ST92T163 ST Microelectronics, ST92T163 Datasheet - Page 51

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ST92T163

Manufacturer Part Number
ST92T163
Description
8/16-BIT FULL SPEED USB MCU FOR COMPOSITE DEVICES WITH 16 ENDPOINTS
Manufacturer
ST Microelectronics
Datasheet

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ARBITRATION MODES (Cont’d)
Example 2
In the second example, (more complex, Figure
23), each interrupt service routine sets Interrupt
Enable with the ei instruction at the beginning of
the routine. Placed here, it minimizes response
time for requests with a higher priority than the one
being serviced.
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-
Figure 23. Complex Example of a Sequence of Interrupt Requests with:
- Concurrent mode selected
- IEN set to 1 during interrupt service routine execution
0
1
2
3
4
5
6
7
CPL is set to 7
Priority Level of
Interrupt Request
MAIN
INT 5
ei
CPL = 7
INT 5
INT 2
INT 3
INT 4
ei
CPL = 7
INT 2
ei
CPL = 7
INT 3
ei
ei
CPL = 7
INT 4
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.
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 iret of 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.
CPL = 7
INT 3
INTERRUP T 2 HAS PRIORITY LEVEL 2
INTERRUP T 3 HAS PRIORITY LEVEL 3
INTERRUP T 4 HAS PRIORITY LEVEL 4
INTERRUP T 5 HAS PRIORITY LEVEL 5
CPL = 7
INT 2
ST92163 - INTERRUPTS
CPL = 7
INT 5
CPL = 7
MAIN
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