CP3BT23 Datasheet, PDF (125/260 Pages) Texas Instruments – CP3BT23 Reprogrammable Connectivity Processor with Bluetooth and Dual CAN Interfaces

Table 58 CAN Module Internal Timing
Task
Cycle Occurrence/
Count
Frame
Table 59 gives examples for the minimum clock frequency in
order to ensure proper functionality at various CAN bus
speeds.
Table 59 Minimum Clock Frequency Requirements
Copy hidden buffer to receive
message buffer
17
0–1
Baud Rate
Minimum Clock
Frequency
Update status from TX_RTR
to TX_ONCE_RTR
3
0–15
1 Mbit/sec
15.25 MHz
Schedule a message for
transmission
2
0–1
The critical path derives from receiving a remote frame,
500 kbit/sec
250 kbit/sec
7.625 MHz
3.81 MHz
which triggers the transmission of one or more data frames. _{18.11.4 Bit Time Logic Calculation Examples}
There are a minimum of four bit times in-between two con-
secutive frames. These bit times start at the validation point
of received frame (reception of 6th EOF bit) and end at the
earliest possible transmission start of the next frame, which
is after the third intermission bit at 100% burst bus load.
The calculation of the CAN bus clocks using CKI = 16 MHz
is shown in the following examples. The desired baud rate
for both examples is 1 Mbit/s.
Example 1
These four bit times have to be set in perspective with the
timing requirements of the CAN module.
te ^{The minimum duration of the four CAN bit times is deter-}
mined by the following Bit Time Logic settings:
PSC = PSCmin = 2
TSEG1 = TSEG1min = 2
TSEG2 = TSEG2min = 1
le Bit time = Sync + Time Segment 1 + Time Segment 2
= (1 + 2 + 1) tq = 4 tq
= (4 tq × PSC) clock cycles
= (4 tq × 2) clock cycles = 8 clock cycles
PSC = PSC[5:0] + 2 = 0 + 2 = 2
TSEG1 = TSEG1[3:0] + 1 = 3 + 1 = 4
TSEG2 = TSEG2[2:0] + 1 = 2 + 1 = 3
SJW = TSEG2 = 3
! Sample point positioned at 62.5% of bit time
! Bit time = 125 ns × (1 + 4 + 3 ± 3) = (1 ± 0.375) µs
! Bus Clock = 16 MHz / (2 × (1 + 4 + 3)) = 1 Mbit/s (nomi-
nal)
Example 2
PSC = PSC[5:0] + 1 = 2 + 2 = 4
TSEG1 = TSEG1[3:0] + 1 = 1 + 1 = 2
For these minimum BTL settings, four CAN bit times take 32
o _{clock cycles.}
The following is an example that assumes typical case:
! Minimum BTL settings
s ! Reception and copy of a remote frame
! Update of one buffer from TX_RTR
! Schedule of one buffer from transmit
As outlined in Table 58, the copy process, update, and
b scheduling the next transmission gives a total of 17 + 3 + 2
= 22 clock cycles. Therefore under these conditions there is
no timing restriction.
The following example assumes the worst case:
O ^{! Minimum BTL settings}
TSEG2 = TSEG2[2:0] + 1 = 0 + 1 = 1
SJW = TSEG2 = 1
! Sample point positioned at 75% of bit time
! Bit time = 250 ns × (1 + 2 + 1 ± 1) = (1 ± 0.25) µs
! Bus Clock = 16 MHz / (2 × (1 + 4 + 3)) = 1Mbit/s (nominal)
18.11.5 Acceptance Filter Considerations
The CAN module provides two acceptance filter masks
GMSK and BMSK, as described in “Acceptance Filtering”
on page 102, “Global Mask Register (GMSKBn/GMSKXn)”
on page 119, and “Basic Mask Register (BMSKBn/BM-
SKXn)” on page 120. These masks allow filtering of up to 32
bits of the message object, which includes the standard
identifier, the extended identifier, and the frame control bits
! Reception and copy of a remote frame
RTR, SRR, and IDE.
! Update of the 14 remaining buffers from TX_RTR
! Schedule of one buffer for transmit
18.11.6 Remote Frames
Remote frames can be automatically processed by the CAN
All these actions in total require 17 + (14 × 3) + 2 = 61 clock _{module. However, to fully enable this feature, the RTR/}
cycles to be executed by the CAN module. This leads to the _{XRTR bits (for both standard and extended frames) within}
limitation of the Bit Time Logic of 61 / 4 = 15.25 clock cycles _{the BMSK and/or GMSK register need to be set to “don’t}
per CAN bit as a minimum, resulting in the minimum clock _{care”. This is because a remote frame with the RTR bit set}
frequencies listed below. (The frequency depends on the _{should trigger the transmission of a data frame with the RTR}
desired baud rate and assumes the worst case scenario _{bit clear and therefore the ID bits of the received message}
can occur in the application.)
need to pass through the acceptance filter. The same ap-
plies to transmitting remote frames and switching to receive
the corresponding data frames.
www.national.com
124