Microcontroller 8051 soft
description
Transcript of Microcontroller 8051 soft
Numerical Bases Used in Programming
• Hexadecimal
• Binary
• BCD
Hexadecimal Basis
• Hexadecimal Digits:
1 2 3 4 5 6 7 8 9 A B C D E F
A=10 B=11 C=12 D=13 E=14 F=15
Decimal, Binary, BCD, & Hexadecimal Numbers
(43)10=
(0100 0011)BCD=
( 0010 1011 )2 =
( 2 B )16
Registers
A
B
R0
R1
R3
R4
R2
R5
R7
R6
DPH DPL
PC
DPTR
PC
Some 8051 16-bit Register
Some 8-bitt Registers of the 8051
SP
Memory mapping in 8051
• ROM memory map in 8051 family
0000H
0FFFH
0000H
1FFFH8751
AT89C51 8752AT89C52
4k 8k
• RAM memory space allocation in the 8051
7FH
30H
2FH
20H
1FH
17H
10H
0FH
07H
08H
18H
00HRegister Bank 0
(Stack )Register Bank 1
Register Bank 2
Register Bank 3
Bit-Addressable RAM
Scratch pad RAM
Addressing Modes
• Register• Direct • Register Indirect• Immediate• Relative• Absolute• Long• Indexed
Register Addressing Mode
MOV Rn, A ;n=0,..,7
ADD A, Rn
MOV DPL, R6
MOV DPTR, A
MOV Rm, Rn
Direct Addressing Mode
Although the entire of 128 bytes of RAM can be accessed using direct addressing mode, it is most often used to access RAM loc. 30 – 7FH.
MOV R0, 40HMOV 56H, AMOV A, 4 ; ≡ MOV A, R4MOV 6, 2 ; copy R2 to R6
; MOV R6,R2 is invalid !
Register Indirect Addressing Mode
• In this mode, register is used as a pointer to the data.
MOV A,@Ri ; move content of RAM loc. where address is held by Ri into A ( i=0 or 1 )
MOV @R1,B
In other word, the content of register R0 or R1 is sources or target in MOV, ADD and SUBB insructions.
Immediate Addressing Mode
MOV A,#65H
MOV R6,#65H
MOV DPTR,#2343H
MOV P1,#65H
Relative, Absolute, & Long Addressing
Used only with jump and call instructions:
SJMP
ACALL,AJMP
LCALL,LJMP
Indexed Addressing Mode
• This mode is widely used in accessing data elements of look-up table entries located in the program (code) space ROM at the 8051
MOVC A,@A+DPTR (A,@A+PC)
A= content of address A +DPTR from ROM Note:
Because the data elements are stored in the program (code ) space ROM of the 8051, it uses the instruction MOVC instead of MOV. The “C” means code.
Some Simple InstructionsMOV dest,source ; dest = source
MOV A,#72H ;A=72HMOV R4,#62H ;R4=62HMOV B,0F9H ;B=the content of F9’th byte of RAM
MOV DPTR,#7634HMOV DPL,#34HMOV DPH,#76H
MOV P1,A ;mov A to port 1
Note 1:MOV A,#72H ≠ MOV A,72HAfter instruction “MOV A,72H ” the content of 72’th byte of RAM will replace in Accumulator.
Note 2:MOV A,R3 ≡ MOV A,3
ADD A, Source ;A=A+SOURCE
ADD A,#6 ;A=A+6
ADD A,R6 ;A=A+R6
ADD A,6 ;A=A+[6] or A=A+R6
ADD A,0F3H ;A=A+[0F3H]
SUBB A, Source ;A=A-SOURCE-C
SUBB A,#6 ;A=A-6
SUBB A,R6 ;A=A+R6
MUL & DIV
• MUL AB ;B|A = A*BMOV A,#25HMOV B,#65HMUL AB ;25H*65H=0E99
;B=0EH, A=99H
• DIV AB ;A = A/B, B = A mod BMOV A,#25MOV B,#10DIV AB ;A=2, B=5
SETB bit ; bit=1CLR bit ; bit=0
SETB C ; CY=1SETB P0.0 ;bit 0 from port 0 =1SETB P3.7 ;bit 7 from port 3 =1SETB ACC.2 ;bit 2 from ACCUMULATOR =1SETB 05 ;set high D5 of RAM loc. 20h
Note:
CLR instruction is as same as SETBi.e.:
CLR C ;CY=0
But following instruction is only for CLR:CLR A ;A=0
DEC byte ;byte=byte-1
INC byte ;byte=byte+1
INC R7
DEC A
DEC 40H ; [40]=[40]-1
RR – RL – RRC – RLC A
EXAMPLE:RR A
RR:
RRC:
RL:
RLC:
C
C
ANL - ORL – XRLBitwise Logical Operations: AND, OR, XOREXAMPLE:
MOV R5,#89HANL R5,#08H
CPL A ;1’s complementExample:
MOV A,#55H ;A=01010101 BL01: CPL A
MOV P1,AACALL DELAYSJMP L01
Stack in the 8051• The register used to
access the stack is called SP (stack pointer) register.
• The stack pointer in the 8051 is only 8 bits wide, which means that it can take value 00 to FFH. When 8051 powered up, the SP register contains value 07.
7FH
30H
2FH
20H
1FH
17H10H
0FH
07H
08H
18H
00HRegister Bank 0
(Stack )Register Bank 1
Register Bank 2
Register Bank 3
Bit-Addressable RAM
Scratch pad RAM
Example:MOV R6,#25HMOV R1,#12HMOV R4,#0F3HPUSH 6PUSH 1PUSH 4
0BH
0AH
09H
08H
Start SP=07H
25
0BH
0AH
09H
08H
SP=08H
F3
12
25
0BH
0AH
09H
08H
SP=08H
12
25
0BH
0AH
09H
08H
SP=09H
LOOP and JUMP Instructions
JZ Jump if A=0
JNZ Jump if A/=0
DJNZ Decrement and jump if A/=0
CJNE A,byte Jump if A/=byte
CJNE reg,#data Jump if byte/=#data
JC Jump if CY=1
JNC Jump if CY=0
JB Jump if bit=1
JNB Jump if bit=0
JBC Jump if bit=1 and clear bit
Conditional Jumps :
DJNZ:
Write a program to clear ACC, then
add 3 to the accumulator ten time
Solution:
MOV A,#0
MOV R2,#10
AGAIN: ADD A,#03
DJNZ R2,AGAIN ;repeat until R2=0 (10 times)
MOV R5,A
LJMP(long jump)LJMP is an unconditional jump. It is a 3-byte instruction. It allows a jump to any memory location from 0000 to FFFFH.
AJMP(absolute jump)In this 2-byte instruction, It allows a jump to any memory location within the 2k block of program memory.
SJMP(short jump)
In this 2-byte instruction. The relative address range of 00-FFH is divided into forward and backward jumps, that is , within -128 to +127 bytes of memory relative to the address of the current PC.
CALL InstructionsAnother control transfer instruction is the CALL instruction, which is used to call a subroutine.
• LCALL(long call)
This 3-byte instruction can be used to call subroutines located anywhere within the 64K byte address space of the 8051.
• ACALL (absolute call)ACALL is 2-byte instruction. the target address of the subroutine must be within 2K byte range.
Example:Write a program to copy a block of 10 bytes from RAM location starting at 37h to RAM location starting at 59h.
Solution:MOV R0,#37h ; source pointerMOV R1,#59h ; dest pointer MOV R2,#10 ; counter
L1: MOV A,@R0MOV @R1,AINC R0INC R1DJNZ R2,L1
. 100's 10's 1's
. 1 5 6
+ 2 4 8
= 4 0 4
Decimal Addition156 + 248
16 Bit Addition
1A44 + 22DB
. 256's 16’s 1's
. 1 A 4 4
+ 2 2 D B
= 3 D 1 F
= 3D1F
Performing the Addition with 8051
. 65536's 256's 1's
. R6 R7
+ R4 R5
= R1 R2 R3
1.Add the low bytes R7 and R5, leave the answer in R3.
2.Add the high bytes R6 and R4, adding any carry from step 1, and leave the answer in R2. 3.Put any carry from step 2 in the final byte, R1.
Steps 1, 2, 3
MOV A,R7 ;Move the low-byte into the accumulator ADD A,R5 ;Add the second low-byte to the accumulator
MOV R3,A ;Move the answer to the low-byte of the result
MOV A,R6 ;Move the high-byte into the accumulator
ADDC A,R4 ;Add the second high-byte to the accumulator, plus carry.
MOV R2,A ;Move the answer to the high-byte of the result
MOV A,#00h ;By default, the highest byte will be zero.
ADDC A,#00h ;Add zero, plus carry from step 2.
MOV R1,A ;Move the answer to the highest byte of the result
The Whole Program;Load the first value into R6 and R7 MOV R6,#1Ah MOV R7,#44h ;Load the first value into R4 and R5 MOV R4,#22h MOV R5,#0DBh ;Call the 16-bit addition routine LCALL ADD16_16
ADD16_16:
;Step 1 of the process MOV A,R7 ;Move the low-byte into the accumulator ADD A,R5 ;Add the second low-byte to the accumulator MOV R3,A ;Move the answer to the low-byte of the result
;Step 2 of the process MOV A,R6 ;Move the high-byte into the accumulator ADDC A,R4 ;Add the second high-byte to the accumulator, plus carry. MOV R2,A ;Move the answer to the high-byte of the result
;Step 3 of the process MOV A,#00h ;By default, the highest byte will be zero. ADDC A,#00h ;Add zero, plus carry from step 2. MOV MOV R1,A ;Move the answer to the highest byte of the result
;Return - answer now resides in R1, R2, and R3. RET
Timer & Port Operations• Example:Write a program using Timer0 to create a 10khz square
wave on P1.0
MOV TMOD,#02H ;8-bit auto-reload modeMOV TH0,#-50 ;-50 reload value in TH0SETB TR0 ;start timer0
LOOP: JNB TF0, LOOP ;wait for overflowCLR TF0 ;clear timer0 overflow flagCPL P1.0 ;toggle port bitSJMP LOOP ;repeatEND
Interrupts
1. Enabling and Disabling Interrupts
2. Interrupt Priority
3. Writing the ISR (Interrupt Service Routine)
Interrupt Enable (IE) Register :
• EA : Global enable/disable.• --- : Undefined.• ET2 :Enable Timer 2 interrupt.• ES :Enable Serial port interrupt.• ET1 :Enable Timer 1 interrupt.• EX1 :Enable External 1 interrupt.• ET0 : Enable Timer 0 interrupt. • EX0 : Enable External 0 interrupt.
Interrupt Vectors
Interrupt Vector AddressSystem Reset 0000H
External 0 0003H
Timer 0 000BH
External 1 0013H
Timer 1 001BH
Serial Port 0023H
Timer 2 002BH
Writing the ISRExample:
Writing the ISR for Timer0 interrupt
ORG 0000H ;reset
LJMP MAIN
ORG 000BH ;Timer0 entry point
T0ISR: . ;Timer0 ISR begins
.
RETI ;return to main program
MAIN: . ;main program
.
.
END
Structure of Assembly language and Running an 8051
programEDITOR
PROGRAM
ASSEMBLERPROGRAM
LINKERPROGRAM
OHPROGRAM
Myfile.asm
Myfile.obj
Other obj fileMyfile.lst
Myfile.hex
Examples of Our Program Instructions
• MOV C,P1.4
JC LINE1
• SETB P1.0
CLR P1.2
8051 Instruction SetACALL: Absolute Call
ADD, ADDC: Add Acc. (With Carry)
AJMP: Absolute Jump
ANL: Bitwise AND
CJNE: Compare & Jump if Not Equal
CLR: Clear Register
CPL: Complement Register
DA: Decimal Adjust
DEC: Decrement Register
DIV: Divide Accumulator by B
DJNZ: Dec. Reg. & Jump if Not Zero
INC: Increment Register
JB: Jump if Bit Set
JBC: Jump if Bit Set and Clear Bit
JC: Jump if Carry Set
JMP: Jump to Address
JNB: Jump if Bit Not Set
JNC: Jump if Carry Not Set
JNZ: Jump if Acc. Not Zero
JZ: Jump if Accumulator Zero
LCALL: Long Call
LJMP: Long Jump
MOV: Move Memory
MOVC: Move Code Memory
MOVX: Move Extended Memory
MUL: Multiply Accumulator by B
NOP: No Operation
ORL: Bitwise OR
POP: Pop Value From Stack
PUSH: Push Value Onto Stack
RET: Return From Subroutine
RETI: Return From Interrupt
RL: Rotate Accumulator Left
RLC: Rotate Acc. Left Through Carry
RR: Rotate Accumulator Right
RRC: Rotate Acc. Right Through Carry
SETB: Set Bit
SJMP: Short Jump
SUBB: Sub. From Acc. With Borrow
SWAP: Swap Accumulator Nibbles
XCH: Exchange Bytes
XCHD: Exchange Digits
XRL: Bitwise Exclusive OR
Undefined: Undefined Instruction