AcroPack Series AP560
CAN Bus Interface Module
- 40 -
Phase Errors (e)
If a bit edge occurs within the Sync Seg as expected, there is no phase error
(e = 0). However, if an edge occurs outside Sync Seg, a phase error is deemed
to have occurred. If the edge occurs after Sync Seg (edge occurs “late”), the
phase error is positive (e > 0), whereas if the edge occurs before Sync Seg
(edge occurs “early”), the phase erro
r is negative (e < 0).
Synchronization
Synchronization is carried out only on recessive-to-dominant bit edges and is
used to ensure the bit times of all nodes on the bus are synchronized. This is
necessary for arbitration and message acknowledgment to function properly.
Only one synchronization can occur per bit time.
Hard synchronization
forces the bit edge to lie within the Sync Seg, regardless
of the phase error. Hard synchronization only occurs on reception of the start
of a frame.
Re-synchronization
results in the shortening or lengthening of the bit time
such that the position of the sample point is shifted with respect to the edge
causing the re-synchronization. For e > 0, Phase Seg 1 is lengthened by the
magnitude of the phase error, up to a maximum of SJW. For e < 0, Phase Seg 2
is shortened by the magnitude of the phase error, up to a maximum of SJW
.
Examples
1)
CAN bit rate (BR) = 50 kHz, f
OSC
= 40 MHz
Assume sample point (at end of TSeg1) will occur at 75% of bit time.
Hence, for Sync Seg = 1Tq, TSeg1 = 11 Tq and TSeg2 = 4Tq. Therefore,
total bit time will be 16Tq. Chose SJW = 1Tq. For 50 kHz, the bit time
needs to be 1/50 kHz = 20
μs.
Hence, 1Tq = 1.25
μ
s. Using
equation (1) => BRP = 25.
2)
CAN bit rate (BR) = 800 kHz, f
OSC
= 32 MHz
Assume sample point (at end of TSeg1) will occur at 75% of bit time.
Hence, for Sync Seg = 1 Tq, TSeg1 = 14 Tq and TSeg2 = 5 Tq.
Therefore, total bit time will be 20 Tq. Chose SJW = 1 Tq. For 800 kHz,
the bit time needs to be 1/800 kHz = 1.25
μs. Hence, 1
Tq = 62.5 ns.
Using equation (1) => BRP = 1.
3)
CAN bit rate (BR) = 1 MHz, f
OSC
= 40 MHz
Assume sample point (at end of TSeg1) will occur at 75% of bit time.
For Sync Seg = 1 Tq, then TSeg1 = 14 Tq and TSeg2 = 5 Tq. Therefore,
total bit time will be 20 Tq. Chose SJW = 1 Tq. For 1 MHz, the bit time
needs to be 1 / 1 MHz =
1μs. Hence, 1
Tq = 50 ns. Using
equation (1) => BRP = 1.
Note:
Choosing the sample point at 75% of the bit time is a requirement of
ARINC 825.
Summary of Contents for AcroPack AP560
Page 28: ...AcroPack Series AP560 CAN Bus Interface Module 28 Figure 2 Standard Data Frame...
Page 30: ...AcroPack Series AP560 CAN Bus Interface Module 30 Figure 3 Extended Data Frame...
Page 32: ...AcroPack Series AP560 CAN Bus Interface Module 32 Figure 4 Remote Frame...
Page 34: ...AcroPack Series AP560 CAN Bus Interface Module 34 Figure 5 Error Frame...
