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ST9 数据表(PDF) 50 Page - STMicroelectronics |
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ST9 数据表(HTML) 50 Page - STMicroelectronics |
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50 / 146 page ![]() 50/146 ST9 USER GUIDE In Concurrent Mode, the CPL is set exclusively by the programmer. You can change it if you need to shift the compromise mentioned above either towards main program efficiency, or to- wards short interrupt latency times. So you can have a high-priority service routine that is re- interrupted by a low-priority interrupt request, if the CPL has been set to a low priority and the IEN bit is set. This mode gives you maximum flexibility, but it is the most difficult to use since you must keep track of every combination of interrupt requests to achieve the efficiency you expect from your program. Figure 19 . Example with Concurrent Interrupts Enabled Concurrent mode does not look like a reasonable way to handle interrupts if you enable inter- rupts in your interrupt service routine. It should be thought of as a way to fully control priorities through programming, if nesting priorities cannot meet your processing requirements. For ex- ample, let us consider the case when a timer produces a periodic interrupt that outputs some data on an external digital to analog converter. The requirement is that the new data be output at the very time of the interrupt, so as to reduce the jitter (or parasitic frequency modulation) that would compromise the spectral purity. Then, once the data is output, the interrupt service routine does some processing to make or get the data for the next interrupt. The latter part of the processing is much less critical in terms of execution time, provided it is finished before the next timer interrupt. Using Concurrent mode, you can assign that interrupt the highest priority so that it will be served immediately, then re-enable the interrupts and, if needed, give it a pri- 7 6 5 4 3 2 1 0 Main program Main Interrupt #5 Interrupt #2 Interrupt #6 Interrupt #2 Interrupt #5 Interrupt #5 is requested Interrupt #2 and #6 are requested. The Interrupt #2 is served first because it’s priority is higher than the interrupt #6 CPL is set to 7 CPL=7 CPL=7 CPL=7 CPL=7 CPL=7 CPL=7 ei ei Interrupt #4 CPL=7 ei Interrupt #4 CPL=7 Interrupt #4 is requested Interrupt #4 is served first because it’s priority is higher than the interrupt #6 Interrupt #2 has priority level 2 Interrupt #4 has priority level 4 Interrupt #5 has priority level 5 Interrupt #6 has priority level 6 Main prog has priority level 7 |
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