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PIC18F4580-I/PT 参数 Datasheet PDF下载

PIC18F4580-I/PT图片预览
型号: PIC18F4580-I/PT
PDF下载: 下载PDF文件 查看货源
内容描述: 28 /40/ 44引脚增强型闪存微控制器与ECAN技术, 10位A / D和纳瓦技术 [28/40/44-Pin Enhanced Flash Microcontrollers with ECAN Technology, 10-Bit A/D and nanoWatt Technology]
分类和应用: 闪存微控制器和处理器外围集成电路时钟
文件页数/大小: 490 页 / 8912 K
品牌: MICROCHIP [ MICROCHIP ]
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PIC18F2480/2580/4480/4580  
The CAN protocol uses a bit-stuffing technique that  
inserts a bit of a given polarity following five bits with the  
opposite polarity. This gives a total of 10 bits transmit-  
ted without re-synchronization (compensation for jitter  
or phase error).  
24.9.1  
EXTERNAL CLOCK, INTERNAL  
CLOCK AND MEASURABLE JITTER  
IN HS-PLL BASED OSCILLATORS  
The microcontroller clock frequency generated from a  
PLL circuit is subject to a jitter, also defined as Phase  
Jitter or Phase Skew. For its PIC18 Enhanced micro-  
controllers, Microchip specifies phase jitter (Pjitter) as  
being 2% (Gaussian distribution, within 3 standard  
deviations, see parameter F13 in Table 28-7) and Total  
Given the random nature of the jitter error added, it can  
be shown that the total error caused by the jitter tends  
to cancel itself over time. For a period of 10 bits, it is  
necessary to add only two jitter intervals to correct for  
jitter-induced error: one interval in the beginning of the  
10-bit period and another at the end. The overall effect  
is shown in Figure 24-5.  
Jitter (Tjitter) as being 2*Pjitter  
.
FIGURE 24-5:  
EFFECTS OF PHASE JITTER ON THE MICROCONTROLLER CLOCK  
AND CAN BIT TIME  
Nominal Clock  
Clock with Jitter  
Phase Skew (Jitter)  
CAN Bit Jitter  
CAN Bit Time  
with Jitter  
Once these considerations are taken into account, it is  
possible to show that the relation between the jitter and  
the total frequency error can be defined as:  
For example, assume a CAN bit rate of 125 Kb/s, which  
gives an NBT of 8 µs. For a 16 MHz clock generated  
from a 4x PLL, the jitter at this clock frequency is:  
1
0.02  
16×106  
Tjitter  
2 × Pjitter  
2% × ------------------- = ----------------- = 1 . 2 5 n s  
Δf = -----------------------= -----------------------  
10 × NBT 10 × NBT  
16 MHz  
and resultant frequency error is:  
where jitter is expressed in terms of time and NBT is the  
Nominal Bit Time.  
2 × (1.25×10–9  
)
--------------------------------------= 3.125×10–5= 0.0031%  
10 × (8×10–6  
)
© 2009 Microchip Technology Inc.  
DS39637D-page 339  
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