SAM9263 Atmel Corporation, SAM9263 Datasheet - Page 658

• Manufacturer Part Number: SAM9263
• Manufacturer: Atmel Corporation

Specifications of SAM9263

• Flash (kbytes): 0 Kbytes
• Pin Count: 324
• Max. Operating Frequency: 240 MHz
• Cpu: ARM926
• Hardware Qtouch Acquisition: No
• Max I/o Pins: 160
• Ext Interrupts: 160
• Usb Transceiver: 3
• Usb Speed: Full Speed
• Usb Interface: Host, Device
• Spi: 2
• Twi (i2c): 1
• Uart: 4
• Can: 1
• Ssc: 2
• Ethernet: 1
• Sd / Emmc: 2
• Graphic Lcd: Yes
• Video Decoder: No
• Camera Interface: Yes
• Resistive Touch Screen: No
• Temp. Sensor: No
• Crypto Engine: No
• Sram (kbytes): 96
• Self Program Memory: NO
• External Bus Interface: 2
• Dram Memory: sdram
• Nand Interface: Yes
• Picopower: No
• Temp. Range (deg C): -40 to 85
• I/o Supply Class: 1.8/3.3
• Fpu: No
• Mpu / Mmu: No / Yes
• Timers: 3
• Output Compare Channels: 3
• Input Capture Channels: 3
• Pwm Channels: 4
• 32khz Rtc: Yes
• Calibrated Rc Oscillator: No

37.6.4.2
6249I–ATARM–3-Oct-11
CAN Bus Synchronization
Two types of synchronization are distinguished: “hard synchronization” at the start of a frame
and “resynchronization” inside a frame. After a hard synchronization, the bit time is restarted
with the end of the SYNC_SEG segment, regardless of the phase error. Resynchronization
causes a reduction or increase in the bit time so that the position of the sample point is shifted
with respect to the detected edge.
The effect of resynchronization is the same as that of hard synchronization when the magnitude
of the phase error of the edge causing the resynchronization is less than or equal to the pro-
grammed value of the resynchronization jump width (t
When the magnitude of the phase error is larger than the resynchronization jump width and
• the phase error is positive, then PHASE_SEG1 is lengthened by an amount equal to the
• the phase error is negative, then PHASE_SEG2 is shortened by an amount equal to the
resynchronization jump width.
resynchronization jump width.
Delay of the bus driver: 50 ns
Delay of the receiver: 30ns
Delay of the bus line (20m): 110ns
The total number of time quanta in a bit time must be comprised between 8
and 25. If we fix the bit time to 16 time quanta:
Tcsc = 1 time quanta = bit time / 16 = 125 ns
=> BRP = (Tcsc x MCK) - 1 = 5
The propagation segment time is equal to twice the sum of the signal’s
propagation time on the bus line, the receiver delay and the output driver
delay:
Tprs = 2 * (50+30+110) ns = 380 ns = 3 Tcsc
=> PROPAG = Tprs/Tcsc - 1 = 2
The remaining time for the two phase segments is:
Tphs1 + Tphs2 = bit time - Tcsc - Tprs = (16 - 1 - 3)Tcsc
Tphs1 + Tphs2 = 12 Tcsc
Because this number is even, we choose Tphs2 = Tphs1 (else we would choose
Tphs2 = Tphs1 + Tcsc)
Tphs1 = Tphs2 = (12/2) Tcsc = 6 Tcsc
=> PHASE1 = PHASE2 = Tphs1/Tcsc - 1 = 5
The resynchronization jump width must be comprised between 1 Tcsc and the
minimum of 4 Tcsc and Tphs1. We choose its maximum value:
Tsjw = Min(4 Tcsc,Tphs1) = 4 Tcsc
=> SJW = Tsjw/Tcsc - 1 = 3
Finally: CAN_BR = 0x00053255
SJW
).
AT91SAM9263
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