One of the 8051s many powerful features is its integrated UART, otherwise known as a serial port. The fact that the 8051 has an integrated serial port means that you may very easily read and write values to the serial port. If it were not for the integrated serial port, writing a byte to a serial line would be a rather tedious process requring turning on and off one of the I/O lines in rapid succession to properly "clock out" each individual bit, including start bits, stop bits, and parity bits.
However, we do not have to do this. Instead, we simply need to configure the serial ports operation mode and baud rate. Once configured, all we have to do is write to an SFR to write a value to the serial port or read the same SFR to read a value from the serial port. The 8051 will automatically let us know when it has finished sending the character we wrote and will also let us know whenever it has received a byte so that we can process it. We do not have to worry about transmission at the bit level--which saves us quite a bit of coding and processing time.
The first thing we must do when using the 8051s integrated serial port is, obviously, configure it. This lets us tell the 8051 how many data bits we want, the baud rate we will be using, and how the baud rate will be determined.
First, lets present the "Serial Control" (SCON) SFR and define what each bit of the SFR represents:
Bit | Name | Bit Addres | Explanation of Function |
7 | SM0 | 9Fh | Serial port mode bit 0 |
6 | SM1 | 9Eh | Serial port mode bit 1. |
5 | SM2 | 9Dh | Mutliprocessor Communications Enable (explained later) |
4 | REN | 9Ch | Receiver Enable. This bit must be set in order to receive characters. |
3 | TB8 | 9Bh | Transmit bit 8. The 9th bit to transmit in mode 2 and 3. |
2 | RB8 | 9Ah | Receive bit 8. The 9th bit received in mode 2 and 3. |
1 | TI | 99h | Transmit Flag. Set when a byte has been completely transmitted. |
0 | RI | 98h | Receive Flag. Set when a byte has been completely received. |
Additionally, it is necessary to define the function of SM0 and SM1 by an additional table:
SM0 | SM1 | Serial Mode | Explanation | Baud Rate |
0 | 0 | 0 | 8-bit Shift Register | Oscillator / 12 |
0 | 1 | 1 | 8-bit UART | Set by Timer 1 (*) |
1 | 0 | 2 | 9-bit UART | Oscillator / 64 (*) |
1 | 1 | 3 | 9-bit UART | Set by Timer 1 (*) |
(*) Note: The baud rate indicated in this table is doubled if PCON.7 (SMOD) is set.
The SCON SFR allows us to configure the Serial Port. Thus, well go through each bit and review its function.
The first four bits (bits 4 through 7) are configuration bits.
Bits SM0 and SM1 let us set the serial mode to a value between 0 and 3, inclusive. The four modes are defined in the chart immediately above. As you can see, selecting the Serial Mode selects the mode of operation (8-bit/9-bit, UART or Shift Register) and also determines how the baud rate will be calculated. In modes 0 and 2 the baud rate is fixed based on the oscillators frequency. In modes 1 and 3 the baud rate is variable based on how often Timer 1 overflows. Well talk more about the various Serial Modes in a moment.
The next bit, SM2, is a flag for "Multiprocessor communication." Generally, whenever a byte has been received the 8051 will set the "RI" (Receive Interrupt) flag. This lets the program know that a byte has been received and that it needs to be processed. However, when SM2 is set the "RI" flag will only be triggered if the 9th bit received was a "1". That is to say, if SM2 is set and a byte is received whose 9th bit is clear, the RI flag will never be set. This can be useful in certain advanced serial applications. For now it is safe to say that you will almost always want to clear this bit so that the flag is set upon reception of any character.
The next bit, REN, is "Receiver Enable." This bit is very straightforward: If you want to receive data via the serial port, set this bit. You will almost always want to set this bit.
The last four bits (bits 0 through 3) are operational bits. They are used when actually sending and receiving data--they are not used to configure the serial port.
The TB8 bit is used in modes 2 and 3. In modes 2 and 3, a total of nine data bits are transmitted. The first 8 data bits are the 8 bits of the main value, and the ninth bit is taken from TB8. If TB8 is set and a value is written to the serial port, the datas bits will be written to the serial line followed by a "set" ninth bit. If TB8 is clear the ninth bit will be "clear."
The RB8 also operates in modes 2 and 3 and functions essentially the same way as TB8, but on the reception side. When a byte is received in modes 2 or 3, a total of nine bits are received. In this case, the first eight bits received are the data of the serial byte received and the value of the ninth bit received will be placed in RB8.
TI means "Transmit Interrupt." When a program writes a value to the serial port, a certain amount of time will pass before the individual bits of the byte are "clocked out" the serial port. If the program were to write another byte to the serial port before the first byte was completely output, the data being sent would be garbled. Thus, the 8051 lets the program know that it has "clocked out" the last byte by setting the TI bit. When the TI bit is set, the program may assume that the serial port is "free" and ready to send the next byte.
Finally, the RI bit means "Receive Interrupt." It funcions similarly to the "TI" bit, but it indicates that a byte has been received. That is to say, whenever the 8051 has received a complete byte it will trigger the RI bit to let the program know that it needs to read the value quickly, before another byte is read.
Once the Serial Port Mode has been configured, as explained above, the program must configure the serial ports baud rate. This only applies to Serial Port modes 1 and 3. The Baud Rate is determined based on the oscillators frequency when in mode 0 and 2. In mode 0, the baud rate is always the oscillator frequency divided by 12. This means if youre crystal is 11.059Mhz, mode 0 baud rate will always be 921,583 baud. In mode 2 the baud rate is always the oscillator frequency divided by 64, so a 11.059Mhz crystal speed will yield a baud rate of 172,797.
In modes 1 and 3, the baud rate is determined by how frequently timer 1 overflows. The more frequently timer 1 overflows, the higher the baud rate. There are many ways one can cause timer 1 to overflow at a rate that determines a baud rate, but the most common method is to put timer 1 in 8-bit auto-reload mode (timer mode 2) and set a reload value (TH1) that causes Timer 1 to overflow at a frequency appropriate to generate a baud rate.
To determine the value that must be placed in TH1 to generate a given baud rate, we may use the following equation (assuming PCON.7 is clear).
- TH1 = 256 - ((Crystal / 384) / Baud)
If PCON.7 is set then the baud rate is effectively doubled, thus the equation becomes:
- TH1 = 256 - ((Crystal / 192) / Baud)
For example, if we have an 11.059Mhz crystal and we want to configure the serial port to 19,200 baud we try plugging it in the first equation:
- TH1 = 256 - ((Crystal / 384) / Baud) TH1 = 256 - ((11059000 / 384) / 19200 ) TH1 = 256 - ((28,799) / 19200) TH1 = 256 - 1.5 = 254.5
As you can see, to obtain 19,200 baud on a 11.059Mhz crystal wed have to set TH1 to 254.5. If we set it to 254 we will have achieved 14,400 baud and if we set it to 255 we will have achieved 28,800 baud. Thus were stuck...
But not quite... to achieve 19,200 baud we simply need to set PCON.7 (SMOD). When we do this we double the baud rate and utilize the second equation mentioned above. Thus we have:
- TH1 = 256 - ((Crystal / 192) / Baud) TH1 = 256 - ((11059000 / 192) / 19200) TH1 = 256 - ((57699) / 19200) TH1 = 256 - 3 = 253
Here we are able to calculate a nice, even TH1 value. Therefore, to obtain 19,200 baud with an 11.059MHz crystal we must:
- 1. Configure Serial Port mode 1 or 3. 2. Configure Timer 1 to timer mode 2 (8-bit auto-reload). 3. Set TH1 to 253 to reflect the correct frequency for 19,200 baud. 4. Set PCON.7 (SMOD) to double the baud rate.
Once the Serial Port has been propertly configured as explained above, the serial port is ready to be used to send data and receive data. If you thought that configuring the serial port was simple, using the serial port will be a breeze.
To write a byte to the serial port one must simply write the value to the SBUF (99h) SFR. For example, if you wanted to send the letter "A" to the serial port, it could be accomplished as easily as:
- MOV SBUF,#A
Upon execution of the above instruction the 8051 will begin transmitting the character via the serial port. Obviously transmission is not instantaneous--it takes a measureable amount of time to transmit. And since the 8051 does not have a serial output buffer we need to be sure that a character is completely transmitted before we try to transmit the next character.
The 8051 lets us know when it is done transmitting a character by setting the TI bit in SCON. When this bit is set we know that the last character has been transmitted and that we may send the next character, if any. Consider the following code segment:
- CLR TI ;Be sure the bit is initially clear MOV SBUF,#A ;Send the letter A to the serial port JNB TI,$ ;Pause until the TI bit is set.
The above three instructions will successfully transmit a character and wait for the TI bit to be set before continuing. The last instruction says "Jump if the TI bit is not set to $"--$, in most assemblers, means "the same address of the current instruction." Thus the 8051 will pause on the JNB instruction until the TI bit is set by the 8051 upon successful transmission of the character.
Reading data received by the serial port is equally easy. To read a byte from the serial port one just needs to read the value stored in the SBUF (99h) SFR after the 8051 has automatically set the RI flag in SCON.
For example, if your program wants to wait for a character to be received and subsequently read it into the Accumulator, the following code segment may be used:
- JNB RI,$ ;Wait for the 8051 to set the RI flag MOV A,SBUF ;Read the character from the serial port
The first line of the above code segment waits for the 8051 to set the RI flag; again, the 8051 sets the RI flag automatically when it receives a character via the serial port. So as long as the bit is not set the program repeats the "JNB" instruction continuously.
Once the RI bit is set upon character reception the above condition automatically fails and program flow falls through to the "MOV" instruction which reads the value.
8051 Microcontroller Interrupts
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:
- JNB TF0,SKIP_TOGGLE CPL P3.0 CLR TF0 SKIP_TOGGLE: ...
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 tests 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 program loop.
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:
- CPL P3.0 RETI
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 normal RET 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.
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 |
Serial | RI/TI | 0023h |
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 |
6 | - | AEh | Undefined |
5 | - | ADh | Undefined |
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:
- MOV IE,#08h
- or
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:
- SETB ET1 SETB EA
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 |
7 | - | - | Undefined |
6 | - | - | Undefined |
5 | - | - | Undefined |
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.
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.
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 | |
EXIT_INT: | RETI |
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.
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:
- PUSH ACC PUSH PSW MOV A,#0FFh ADD A,#02h POP PSW POP ACC
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:
- PSW
- DPTR (DPH/DPL)
- PSW
- ACC
- B
- 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
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:
- PUSH 00h
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.
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.
- 256 bytes of Internal RAM (compared to 128 in the standard 8051).
- A third 16-bit timer, capable of a number of new operation modes and 16-bit reloads.
- Additional SFRs to support the functionality offered by the third timer.
The 8052's additional Internal RAM may only be referred by Indirect Addressing. Indirect addressing always refers to Internal RAM, never to an SFR.
Thus, to read the value contained in Internal RAM address 90h, the developer would need to code something along the lines of the following:
In addition to the 8051's standard 21 SFRs, the 8052 adds an additional 5 SFRs related to the 8052's third timer. All of the original 8051 SFRs function exactly as they do in the 8051-the 8052 simply adds new SFRs, it doesn't change the definition of the standard SFRs.
The five new SFRs are in the range of C8h to CDh (SFR C9h is not defined).
BIT | NAME | BIT ADDRESS | DESCRIPTION |
7 | TF2 | CFh | Timer 2 Overflow. This bit is set when T2 overflows. When T2 interrupt is enabled, this bit will cause the interrupt to be triggered. This bit will not be set if either TCLK or RCLK bits are set. |
6 | EXF2 | CEh | Timer 2 External Flag. Set by a reload or capture caused by a 1-0 transition on T2EX (P1.1), but only when EXEN2 is set. When T2 interrupt is enabled, this bit will cause the interrupt to be triggered. |
5 | RCLK | CDh | Timer 2 Receive Clock. When this bit is set, Timer 2 will be used to determine the serial port receive baud rate. When clear, Timer 1 will be used. |
4 | TCLK | CCh | Timer 2 Receive Clock. When this bit is set, Timer 2 will be used to determine the serial port transmit baud rate. When clear, Timer 1 will be used. |
3 | EXEN2 | CBh | Timer 2 External Enable. When set, a 1-0 transition on T2EX (P1.1) will cause a capture or reload to occur. |
2 | TR2 | CAh | Timer 2 Run. When set, timer 2 will be turned on. Otherwise, it is turned off. |
1 | C/T2 | C9h | Timer 2 Counter/Interval Timer. If clear, Timer 2 is an interval counter. If set, Timer 2 is incremented by 1-0 transition on T2 (P1.0). |
0 | CP/RL2 | C8h | Timer 2 Capture/Reload. If clear, auto reload occurs on timer 2 overflow, or T2EX 1-0 transition if EXEN2 is set. If set, a capture will occur on a 1-0 transition of T2EX if EXEN2 is set. |
With the standard 8051, Timer 1 is the only timer which may be used to determine the baud rate of the serial port. Additionally, the receive and transmit baud rate must be the same.
With the 8052, however, the user may configure the serial port to receive at one baud rate and transmit with another. For example, if RCLK is set and TCLK is cleared, serial data will be received at the baud rate determined by Timer 2 whereas the baud rate of transmitted data will be determined by Timer 1.
Determining the auto-reload values for a specific baud rate is discussed in Serial Port Operation; the only difference is that in the case of Timer 2, the auto-reload value is placed in RCAP2H and RCAP2L, and the value is a 16-bit value rather than an 8-bit value.
- NOTE: When Timer 2 is used as a baud rate generator (either TCLK or RCLK are set), the Timer 2 Overflow Flag (TF2) will not be set.
The first mode in which Timer 2 may be used is Auto-Reload. The auto-reload mode functions just like Timer 0 and Timer 1 in auto-reload mode, except that the Timer 2 auto-relaod mode performs a full 16-bit reload (recall that Timer 0 and Timer 1 only have 8-bit reload values). When a reload occurs, the value of TH2 will be reloaded with the value contained in RCAP2H and the value of TL2 will be reloaded with the value contained in RCAP2L.
To operate Timer 2 in auto-reload mode, the CP/RL2 bit (T2CON.0) must be clear. In this mode, Timer 2 (TH2/TL2) will be reloaded with the reload value (RCAP2H/RCAP2L) whenever Timer 2 overflows; that is to say, whenever Timer 2 overflows from FFFFh back to 0000h. An overflow of Timer 2 will cause the TF2 bit to be set, which will cause an interrupt to be triggered, if Timer 2 interrupt is enabled. Note that TF2 will not be set on an overflow condition if either RCLK or TCLK (T2CON.5 or T2CON.4) are set.
Additionally, by also setting EXEN2 (T2CON.3), a reload will also occur whenever a 1-0 transition is detected on T2EX (P1.1). A reload which occurs as a result of such a transition will cause the EXF2 (T2CON.6) flag to be set, triggering a Timer 2 interrupt if said interrupt has been enabled.
TIMER 2 INTERRUPTAs is the case with the other two timers, timer 2 can be configured to trigger and interrupt. In fact, the text above indicates a number of situations that can trigger a timer 2 interrupt.To enable Timer 2 interrupt, set ET2 (IE.5). This bit of IE is only valid on an 8052. Similarly, the priority of Timer 2 interrupt can be configured using PT2 (IP.5). As always, be sure to alsoset EA (IE.7) when enabling any interrupt.
Once Timer 2 interrupt has been enabled, a Timer 2 interrupt will be triggered whenever TF2 (T2CON.7) or EXF2 (T2CON.6) are set. The Timer 2 Interrupt routine must be placed at 002Bh in code memory.
NOTE: Like the Serial Interrupt, Timer 2 interrupt does not clear the interrupt flag that triggered the interrupt. Since there are two conditions that can trigger a Timer 2 interrupt, either TF2 or EXF2 being set, the microcontroller leaves the flags alone so that your interrupt routine can determine the source of the interrupt and act accordingly. It is possible (and even probable!) that you will want to do one thing when the timer overflows and something completely different when a capture or reload is triggered by an external event. Thus, be sure to always clear TF2 and EXF2 in your Timer 2 Interrupt. Failing to do so will cause the interrupt to be triggered repeatedly until the bits are cleared.
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