Datasheet

452
32142D–06/2013
ATUC64/128/256L3/4U
The oversampling mode (OVER=0 => 16x, or OVER=1 => 8x)
The following formula is used to calculate synchronization deviation, where F
SLAVE
is the real
slave node clock frequency, and F
TOL_UNSYNC
is the difference between F
Nom
and F
SLAVE
Accord-
ing to the LIN specification, F
TOL_UNSYNCH
may not exceed ±15%, and the bit rates between two
nodes must be within ±2% of each other, resulting in a maximal BaudRate_deviation of ±1%.
Minimum nominal clock frequency with a fractional part:
Examples:
Baud rate = 20 kbit/s, OVER=0 (Oversampling 16x) => F
Nom
(min) = 2.64 MHz
Baud rate = 20 kbit/s, OVER=1 (Oversampling 8x) => F
Nom
(min) = 1.47 MHz
Baud rate = 1 kbit/s, OVER=0 (Oversampling 16x) => F
Nom
(min) = 132 kHz
Baud rate = 1 kbit/s, OVER=1 (Oversampling 8x) => F
Nom
(min) = 74 kHz
If the fractional part is not used, the synchronization accuracy is much lower. The 16 most signif-
icant bits, added with the first least significant bit, becomes the new clock divider (CD). The
equation of the baud rate deviation is the same as above, but the constants are:
Minimum nominal clock frequency without a fractional part:
Examples:
Baud rate = 20 kbit/s, OVER=0 (Oversampling 16x) => F
Nom
(min) = 19.12 MHz
Baud rate = 20 kbit/s, OVER=1 (Oversampling 8x) => F
Nom
(min) = 9.71 MHz
Baud rate = 1 kbit/s, OVER=0 (Oversampling 16x) => F
Nom
(min) = 956 kHz
Baud rate = 1 kbit/s, OVER=1 (Oversampling 8x) => F
Nom
(min) = 485 kHz
20.6.5.8 Identifier Parity
An identifier field consists of two sub-fields; the identifier and its parity. Bits 0 to 5 are assigned
to the identifier, while bits 6 and 7 are assigned to parity. Automatic parity management is dis-
abled by writing a one to the Parity Disable bit in the LIN Mode register (LINMR.PARDIS).
BaudRate_deviation 100
8 2 OVER+ BaudRate
8F
SLAVE
---------------------------------------------------------------------------------------------------


%=
BaudRate_deviation 100
8 2 OVER+ BaudRate
8
F
TOL_UNSYNC
100
----------------------------------- -


xF
Nom
---------------------------------------------------------------------------------------------------





%=
0.5 +0.5 -1 +1
F
Nom
min 100
0.5 8 2 OVER 1+BaudRate
8
15
100
--------- - 1+


1%
------------------------------------------------------------------------------------------------------ -





Hz=
4 +4 -1 +1
F
Nom
min 100
48 2OVER 1+Baudrate
8
15
100
--------- - 1+


1%
-----------------------------------------------------------------------------------------------





Hz=