ECET 410 Lab 3
Delay routines using both assembly and C programs
Delay functions and subroutines are important function in real time programming.
In this lab you will exercise both assembly and C programs to implement delays.
In this lab we will use the assembler. In the next lab we will use the C compiler to
demonstrate similar learning objectives.
Lab Objectives:
1- Learn and understand how to use I/O port registers
2- Learn and understand how to use timer functions and the underlying
mechanism for setting up real time counters.
3- Understand the basic idea of interrupts
4- Learn how to work in C environment
5- Learn how to compile and download C programs to the development board
6- Learn how to use functions in C programs.
Pre-lab activities:
1. Scan the file reg9s12.h to get familiar with the assembly definitions of
the HCS12 registers.
2. Read the I/O section relevant to the program from chapter 7
3. Examine the vector.h header file supplied with your textbook CD.
4. Provide a more detailed explanation of the program below before
using it.
Lab activities:
1. Assemble and down load the following program
2. Use the software interrupt instruction in the proper place in the program
in order to get the > prompt back to the hyper terminal, so you can
continue debugging your program.
3. Modify the program such that it runs continually or until you hit the
abort or the reset button on the board.
4. Modify the program to create a chaser light ( where LED 1 is lighted
first, 5 seconds later LED1 is turned off and LED2 is lighted for 5 second
and so on) .
5. Change the delay in your program: one time use to use 10 seconds instead
of 5 and repeat the program.
A 5 seconds delay assembly program
; 5s_delay.asm 5 second delay timer for DRAGON12 Rev. E board
;
(c)2002, EVBplus.com, written by Wayne Chu
;
; Function: 5 second delay routine using output comparator 6
;
The PB0 LED will be turned on immediately after running
;
this program. It will be turned off after 5 second delay.
;
Change the DELAY_TIME to 36000 will delay 3 minutes.
PB0: equ
DB6: equ
;
TB1MS:
;
1
$40
;DELAY_TIME:
DELAY_TIME:
REGBLK:
#include
org
equ
24000 ; 1ms time base of 24,000 instruction cycles
; 24,000 x 1/24MHz = 1ms at 24 MHz bus speed
equ
equ
36000 ; 36000 X 5 ms= 180 sec = 3 min
5000 ; 5000 X 1 ms= 5 sec
equ
reg9s12.h
$0
; include register equates
$1000
;
flag_5s:
cnt_5s:
rmb
rmb
1
2
STACK:
equ
$2000
org
$2000
lds
ldx
stx
#STACK
#timer6
$3E62
; initialize the int vetctor
ldx
ldaa
staa
staa
staa
staa
#REGBLK
#$ff
ddrb,x
ddrp,x
ptp,x
ddrj,x
; make port B an output port
; make port P an output port
; turn off 7-segment LED display
; make port J an output port
start:
clr
ptj,x
ldaa
staa
ldaa
staa
staa
#$80
tscr,x
#DB6
tios,x
tmsk1,x
; make PJ1 low to enable LEDs
; enable timer
; select t6 as an output compare
cli
bset portb,x PB0 ; turn on LED PB0
jsr
delay_5s
bclr portb,x PB0 ; turn off LED PB0
stp:
jmp
delay_5s:
clr
clr
clr
delay: ldaa
beq
rts
stp
cnt_5s
cnt_5s+1
flag_5s
flag_5s
delay
timer6:
ldx
inx
stx
cpx
bne
clr
clr
ldaa
staa
rti
tmr6: ldx
ldd
addd
std
ldaa
staa
rti
org
cnt_5s
cnt_5s
#DELAY_TIME
tmr6
cnt_5s
cnt_5s+1
#1
flag_5s
#REGBLK
#TB1MS
tc6,x
tc6,x
#DB6
tflg1,x
$3E62
; reload the count for 1 ms time base
; clear flag
fdb
end
timer6
A simpler delay program
This program is from your text chapter 2, it depends on the time an instruction
takes to execute. The following sequence of instruction takes 10 seconds to execute.
If you have hard time with above program, you can start with this delay routine to
get your lab completed on time. In the mean time work on understanding the
mechanics behind the timer interrupt driven program above.
ldab
#100
out_loop ldx
in_loop psha
pula
psha
pula
psha
pula
psha
pula
psha
pula
psha
pula
psha
pula
nop
nop
dbne
dbne
#20000
; 2 E cycles
; 3 E cycles
x,in_loop
b,out_loop
; 1 E cycle
; 1 E cycle
; 3 E cycles
; 3 E cycles