MCP2515: Listen-only geht, normal geht nicht

void can_prepare(void) {

  /* IRQ pin as input with pullup */

  DDRB &= ~_BV(PB1);

  PORTB |= _BV(PB1);

  /* Reset */

  can_command(CAN_RESET);

  _delay_ms(100);

  /* Tie BF pins */

  can_write_register(BFPCTRL,

    _BV(B1BFE) | _BV(B1BFM) | _BV(B0BFE) | _BV(B0BFM)

  );

  /* Disable RTS pins */

  can_write_register(TXRTSCTRL, 0);

  /* Configure baud rate to 100kbps.

    Given by the quartz oscillator used in the circuit:

      Fosc = 8 MHz -> Tosc = 125 ns

    The nominal baud rate 'NBR' of 100 kbps is also the

    nominal bit rate. This yields a nominal bit time of

      Tbit = 1/NBR = 10 us

    The bit time is divided into time quants 'Tq' by the can

    bus controller and the bit time is composed out of time

    quants as

      Tbit = Tsync + Tprop + Tpha1 + Tpha2

    where

      . Tsync is fixed to 1*Tq for synchronization to the bus,

      . Tprop is the physical propagation delay on the bus,

      . Tpha1 and Tpha2 divide the remaining bit time into

        two phase segments. The bit is sampled between these

        phase segments.

    The time quant must be choosen short enough to allow for

    a sensible choice of the propagation sement Tprop and large

    enough to be able to divide the total bit time at a ratio

    of about 60..70% : 40..30%.

    The time quant is determined by a prescaler 'BRP'.

      Tq = 2*(BRP+1) * Tosc

    Choosing 'BRP' = 2 yields Tq = 1 us.

    The propagation delay is programmed by 'PRSEG' as

      Tprop = (PRSEG+1) * Tq

    and 'PRSEG' = 1 gives a propagation delay of 2 us, which

    is enough for a tiny bus.

    After Tsync and Tprop, there are 6 time quants left.

    Splitting the total bit time at a ratio of 70% : 30% makes

    7 time quants before and 3 after the sampling point. As

    3 time quants have already been used up by synchronization

    and propagation, there are 4 time quants left for the first

    phase segment and another 3 for the second one.

      Tpha1 = (PHSEG1+1) * Tq

      Tpha2 = (PHSEG2+1) * Tq

    Choosing 'PHSEG1' = 3 yields Tpha1 = 4 us and choosing

    'PHSEG2' = 2 yields Tpha2 = 3 us.

    Summarizing:

      Tq = 1 us

      Tbit =  Tsync + Tprop + Tpha1 + Tpha2

           = (  1   +   2   +   4   +   3  ) * Tq = 10 us

                                    ^

                             sampling point

  */

  can_write_register(CNF1,

    _BV(SJW0) |

    _BV(BRP1) | _BV(BRP0) /* 'BRP' = 3 */

  );

  can_write_register(CNF2,

    _BV(SAM) |

    _BV(BTLMODE) |

    _BV(PHSEG11) | _BV(PHSEG10) | /* 'PHSEG1' = 3 */

    _BV(PRSEG0) /* 'PRSEG' = 1 */

  );

  can_write_register(CNF3,

    _BV(PHSEG21) /* 'PHSEG2' = 2 */

  );

  /* Interrupts on reception into any of the two RX buffers and on

  completing transmission of any of the three TX buffers. */

  can_write_register(CANINTE,

    _BV(RX1IE) | _BV(RX0IE) |

    _BV(TX2IE) | _BV(TX1IE) | _BV(TX0IE)

  );

  /* Accept anything in first buffer */

  can_write_register(RXB0CTRL, _BV(RXM1) | _BV(RXM0));

  /* Leave configuration mode and speed up clock output. */

  can_write_register(CANCTRL, _BV(CLKEN));

}