AN2575 Freescale Semiconductor / Motorola, AN2575 Datasheet - Page 13

• Manufacturer Part Number: AN2575
• Description: MC68HC908EY16 ESCI LIN Drivers
• Manufacturer: Freescale Semiconductor / Motorola

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References
MOTOROLA
[1] LIN Protocol Specification, Version 1.3, 12 December 2002.
[2] AN2498/D, Initial trimming of the MC68HC908 ICG.
[3] MC68HC908EY16/D, MC68HC908EY16 Technical Data Sheet.
[4] AN2264/D, LIN Node Temperature Display.
[5] AN2344/D, HC908EY16 EMI Radiated Emissions Results.
[6] AN2573/D, LINkits LIN Evaluation Boards.
1. In the current implementation, FD reverts back to 4 before the reception
2. To incorporate other bus frequencies and baud rates, leave BPD at 1
3. If the methods described in 1 and 2 above are both used, FD should be
4. Avoid uncontrolled overflow of the arbiter counter. Although the overflow
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of each message. As an alternative, it would be possible to retain the
adjusted value of FD and change it only when required (e.g., due to
temperature change causing a drift in the ICG frequency). To do this
without risk of getting stuck on an incorrect value, the range of FD should
be limited to ±14% of its nominal value (i.e., if the nominal is 4., the
acceptable FD range is between 3
and select the most appropriate values of BD and nominal FD to suit the
required combination.
Let R = bus clock ÷ (64 × baud rate), divide R by 2
Then BD = N and FD
Calculate x = (R ÷ 2
FD
When the arbiter count is being transferred to SCPCS, it should be
shifted to the right N times.
limited to the following values:
SCPSC
SCPSC
bit can be used as a tenth bit, it is safer to limit the maximum acceptable
count to 511 (nine bits) and leave the overflow bit for use as an indication
of an error. The maximum count = 1.1 × nominal f
9600 baud, this gives a count of 458 at a bus clock of 8 MHz. This is
acceptable, but is near the limit. Therefore, the use of lower baud rates
necessarily implies the use of lower bus speeds.
MC68HC908EY16 ESCI LIN Drivers
nom
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= PS + 1 = 2
min
max
= INT (0.86 × (SCPSC
= INT (1.14 × (SCPSC
N
x
nom
/
– 2) ÷ 32 to nearest integer to get
32
= R ÷ 2
N
nom
nom
14
/
32
+ 32) – 32)
+ 32) – 32)
and 4
18
/
Bus
32
N
).
so that 2<R÷2
÷ 2 × baud rate. At
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AN2575/D
N
<4
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