INSULATOR 1P6452 - Caterpillar

Limited Modbus Master
The PL1000E can act as a "Modbus Master" in a limited fashion. As a "Modbus Master", the features of the PL1000E are limited to translating data from CDL or J1939 to "Modbus". The PL1000E cannot act as a full "Modbus Master".Considerations for Designing a Robust System
Modbus is a "Point-to-Point" protocol. Therefore, the PL1000E will send a request message on Modbus and then will wait for a response. In a "Point-to-Point" protocol system that operates properly, the process to send a request and wait for a response is not an issue. If the PL1000E does not receive a response, the PL1000E will wait for 250ms, and then move on to the next request. If some slaves do not respond to the requests of the PL1000E, then system operation will slow due to the wasted time. If this happens with too many messages, then the data may not be received in a timely fashion.There is another issue that must be considered with the "Point-to-Point" protocol. If too many parameters are being requested within a given time period, there is possibility that the transmit buffer on the PL1000E may become full. If the transmit buffer becomes full, then transmit messages may be dropped. The equation below is a guideline in order to determine the load on the Modbus Link. The equation calculates the total time the system takes to send one request and receive one response.
Illustration 1 g01412383
LengthReqBit = Total Length of the Request Message in Bits.BaudRate = The Baud rate you system configured.TotalPropagationTime = Time that it takes the message to reach the destination.SlaveResponseTime = Time it takes the Modbus slave to respond to the request from PL1000E.LengthRespBits = Total Length of the Response Message in Bits.Modbus Load Equation Example
The following values are used in order to provide an example of how to use the equation.
The PL1000E has been configured to request 10 registers at rate of 500ms.
The PL1000E has been configured for a baud rate of 9,600 bps.
The PL1000E can only request one register at a time, so the total length of a Modbus request message is 7 bytes, which is equal to 56 bits.
Since the PL1000E can only request one register at a time as a "Master". The response is a total of 6 bytes, which is equal to 48 bits.
The PL1000E is close to the Modbus slave that the PL1000E is requesting information from, therefore, for this example, the propagation time is 0
The response from the Modbus slave takes 100ms.The equation would look like this:
Illustration 2 g01412389
The total time that is required in order to send one request and receive one response from the system is 110.8ms. In theory you should be able to request 9 messages per second. In the example, 20 requests a second are being sent to the transmit buffer, but only 9 messages per second can be sent out. The buffer for the PL1000E will eventually become full because the amount of messages that are being queued per second is more than what