As the name implies, an interrupt is some event which interrupts normal program execution.
As stated earlier, program flow is always sequential, being altered only by those instructions which expressly cause program flow to deviate in some way. However, interrupts give us a mechanism to "put on hold" the normal program flow, execute a subroutine, and then resume normal program flow as if we had never left it. This subroutine, called an interrupt handler, is only executed when a certain event (interrupt) occurs. The event may be one of the timers "overflowing," receiving a character via the serial port, transmitting a character via the serial port, or one of two "external events." The 8051 may be configured so that when any of these events occur the main program is temporarily suspended and control passed to a special section of code which presumably would execute some function related to the event that occured. Once complete, control would be returned to the original program. The main program never even knows it was interrupted.
The ability to interrupt normal program execution when certain events occur makes it much easier and much more efficient to handle certain conditions. If it were not for interrupts we would have to manually check in our main program whether the timers had overflown, whether we had received another character via the serial port, or if some external event had occured. Besides making the main program ugly and hard to read, such a situation would make our program inefficient since wed be burning precious "instruction cycles" checking for events that usually dont happen.
For example, lets say we have a large 16k program executing many subroutines performing many tasks. Lets also suppose that we want our program to automatically toggle the P3.0 port every time timer 0 overflows. The code to do this isnt too difficult:
Luckily, this isnt necessary. Interrupts let us forget about checking for the condition. The microcontroller itself will check for the condition automatically and when the condition is met will jump to a subroutine (called an interrupt handler), execute the code, then return. In this case, our subroutine would be nothing more than:
Thus, every 65536 instruction cycles we execute the CPL instruction and the RETI instruction. Those two instructions together require 3 instruction cycles, and weve accomplished the same goal as the first example that required 1312 instruction cycles. As far as the toggling of P3.0 goes, our code is 437 times more efficient! Not to mention its much easier to read and understand because we dont have to remember to always check for the timer 0 flag in our main program. We just setup the interrupt and forget about it, secure in the knowledge that the 8051 will execute our code whenever its necessary.
The same idea applies to receiving data via the serial port. One way to do it is to continuously check the status of the RI flag in an endless loop. Or we could check the RI flag as part of a larger program loop. However, in the latter case we run the risk of missing characters--what happens if a character is received right after we do the check, the rest of our program executes, and before we even check RI a second character has come in. We will lose the first character. With interrupts, the 8051 will put the main program "on hold" and call our special routine to handle the reception of a character. Thus, we neither have to put an ugly check in our main code nor will we lose characters.
We can configure the 8051 so that any of the following events will cause an interrupt:
- Timer 0 Overflow.
- Timer 1 Overflow.
- Reception/Transmission of Serial Character.
- External Event 0.
- External Event 1.
Obviously we need to be able to distinguish between various interrupts and executing different code depending on what interrupt was triggered. This is accomplished by jumping to a fixed address when a given interrupt occurs.
|Interrupt||Flag||Interrupt Handler Address|
By consulting the above chart we see that whenever Timer 0 overflows (i.e., the TF0 bit is set), the main program will be temporarily suspended and control will jump to 000BH. It is assumed that we have code at address 000BH that handles the situation of Timer 0 overflowing.
By default at powerup, all interrupts are disabled. This means that even if, for example, the TF0 bit is set, the 8051 will not execute the interrupt. Your program must specifically tell the 8051 that it wishes to enable interrupts and specifically which interrupts it wishes to enable.
Your program may enable and disable interrupts by modifying the IE SFR (A8h):
|Bit||Name||Bit Address||Explanation of Function|
|7||EA||AFh||Global Interrupt Enable/Disable|
|4||ES||ACh||Enable Serial Interrupt|
|3||ET1||ABh||Enable Timer 1 Interrupt|
|2||EX1||AAh||Enable External 1 Interrupt|
|1||ET0||A9h||Enable Timer 0 Interrupt|
|0||EX0||A8h||Enable External 0 Interrupt|
As you can see, each of the 8051s interrupts has its own bit in the IE SFR. You enable a given interrupt by setting the corresponding bit. For example, if you wish to enable Timer 1 Interrupt, you would execute either:
However, before Timer 1 Interrupt (or any other interrupt) is truly enabled, you must also set bit 7 of IE. Bit 7, the Global Interupt Enable/Disable, enables or disables all interrupts simultaneously. That is to say, if bit 7 is cleared then no interrupts will occur, even if all the other bits of IE are set. Setting bit 7 will enable all the interrupts that have been selected by setting other bits in IE. This is useful in program execution if you have time-critical code that needs to execute. In this case, you may need the code to execute from start to finish without any interrupt getting in the way. To accomplish this you can simply clear bit 7 of IE (CLR EA) and then set it after your time-criticial code is done.
So, to sum up what has been stated in this section, to enable the Timer 1 Interrupt the most common approach is to execute the following two instructions:
The 8051 automatically evaluates whether an interrupt should occur after every instruction. When checking for interrupt conditions, it checks them in the following order:
- External 0 Interrupt
- Timer 0 Interrupt
- External 1 Interrupt
- Timer 1 Interrupt
- Serial Interrupt
The 8051 offers two levels of interrupt priority: high and low. By using interrupt priorities you may assign higher priority to certain interrupt conditions.
For example, you may have enabled Timer 1 Interrupt which is automatically called every time Timer 1 overflows. Additionally, you may have enabled the Serial Interrupt which is called every time a character is received via the serial port. However, you may consider that receiving a character is much more important than the timer interrupt. In this case, if Timer 1 Interrupt is already executing you may wish that the serial interrupt itself interrupts the Timer 1 Interrupt. When the serial interrupt is complete, control passes back to Timer 1 Interrupt and finally back to the main program. You may accomplish this by assigning a high priority to the Serial Interrupt and a low priority to the Timer 1 Interrupt.
Interrupt priorities are controlled by the IP SFR (B8h). The IP SFR has the following format:
|Bit||Name||Bit Address||Explanation of Function|
|4||PS||BCh||Serial Interrupt Priority|
|3||PT1||BBh||Timer 1 Interrupt Priority|
|2||PX1||BAh||External 1 Interrupt Priority|
|1||PT0||B9h||Timer 0 Interrupt Priority|
|0||PX0||B8h||External 0 Interrupt Priority|
When considering interrupt priorities, the following rules apply:
- Nothing can interrupt a high-priority interrupt--not even another high priority interrupt.
- A high-priority interrupt may interrupt a low-priority interrupt.
- A low-priority interrupt may only occur if no other interrupt is already executing.
- If two interrupts occur at the same time, the interrupt with higher priority will execute first. If both interrupts are of the same priority the interrupt which is serviced first by polling sequence will be executed first.
When an interrupt is triggered, the following actions are taken automatically by the microcontroller:
- The current Program Counter is saved on the stack, low-byte first.
- Interrupts of the same and lower priority are blocked.
- In the case of Timer and External interrupts, the corresponding interrupt flag is cleared.
- Program execution transfers to the corresponding interrupt handler vector address.
- The Interrupt Handler Routine executes.
An interrupt ends when your program executes the RETI (Return from Interrupt) instruction. When the RETI instruction is executed the following actions are taken by the microcontroller:
- Two bytes are popped off the stack into the Program Counter to restore normal program execution.
- Interrupt status is restored to its pre-interrupt status.
Serial Interrupts are slightly different than the rest of the interrupts. This is due to the fact that there are two interrupt flags: RI and TI. If either flag is set, a serial interrupt is triggered. As you will recall from the section on the serial port, the RI bit is set when a byte is received by the serial port and the TI bit is set when a byte has been sent.
This means that when your serial interrupt is executed, it may have been triggered because the RI flag was set or because the TI flag was set--or because both flags were set. Thus, your routine must check the status of these flags to determine what action is appropriate. Also, since the 8051 does not automatically clear the RI and TI flags you must clear these bits in your interrupt handler.
A brief code example is in order:
|INT_SERIAL:||JNB RI,CHECK_TI||;If the RI flag is not set, we jump to check TI|
|MOV A,SBUF||;If we got to this line, its because the RI bit *was* set|
|CLR RI||;Clear the RI bit after weve processed it|
|CHECK_TI:||JNB TI,EXIT_INT||;If the TI flag is not set, we jump to the exit point|
|CLR TI||;Clear the TI bit before we send another character|
|MOV SBUF,#A||;Send another character to the serial port|
One very important rule applies to all interrupt handlers: Interrupts must leave the processor in the same state as it was in when the interrupt initiated.
Remember, the idea behind interrupts is that the main program isnt aware that they are executing in the "background." However, consider the following code:
CLR C ;Clear carry
MOV A,#25h ;Load the accumulator with 25h
ADDC A,#10h ;Add 10h, with carry
After the above three instructions are executed, the accumulator will contain a value of 35h.
But what would happen if right after the MOV instruction an interrupt occured. During this interrupt, the carry bit was set and the value of the accumulator was changed to 40h. When the interrupt finished and control was passed back to the main program, the ADDC would add 10h to 40h, and additionally add an additional 1h because the carry bit is set. In this case, the accumulator will contain the value 51h at the end of execution.
In this case, the main program has seemingly calculated the wrong answer. How can 25h + 10h yield 51h as a result? It doesnt make sense. A programmer that was unfamiliar with interrupts would be convinced that the microcontroller was damaged in some way, provoking problems with mathematical calculations.
What has happened, in reality, is the interrupt did not protect the registers it used. Restated: An interrupt must leave the processor in the same state as it was in when the interrupt initiated.
What does this mean? It means if your interrupt uses the accumulator, it must insure that the value of the accumulator is the same at the end of the interrupt as it was at the beginning. This is generally accomplished with a PUSH and POP sequence. For example:
The guts of the interrupt is the MOV instruction and the ADD instruction. However, these two instructions modify the Accumulator (the MOV instruction) and also modify the value of the carry bit (the ADD instruction will cause the carry bit to be set). Since an interrupt routine must guarantee that the registers remain unchanged by the routine, the routine pushes the original values onto the stack using the PUSH instruction. It is then free to use the registers it protected to its hearts content. Once the interrupt has finished its task, it pops the original values back into the registers. When the interrupt exits, the main program will never know the difference because the registers are exactly the same as they were before the interrupt executed.
In general, your interrupt routine must protect the following registers:
- DPTR (DPH/DPL)
- Registers R0-R7
Remember that PSW consists of many individual bits that are set by various 8051 instructions. Unless you are absolutely sure of what you are doing and have a complete understanding of what instructions set what bits, it is generally a good idea to always protect PSW by pushing and popping it off the stack at the beginning and end of your interrupts.
Note also that most assemblers (in fact, ALL assemblers that I know of) will not allow you to execute the instruction:
- PUSH R0
Thus, if you are using any "R" register in your interrupt routine, you will have to push that registers absolute address onto the stack instead of just saying PUSH R0. For example, instead of PUSH R0 you would execute:
Interrupts are a very powerful tool available to the 8051 developer, but when used incorrectly they can be a source of a huge number of debugging hours. Errors in interrupt routines are often very difficult to diagnose and correct.
If you are using interrupts and your program is crashing or does not seem to be performing as you would expect, always review the following interrupt-related issues:
- Register Protection: Make sure you are protecting all your registers, as explained above. If you forget
to protect a register that your main program is using, very strange results may occur. In our example above we saw
how failure to protect registers caused the main program to apparently calculate that 25h + 10h = 51h. If you witness
problems with registers changing values unexpectedly or operations producing "incorrect" values, it is very likely that
you've forgotten to protect registers. ALWAYS PROTECT YOUR REGISTERS.
- Forgetting to restore protected values: Another common error is to push registers onto the stack to
protect them, and then forget to pop them off the stack before exiting the interrupt. For example, you may push ACC,
B, and PSW onto the stack in order to protect them and subsequently pop only ACC and PSW off the stack before exiting.
In this case, since you forgot to restore the value of "B", an extra value remains on the stack. When you execute the
RETI instruction the 8051 will use that value as the return address instead of the correct value. In this case, your
program will almost certainly crash. ALWAYS MAKE SURE YOU POP THE SAME NUMBER OF VALUES OFF THE STACK AS YOU
PUSHED ONTO IT.
- Using RET instead of RETI: Remember that interrupts are always terminated with the RETI instruction. It is
easy to inadvertantly use the RET instruction instead. However, the RET instruction will not end your interrupt.
Usually, using a RET instead of a RETI will cause the illusion of your main program running normally, but your
interrupt will only be executed once. If it appears that your interrupt mysteriously stops executing, verify that
you are exiting with RETI.
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