Datasheet

609
SAM9G45 [DATASHEET]
Atmel-6438O-ATARM-SAM9G45-Datasheet_08-Dec-15
If the fractional baud rate is not used, the accuracy of the synchronization becomes much lower. When the counter
is stopped, the 16 most significant bits of the counter (value divided by 8) gives the new clock divider (CD). This
value is rounded by adding the first insignificant bit. The equation of the Baudrate deviation is the same as given
above, but the constants are as follows:
It follows from that, a minimum value for the nominal
clock frequency:
Examples:
 Baudrate = 20 kbit/s, OVER = 0 (Oversampling 16X) => f
Nom
(min) = 19.12 MHz
 Baudrate = 20 kbit/s, OVER = 1 (Oversampling 8X) => f
Nom
(min) = 9.71 MHz
 Baudrate = 1 kbit/s, OVER = 0 (Oversampling 16X) => f
Nom
(min) = 956 kHz
 Baudrate = 1 kbit/s, OVER = 1 (Oversampling 8X) => f
Nom
(min) = 485 kHz
32.7.8.8 Identifier Parity
A protected identifier consists of two sub-fields; the identifier and the identifier parity. Bits 0 to 5 are assigned to the
identifier and bits 6 and 7 are assigned to the parity.
The USART interface can generate/check these parity bits, but this feature can also be disabled. The user can
choose between two modes by the PARDIS bit of the LIN Mode register (US_LINMR):
 PARDIS = 0
 During header transmission, the parity bits are computed and sent with the 6 least significant bits of
the IDCHR field of the LIN Identifier register (US_LINIR). The bits 6 and 7 of this register are
discarded.
 During header reception, the parity bits of the identifier are checked. If the parity bits are wrong, an
Identifier Parity error occurs (see Section 32.7.3.9). Only the 6 least significant bits of the IDCHR field
are updated with the received Identifier. The bits 6 and 7 are stuck at 0.
 PARDIS = 1
 During header transmission, all the bits of the IDCHR field of the LIN Identifier register (US_LINIR) are
sent on the bus.
 During header reception, all the bits of the IDCHR field are updated with the received Identifier.
4– α +4 -1 β +1<<≤≤
f
Nom
(min) 100
4 8 2 Over–()×× 1+[]Baudrate×
8
15–
100
---------
1+


× 1%×
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×





Hz=