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I am trying to connect CAN to the LPC1768, like here

Can Example

Expect that I don't understand the picture or the meaning of "you need two 120 ohm terminating resistors at either end of the bus".

Does it mean I am suppose to connect a resistor of value 120 ohm between CANL and CANH (between pin 6 and pin 7) on both of the MCP2551s? For me this schematic shows two resistors connected together.

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If the bus is short enough you can probably get away with a single resistor of 56R or so but because CAN is current driven the ideal layout is to connect one resistor of 120R at each end of the bus. In this case this would directly between and as close as possible to pins 6 & 7 on each of the two ICs.

This is to allow for the inductance of the wires and match impedance at either end.

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Yes, you have to connect 120 Ohms between CANL and CANH on both of the MCP2551s. In case you have many devices you don't have to put a resistor for every device. enter image description here

Many devices: enter image description here

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    \$\begingroup\$ you'd improve this answer greatly if you emphasized 'one resistor at each end' of the transmission line \$\endgroup\$ – Neil_UK Jan 29 '19 at 14:48
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The resistors are in parallel, but physically far apart at either end of the bus. So yes, one resistor at each device.

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  • \$\begingroup\$ Long wires can act like inductors. The terminating resistors are there to match the impedance of the wires to prevent distortion of signals. \$\endgroup\$ – daniel Jan 29 '19 at 14:16
  • \$\begingroup\$ Long wires also have substantial capacitance between the two wires of the twisted pairs, or for coax, between the center and the shield. Using math Zo = sqrt( L / C ), if C= 50pF/meter and Zo = 120, then L = Zo * Zo * C (farad/meter), and L= 15,000 *50pF = 750 nanoHenry or 0.75 microHenry. Per Meter. \$\endgroup\$ – analogsystemsrf Jan 30 '19 at 3:58
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You have, I suspect, not really understood CAN systems. The important part is that all communication is conducted on a buss - that is, a single twisted pair which runs from one end of the system to the other. All devices connected to the buss do so as stubs which should be as short as possible (0.1 meter is the standard max, and shorter is better). No branches are allowed. In general, the impedance of commercially available twisted pair is about 120 ohms.

For proper operation, each physical end of the buss should have a terminating resistor of (nominally) 120 ohms. These two resistors are what you have included in your schematic.

The two resistors on your schematic will not appear on a single pcb. For each device, there will be one or zero, depending on where the device sits on the buss.

Another approach is not to include the terminating resistors on the device cards at all, but to install them at each end of the buss, then install transceivers where appropriate.

TLDR; The schematic is wrong. Make the board with room for one resistor. If you have only two units on the buss, install a resistor on each. If you have more than two on the buss, install the resistor only on the two boards at each end of the twisted pair (and make sure the wires do not extend past the board). Do not install the resistor on the intermediate units.

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