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ST90135M6 参数 Datasheet PDF下载

ST90135M6图片预览
型号: ST90135M6
PDF下载: 下载PDF文件 查看货源
内容描述: 8月16号位微控制器( MCU ), 16至64K的ROM 。 OTP或EPROM 。 512 2K的RAM - ST9 +系列\n [8/16-BIT MICROCONTROLLER (MCU) WITH 16 TO 64K ROM. OTP OR EPROM. 512 TO 2K RAM - ST9 + FAMILY ]
分类和应用: 微控制器可编程只读存储器电动程控只读存储器
文件页数/大小: 199 页 / 2805 K
品牌: ETC [ ETC ]
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ST90158 - INTERRUPTS  
ARBITRATION MODES (Cont’d)  
Example 2  
In the second example, (more complex, Figure  
22), each interrupt service routine sets Interrupt  
Enable with the eiinstruction at the beginning of  
the routine. Placed here, it minimizes response  
time for requests with a higher priority than the one  
being serviced.  
ice routines being executed in the opposite order  
of their priority.  
It is therefore recommended to avoid inserting  
the ei instruction in the interrupt service rou-  
tine in Concurrent mode. Use the ei instruc-  
tion only in nested mode.  
The level 2 interrupt routine (with the highest prior-  
ity) will be acknowledged first, then, when the ei  
instruction is executed, it will be interrupted by the  
level 3 interrupt routine, which itself will be inter-  
rupted by the level 4 interrupt routine. When the  
level 4 interrupt routine is completed, the level 3 in-  
terrupt routine resumes and finally the level 2 inter-  
rupt routine. This results in the three interrupt serv-  
WARNING: If, in Concurrent Mode, interrupts are  
nested (by executing ei in an interrupt service  
routine), make sure that either ENCSR is set or  
CSR=ISR, otherwise the iretof the innermost in-  
terrupt will make the CPU use CSR instead of ISR  
before the outermost interrupt service routine is  
terminated, thus making the outermost routine fail.  
Figure 22. Complex Example of a Sequence of Interrupt Requests with:  
- Concurrent mode selected  
- IEN set to 1 during interrupt service routine execution  
0
Priority Level of  
Interrupt Request  
INTERRUPT 2 HAS PRIORITY LEVEL 2  
INTERRUPT 3 HAS PRIORITY LEVEL 3  
INTERRUPT 4 HAS PRIORITY LEVEL 4  
INTERRUPT 5 HAS PRIORITY LEVEL 5  
1
2
3
4
5
6
7
INT 2  
INT 2  
CPL = 7  
CPL = 7  
INT 3  
CPL = 7  
INT 3  
CPL = 7  
ei  
INT 2  
INT 3  
INT 4  
ei  
ei  
INT 4  
CPL = 7  
INT 5  
INT 5  
CPL = 7  
CPL = 7  
ei  
ei  
INT 5  
MAIN  
MAIN  
CPL is set to 7  
CPL = 7  
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