the Voltages section; On the RS-485 network there will be times when the two wires are not driven by a transmitter. This means there will be times when no driver is connected to the wires. When an RS-485 driver gets its data directly from a UART (with no added inversion), you would expect the "A" and "B" wires to match the voltages in the RS-485 standard for the voltages on the wires, but they will not (unless the driver inverts its input). When the input is high the inverting output goes low, and when the input is low the inverting output goes high. 0.2V the receivers output switches to a 1 and when the voltage goes below -0.2V the receivers output switches to a 0. (or vice versa if there is inversion on the receiver) Since the last bit from a UART will be the stop bit (1), then the transmitter is turned off (the differential voltage goes to 0V, but not having gone less than -0.2V), this should leave the receiver with a 1 being output to the receiving UART.


The logic function of the driver and receiver are not defined, only the binary state of the differential voltages on the wires. Even though the logic function of the generator and receiver are not defined by RS-485, it makes sense to many engineers to have a binary 1 appear on the RS-485 wires when a binary 1 is being transmitted. Digital logic usually uses the same convention (but not the same voltage) as the lamp example. The exact voltage of high and low are not usually specified, rather a voltage difference between the two outputs is specified. The driver's impedance (when active) is not specified, but the driver needs to be capable of driving 32 unit loads and a termination resistance as low as 60 Ω. So are termination resistors required? If you've been paying attention, you may notice that the last two points are contradictory. If the data bit is sampled while there is ringing it may be sampled as an incorrect value. We may already have the solution.



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