Expansion of Parallel I/O Port in Single Chip Microcomputer Development

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 目录

一、题目

二、要求

三、Process and result record

1. 在ProteusThe principle diagram of the environment to establish,如图1所示,并保存为expandIO.DSN文件.

2. Will test code into a text document,并保存为expandIO .asm文件.

3. 输入源程序,Will join the rest of the system source code and compile the source code.

4. 双击芯片,将编译得到的hexFile downloaded to the chip

5. Perform simulation process to observe all directions of traffic lights indicate,Check the function program is correct.

四、思路

6. 对于状态1：东西红灯,南北绿灯

7. 对于状态2：东西红灯,南北黄灯

8. 对于状态3：东西绿灯,南北红灯

9. 对于状态4：东西黄灯,南北红灯

五、源程序

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一、题目

3.2 拓展并行I/O口实验

二、要求

1、Draw a process flow diagram that experiment

2、Write the source program and annotate

3、Document the process

4、Document the program runs results screenshot

三、Process and result record

According to the ideas to build circuit diagram1如下：

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图1：实验电路图

实现功能：Control traffic lights

The simulation traffic light control function.

Control the order：

① 南北绿灯亮,At the same time things red light 10s;

② 南北黄灯亮,At the same time things red light 2s;

③ 南北红灯亮,At the same time something green light 10s;

④ 东西黄灯亮,At the same time the north and south red light 2s;

⑤ 重复①～④.

实验过程：

1. 在ProteusThe principle diagram of the environment to establish,如图1所示,并保存为expandIO.DSN文件.

2. Will test code into a text document,并保存为expandIO .asm文件.

3. 输入源程序,Will join the rest of the system source code and compile the source code.

4. 双击芯片,将编译得到的hexFile downloaded to the chip

5. Perform simulation process to observe all directions of traffic lights indicate,Check the function program is correct.

The diagram below with the results of the experiment2~5所示：

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图2：南北绿灯亮,At the same time things red light10s

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图3：南北黄灯亮,At the same time things red light2s

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图4：南北红灯亮,At the same time something green light 10s

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图5：东西黄灯亮,At the same time the north and south red light 2s

四、思路

The flow chart is as follows：

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图6：实验流程

1. 进行初始化工作,Including set the stack pointerSP,将两个373 All bits are set to the output1,Make all leds put out.

2. Define two latch address: U4的地址是0FE00H, U5的地址是0FD00H

3. 分析4 A state of two373 The output of the data values：假定“南北绿灯亮,At the same time things red light”为状态1,即：Stat1;“南北黄灯亮,At the same time things red light”为状态2,即：Stat2;“南北红灯亮,At the same time something green light”为状态3,即：Stat3;“东西黄灯亮,At the same time the north and south red light”为状态4,即：Stat4.Can analyze its the output of each state.

6. 对于状态1：东西红灯,南北绿灯

    ① 那么对于U4,There are only two red light,A red light the address of the corresponding place0即可.对应的控制信号是11110011即0F3H

② 对于U5,绿灯亮,Only need to put the corresponding green light0 即可,For high four does not use the port,Also put together0.控制信号为00001100即0CH

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图7：状态1下U4和U5的输出

7. 对于状态2：东西红灯,南北黄灯

    ① 对于U4：The north and the south yellow light、What red light corresponding port place0.So the control signals for the11000011即0C3H

② 对于U5：There is no green light,A useful（低四位）置1,In less than the bit or set of0.So the control signals for the00001111即0FH

​编辑图8：状态2下U4和U5的输出

8. 对于状态3：东西绿灯,南北红灯

    ① 对于U4：To the north and the south red light corresponding port place0,其余位为0.So the control signals for the11111100即0FCH

② 对于U5: To put something green corresponding port0,其余位为1.Pay attention to not use a zero.So the control signals for the00000011即03H

​编辑图9：状态3下U4和U5的输出

9. 对于状态4：东西黄灯,南北红灯

    ① 对于U4：Yellow light things、The north and the south red light corresponding port place0,其余位置1,.So the control signals for the00111100即3CH

    ② 对于U5：Did not involve green light,So will useful position1即可.So the control signals for the00001111即0FH

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图10：状态4下U4和U5的输出

五、源程序

ORG 0000H     ; 在内存的0Address is forced to turn to the main program,Bypass the interrupt routineAJMP MAIN     ; Unconditionally transferred to the main programORG 0030H    MAIN:				;定义主程序  MOV SP,#60H			;设置栈指针  CLR P3.6				; 使用CLRAn operating instruction will beP3.6口清零  MOV P0,#0FFH			;设置P0口,Make both output a high level,此时所有的LED灯灭STATUS1:				; 状态1   MOV DPTR,#0FE00H	    ;数据指针指向U4  MOV A,#0F3H		    ;使U4输出0F3H  MOVX @DPTR,A     ; 让ARegister access memory,Let things red light  MOV DPTR,#0FD00H	    ;数据指针指向U5  MOV A,#0CH			;使U5输出0CH  MOVX @DPTR,A   ; 让ARegister access memory,To the north and south green light  ACALL DELAY10	 ; 调用延时10S子程序STATUS2:			; 状态2   MOV DPTR,#0FE00H		;数据指针指向U4  MOV A,#0C3H		   ;使U4输出03CH  MOVX @DPTR,A    ; 使用ARegister access memory,Make something red light,南北黄灯  MOV DPTR,#0FD00H		;数据指针指向U5  MOV A,#0FH			; 使U5输出0FH  MOVX @DPTR,A    ;置1,The green light completely destroyed  ACALL DELAY2			; 调用延时2S子程序STATUS3:			;  状态3  MOV DPTR,#0FE00H	    ;数据指针指向U4  MOV A,#0FCH		; U4输出0FCH  MOVX @DPTR,A    ; 借ARegister indirect access memory,To the north and the south red light  MOV DPTR,#0FD00H	    ;数据指针指向U5  MOV A,#03H	  ; U5输出0FD00H  MOVX @DPTR,A     ;东西绿灯亮  ACALL DELAY10		; 调用延时10S子程序STATUS4:			; 状态4  MOV DPTR,#0FE00H	   ;DPTR指向U4  MOV A,#3CH	   ; U4输出3CH  MOVX @DPTR,A  ; 寄存器A访问内存,南北红灯,东西黄灯  MOV DPTR,#0FD00H	从   ;使FPTR指向U5  MOV A,#0FH		;  使U5输出0FH  MOVX @DPTR,A    ;置1,The green light completely destroyed  ACALL DELAY2		; 调用延时2S子程序ACALL DELAY1	   ; 调用延时1S子程序AJMP STATUS1	  ; 跳转到状态1,形成循环DELAY1:				  ;定义延时1S子程序MOV R7,#10     ;Define the outermost loop,循环10次DL2:MOV R6,#200    ; Define the middle tier cycle200次DL1:MOV R5,#250    ; 定义内层循环250次     DJNZ R5,$      ;等待,And the instruction of two clock cycles	DJNZ R6,DL1    ;Of two clock cycles	DJNZ R7,DL2    ;Calculate the total delay time（（1μs*2*250+2+1）*200+1+2）*10+1=1sRET     ; 子程序返回DELAY2:					;定义延时2S子程序MOV R7,#20           DL4:MOV R6,#200DL3:MOV R5,#250    DJNZ R5,$	DJNZ R6,DL3	DJNZ R7,DL4RET       ; 子程序返回DELAY10:				;延时10S子程序MOV R7,#100         ; 延时10SThan the time delay1SThe application of the outermost loop more10倍DL6:MOV R6,#200    DL5:MOV R5,#250    DJNZ R5,$	DJNZ R6,DL5	DJNZ R7,DL6RET           ; 子程序返回END						;程序结束

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