作业帮求微机原理实现A/D转换的程序?
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/09 12:07:42
求微机原理实现A/D转换的程序?
求微机原理实现A/D转换的程序?
求微机原理实现A/D转换的程序?
CODE SEGMENT ;H0809.ASM
ASSUME CS:CODE
ADPORT EQU 0FF80h
PA EQU 0FF20H ;字位口
PB EQU 0FF21H ;字形口
PC EQU 0FF22H ;键入口
ORG 1000H
START:JMP START0
BUF DB ,,,,,
data1:
db0c0h,0f9h,0a4h,0b0h,99h,92h,82h,0f8h,80h,90h,88h,83h,0c6h,0a1h
db 86h,8eh,0ffh,0ch,89h,0deh,0c7h,8ch,0f3h,0bfh,8FH
START0:CALL BUF1
ADCON:MOV AX,00
MOV DX,ADPORT
OUT DX,AL
MOV CX,0500H
;DELAY:LOOP DELAY
MOV DX,ADPORT
IN AL,DX
CALL CONVERS
CALL DISP
JMP ADCON
CONVERS:MOV AH,AL
AND AL,0FH
MOV BX,OFFSET BUF
MOV [BX+5],AL
MOV AL,AH
AND AL,0F0H
MOV CL,04H
SHR AL,CL
MOV [BX+4],AL
RET
DISP:MOV AL,0FFH ;00H
MOV DX,PA
OUT DX,AL
MOV CL,0DFH ;20H ;显示子程序 ,5ms
MOV BX,OFFSET BUF
DIS1:MOV AL,[BX]
MOV AH,00H
PUSH BX
MOV BX,OFFSET DATA1
ADD BX,AX
MOV AL,[BX]
POP BX
MOV DX,PB
OUT DX,AL
MOV AL,CL
MOV DX,PA
OUT DX,AL
PUSH CX
DIS2:MOV CX,00A0H
LOOP $
POP CX
CMP CL,0FEH ;01H
JZ LX1
INC BX
ROR CL,1 ;SHR CL,1
JMP DIS1
LX1:MOV AL,0FFH
MOV DX,PB
OUT DX,AL
RET
BUF1:MOV BUF,00H
MOV BUF+1,08H
MOV BUF+2,00H
MOV BUF+3,09H
MOV BUF+4,00H
MOV BUF+5,00H
RET
CODE ENDS
END START