SAM9M10 Atmel Corporation, SAM9M10 Datasheet - Page 541

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SAM9M10

Manufacturer Part Number
SAM9M10
Description
Manufacturer
Atmel Corporation
Datasheets

Specifications of SAM9M10

Flash (kbytes)
0 Kbytes
Pin Count
324
Max. Operating Frequency
400 MHz
Cpu
ARM926
Hardware Qtouch Acquisition
No
Max I/o Pins
160
Ext Interrupts
160
Usb Transceiver
3
Usb Speed
Hi-Speed
Usb Interface
Host, Device
Spi
2
Twi (i2c)
2
Uart
5
Lin
4
Ssc
2
Ethernet
1
Sd / Emmc
2
Graphic Lcd
Yes
Video Decoder
Yes
Camera Interface
Yes
Adc Channels
8
Adc Resolution (bits)
10
Adc Speed (ksps)
440
Resistive Touch Screen
Yes
Temp. Sensor
No
Crypto Engine
No
Sram (kbytes)
64
Self Program Memory
NO
External Bus Interface
2
Dram Memory
DDR2/LPDDR, SDRAM/LPSDR
Nand Interface
Yes
Picopower
No
Temp. Range (deg C)
-40 to 85
I/o Supply Class
1.8/3.3
Operating Voltage (vcc)
0.9 to 1.1
Fpu
No
Mpu / Mmu
No / Yes
Timers
6
Output Compare Channels
6
Input Capture Channels
6
Pwm Channels
4
32khz Rtc
Yes
Calibrated Rc Oscillator
No
31.7.8.25
6355D–ATARM–7-Sep-11
Wake-up Request
Any node in a sleeping LIN cluster may request a wake-up.
In the LIN 2.0 specification, the wakeup request is issued by forcing the bus to the dominant
state from 250 μs to 5 ms. For this, it is necessary to send the character 0xF0 in order to impose
5 successive dominant bits. Whatever the baud rate is, this character respects the specified
timings.
In the LIN 1.3 specification, the wakeup request should be generated with the character 0x80 in
order to impose 8 successive dominant bits.
The user can choose by the WKUPTYP bit in the LIN Mode register (US_LINMR) either to send
a LIN 2.0 wakeup request (WKUPTYP=0) or to send a LIN 1.3 wakeup request (WKUPTYP=1).
A wake-up request is transmitted by writing the Control Register (US_CR) with the LINWKUP bit
at 1. Once the transfer is completed, the LINTC flag is asserted in the Status Register (US_SR).
It is cleared by writing the Control Register (US_CR) with the RSTSTA bit at 1.
• Baud rate min = 1 kbit/s -> Tbit = 1ms -> 5 Tbits = 5 ms
• Baud rate max = 20 kbit/s -> Tbi t= 50 μs -> 5 Tbits = 250 μs
SAM9M10
541

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