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:
Since the TF0 flag is set whenever timer 0 overflows, the above code will toggle P3.0 every time timer 0
overflows. This accomplishes what we want, but is inefficient. The JNB
instruction consumes 2
instruction cycles to determine that the flag is not set and jump over the unnecessary code. In the
event that timer 0 overflows, the CPL and CLR instruction require 2 instruction cycles to execute. To
make the math easy, lets say the rest of the code in the program requires 98 instruction cycles. Thus,
in total, our code consumes 100 instruction cycles (98 instruction cycles plus the 2 that are executed
every iteration to determine whether or not timer 0 has overflowed). If were in 16-bit timer mode, timer
0 will overflow every 65,536 machine cycles. In that time we would have performed 655 JNB
for a total of 1310 instruction cycles, plus another 2 instruction cycles to perform the code. So to
achieve our goal weve spent 1312 instruction cycles. So 2.002% of our time is being spent just checking
when to toggle P3.0. And our code is ugly because we have to make that check every iteration of our main
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:
First, youll notice the CLR TF0 command has disappeared. Thats because when the 8051 executes our
"timer 0 interrupt routine," it automatically clears the TF0 flag. Youll also notice that instead of a
instruction we have a RETI
instruction. The RETI instruction does the same thing
as a RET instruction, but tells the 8051 that an interrupt routine has finished. You must always end
your interrupt handlers with RETI.
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.
What Events Can Trigger Interrupts, and where do they go?
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.
In other words, we can configure the 8051 so that when Timer 0 Overflows or when a character is sent/received,
the appropriate interrupt handler routines are called.
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|
|External 0||IE0 ||0003h|
|Timer 0 ||TF0 ||000Bh|
|External 1||IE1 ||0013h|
|Timer 1 ||TF1 ||001Bh|
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.
Setting Up Interrupts
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:
Both of the above instructions set bit 3 of IE, thus enabling Timer 1 Interrupt. Once Timer 1 Interrupt is
enabled, whenever the TF1 bit is set, the 8051 will automatically put "on hold" the main program and execute
the Timer 1 Interrupt Handler at address 001Bh.
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:
Thereafter, the Timer 1 Interrupt Handler at 01Bh will automatically be called whenever the TF1 bit is set (upon
Timer 1 overflow).
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
This means that if a Serial Interrupt occurs at the exact same instant that an External 0 Interrupt occurs,
the External 0 Interrupt will be executed first and the Serial Interrupt will be executed once the External
0 Interrupt has completed.
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.
What Happens When an Interrupt Occurs?
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.
Take special note of the third step: If the interrupt being handled is a Timer or External interrupt, the
microcontroller automatically clears the interrupt flag before passing control to your interrupt handler
routine. This means it is not necessary that you clear the bit in your code.
What Happens When an Interrupt Ends?
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|
As you can see, our code checks the status of both interrupts flags. If both flags were set, both sections of
code will be executed. Also note that each section of code clears its corresponding interrupt flag. If you forget
to clear the interrupt bits, the serial interrupt will be executed over and over until you clear the bit. Thus it
is very important that you always clear the interrupt flags in a serial interrupt.
Important Interrupt Consideration: Register Protection
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:
This is due to the fact that depending on which register bank is selected, R0 may refer to either internal ram address
00h, 08h, 10h, or 18h. R0, in and of itself, is not a valid memory address that the PUSH and POP instructions can use.
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:
Of course, this only works if youve selected the default register set. If you are using an alternate register set,
you must PUSH the address which corresponds to the register you are using.
Common Problems with Interrupts
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
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.