mpi lab manual

G.N.I.T.S-ECE DEPARTMENT:MPI LAB

Exercise No:M.1.1 Page No: 1 Class:3/4 ECE,ETM,ICE –2nd sem & 4/4 EEE – 1st sem

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G NARAYANAMMA INSTITUTE OF TECHNOLOGY & SCIENCE (FOR WOMEN)

SHAIKPET, HYDERABAD – 500008

MICROPROCESSORS LABORATORY MANUAL

DEPARTMENT OF

ELECTRONICS & COMMUNICATION ENGINEERING MARCH - 2006



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G NARAYANAMMA INSTITUTE OF TECHNOLOGY & SCIENCE (FOR WOMEN)

SHAIKPET, HYDERABAD – 500008

GNITS GNITS-D / ECE / LLM / 023(f) / 01 LAB WISE-LABMANUALS DEPARTMENT : ECE

MICROPCROCESSORS LABORATORY MANUAL

DEPARTMENT OF

ELECTRONICS & COMMUNICATION ENGINEERING MARCH - 2006



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TABLE OF CONTENTS

S.NO NAME OF THE EXERCISE PAGE NO

I. MICROPROCESSOR 8086

M.1

1. Introduction 5 2. Using Turbo – Assembler-Linker-Debugger 8 (TASM,TLINK,TD) 3. Introduction to the trainer & Instructions to user 10 M.2

1.a). 8bit addition 18 b). 16bit addition 20 2.a). 8bit subtraction 22 b). 16bit subtraction 23 3.a). 8bit multiplication 24 b). 16bit multiplication 25 4.a). 8bit division 26 b). 16bit division 28 5. Multi byte addition 29 6. Multi byte subtraction 31 7. Signed multiplication 33 8. Signed division 34 9. ASCII addition 35 10. ASCII subtraction 36 11. ASCII multiplication 37 12. ASCII division 38 M.3 1. Packed to unpacked BCD 39 2. BCD to ASCII conversion 41



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S.NO NAME OF THE EXERCISE PAGE NO M.4 1. Block Transfer 42 2. String Reversal 43 3. String insertion 44 4. String deletion 45 5. Length of string 46 6. String comparison 47 M.5 1. Program design example (calculator) 48 M.6 1. DOS interrupts 56

II. INTERFACING 1. i8279 Keyboard/Display Interface 58 2. i8086 LED’s & switch interface 67

III. MICROCONTROLLER 8051 1. i8051 LED’s & switch interface 73 2. Transferring a block of data from 77

internal ROM to internal RAM 3. a) Internal ROM to external RAM 78

b) Internal ROM to external RAM 79 4. Understanding the three memory 80 areas of 00-FF



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INTRODUCTION EDITOR An editor is a program, which allows you to create a file containing the assembly language statements for your program. As you type in your program, the editor stores the ASCII codes for the letters and numbers in successive RAM locations. When you have typed in all of your programs, you then save the file on a floppy of hard disk. This file is called source file. The next step is to process the source file with an assembler. In the TASM assembler, you should give your source file name the extension, .ASM ASSEMBLER An assembler program is used to translate the assembly language mnemonics for instructions to the corresponding binary codes. When you run the assembler, it reads the source file of your program the disk, where you saved it after editing on the first pass through the source program the assembler determines the displacement of named data items, the offset of labels and pails this information in a symbol table. On the second pass through the source program, the assembler produces the binary code for each instruction and inserts the offset etc that is calculated during the first pass. The assembler generates two files on floppy or hard disk. The first file called the object file is given the extension. OBJ. The object file contains the binary codes for the instructions and information about the addresses of the instructions. The second file generated by the assembler is called assembler list file. The list file contains your assembly language statements, the binary codes for each instructions and the offset for each instruction. In TASM assembler, TASM source file name ASM is used to assemble the file. Edit source file name LST is used to view the list file, which is generated, when you assemble the file. LINKER A linker is a program used to join several object files into one large object file and convert to an exe file. The linker produces a link file, which contains the binary codes for all the combined modules. The linker however doesn’t assign absolute addresses to the program, it assigns is said to be relocatable because it can be put anywhere in memory to be run. In TASM, TLINK source filename is used to link the file. DEBUGGER A debugger is a program which allows you to load your object code program into system memory, execute the program and troubleshoot are debug it the debugger allows you to look at the contents of registers and memory locations after your program runs. It allows you to change the contents of register and memory locations after your program runs. It allows you to change the contents of register and memory locations and return the program. A debugger also allows you to set a break point at any point in the program. If you inset a breakpoint the debugger will run the program upto the instruction where the breakpoint is set and stop execution. You can then examine register and memory contents to see whether the results are correct at that point. In TASM, td filename is issued to debug the file.



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DEBUGGER FUNCTIONS:

1. Debugger allows to look at the contents of registers and memory locations. 2. We can extend 8-bit register to 16-bit register which the help of extended register option. 3. Debugger allows to set breakpoints at any point with the program. 4. The debugger will run the program upto the instruction where the breakpoint is set and then

stop execution of program. At this point, we can examine registry and memory contents at that point.

5. With the help of dump we can view register contents. 6. we can trace the program step by step with the help of F7. 7. We can execute the program completely at a time using F8.

DEBUGGER COMMANDS ASSEMBLE: To write assembly language program from the given address. A starting address <cr> Eg: a 100 <cr> Starts program at an offset of 100. DUMP: To see the specified memory contents D memory location first address last address (While displays the set of values stored in the specified range, which is given above) Eg: d 0100 0105 <cr> Display the contents of memory locations from 100 to 105(including). ENTER: To enter data into the specified memory locations(s). E memory location data data data data data …<cr> Eg: e 1200 10 20 30 40 …. Enters the above values starting from memory locations 1200 to 1203, by loading 10 into 1200,20 into 1201 and soon. GO: To execute the program G: one instruction executes (address specified by IP) G address <cr>: executes from current IP to the address specified G first address last addresses <cr>: executes a set of instructions specified between the given addresses. MOVE: Moves a set of data from source location to destination location M first address last address destination address Eg: m100 104 200 Transfers block of data (from 100 to 104) to destination address 200.



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QUIT: To exit from the debugger. Q <cr> REGISTER: Shows the contents of Registers R register name Eg: r ax Shows the contents of register. TRACE: To trace the program instruction by instruction.

T = 0100 <cr>: traces only the current instruction. (Instruction specified by IP) T = 0100 02 <cr>: Traces instructions from 100 to 101, here the second argument specifies the

number of instructions to be traced.

UNASSEMBLE: To unassembled the program. Shows the opcodes along with the assembly language program. U 100 <cr>: unassembled 32 instructions starting from 100th location. U 0100 0109 <cr>: unassebles the lines from 100 to 104



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Using Turbo – Assembler – Linker – Debugger

(TASM, TLINK, TD) 1. Open an MSDOS window. 2. Set the PATH so that the TASM programs are available

The TASM programs are on the C drive; set the path so that DOS can find them. This only needs to be done once each time you open an MSDOS prompt. set PATH=%PATH%;C:\TASM\BIN 3. Use a Text Editor to Edit the .ASM File Create your file using one of the following programs: notepad proj.asm wordpad proj.asm edit proj.asm 4. Compile the source code to create an object module. tasm/z/zi proj.asm The /z switch causes TASM to display the lines that generate compilation errors. The /zi switch enables information needed by the debugger to be included in the .OBJ file. Note that you should use "real mode" assembler, TASM.EXE. Do not use the "protected mode" assembler TASM32.EXE for the assignments that will be given in class 5. Run Linker TLINK.EXE- generate .EXE file from the .OBJ file tlink/v proj 6. Run the Program Your final program (if there were no errors in the previous step) will have an .EXE ending. To just run it, type: proj If you want to use the debugger to examine the instructions, registers, etc., type: td proj This brings up the regular full-screen version of the Turbo debugger. 1. Tracing the Program's Execution The Turbo debugger first starts, a Module Window which displays the . Executable lines of program code, marked with a bullet in the left column of the window. You can set breakpoints or step to any of these lines of code. An arrow in the first column of the window indicates the location of the instruction pointer. This always points to the next statement to be executed. To execute just that instruction use one of the two methods listed under the Run menu item: o Trace into (can use F7 key): executes one instruction; traces "into" procedures. o Step over (can use F8 key): executes one instruction; skips (does not trace into) procedures. Hitting either of these executes the instruction, and moves the arrow to the next instruction. As each instruction executes, the effects might be visible in the Registers Window and Watches Window



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o Trace into (can use F7 key): executes one instruction; traces "into" procedures. o Step over (can use F8 key): executes one instruction; skips (does not trace into) procedures. Hitting either of these executes the instruction, and moves the arrow to the next instruction. As each instruction executes, the effects might be visible in the Registers Window and Watches Window 2. Setting and Removing Breakpoints To set a breakpoint, position the cursor on the desired line of source code and press F2. The line containing the breakpoint will turn red. Pressing F2 again removes the breakpoint. To execute all instructions from the current instruction pointer up to the next encountered breakpoint, choose Run (can use F9 key) from the Run menu item. 3. Examining Registers Another window, the Registers Window, can be opened to examine the current value of the CPU registers and flags. The View menu can be used to open this Registers Window. The registers and flags might change as each instruction is executed. 4. Examining Memory To examine memory, you will need to open an Inspector window. An Inspector window shows the contents of a data structure (or simple variable) in the program you are debugging. It also allows you to modify the contents of the data structure or variable. To open an Inspector window, place the cursor on what you want to inspect and press CTRL-I. After you've examined the data item, press the ESC key to remove the Inspector window. 5. Viewing the Program's Output Output written to the screen by a program is not immediately visible, since the main purpose of using a debugger is to examine the internal operation of the program. To observe what the user would see, press ALT-F5. The entire screen will change to a user-view showing the program's input and output (and possibly that of previous programs as well). Press any key to return to the debugger screen.



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Introduction to the trainer & Instructions to user INTRODUCTION

ESA 86/88 –2 is a powerful, general-purpose microcomputer system, which can be operated either with 8086 CPU or with 8086CPU. It is generally supplied with 8086 CPU. To change it to 8088, user has to just remove the 8086, insert 8088 into that socket and set a DIP switch. The 8086 and 8088 are third generation CPU is from INTEL that differ primarily in their external data paths. 8088 uses and 8-bit wide data bus while 8086 uses a 16-bit wide bus. ESA86/88-2 can be operated with either CPU and the only possible difference would be in the speed of execution (with 8088 CPU, a small speed degradation occurs because of the 8-bit wide data bus). In either case, the CPU is operated in the maximum mode. Following are the system capabilities

• Examine and optionally modify the contents of memory (byte or word format) • Examine and optionally modify the processor registers. • Assemble and Disassemble 8086/8088 instructions (via line assembler, disassembler). • Perform fast numerical computations using the optional 8087 Numeric data processor. • Execute the user program at full speed. • Debug user program through single step and Breakpoint facilities. • Write or read data to or from I/O ports (byte or word format). • Move a block of data or program within the memory • Download user programs into ESA 86/88-2 from a host computer system.

SPECIFICATIONS:

Central processor 8086 CPU, operating at 8MHz in maximum mode. (Supplied with 8086 CPU). (Memory cycles have zero wait states and I/O cycles have one wait state).

Co-Processor On-board 8087 Numeric Data processor (optional)

Memory EPROM: 4 JEDEC compatible slots offer the following options: 64K bytes using 27128s or, 128K bytes using 27256s or, 256K bytes using 27512s

(System firmware is supplied in 2x27256s. The other two sockets are for user expansion).



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RAM: 4 JEDEC Compatible offer the following: 128K bytes using 6255s (64K bytes supplied using 2x62556s. The other two sockets are for user expansion). RAM has battery backup facility. Peripherals and Controllers 8251A: Programmable Communication Interface for serial communication supporting all standard from 110-19,200 (Baud is selected through on-board Dipswitch). 8253-5: (2 Nos.) programmable peripheral Interface devices provide 48 Programmable I/O lines. 8259A: Programmable interrupt Controller provides interrupt vectors for 8 sources 8288: Bus controller used for generating control signals. Interrupts External: NMI for INTR Keyboard.

INTER controlled through 8259A, on-board Interrupt Controller: provides interrupt vectors for eight sources. Complete flexibility in selecting off-board or on-board interrupt sources. On-board interrupt sources # 8251 (TxRDY and RxRDY) # 8253-5 (OUT1 and OUT2) # 8255A (PC0 and PC3 in handshake mode) # 8087 (NDP INT)

Internal: Interrupt vectors 1(single step) and 3 (breakpoint) reserved for monitor. Interface Signals CPU Bus: Demultiplexed and fully buffered, TTL compatible, Address, Data & Control signals are available on two 50-pin ribbon cable connectors. Parallel I/O: 48 Programmable parallel I/O lines (TTL compatible) through two 26 pin ribbon cable connectors. (Connector details compatible to our other microcomputer trainers). Serial I/O: RS 232 C through on-board 9pin D-type female connector. Power supply (optional): +5V @ 3.0Amp

CONFIGURATION AND INSTALLATION Configuration ESA 86/88-2 ESA 86/88-2-microcomputer trainer is versatile and can be configured in a number of ways, as determined by the setting of a DIPswitch and other jumpers. (Refer to the component layout diagram in appendix C to locate the DIPswitch and the jumpers). This chapter describes all the configuration options and the installation procedures. Operational mode selection ESA 86/88-2 can be operated either in the serial mode or in Hexadecimal keypad mode. In the serial mode, the trainer is connected to a CRT terminal or to a host computer system (like PC compatible) through an RS 232 C interface. In the keypad mode, the trainer is operated through Hexadecimal keypad.



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SW4 of the DIP switch Operational mode OFF Serial mode ON Hexadecimal keypad mode* (*Factory installed Option) Printer Enable/Disable ESA 86/88-2 firmware includes the driver program fro centronics compatible parallel printer interface. This driver can be enabled/disabled as shown below: SW5 of the DIP Switch Printer Driver OFF Disabled* ON Enabled (*Factory installed Option) Baud rate selection In the serial mode of operation, ESA 86/88-2 configures an 8251A USART as follows:

Asynchronous mode 8-bit character length 2 stop bits No parity Baud rate factor of 16X

Timers 0 of an 8253 provide the Transmit and receive baud clocks for the USART. (Refer to chapter 5 for a detailed discussion of the Hardware).This timer is initialized by the system firmware to provide proper baud clock based on the settings of the DIP Switch as shown below. DIP SWITCH SW3 SW2 SW1 Baud rate OFF OFF ON 9,600*

Memory selection: ESA 86/88-2 has four sockets, labeled U9, U8, U7, U6 for RAM. These sockets are configured for 62256(32X 4) devices. Two of these sockets are populated (providing 64K Bytes of RAM) and two are for user expansion. DEVICE DIP SWITCH JUMPER SW7 SW6 27256 ON OFF JP10 – 1-2



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Hexadecimal keypad Legend Interpretation Hexadecimal Acronym Name Acronym Name 0 EB/AX 1 ER/BX 2 GO/CX 3 ST/DX 4 IB/SP 5 OB/BP 6 MV/SI 7 EW/DI 8 IW/CS 9 OW/DS A /SS B /ES C /IP D /FL E F

EB ER GO ST IB OB MV EW IW OW None None None None None None

Examine Byte Examine Register Go Single Step Input Byte Output Byte Move Examine Word Input word Output Word N/A N/A N/A N/A N/A N/A

AX BX CX DX SP BP SI DI CS DS SS ES IP FL none none

Accumulator Base Register Count Register Data Register Stack Pointer Base Pointer Source Index Destination index Code Segment Data Segment Stack Segment Extra Segment Instruction Pointer Flag Register N/A N/A



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Function Key Operation Function Key

Operation RESET KB INT + - : REG NEXT (,) PREV EXEC (.) NOTE:

The RESET key allows you to terminate any present activity and to return your ESA 86/88-2 to an initialize state. When pressed, the sign-on message appears in the display and the monitor is ready for command entry. The INTR (interrupt) key is used to generate an immediate non-maskable type 2 interrupt (NM). The NMI interrupt vector is initialized on power up or system reset to point to a routine within the monitor which causes all of the 8086/8088’s registers to be saved. Control is returned to the monitor for subsequent command entry. The + (plus) key allows you to add two hexadecimal values. This function simplifies relative addressing by allowing you to readily calculate an address location relative to a base address. The – (minus) key allows you to subtract one hexadecimal value from another. The : (colon) key is used to separate an address to be entered into two parts; a segment value and an offset value. The REG (register) key allows you to use the contents of any of the 8086/8088’s registers as an address or data value. The NEXT key is used to separate keypad entries and To increment the address field to the next consecutive memory location. The PREV key is used to decrement the address field to previous memory location. The dot key is the command terminator. When Pressed, the current command is executed.

1) NEXT or, means the same operation 2) EXEC or, means the same operation



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Summary of Monitor commands Command Group Command Function/Format Examine/modify Input/Output Execution Block Move

Examine Byte Displays/modifies memory byte locations EB <address>,[[<data>] NEXT or PREV]. Examine word Displays/modifies memory word locations EW <address>,[[<data>],]NEXT of PREV]. Examine Register Displays/modifies processor register contents

ER<reg key>[[<data>] NEXT] [.] Input Byter Displays the data byte at the input port. IB <port address> NEXT [NEXT]. Input word Displays the data word at the input port. IW <port address> NEXT [NEXT]. Output byte outputs the data byte to the output port.

OB <port address>NEXT <data> [NEXT <data>] Output word outputs the data word to the output port. OW <port address>NEXT <data> [NEXT<data>]. Step Executes one single instruction.

ST [<Start address>NEXT [[<start address>] NEXT] Go Transfers control from monitor to user program GO [<address>] [NEXT<breakpoint address>]. Move Moves block of data within memory MV <start address> NEXT <end address> NEXT <destination address>.

EXAMINE BYTE AND EXAMINE WORD COMMANDS Function: The Examine byte (EB) and Examine Word (EW) commands are used to examine the contents of selected memory locations. If the memory location can be modified (e.g. a location in RAM), the contents optionally be modified. Format EB <address> NEXT [[<data>] PREV/NEXT]. EW <address> NEXT [[<data>] PREV/NEXT].



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Operation

1. Both the commands operate in a similar fashion. To use these commands, press the EB key or EW key when prompted for a command.

2. When either key is pressed, a dot appears at the right edge of the address field indicating that an address entry is required.

3. Enter the memory address of the byte (for EB) of word (for EW) to be examined. (The entry of address values is discussed in detail in section 3.2)

4. After entering the address value, press the “,” key. (i.e. the NEXT key). 5. The data byte of word contents of the addressed memory location will be displayed in the

data field and a decimal point (a dot) appears at the right edge of data field indicating that the data can be updated. Note that when using the Examine word command, the byte contents of the displayed updated memory location appear in the two least-significant digits of the field and the byte contents of the next consecutive memory location (i.e. entered memory address + 1) appear in the two most significant digits of the data field.

6. If the contents of the memory location addressed are only to be examined, press the ‘’.’’ Key to terminate the command, of press the ‘’,’’ (NEXT) key to examine the next consecutive memory location (Examine Byte Command) or the next two consecutive memory locations (Examine Word Command) or press the ‘’PREV’’ key to examine previous byte or word location.

7. To modify the contents of an addressed memory location, enter the new data from the hexadecimal keyboard (entering the data values is discussed in detail in section 3.2).

8. The data displayed is not updated in memory until either the ‘’,’’ or ‘’.’’ Key is pressed.

EXAMINE BYTE AND EXAMINE WORD COMMANDS Function: The Examine Byte (EB) and Examine Word (EW) commands are used to examine the contents of selected memory locations. If the memory location can be modified (e.g. a location in RAM), the contents optionally be modified.

Format EB <address> NEXT [[<data>] PREV/NEXT]. EW <address> NEXT [[<data>] PREV/NEXT]. EXAMINE REGISTER COMMAND

Function The Examine Register (ER) command is used to examine and optionally modify the contends of nay of the 8086/8088’s registers.

Format ER <reg key> [[<data>],] [.]



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INPUT/OUTPUT COMMANDS There are 4 commands available for Input/Output of Byte/Word data form/to a specified port. In entering the port address (in any of these four commands), it should be noted that 8086/8088 I/O addressing is limited to 64K (maximum address is FFFFH). Thus no segment value is permitted with the port address. INPUT BYTE AND INPUT WORD COMMANDS Function The Input Byte (IB) and Input Word (IW) commands are used to input (accept) an 8-bit byte or 16-bit word from an input port. Format IB <port address > , [,]. IW <port address> , [,]. STEP COMMAND Function This command is used for single step execution of a program. In other words, this step (ST) command permits program instructions in memory to be executed individually. With each instruction executed, control is returned to the monitor from the program being executed, Format ST [< start address>], [[<start address>,. GO COMMAND Function The GO command is used to transfer control of the 8086/8088 from the keyboard monitor program to user’s program. Format

GO [<address>] [, <breakpoint address>] MOVE COMMAND Function This command (MV) can be used to move a block of data from one portion of the memory to another potion of the memory. Format MV <Start address> NEXT <end address> NEXT <destination address>.



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PROGRAM: 8BIT ADDITION .model tiny .stack 32h .code org 2000h start: mov ax,cs mov ds,ax mov ax,00 mov al,num1 mov bl,num2 add al,bl mov result,al int 3 mov ah,4ch int 21h org 20f0h num1 db 03 num2 db 08 result db 00 end start Result: 03h 08h 0Bh



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Exercise:

1. What is the significance of model tiny? 2. How many model assignment are the name them? 3. What is a directive? 4. What is a pseudo operation? 5. ORG 2000H implies what? 6. At register is used why not AX? 7. What is the purpose of INT3 in the program?

8. What is the purpose of MOV AH, 4CH, INT 21H in the program?.



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PROGRAM: 16BIT ADDITION .model tiny .stack 32h .code org 2000h start: mov ax,cs mov ds,ax mov ax,00 mov ax,num1 mov bx,num2 add ax,bx mov result,ax int 3 mov ah,4ch int 21h org 20f0h num1 dw 0ffffh num2 dw 0ffffh result dw 00 end start

Result: 0FFFFh 1111 1111 1111 1111 1111 0FFFFh 1111 1111 1111 1111 1111 1FFFEh 1,1111 1111 1111 1111 1110

Carry



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Exercise: 1. Stack 32 implies what? 2. Can ORG have other numbers instead of 2000h? 3. What is the purpose of MOV AX,CS? MOV DS,AX? 4. Why AX register is used and not AL? 5. What is the purpose of DW? 6. What happens if the result is greater than 16bit when result is declared an DW.?



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PROGRAM: 8BIT SUBTRACTION

.model tiny .stack 32h .code org 2000h start: mov ax,cs mov ds,ax mov ax,00 mov al,num1 mov bl,num2 sub al,bl mov result,al int 3 mov ah,4ch int 21h org 20f0h num1 db 0ffh num2 db 0aah result db 00 end start Result: 0ffh 0aah 055h Exercise:

1. What AL has been used and not AX ? 2. What happens if num1 contains 0AAH and num2 contains 0FFH. ?

3. How do you account for the difference obtained in previous question?



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PROGRAM: 16BIT SUBTRACTION

.model tiny .stack 32h .code org 2000h start: mov ax,cs mov ds,ax mov ax,00 mov ax,num1 mov bx,num2 sub ax,bx mov result,ax int 3 mov ah,4ch int 21h org 20f0h num1 dw 0ffffh num2 dw 0eabch result dw 00 end start Result: 0ffffh 0eabch 01543h Exercise:

1. Why should AX be used not AL. ? 2. What happens if num1 and num2 values are interchanged? 3. If carry is set to 1 before subtraction what is the instruction to be used?



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PROGRAM: 8BIT MULTIPLICATION

.model tiny .stack 32h .code org 2000h start: mov ax,cs mov ds,ax mov ax,00 mov al,num1 mov bl,num2 mul bl mov result,al mov result1,ah int 3 mov ah,4ch Org 20f0h num1 db 0ffh num2 db 0aah result db 00 result1 db 00 end start Result: 0ffh 0aah a956h Exercise:

1. What is an extended accumulator? 2. AL and BL are used for multiplying why not AX & BX?

3. Instead of using MOV BL is it not possible to MUL num2?

4. What is the instruction used for signed multiplication?



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PROGRAM: 16BIT MULTIPLICATION .model tiny .stack 32h .code org 2000h start: mov ax,cs mov ds,ax mov ax,00 mov ax,num1 mov bx,num2 mul bx mov result,ax mov result1,dx int 3 mov ah,4ch int 21h org 20f0h num1 dw 0ffffh num2 dw 0ffffh result dw 00 result1 dw 00 end start Result: 0ffffh 0ffffh fffe0001h Exercise:

1. Why AL & BL are not used in this? 2. If result exceeds 32 bit where is it stored?

3. What is the name given to the register combination DX:AX?



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PROGRAM: 8BIT DIVISION

.model tiny .stack 32h .code org 2000h start: mov ax,cs mov ds,ax mov ax,00 mov dx,00 mov al,num1 mov bl,num2 div bl mov Quotient,al mov remainder,ah int 3 mov ah,4ch int 21h org 20f0h num1 db 0ffh num2 db 0aah Quotient db 00 remainder db 00 end start Result: 0ffh 0aah 5501h Quotient: 01h Remainder r: 55h



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Exercise:

1. Why is the registers DX & AX made zero in the above program? 2. The above program?

3. Where is the remainder in 8 bit division?

4. Where is the quotient in 8 bit division?

5. If AH contains a non-zero value, what will be the result of the division?

6. Which interrupt is used when a divide overflow error occurs?



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PROGRAM: 16BIT DIVISION

.model tiny .stack 32h .code org 2000h start: mov ax,cs mov ds,ax mov ax,00 mov dx,00 mov ax,num1 mov bx,num2 div bx mov Quotient,ax mov remainder,dx int 3 mov ah,4ch int 21h org 20f0h num1 dw 0ffffh num2 dw 0aaaah Quotient dw 00 remainder dw 00 end start Result: 0ffffh 0aaaah 55550001h Quotient: 0001h Remainder: 5555h Exercise:

1. What happens if DX register contains a nonzero value before DIV instruction? 2. What is the instruction used for signed division?

3. In the above program instead of DIV BX is it possible to use DIV num2?



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PROGRAM: MULTY BYTE ADDITION .model tiny data segment dp1 dd 11223344h dp2 dd 55667788h res dd 00000000h data ends code segment assume cs:code,ds:data start: mov ax,data mov ds,ax mov si,offset dp1 mov di,offset dp2 mov bx,offset res mov cx,03 sub ax,ax mov al,[si] mov dl,[di] add al,dl mov [bx],al back: inc si inc di inc bx mov al,[si] mov dl,[di] adc al,dl mov [bx],al loop back nop mov ah,4ch int 21h code ends end start



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EXERCISE:

1. Why is add with carry instruction adc is used in the loop? 2. what is the purpose served by BX register?

3. Why addition is done with AL register why not with AX?

4. What is the other instruction which can be used instead of MOV SI,offset dp1.?



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PROGRAM: MULTY BYTE SUBTRACTION

.model tiny data segment dp2 dd 55667788h dp1 dd 11223344h res dd 00000000h data ends code segment assume cs:code,ds:data start: mov ax,data mov ds,ax mov si,offset dp2 mov di,offset dp1 mov bx,offset res mov cx,03 sub ax,ax mov al,[si] mov dl,[di] sub al,dl mov [bx],al back: inc si inc di inc bx mov al,[si] mov dl,[di] sbb al,dl mov [bx],al loop back nop mov ah,4ch int 21h code ends end start



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EXERCISE:

1. Why is subtract with carry instruction is used in the loop? 2. What is the purpose served by BX register?

3. Why subtraction is done with AL register why not with AX ?

4. What is the other instruction which can be used instead of

MOV DI, offset dp2. ?



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PROGRAM: SIGNED MULTIPLICATION

.model tiny data segment dp1 db 0a5h dp2 db 20h res dw 00h data ends code segment assume cs:code,ds:data start: mov ax,data mov ds,ax sub ax,ax mov al,dp1 mov bl,dp2 imul bl mov res,ax nop mov ah,4ch int 21h code ends end start EXERCISE:

1. What is the difference between IMUL and MUL? 2. What is the use of instruction CBW?

3. What is the use of instruction CWD?



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PROGRAM: SIGNED DIVISION

.model tiny data segment dp1 db 0d5h dp2 db 20h quo db 0h rem db 0h data ends code segment assume cs:code,ds:data start: mov ax,data mov ds,ax xor ax,ax mov al,dp1 cbw mov bl,dp2 idiv bl mov quo,al mov rem,ah nop mov ah,4ch int 21h code ends end start EXERCISE:

1. What is the purpose of SUB AX,AX? 2. What is the difference between IDIV and DIV?

3. What is the use of instruction CBW & CWD?



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PROGRAM: ASCII ADDITION

.model tiny data segment ip1 db '9',0dh,0ah ip2 db '6',0dh,0ah res db 0 data ends code segment assume cs:code,ds:data start: mov ax,data mov ds,ax xor ax,ax mov al,ip1 add al,ip2 aaa mov res,al mov ah,4ch int 21h code ends end start

EXERCISE:

1. What is the purpose of ASCII addition? 2. What is the instruction used for ASCII addition?

3. Why do we make use of instruction ORL AX,3030H ?

4. Why is aaa after addition?



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PROGRAM: ASCII SUBTRACTION

.model tiny data segment ip1 db '9',0dh,0ah ip2 db '6',0dh,0ah res db 0 data ends code segment assume cs:code,ds:data start: mov ax,data mov ds,ax xor ax,ax mov al,ip1 sub al,ip2 aas mov res,al mov ah,4ch int 21h code ends end start

EXERCISE:

1. What is the purpose of ASCII subtraction? 2. What is the instruction used for ASCII subtraction?



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PROGRAM: ASCII MULTIPLICATION

.model tiny data segment ip1 db 4 ip2 db 6 res db 0 data ends code segment assume cs:code,ds:data start: mov ax,data mov ds,ax xor ax,ax mov al,ip1 mul ip2 aam mov res,al mov ah,4ch int 21h code ends end start EXERCISE:

1. What is the purpose of XCHG instruction is ASCII adjust after multiplication. ? 2. Why is ASCII adjust after multiply cab be called as 1 byte binary to bcd

conversion?

3. What does AL & AH contains?



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PROGRAM: ASCII DIVISION

.model tiny data segment ip1 db 6 ip2 db 4 res db 0 data ends code segment assume cs:code,ds:data start: mov ax,data mov ds,ax xor ax,ax mov al,ip1 aad div ip2 mov res,al mov ah,4ch int 21h code ends end start EXERCISE:

1. What is the ASCII instruction, which is used before the arithmetic operation? 2. Why is ASCII adjust before division is done before actual division?

3. What does AL & AH contains?



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PROGRAM: PACKED TO UNPACKED BCD

.model tiny data segment bcdip db 56h ubcdop dw 0 data ends code segment assume cs:code,ds:data start: mov ax,data mov ds,ax xor ax,ax mov al, bcdip mov dl,al and al,0f0h mov cl,4 ror al,cl mov bh,al and dl,0fh mov bl,dl mov ubcdop,bx mov ah,4ch int 21h code ends end start



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EXERCISE:

1. What is the purpose of the instruction ROR AL, CL? 2. What is the purpose of the instruction AND AL, 0FH & AND AL,0F0H in the

program.?

3. What is the expansion of UPBCD?

4. What is the use of DAA instruction?

5. What is the reason for packing unpacked BCD?

6. What is common between unpacked BCD and ASCII?



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PROGRAM: BCD to ASCII CONVERSION

.model tiny data segment bcdip db 56h ubcdop dw 0 data ends code segment assume cs:code,ds:data start: mov ax,data mov ds,ax xor ax,ax mov al, bcdip mov dl,al and al,0f0h mov cl,4 ror al,cl mov bh,al and dl,0fh mov bl,dl mov ubcdop,bx add bx,3030h mov ah,4ch int 21h code ends end start EXERCISE:

1. What is the difference between adding 30h and OR 30H to a BCD number to conversion to ASCII?

2. Why unpacking is necessary during the conversion?

3. What is the ASCII character for symbol A?

4. What is the ASCII character for symbol zero ‘0’?



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PROGRAM: BLOCK TRANSFER ;.model tiny data segment srcdata db 'Empty vessels make much noise',24h data ends extra segment dstdata db 12 dup(0) extra ends code segment assume cs:code,ds:data,es:extra start: mov ax,data mov ds,ax mov ax,extra mov es,ax mov si,offset srcdata mov di,offset dstdata cld mov cx,29 rep movsb nop mov ah,4ch int 21h code ends end start EXERCISE:

1. If the DF=1, will the SI and DI register decremented? 2. The destination memory is pointed by which register combination?

3. The source is pointed to by which register combination?



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PROGRAM: STRING REVERSAL

.model tiny data segment string1 db 'Empty' strlen equ ($-string1) string2 db 5 dup(0) data ends code segment assume cs:code,ds:data,es:data start: mov ax,data mov ds,ax mov es,ax mov bx,offset string1 mov si,bx mov di,offset string2 add di,5 cld mov cx,strlen a1: mov al,[si] mov es:[di],al inc si dec di loop a1 ;rep movsb ;nop mov ah,4ch int 21h code ends end start EXERCISE:

1. Why BX register is added with ‘5’? 2. Why MOVS instruction is not used?

3. What is the function of LODS and STOS instructions?



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PROGRAM: STRING INSERTION .model tiny data segment

string1 db 'Empty vessels more noise$' strlen equ ($-string1)

data ends extra segment

string2 db strlen+5 dup(0) extra ends code segment

assume cs:code,ds:data,es:extra

start: mov ax,data mov ds,ax mov si,offset string1 mov di,offset string2 cld mov cx,14 rep movsb mov dl,5

back: mov ah,01 int 21h stos string2 dec dl jnz back mov cx,11 rep movsb nop mov ah,4ch int 21h

code ends end start

EXERCISE: 1. Why register ‘DI’ is loaded with 5? 2. What is the function of rep movsb?

3. What is the purpose of mov ah,01h / int 21h?



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PROGRAM: STRING DELETION .model tiny data segment string1 db 'Empty vessels make more noise$' strlen equ ($-string1) data ends extra segment string2 db strlen-5 dup(0) extra ends code segment assume cs:code,ds:data,es:extra start: mov ax,data mov ds,ax mov ax,extra mov es,ax mov si,offset string1 mov di,offset string2 cld mov cx,13 rep movsb cld mov si,18 mov cx,12 rep movsb mov ah,4ch int 21h code ends end start EXERCISE:

1. What is the purpose of string length? 2. What does ‘equ’ stands for?

3. What is the purpose of label start after the end directive?



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PROGRAM: LENGTH OF THE STRING .model tiny data segment string1 db 'Empty vessels make more noise$' strlen equ ($-string1) res db 0 cort db 'strlength found correct$' incort db 'strlength found incorrect$' data ends code segment assume cs:code,ds:data start: mov ax,data mov ds,ax sub cl,cl mov bl,strlen mov si,offset string1 back:lodsb inc cl cmp al,'$' jnz back mov res,cl cmp cl,bl jz correct mov dx,offset incort mov ah,09 int 21h correct:mov dx,offset cort mov ah,09 int 21h mov ah,4ch int 21h code ends end start EXERCISE:

1. What is the operation performed by the instruction cmp al,$ ? 2. What is function 09h / int 21h performed?

3. Why SI is not been incremented is the program?



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PROGRAM: STRING COMPARISION .model tiny data segment string1 db 'Empty' strlen equ ($-string1) notsful db 'strings are unequal$' sful db 'strings are equal$' data ends extra segment string2 db 'Empty' extra ends code segment assume cs:code,ds:data,es:extra start: mov ax,data mov ds,ax mov ax,extra mov es,ax mov si,offset string1 mov di,offset string2 cld ;mov cx,length string1 mov cx,strlen rep cmpsb jz forw mov ah,09h mov dx,offset notsful int 21h jmp exitp forw:mov ah,09h mov dx,offset sful int 21h exitp: nop mov ah,4ch int 21h code ends end start EXERCISE:

1. What is the significance of CLD? 2. How does CMPSB perform the comparison?



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Program design example (calculator) EXTRN ADDER:near, MULT:near, DIVD:near, SUBB:near .model tiny printf macro string mov ah,09h mov dx, offset string int 21h endm getch macro mov ah,01 int 21h endm .stack 160 .data menu db 0ah,0dh,' A Simple Calculator' db 0ah,0dh,' 1. addition' db 0ah,0dh,' 2. subtraction' db 0ah,0dh,' 3. multiplication' db 0ah,0dh,' 4. division' db 0ah,0dh,' X. exit the program' db 0ah,0dh,' input your choice ?$' addproc db 0ah,0dh,' addition procedure?$' subproc db 0ah,0dh,' subtraction procedure?$' divproc db 0ah,0dh,' division procedure?$' mulproc db 0ah,0dh,' multiplication procedure?$' num1 dw 0 num2 dw 0 PRESSANYKEY DB 0AH,0DH,'PRESS ANY KEY $' binnum dw 0 ;accessed by asc2bin overflowdmsg db 0dh,0ah, 'overflow error $'; invalidmsg db 0dh,0ah, 'invalid error $'; enternum db 0dh,0ah, 'please type a number less than 32000 $';from asciinum db '00000$';getnum binout dw 0ffffh RESULT DB 0AH,0DH,0DH,0AH,0AH,0DH,0DH,0AH,0AH,0DH,0DH,0AH,'THE RESULT IS ' ascout db '00000',0AH,0DH,0DH,0AH,0AH,0DH,0DH,0AH,0AH,0DH,0DH,0AH,'$'



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.code .startup

begin: mov ax,@data mov ds,ax menuagain: printf menu getch cmp al,'1' jne nxt1 call addition jmp menuagain nxt1: cmp al,'2' jne nxt2 call subtract jmp menuagain nxt2: cmp al,'3' jne nxt3 call multiply jmp menuagain nxt3: cmp al,'4' jne nxt4 call divide jmp menuagain nxt4: cmp al,'X' jne menuagain .exit



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asc2bin proc mov cx,5 ;put number of digits in cx mov di,10 ;put base 10 in di to multiply acc with mov bh,'9' ;bh to check if number greater than 9 mov bl,'0' ;bl checks lower limit of number mov si,offset asciinum xor ax,ax ; make accumulator zero A2Bagain:mul di jc overflow mov dx,ax ; save multiply accumulate reister AX mov al,[si] cmp al,bl jl invalid cmp al,bh jg invalid sub al,30h cbw add ax,dx jc overflow inc si loop A2Bagain over: mov binnum,ax ret invalid: PRINTF invalidmsg PRINTF PRESSANYKEY GETCH jmp BEGIN overflow: PRINTF overflowdmsg PRINTF PRESSANYKEY GETCH jmp BEGIN asc2bin endp bin2asc proc xor si,si mov di,offset ascout add di,4 mov ax,binOUT mov dx,0 mov bx,10 mov cx,5 ba: div bx add dl,30h mov [di],dl dec di xor dx,dx loop ba ret bin2asc endp



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getnum proc mov ah,9 mov dx,offset enternum int 21h mov si,offset asciinum mov cx,5 GNagain: mov ah,01 int 21h mov [si],al inc si loop GNagain ret getnum endp multiply proc call getnum call asc2bin mov ax,binnum push ax call getnum call asc2bin mov ax,binnum push ax call mult pop ax pop bx add ax,bx mov binout,ax call bin2asc PRINTF RESULT ret multiply endp subtract proc call getnum call asc2bin mov ax,binnum push ax call getnum call asc2bin mov ax,binnum push ax call subb pop ax pop bx add ax,bx mov binout,ax call bin2asc PRINTF RESULT ret subtract endp



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divide proc call getnum call asc2bin mov ax,binnum push ax call getnum call asc2bin mov ax,binnum push ax call divd pop ax pop bx add ax,bx mov binout,ax call bin2asc PRINTF RESULT ret divide endp addition proc call getnum call asc2bin push binnum call getnum call asc2bin push BINNUM CALL ADDER pop ax pop bx mov binout,ax call bin2asc PRINTF RESULT ret addition endp END



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public adder .model tiny .code adder proc near push bp mov bp,sp add bp,6 mov ax,[bp] dec bp dec bp mov bx,[bp] add ax,bx mov [bp],ax inc bp inc bp mov ax,0 mov [bp],ax pop bp ret adder endp end public mult .model tiny .code mult proc near push bp mov bp,sp add bp,6 mov ax,[bp] dec bp dec bp mov bx,[bp] mul bx mov [bp],ax inc bp inc bp mov ax,0 mov [bp],ax pop bp ret mult endp end



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public subb .model tiny .code subb proc near push bp mov bp,sp add bp,6 mov ax,[bp] dec bp dec bp mov bx,[bp] sub ax,bx mov [bp],ax inc bp inc bp mov ax,0 mov [bp],ax pop bp ret subb endp end public divd .model tiny .code divd proc near push bp mov bp,sp add bp,6 mov ax,[bp] dec bp dec bp mov bx,[bp] mov dx,00 div bx mov [bp],ax inc bp inc bp mov ax,0 mov [bp],ax pop bp ret divd endp end



Lab incharge HOD

EXERCISE:

1. What is the difference between a macro and a procedure? 2. What is the purpose of having 0ah & 0dh in message statement?

3. What is the associated directive to the EXTRN directive?

4. The menu in the instruction printf menu represents what?

5. What is performed by the directive startup?

6. Convert one of the procedure into near procedure?

7. Convert another procedure into far procedure in ASCII to binary procedure, a

famous algorithm is used. What is the name of the algorithm [refer Gibbson’s book] ?

8. What is the significance of public in the file named adder?

9. What is BP used for in the adder procedure?

10. If the variables are RXCD and passed by the adder is it been by done by the

value or reference?



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PROGRAM: DOS INTERRUPTS

.model tiny data segment msg1 db 0ah,0dh,'Enter the first number: $' msg2 db 0ah,0dh,'Enter the second number: $' msg3 db 0ah,0dh,'Result: $' res db 00 data ends code segment assume cs:code,ds:data start: mov ax,data mov ds,ax lea dx,msg1 mov ah,09h int 21h mov ah,01h int 21h sub al,30h mov bl,al lea dx,msg2 mov ah,09h int 21h mov ah,01h int 21h sub al,30h add al,bl add al,30h mov res,al lea dx,msg3 mov ah,09h int 21h mov dl,res mov ah,02h int 21h mov ah,4ch int 21h code ends end start



Lab incharge HOD

EXERCISE:

1. After executing the instructions mov ah,01h / int 21h, what are contents of AL ? 2. What are the contents of AH for displaying a sting in the screen?

3. What are the contents of AH for accepting a number from keyboard?

4. Which register contains the address of the string to be displayed on screen?



Lab incharge HOD

Programmable Keyboard/Display Interface – 8279

• A programmable keyboard and display interfacing chip. o Scans and encodes up to a 64-key keyboard. o Controls up to a 16-digit numerical display.

• Keyboard has a built-in FIFO 8 character buffer.

• The display is controlled from an internal 16x8 RAM that stores the coded

display information. Pinout Definition 8279

• A0: Selects data (0) or control/status (1) for reads and writes between micro and 8279. • BD: Output that blanks the displays. • CLK: Used internally for timing. Max is 3 MHz. • CN/ST: Control/strobe, connected to the control key on the keyboard. • CS: Chip select that enables programming, reading the keyboard, etc. • DB7-DB0: Consists of bidirectional pins that connect to data bus on micro. • IRQ: Interrupt request, becomes 1 when a key is pressed, data is available. • OUT A3-A0/B3-B0: Outputs that sends data to the most significant/least significant nibble of

display. • RD(WR): Connects to micro's IORC or RD signal, reads data/status registers. • RESET: Connects to system RESET. • RL7-RL0: Return lines are inputs used to sense key depression in the keyboard matrix. • Shift: Shift connects to Shift key on keyboard. • SL3-SL0: Scan line outputs scan both the keyboard and displays.

Keyboard Interface of 8279 • The keyboard matrix can be any size from 2x2 to 8x8.

• Pins SL2-SL0 sequentially scan each column through a counting operation.

o The 74LS138 drives 0's on one line at a time. o The 8279 scans RL pins synchronously with the scan. o RL pins incorporate internal pull-ups, no need for

external resistor pull-ups.

Unlike the 82C55, the 8279 must be programmed first. D7 D6 D5 Function Purpose

0 0 0 Mode set Selects the number of display positions, type of key scan...

0 0 1 Clock Programs internal clk, sets scan and debounce times.

0 1 0 Read FIFO Selects type of FIFO read and address of



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the read.

0 1 1 Read Display Selects type of display read and address of the read.

1 0 0 Write Display Selects type of write and the address of the write.

1 0 1 Display write inhibit Allows half-bytes to be blanked.

1 1 0 Clear Clears the display or FIFO

1 1 1 End interrupt Clears the IRQ signal to the microprocessor.

The first 3 bits of # sent to control port selects one of 8 control words.

First three bits given below select one of 8 control registers (opcode).

• 000DDMMM o Mode set: Opcode 000.

DD sets displays mode. MMM sets keyboard mode.

o DD field selects either:

• 8- or 16-digit display • Whether new data are entered to the rightmost or leftmost display position.

DD Function 00 8-digit display with left entry 01 16-digit display with left entry 10 8-digit display with right entry 11 16-digit display with right entry

Keyboard Interface of 8279

MMM field:

Encoded: Sl outputs are active-high, follow binary bit pattern 0-7 or 0-15. Decoded: SL outputs are active-low (only one low at any time).

MMM Function 000 Encoded keyboard with 2-key lockout 001 Decoded keyboard with 2-key lockout 010 Encoded keyboard with N-key rollover 011 Decoded keyboard with N-key rollover 100 Encoded sensor matrix 101 Decoded sensor matrix 110 Strobed keyboard, encoded display scan111 Strobed keyboard, decoded display scan



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Pattern output: 1110, 1101, 1011, 0111. Strobed: An active high pulse on the CN/ST input pin strobes data from the RL pins into an internal FIFO for reading by micro later.

o 2-key lockout/N-key rollover: Prevents 2 keys from being recognized

if pressed simultaneously/Accepts all keys pressed from 1st to last.

• 001PPPPP o The clock command word programs the internal clock driver. o The code PPPPP divides the clock input pin (CLK) to achieve the

desired operating frequency, e.g. 100KHz requires 01010 for a 1 MHz CLK input.

• 010Z0AAA

o The read FIFO control word selects the address (AAA) of a keystroke from the FIFO buffer (000 to 111).

o Z selects auto-increment for the address.

• 011ZAAAA o The display read control word selects the read address of one of the

display RAM positions for reading through the data port.

• 100ZAAAA o Selects write address -- Z selects auto-increment so subsequent writes

go to subsequent display positions.

• 1010WWBB o The display write inhibit control word inhibits writing to either the

leftmost 4 bits of the display (left W) or rightmost 4 bits. o BB works similarly except that they blank (turn off) half of the output

pins. • 1100CCFA

o The clear control word clears the display, FIFO or both o Bit F clears FIFO and the display RAM status, and sets address

pointer to 000. If CC are 00 or 01, all display RAM locations become 00000000. If CC is 10, --> 00100000, if CC is 11, --> 11111111.

• 1110E000 o End of Interrupt control word is issued to clear IRQ pin in sensor

matrix mode.

• 1) Clock must be programmed first. If 3.0 MHz drives CLK input, PPPPP is programmed to 30 or 11110.



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• 2) Keyboard type is programmed next.

o The previous example illustrates an encoded keyboard, external decoder used to drive matrix.

• 3) Program the FIFO.

• Once done, a procedure is needed to read data from the keyboard.

o To determine if a character has been typed, the FIFO status register is checked.

o When this control port is addressed by the IN instruction, the contents of the FIFO status word is copied into register AL:

• Data returned from 8279 contains raw data that need to be translated to ASCII:

o Row and column number are given the rightmost 6 bits (scan/return).

o This can be converted to ASCII using the XLAT instruction with an ASCII code lookup table.

o The CT and SH indicate whether the control or shift keys were pressed.

The Strobed Keyboard code is just the state of the RLx bits at the time a 1 was `strobed' on the strobe input pin.



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PROGRAM: KEYBOARD DISPLAY RIGHT ENTRY

cmdreg equ 0ffebh datareg equ 0ffe9h

.8086 .model tiny .stack 32 .data .code org 2000h start:

mov ax,cs mov ds,ax mov al,0H ; ENCODED SCAN-8CHAR 8 BIT LEFT ENTRY mov dx,cmdreg out dx,al mov al,090h ;WRITE RAM AUTO INCREMENT out dx,al mov cx,08 clear: mov al,00 MOV DX,datareg out DX,al loop clear back: mov dx,cmdreg in al,dx ; READ THE COMMAND R4EGISTER TO GET ; THE 8279 STATUS and al,07 ; NUMBER OF KEYS PRESSED IS IN THE ; LOWER 3 BITS ; MASK THESE AND CHECK IF NOT ZERO. IF ; ZERO NO KEY PRESSED jz back mov bx,offset sscharlut mov al,040h ; READ THE FIRST RAM ADDRESS OF FIFO out dx,al mov Dx,datareg in al, dx and al,01fh



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xlat ; CONVERT THE KEY READ TO THE SS CODE mov dx,datareg out dx,al ; DISPLAY AT THE CURRENT DIGIT ; POSITION jmp back sscharlut: SS0 DB 3FH ;0 SS1 DB 6H ;1 SS2 DB 5BH ;2 SS3 DB 4FH ;3 SS4 DB 66H ;4 SS5 DB 6DH ;5 SS6 DB 7DH ;6 SS7 DB 07H ;7 SS8 DB 7FH SS9 DB 6FH SSA DB 77H SSB DB 7cH SSC DB 39H SSD DB 5eH SSE DB 79H SSF DB 71H end start



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PROGRAM: KEYBOARD DISPLAY LEFT ENTRY cmdreg equ 0ffebh datareg equ 0ffe9h

.8086 .model tiny .stack 32 .data .code org 2000h start: mov ax,cs mov ds,ax mov al,10H ; ENCODED SCAN-8CHAR 8 BIT LEFT ENTRY mov dx,cmdreg out dx,al mov al,090h ;WRITE RAM AUTO INCREMENT out dx,al mov cx,08 clear: mov al,00 MOV DX,datareg out DX,al loop clear back: mov dx,cmdreg in al,dx ; READ THE COMMAND R4EGISTER TO GET ; THE 8279 STATUS and al,07 ; NUMBER OF KEYS PRESSED IS IN THE ; LOWER 3 BITS ; MASK THESE AND CHECK IF NOT ZERO. IF ; ZERO NO KEY PRESSED jz back mov bx,offset sscharlut mov al,040h ; READ THE FIRST RAM ADDRESS OF FIFO out dx,al mov Dx,datareg; in al, dx and al,01fh xlat ; CONVERT THE KEY READ TO THE SS CODE mov dx,datareg



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out dx,al ; DISPLAY AT THE CURRENT DIGIT

; POSITION jmp back sscharlut: SS0 DB 3FH ;0 SS1 DB 6H ;1 SS2 DB 5BH ;2 SS3 DB 4FH ;3 SS4 DB 66H ;4 SS5 DB 6DH ;5 SS6 DB 7DH ;6 SS7 DB 07H ;7 SS8 DB 7FH SS9 DB 6FH SSA DB 77H SSB DB 7cH SSC DB 39H SSD DB 5eH SSE DB 79H SSF DB 71H end start EXERCISE:

1. How many puns does 8279 IC have? 2. How many scan lines are available on 8279?

3. How many row-sense pins are there on 8279?

4. How many registers are there in 8279?

5. What is the purpose of command & data registers?

6. What are the first 4 bits D7 – D4 in the command register used for?



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7. If 16 character 8 bit left entry LED display is to be programmed. What is the command code?

8. What is the use of org 2000h? can the 2000h changed to another value?

9. What are the data values of ‘SSCHARLUT’ represent?

10. Explain the use of XLAT and what default registers it use?

11. Out dx,al the operand 1 is 16 bit and operand 2 is 16 bit. How does it work?

12. What is the interrupt used for inputting data from i/o port?

13. What is the states register address of 8279?

14. In 8279, if consists of lots of registers, how is it only 2 addresses are used?

15. How do you change the right entry to left entry mode is LED character display?

16. Explain the sensor mode of 8279?

17. Can the addressed RAM be automatically incremented, if so how it is done?

18. When the states register is read, the lower 32bits represents what?

19. How are the shift and ctrl keys are identified in 8279?

20. Write down the control word required for 16 characters 8bit left entry, 16 keys

N-key roll over?

21. What is the meaning N-key roll over?

22. What is 2-key lock out?

23. What is the circuit which processes the clock given to 8279 and steps it down programmable to the required rate called?



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LEDS & SWITCH INTERFACE(8086)

This experimental board has been developed for the purpose of understanding the operation of LEDS & SWITCH. This experiment makes use of programmable i/o port called programmable peripheral interface (PPI) which is numbered as i8255 has three 8086 on the trainer kit. 8255 has three i/o ports and they are grouped as group A & groupB. Group A contains port A & port B & lower nibble of port C. The 8255 can be operated in three modes. Mode 0 is for simple I/O mode 1 is for handshake i/o. Mode 2 is for bi-directional multiprocessor interfacing. The i/o ports of 8255 have a fan in of 10 & fanout of 1 the operations of8255 are controlled by a control register which can be programmed to set the mode & i/p o/p functions of ports. All the port pins are collectively programmable & not individually programmable. For example 1 port A pins PA0 to PA7 have to be all inputs or all outputs. They cannot be programmed as PA0 in & PA0 out individually as 2 four bit ports which can be programmed as a group for i/p of o/p. PC0 to PC3 which are the lower nibble of port C can be programmed as input or output while upper nibble of port C PC4 to PC7 can be programmed as input or output. The control word format is as follows.

D7 D6 D5 D4 D3 D2 D1 D0 Active/BSR Group A PA PCH GroupB PB PCL Mode i/p of o/p Mode D7- When set it is in the action operation where the i/o s can be programmed. If D7 is clear it is in bit set reset mode. Which is used to set reset port C bits. This mode is not of great significance in the use of 8255. D7 is one as 8255 is used in the active mode. D6,D5 are the bits used far setting the mode of group A ports.

00 Mode 0 01 Mode 1 10 Mode 2 11 Mode 3

D4 sets the port A as input or output depending upon whether it is 1 or 0 respectively. D3 sets the port C higher as input or output depending upon whether it is 1 of 0 respectively. D2 this bit is for setting the mode of group B which has only 2 modes mode0 and mode1. D1 sets the port B as i/p or o/p depending upon whether it is 1 or 0 respectively. D0 sets the port C lower as i/p or o/p depending upon whether it is 1 or 0 respectively. To program 8255 for lighting the LED lamps it is important for us to know the LED specifications.



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Green Red Forward voltage drop 1.9V 1.8V Typical operating current 5ma 5ma Max current 15ma 15ma Dimensions 5mm φ 5mm φ For further details refer website http:\\ crsarma.tripod.com. A current of 10ma was passed through lamp for lighting. The output voltage, used to drive it was form a TTL port. So the balance of voltage 3.4V has to be dropped across this resistor while passing 10ma of current therefore resistor value is 5V-1.6V = 340 Ω is not available in the standard range of resistors. 10ma So in the standard range which ever is close to 340 Ω is chosen. Port A was connected to four RED & four GREEN LEDs arranged alternately. SWITCHES Port C was programmed for2 ports of nibble each. So the upper nibble (PCH PC4….PC7) could be used for i/p or o/p subsequently. An i/p port papers in a tri-stated condition which results in a floating voltage. So inputs have to be given a pull-up resistor which makes at logic 1 level. A toggle switch is then connected to port pin with its other end connected to ground. If the toggle switches open, the particular bit is at logic 1 otherwise logic 0. The switches are connected to PC0 & PC1.



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PROGRAM: 8255(LED’S USING SWITCHES)

.8086 .model tiny .stack 16 .code org 2000h start: mov ax,cs mov ds,ax

mov dx,0ffe7h mov al,81h out dx,al h1: mov dx,0ffe5h in al,dx test al,03 jnz nxt1 call alternet jmp h1 nxt1: test al,01 jnz nxt2 call r2l jmp h1 nxt2: test al,02 jnz h1 call l2r jmp h1 alternet proc mov dx,0ffe1h BACK: mov al,0aah out dx,al call delay mov al,055h out dx,al call delay ret alternet endp



Lab incharge HOD

l2r proc mov cx,8 mov al,01h mov dx,0ffe1h BACK2: out dx,al call delay rol al,01h loop back2 ret l2r endp delay proc push cx mov cx,0ffffh

s1: loop s1 pop cx ret delay endp r2l proc mov cx,8 mov al,80h MOV DX,0FFE1H BACK3: out dx,al call delay ror al,01h loop back3 ret r2l endp end start



Lab incharge HOD

LED’S USING SWITCHES

Exercise:

1. What is the control word format for 8255?

2. How is the choice of resistor connected to the led made?

3. What is the forward drop of green leds?

4. What is the forward drop of red leds?

5. What is the current drown by LED’s(8)?



Lab incharge HOD

MCS 51 DEVELOPMENT TOOLS

Using Command Line

1. Edit the program using ‘EDIT’ editor 2. Save as name .A51 3. Assemble using A51.EXE

C: A51 name .A51 Result: name .LST, name .OBJ

4. LINK & Convert to Hex format using OH51.EXE C: OH51 name .OBJ Result: name .HEX

5. Simulate Using SIM31.EXE

C: SIM31

Using Integrated Design Environment (IDE) KEIL µVision – (Windows)

An Integrated design environment consists of all the tools needed for the development of software for the MCS 51 family. These tools are linked together in one environment. The main feature of IDE is the BUILD process. The BUILD ensures that the target file is current. In order to keep modular development the PROJECT feature is provided.

1. Select START\PROGRAMS\KEIL <Left Click> 2. Create New project

Select PROJECT\NEW\<Left Click> “enter name’<Left Click> Choose TARGET\INTEL\8051 <Left Click> Choose BUILD OPTIONS Check Box HEX <OK><Left Click>

3. Create File <ALT><F><N> ‘name of file .ASM’ <Left Click> Edit File & Save

4. SELECT PROJECT WINDOW – ADD FILE – SELECT - <Left Click> 5. PROJECT\BUILD\ <Left Click>

If no errors goto 7 else 3 DEBUG\START\ <Left Click>



Lab incharge HOD

LEDS & SWITCH INTERFACE(8051)

This experimental board has been developed for the purpose of understanding the operation of LEDS & SWITCH. This experiment makes use of programmable i/o port called programmable peripheral interface (PPI) which is numbered as i8255 has three 8086 on the trainer kit. 8255 has three i/o ports and they are grouped as group A & groupB. Group A contains port A & port B & lower nibble of port C. The 8255 can be operated in three modes. Mode 0 is for simple I/O mode 1 is for handshake i/o. Mode 2 is for bi-directional multiprocessor interfacing. The i/o ports of 8255 have a fan in of 10 & fanout of 1 the operations of8255 are controlled by a control register which can be programmed to set the mode & i/p o/p functions of ports. All the port pins are collectively programmable & not individually programmable. For example 1 port A pins PA0 to PA7 have to be all inputs or all outputs. They cannot be programmed as PA0 in & PA0 out individually as 2 four bit ports which can be programmed as a group for i/p of o/p. PC0 to PC3 which are the lower nibble of port C can be programmed as input or output while upper nibble of port C PC4 to PC7 can be programmed as input or output. The control word format is as follows.

D7 D6 D5 D4 D3 D2 D1 D0 Active/BSR Group A PA PCH GroupB PB PCL Mode i/p of o/p Mode D7- When set it is in the action operation where the i/o s can be programmed. If D7 is clear it is in bit set reset mode. Which is used to set reset port C bits. This mode is not of great significance in the use of 8255. D7 is one as 8255 is used in the active mode. D6,D5 are the bits used far setting the mode of group A ports.

02 Mode 0 03 Mode 1 12 Mode 2 13 Mode 3

D4 sets the port A as input or output depending upon whether it is 1 or 0 respectively. D3 sets the port C higher as input or output depending upon whether it is 1 of 0 respectively. D2 this bit is for setting the mode of group B which has only 2 modes mode0 and mode1. D1 sets the port B as i/p or o/p depending upon whether it is 1 or 0 respectively. D0 sets the port C lower as i/p or o/p depending upon whether it is 1 or 0 respectively. To program 8255 for lighting the LED lamps it is important for us to know the LED specifications.



Lab incharge HOD

Green Red Forward voltage drop 1.9V 1.8V Typical operating current 5ma 5ma Max current 15ma 15ma Dimensions 5mm φ 5mm φ For further details refer website http:\\ crsarma.tripod.com. A current of 10ma was passed through lamp for lighting. The output voltage, used to drive it was form a TTL port. So the balance of voltage 3.4V has to be dropped across this resistor while passing 10ma of current therefore resistor value is 5V-1.6V = 340 Ω is not available in the standard range of resistors. 10ma So in the standard range which ever is close to 340 Ω is chosen. Port A was connected to four RED & four GREEN LEDs arranged alternately. SWITCHES Port C was programmed for2 ports of nibble each. So the upper nibble (PCH PC4….PC7) could be used for i/p or o/p subsequently. An i/p port papers in a tri-stated condition which results in a floating voltage. So inputs have to be given a pull-up resistor which makes at logic 1 level. A toggle switch is then connected to port pin with its other end connected to ground. If the toggle switches open, the particular bit is at logic 1 otherwise logic 0. The switches are connected to PC0 & PC1.



Lab incharge HOD

LED ROTATE RIGHT AND ROTATE LEFT PROGRAM

PORTA EQU 2808H PB EQU 2809H PORTC EQU 280AH PCW EQU 280BH DELAY EQU 0114H FIRST EQU 170 SECOND EQU 85 ORG 6500H MOV DPTR,#PCW ; DPTR HOLDS THE CONTROL REGISTER ADDRESS MOV A,#080H MOVX @DPTR,A RIGHTLED: MOV A,#01 MOV DPTR,#PORTA RIGHTLED1: MOVX @DPTR,A MOV R2,#0FFH MOV R1,#0FFH LCALL DELAY RR A JMP RIGHTLED1 LEFTLED: MOV A,#01 MOV DPTR,#PORTA LEFTLED1: MOVX @DPTR,A MOV R2,#0FFH MOV R1,#0FFH LCALL DELAY RL A JMP LEFTLED1 AGAIN: MOV DPTR,#PORTA MOV A,#FIRST MOVX @DPTR,A MOV R2,#0FFH MOV R1,#0FFH LCALL DELAY MOV A,#SECOND MOVX @DPTR,A MOV R2,#0FFH MOV R1,#0FFH LCALL DELAY JMP AGAIN END



Lab incharge HOD

LED’S AND SWITCH INTERFACE

Exercise: 1. What is the control word format for 8255?

2. How is the choice of resistor connected to the led made?

3. What is the forward drop of green LED’s?

4. What is the forward drop of red LED’s?

5. What is the current drown by LED’s(8)?



Lab incharge HOD

PROGRAM: Transferring a block of data from internal

ROM to internal RAM org 0000h

ljmp start org 000bh reti org 0013h reti org 001bh reti org 0023h reti org 0023h reti org 0026h reti org 200h

start: mov dptr,#message mov r0,#20h mov r7,#11

rep: mov a,#00h movc a,@a+dptr mov @r0,a inc r0 inc dptr djnz r7,rep message: db "hello world" end Exercise:

1. What are the two registers in 8051, which are used for indirect addressing? 2. What are the 16 bit registers available in 8051?

3. Which register cannot be decremented?

4. 8051 is a ------- bit microcontroller?



Lab incharge HOD

PROGRAM: (a). Internal ROM to external RAM

ORG 0000H START: mov a,#00 mov sp,#07h mov r3,#11 mov dptr,#200h push dpl push dph rep: mov dptr,#message mov r2,a movc a,@a+dptr pop dph pop dpl movx @dptr,a inc dptr push dpl push dph inc r2 mov a,r2 djnz r3,rep message: db "hello word" end Exercise:

1. What does “djnz” instruction do? 2. What is DPTR relative addressing? 3. How many 8 bit registers are available in 8051?

4. What is the purpose of R2?




Lab incharge HOD

PROGRAM: b). Internal ROM to external RAM

org 200h start: mov r0,#40h mov a,#30h mov r6,#10 rep: mov @r0,a inc a inc r0 djnz r6,rep mov r1,#40h mov dptr,#400h mov r7,#00h rep1: mov a,@r0 movx @dptr,a inc dptr inc r1 inc r7 cjne r7,#10,rep1 end Exercise:

6. What is DPTR relative addressing? 7. What is the purpose of R7?


9. What does “cjne” instruction do?

6. The stack pointer of 8051 grows upwards or downwards



Lab incharge HOD

PROGRAM: Understanding three memory areas of 00-FF

ORG 0000H ljmp start org 200h start: mov a,#41h mov 20h,a mov a,#42h mov r0,#20h mov @r0,a mov a,#43h mov 80h,a mov a,#44h mov r0,#80h mov @r0,a end Exercise:

1. What is the instruction used for loading accumulator from code memory? 2. What is the purpose of register R0?

3. What are the two registers in 8051 which are used for indirect addressing?

mpi lab manual

Documents

assembler program

source program

assembler list file

tasm source file

link file

exe file

program upto

large object file