Datasheet

()
(
)
fnomdffoscfnomdf
×
+
≤
≤
×
− 11
(
)
()
2_132
min2_,1_
SegPhasetbit
segPhasesegPhase
df
−××
≤
f
no
m
d
fdf
×
×
=
2
max
()
(
)
fnomdffoscfnomdf
×
+
≤
≤
×
− 11
(
)
()
2_132
min2_,1_
SegPhasetbit
segPhasesegPhase
df
−××
≤
f
no
m
d
fdf
×
×
=
2
max
where:
■ Phase1 and Phase2 are from Table 23-3 on page 1549
■ tbit = Bit Time
■ dfmax = Maximum difference between two oscillators
If more than one configuration is possible, that configuration allowing the highest oscillator tolerance
range should be chosen.
CAN nodes with different system clocks require different configurations to come to the same bit
rate. The calculation of the propagation time in the CAN network, based on the nodes with the
longest delay times, is done once for the whole network.
The CAN system's oscillator tolerance range is limited by the node with the lowest tolerance range.
The calculation may show that bus length or bit rate have to be decreased or that the oscillator
frequencies' stability has to be increased in order to find a protocol-compliant configuration of the
CAN bit timing.
23.3.16.1 Example for Bit Timing at High Baud Rate
In this example, the frequency of CAN clock is 25 MHz, and the bit rate is 1 Mbps.
bit time = 1 µs = n * t
q
= 5 * t
q
t
q
= 200 ns
t
q
= (Baud rate Prescaler)/CAN Clock
Baud rate Prescaler = t
q
* CAN Clock
Baud rate Prescaler = 200E-9 * 25E6 = 5
tSync = 1 * t
q
= 200 ns \\fixed at 1 time quanta
delay of bus driver 50 ns
delay of receiver circuit 30 ns
delay of bus line (40m) 220 ns
tProp 400 ns = 2 * t
q
\\400 is next integer multiple of t
q
bit time = tSync + tTSeg1 + tTSeg2 = 5 * t
q
bit time = tSync + tProp + tPhase 1 + tPhase2
tPhase 1 + tPhase2 = bit time - tSync - tProp
tPhase 1 + tPhase2 = (5 * t
q
) - (1 * t
q
) - (2 * t
q
)
tPhase 1 + tPhase2 = 2 * t
q
tPhase1 = 1 * t
q
tPhase2 = 1 * t
q
\\tPhase2 = tPhase1
1551December 13, 2013
Texas Instruments-Advance Information
Tiva
™
TM4C129XNCZAD Microcontroller