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AU5790D14-T PDF预览

AU5790D14-T

更新时间: 2024-01-29 02:21:58
品牌 Logo 应用领域
飞利浦 - PHILIPS 电信光电二极管电信集成电路
页数 文件大小 规格书
25页 127K
描述
CAN Transceiver, 1-Trnsvr, PDSO14,

AU5790D14-T 技术参数

是否Rohs认证: 符合生命周期:Transferred
包装说明:SOP, SOP14,.25Reach Compliance Code:unknown
风险等级:5.81Is Samacsys:N
数据速率:100 MbpsJESD-30 代码:R-PDSO-G14
JESD-609代码:e4端子数量:14
收发器数量:1最高工作温度:125 °C
最低工作温度:-40 °C封装主体材料:PLASTIC/EPOXY
封装代码:SOP封装等效代码:SOP14,.25
封装形状:RECTANGULAR封装形式:SMALL OUTLINE
电源:12 V认证状态:Not Qualified
子类别:Network Interfaces最大压摆率:85 mA
标称供电电压:12 V表面贴装:YES
电信集成电路类型:CAN TRANSCEIVER温度等级:AUTOMOTIVE
端子面层:Nickel/Palladium/Gold (Ni/Pd/Au)端子形式:GULL WING
端子节距:1.27 mm端子位置:DUAL
Base Number Matches:1

AU5790D14-T 数据手册

 浏览型号AU5790D14-T的Datasheet PDF文件第1页浏览型号AU5790D14-T的Datasheet PDF文件第2页浏览型号AU5790D14-T的Datasheet PDF文件第3页浏览型号AU5790D14-T的Datasheet PDF文件第5页浏览型号AU5790D14-T的Datasheet PDF文件第6页浏览型号AU5790D14-T的Datasheet PDF文件第7页 
Philips Semiconductors  
Application note  
AU5790 Single wire CAN transceiver  
AN2005  
Within CAN each node must synchronize to each other’s message on the first recessive to dominant edge of the message and all the other  
recessive to dominant edges in the message waveform. Because each node has its own clock reference, the oscillator tolerance, f, will affect  
the bit time and the sample time, so f has big impact on the synchronization. Meanwhile, CAN supports arbitration and in-frame  
acknowledgment, which means after sending out a data bit the transceiver needs to read back the bus level, so the propagation delay between  
nodes in the network must be limited to guarantee synchronization.  
The propagation delay from node A to node B includes all the device delays in the transmission path from A to B, CAN controller A delay time,  
transceiver A transmit delay, transceiver B receive delay, and bus line delay, etc. Since all nodes must receive each other’s signal, and  
synchronize to it, then send them back during arbitration, the total propagation delay in the network should be the round trip delay.  
Dietmayer and Overberg analyzed CAN bit timing requirements in detail in their SAE technical paper #970295[1]. By summarizing their analysis,  
we can find that in order to guarantee CAN bit time requirement, the total propagation delay has to satisfy following equations:  
t
(max) < t – t  
– ( 25t – t  
)* f (1)  
seg2  
prop  
bit  
seg2  
bit  
t
(max) < t – t  
– ( 25t – t  
)* f + t  
(min)/2 – t (1- f) (2)  
prop Q  
prop  
bit  
seg2  
bit  
seg2  
The requirement on Equation (1) is more severe than that on Equation (2) if the minimum propagation delay is larger than 2* t .  
Q
2.1.2 Arbitration  
If no device is transmitting a message, the network bus is in a recessive state, and any device may start to transmit a message. If more than  
one device starts to transmit a message at the same time, only one device gets bus access successfully by bit arbitration using the identifier.  
All devices on the bus are connected to the bus in a wired OR configuration. During arbitration, every device compares the read-back bus level  
with the transmitted data level. If these levels are the same, the transmission continues. If a device sends a recessive level, and reads back a  
dominant level, it has lost arbitration and has to stop sending any more bits, and becomes a receiver.  
The following figure shows an arbitration example. Node 1, 2, and 3 start to send out message at the same time. At bit ID-23, node 2 sends a  
recessive level, but the readback bus level is dominant, thus node 2 loses arbitration and becomes a receiver. Node 1 loses its arbitration at bit  
ID-20. Node 3 finally wins bus access and continues message transmission.  
ARBITRATION  
CONTROL  
DATA  
ID 28 ... 18  
R
T
R
S
O
F
RECEIVE  
*
NODE 1  
RECEIVE  
NODE 2  
*
NODE 3  
DOMINANT  
RECESSIVE  
BUS-LEVEL  
= ARBITRATION LOSS  
BITS NOT SENT  
*
SL01259  
Figure 3. CAN bus arbitration  
3
2001 Apr 16  

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