1 / 20

Basic

Basic. Controller PLC / PC. Dupline (or other Fieldbus) Installation. Traditional Installation. Field Devices Sensors, Switches, Valves, Contactors, Drives etc. Typical Automation System. General Fieldbus Advantages. Reduced Installation Cost Reduced Cable Cost

tirzah
Download Presentation

Basic

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Basic Jens Neigaard

  2. Controller PLC / PC Dupline (or other Fieldbus) Installation Traditional Installation Field Devices Sensors, Switches, Valves, Contactors, Drives etc. Typical Automation System Jens Neigaard

  3. General Fieldbus Advantages • Reduced Installation Cost • Reduced Cable Cost • Reduction in maintenance cost • Increased flexibility • Capacity for more signals Jens Neigaard

  4. Basic components: The twisted pair Signal Common Jens Neigaard

  5. Basic Components: The Channel Generator Jens Neigaard

  6. Basic Components: The Transmitter Jens Neigaard

  7. Basic Components: The Receiver Jens Neigaard

  8. Dupline Addressing Scheme Channels: 1 ... 8 I/OAddresses: A1, A2, A3 .. A8, B1, B2 .. P7, P8 Groups: A ... P Jens Neigaard

  9. Dupline Address Coding Device: GAP1605 Jens Neigaard

  10. Dupline used for wire replacement A1 A2 A4 A3 A2 A4 A1 A3 A1 A4 A3 A3 A2 FMK 8 Channel Generator A1 A2 A4 Jens Neigaard

  11. Receiver Transmitter A1 A5 A1 A5 A1 A5 A1 A5 Transmitter Transmitter n-Remote Transmitters have Effect on One Central Receiver Channel Generator 64 Jens Neigaard

  12. Transmitter Receiver A1 A5 A1 A5 A1 A5 A1 A5 Receiver Receiver A Central Transmitter has Effect on n-Remote Receivers Channel Generator 64 Jens Neigaard

  13. Carlo Gavazzi Dupline Carlo Gavazzi Dupline Dupline Standard Housings Jens Neigaard

  14. Dupline Digital I/O-modules • Different input Types • Contact /NPN Transistor • Voltage • Different output Types • Relay • NPN Transistor • PNP Transistor • Different number of I/O’s • H-Housings, D-Housings and Open PCB • Different supply voltages Jens Neigaard

  15. MODE: DIGITAL 1 GROUP 1 3 7 8 7 Dupline Test Unit: GTU8 Monitoring of Channel Status’ in a Selected Channel Group Dupline Test Unit GTU 8 Display of Channel- Status’ A 1 - 3 - - - - 8 Mode Enter Select Channel- Group 8 Force Channels to ON (monostable) 4 5 6 1 2 3 Jens Neigaard

  16. Dupline has free Topology Jens Neigaard

  17. Dupline Cables • Single (twisted)pair of wires • Shield not required • Recommended min. cross sections • Long distance: 1.5mm2 • Medium distance:0.8 - 1.0 mm2 • Short distance:0.35 mm2 • Low capacitance for longdistances • (preferably < 80 nF/km) Jens Neigaard

  18. Dupline Installation • Never connect DuplineCommon • to Ground • Treat the Dupline cable as a signal wire • Use DT01 termination unit for • distances > 1.5 km • Check correct wiring during • installation Jens Neigaard

  19. Cable Splicing / Isolation • Use soldering splices • Splices must be water tight • No leakage or connection between wires or • fromwires to ground Jens Neigaard

  20. Rules for Dupline-powered Devices • The total current consumption must not exceed max current output of the channel-generator: • G34900000:Max. 70 mA • G38000015: Max. 130 mA • b) The total current consumption in the network multiplied bythe cable loop resistance must not exceed 2 V • An example: 2 km 1.0 sq.mm twisted pair is used. G50101106 is used both with input and LED • The resistance of a 1.0 sq.mm.core is app. 20 Ohms/km. • With a wire length of 2 km the resistance per wire is 40 Ohms • With the 2 Dupline wires the total resistance is 2 x 40 ohms = 80 Ohms. • This means that the max total current is 2 V / 80 Ohms = 25 mA. • One G50101106 with LED function uses 1.2 mA • Max Number of G50101106 = 25 mA / 1.2 mA = 20 units • If the LED is not used the current per G50101106 is only 0.15 mA so in this case the result is: • Max number of G50101106's = 25 mA / 0.15 mA = 166 units. Jens Neigaard

More Related