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Remember these times are irrespective of your external clock frequency. Think of these times as real time, rather than clock times. To help make this clear, let us suppose we want the WDT to reset our PIC after about half a second as a failsafe. The nearest we have is 576mS, or 0.576 seconds. All we do is send b’101’ to our OPTION register, as follows: movlw b’101’
;This is 0x05 in Hex Simple, really. Now, there is a catch. By default the prescaler is assigned to the other internal timer. This means that we have to change the prescaler over to the WDT. First, we have to reset the other counter to 0 first. We then have to change to Bank 1 to assign the prescaler to the WDT and to set up the time, and then come back to Bank 0. The code is below, where xx is the prescaler time: bcf
STATUS,0 ;make sure we are in bank 0 The CLRWDT command above is how we clear the WDT before it resets the PIC. So, all we need to do is calculate where in our program the WDT will time out, and then enter the CLRWDT command just before this point to ensure the PIC doesn’t reset. If your program is long, bear in mind that you may need more than one CLRWDT. For example, if we use the default time of 18mS, then we need to make sure that the program will see CLRWDT every 18mS. So now we come to the point where we need to work out how long our code takes in real time. The principle is very simple, but could cause you to pull your hair out!
Instruction Timing As you are probably already aware, the PIC takes the external clock timing and divides it by 4. This internal time is called an instruction cycle. Now if we have, say, a 4MHz xtal connected to the PIC, internally the PIC will run at 1MHz. In timing terms, this is 1/(4MHz/4) = 1uS. Now, some instructions take just one instruction cycle to complete, i.e. 1uS using a 4MHz crystal, while others take two cycles – 2uS – to complete. The data sheet tells us how many cycles each instruction takes. The easiest way to remember this is quite simple. Assume ALL instructions take 1 cycle. But, if an instruction causes the program to go somewhere else, then it will take 2 cycles. Let me give you a couple of examples. The movwf command takes only one cycle, because it is only moving data from one place to another. The goto command takes 2 cycles, because it is causing the Program Counter (PC) to go elsewhere in the program. The RETURN command takes 2 cycles, because it is causing the PC to go back in the program. I think you can see the pattern here. However, there are four commands which can take 1 or 2 cycles. These are DECFSZ, INCFSZ, BTFSC and BTFSS. These commands have one thing in common. They will skip the next instruction is a certain condition is met. If that condition is not met, then the next instruction will be carried out. For example, the DECFSZ command will decrement the value stored in the F register by 1. If the result is not 0, then the next instruction will be executed. This instruction therefore takes 1 cycle. If the result is 0, then the next instruction will be skipped, and the one following that will be executed. In this instance the instruction takes 2 cycles. The reason is that the instruction alters the value of the PC. It needs one cycle to carry out the function, and it will need another to alter the PC by an extra one. To clarify this, let us look at a sample code, and work out how many instruction cycles it takes.
movlw 02 Our first instruction simply moves the value 02 into w. This does not cause the program to off course, therefore it is only 1 cycle. The next instruction is similar, in as much that it moves the contents of the w register into COUNT. Again, this will be 1 cycle. Now, the next instruction will first decrement COUNT by 1. This is 1 cycle. It will then do a test to see if COUNT is equal to 0. At this stage it doesn’t, and so we move onto the next instruction. The next instruction is a goto statement, and so is 2 cycles long. We come back to our decfsz instruction, which decrements COUNT by 1 again. This is another instruction cycle. It does a test to see if COUNT is equal to 0. This time it does, and so the next instruction is skipped. To skip the next instruction requires another cycle. We reach the end of the program. So in total, with the value 02 placed into COUNT, this program will take a total of 7 cycles. If we were using a 4MHz crystal for our clock, then the program will take: 1/(4MHz/4) = 1uS per cycle, therefore 7 cycles takes 7 x 1uS = 7uS. So you can see that it can get a little confusing when you have instructions like DECFSZ.
Programmer Software Inside the PIC there are things called ‘Fuses’. These are not the same as the fuses you would find in a mains plug, but electronic switches which are ‘blown’ by the programmer. Now, one of these fuses has to be ‘blown’ in order for the WDT to operate. There are two ways of doing this. One way is to write a couple of lines at the beginning of your program to tell the PIC programming software to enable or disable certain fuses. The other way is to tell the PIC programming software manually which fuses to enable. We will look at getting your program to instruct the programming software in a later tutorial, when we look at including other files and macros. To tell the programming software manually, varies from program to program. The documentation that came with the programmer should tell you how to do this. As I am using the PICALLW software, which is linked on my main page, I will explain how to do change fuses within this program. The fuses are configured by pressing the F3 key, or clicking on the ‘Config’ button. Then you can select the fuse you want enabled, in this case the WDT, by clicking on the box next to it.
Sample Program Let us write a program, where we will turn on the WDT, and let the PIC perform a function. We will first of all periodically clear the WDT, to show that the program works, and then remove the CLRWDT command to show that the PIC will indeed reset. The program I have chosen is the one used in tutorial 9 where we cause a row of LEDs to light up one at a time from left to right, then right to left. The circuit is shown below, and with the RC values shown will give us a clock frequency of 8KHz. This clock speed will allow us to actually see the LEDs moving one by one. I chose this program because it is slow enough for us to play with the WDT, and you can easily see when the PIC is reset. I have removed the original comments, and I have replaced them with a description of the WDT lines, a running total of the time from the start (assuming a 8KHz clock), and the number of clock cycles at each line.
TIME
equ 9FH ; Variable for the delay loop. bsf
STATUS,5 ; 1 cycle, 0.5mS ; Start of main program RUN
movlw 01H
; 1 cycle, 4.5mS ; Move the bit on Port B left, then pause.
rlf PORTB,1 ; 1 cycle, 967.5mS ; Now move onto Port A, and move the bit left.
rlf PORTA,1 ;
1 cycle, 7.70S ; Move the bit back on Port A rrf
PORTA,1 ; 1 cycle, 11.55S ; Now move the bit back on Port B rrf
PORTB,1 ; 1 cycle, 14.44S ; Subroutine to give a delay
between bit movements. DELAY END ;
With an 8KHz clock, it takes just under 1 second for the next LED illuminates, and it takes a total of about 21 seconds to run from one end to the other and back again i.e. to go through the routine once only. The delay routine takes 480mS, and we are calling it twice before moving the bit on the ports. Now, we need to periodically reset the WDT. The largest time we can set the WDT is 2.3 seconds, and the next one down form this is 1.1 seconds. We have two options here. We could make a call to a subroutine to clear the WDT after the two delays have finished, or we could incorporate the CLRWDT within the delay itself. I have decided, for no real reason at all, to incorporate the CLRWDT within the delay loop.
TIME
equ 9FH ; Variable for the delay loop. OPT equ
81h ; Option Register to control the WDT ;*************Set up the ports, WDT and prescaler******************
clrf 01h
;Clear TMR0
movlw 00H ;
;*************Start of main program***************************** RUN
movlw 01H ; ; *************Move the bit on Port B left, then pause.**************
rlf PORTB,1 ; ; *************Now move onto Port A, and move the bit left.***********
rlf PORTA,1 ; ;************** Move the bit back on Port A************************
rrf PORTA,1 ; ;****************** Now move the bit back on Port B******************
rrf PORTB,1 ;
; ******************Subroutine to give a delay between bit movements.****** DELAY
movlw TIME ;
LOOP2 ; clrwdt ;This simply resets the WDT. return ; END ;
If you comment out, or remove the CLRWDT command, you will find that the PIC will not go past lighting the second LED. This is because the WDT is resetting the PIC. With the CLRWDT in place, the program works as it should.
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This site was last updated 08/20/02
