8051 chap3 instruction ver 02

7/31/2019 8051 Chap3 Instruction Ver 02

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H Bch Khoa TP.HCM L Ch Thng

www.tinyurl.com/thongchile 1

Chapter 3

Instruction Set

How to write a program

The 8051 Microcontroller

L Ch Thng 1

L Ch Thng

[email protected]

Ho Chi Minh City University of Technology

Instruction Set

255 instructions

1-byte instructions: 139



Instruction set summary (pdf)

Full instruction set (pdf)

L Ch Thng 2


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Internal RAM

128 locations from address 00H to 7FH (256 locations for 8052)

The content of one location is 8 bit.

Register banks: address 00H to 1FH

Bit Addressable RAM: address 20H to 2FH

General purpose RAM: address from 30H to 7FH

L Ch Thng 5

Bit-

addressable

RAM(20H-2FH)

Register

banks

(00H-1FH)

General

purpose

RAM

(30H-7FH)Special

function

registers

(SFRs)(80H-FFH)

L Ch Thng 6


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Address from 30H to 7FH


Ex: MOV A,30H

This instruction moves (reads) the content of location 30H

(NOT data 30H) to register A.

Ex:MOV 31H,R4

This instruction moves (writes) the content of register R4

to location 31H.

LCh Thng

General Purpose RAM

A 30H

31H R4

7

Bit-

addressable

RAM(20H-2FH)

Register

banks

(00H-1FH)

General

purpose

RAM

(30H-7FH)Special

function

registers

(SFRs)(80H-FFH)

L Ch Thng 8


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Address from 20H to 2FH


Can read/write a byte or a bit

Ex: MOV 20H, A ; writes the content of register A to location 20H.

Ex: SETB 20H.0

or SETB 00H

Ex: MOV C,31H

or MOV C, 26H.1

LCh Thng

Bit Addressable RAM

20H A

20H 1

Not affected

26HCY

9

Bit-

addressable

RAM(20H-2FH)

Register

banks

(00H-1FH)

General

purpose

RAM

(30H-7FH)Special

function

registers

(SFRs)(80H-FFH)

L Ch Thng 10


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4 banks: Bank 0 (default), Bank 1, Bank 2, and Bank 3

Change register bank by selecting bits RS1 and RS0 (in register PSW)

One bank includes 8 registers: R0 through R7

o R0 of Bank 0 is location 00H

o



o

o R7 of Bank 1 is location 0FH

oR0 of Bank 2 is location 10H

o



o

o R7 of Bank 3 is location 1FH L Ch Thng

Register Banks

11

Ex: use default bank (Bank 0)

MOV A, R5 = MOV A, 05H

MOV R0, A = MOV 00H, A

L Ch Thng

Register Banks

R5 05H

Read the contents of location 05H

into the accumulator

Read the contents of register R5

into the accumulator

Read the contents of register A

into register R0

Read the contents of register A

into location 00H

12


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L Ch Thng

Register PSW (Program Status Word)

13

The active bank is selected by bit RS1 (PSW.4) & RS0 (PSW.3)

Ex: MOV PSW, #00010000B ; select bank 2

or

SETB RS1

CLR RS0

L Ch Thng

Selecting Register Bank

RS1 RS0 Bank

0 0 0

0 1 1

1 0 2

1 1 3

14


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Bit-

addressable

RAM

(20H-2FH)

Register

banks

(00H-1FH)

General

purpose

RAM

(30H-7FH)Special

function

registers

(SFRs)

(80H-FFH)

L Ch Thng 15

L Ch Thng

Special Function Registers (SFRs)

SFRs include register A, B, PSW, P0, P1, P2, P3, DPTR,TMOD, SCON

All SFRs are accessible by name and direct address.

Both of them must be coded as direct addressEx: MOV P1, A ; Accumulator Port 1

MOV 90H, A ; same

Same opcode: F5 90Ex: MOV R1, ACC

MOV R1, 0E0H

Same opcode: A9 E0Ex: MOV R1, ACC

MOV R1, A

Same function BUT different opcode (A9 E0 vs. F9)

16


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Immediate Addressing

Register Addressing

Direct Addressing

Indirect Addressing

Relative Addressing

Absolute Addressing

Long Addressing

Indexed Addressing

Addressing Modes

L Ch Thng 17

L Ch Thng

Coding format- Addressing modes

18


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- Source operand is constant- Use # sign

Ex 3.1:

MOV A,#12

MOV A,#0C4H

MOV R1,#1110B

ADD A,#11110001B

MOV DPTR,#2000H

Immediate Addressing

L Ch Thng

A 0 0 0 0 1 1 0 0

A 1 1 0 0 0 1 0 0

R1 0 0 0 0 1 1 1 0

A 1 0 1 1 0 1 0 1CY 1

DPH 0 0 1 0 0 0 0 0 DPL 0 0 0 0 0 0 0 0

19

Ex: MOV R1,#36H

MOV A,R1

MOV R7,#0FH

ANL A,R7

INC A

DEC A

MOV DPTR,#2000

INC DPTR

L Ch Thng

Register Addressing

R1 0 0 1 1 0 1 1 0

A 0 0 1 1 0 1 1 0

R7 0 0 0 0 1 1 1 1

A 0 0 1 1 0 0 0 0

A 0 0 1 1 0 0 0 1

A 0 0 1 1 0 0 0 0

DPH 0 0 1 0 0 0 0 0 DPL 0 0 0 0 0 0 0 1

20


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Direct Addressing

L Ch Thng

Access internal RAM and SFRs

MOV A,70H ; copy contents of RAM at 70H to A

MOV R0,40H ; copy contents of RAM at 40H to A

MOV 56H,A ; put contents of A at 56H

MOV 0D0H,A ; put contents of A into PSW

MOV PSW,A ; same

21

Direct Addressing vs. Immediate Addressing

L Ch Thng

MOV A,30H MOV A,#30H

Direct addressing Immediate addressing

A 30H A 0 0 1 1 0 0 0 0

22


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Play with R0-R7 (bank 0) by Direct Addressing

L Ch Thng

MOV A,4 MOV A,R4

MOV A,7 MOV A,R7

MOV 7,6 MOV R7,R6

MOV R2,#05 ; Put 5 in R2

MOV R2,5 ; Put content of RAM at address 5 in R2

R2 05H R2 0 0 0 0 0 1 0 1

23

MOV R2,5 MOV R2,#5

The address of the source or destination is specified in registers.

Use registers R0 or R1 for 8-bit address (internal or external

RAM)

Use @ sign to access the content of the memory location: @R0,@R1

Ex: MOV R0,#30H ;R0 30H

MOV A,@R0 ;A(R0): read content ofinternal RAM at; address specified by R0 to A

Indirect Addressing

L Ch Thng

A

7FH

30H

00H

R0 0 0 1 1 0 0 0 0Internal RAM

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Ex: MOV R0,#30H ;R0 30H

MOVX A,@R0 ;A(R0): read content ofeXternal RAM at; address specified by R0 to A

Indirect Addressing

L Ch Thng

A 30H

00H

External RAM

R0 0 0 1 1 0 0 0 0

25

Uses DPTR register for 16-bit addresses (external memory)

Use @ sign to access the content of the memory location:@DPTR

Ex: MOV DPTR,#4000H

MOVX A,@DPTR ; read content of external RAM at

; address 4000H to A

Indirect Addressing

LCh Thng

A 4000H

00H

External RAMDPTR 40H 00H

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Ex: MOV DPTR,#31FFH

MOVX @DPTR,A ; write content of A to external RAM

; address 31FFH

Indirect Addressing

A 31FFH

00H

External RAM

DPTR 31H FFH

L Ch Thng 27

Register Addressing vs. Indirect Addressing

L Ch Thng

MOV A,R1 MOV A,@R1

Register addressing

Indirect addressing

A R1

28

A

7FH

00H

R1Internal RAM


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Ex: Implementation of array

MOV R0, #60H

LOOP: MOV @R0, #0

INC R0

CJNE R0, #80H, LOOP

Example of Indirect Addressing

L Ch Thng 29

R0 60H

(R0) 0

R0 R0+1

R0= 80h?N

Y

Used in SJMP instruction

A relative address (or offset) is an 8-bit signed value.

It is added to PC to form a new value for PC.

Range: -128 ~ +127

Usually used with label

Ex: SJMP LABEL1 is in memoryat locations 0100H and 0101H

PC = 0102H

If LABEL1 is the label representingan instruction at location 0107H

Relative offset is0107H 0102H = 5

Opcode of SJMP LABEL1 is 80 05

Relative Addressing

L Ch Thng 30


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Ex: SJMP LABEL2 is in memoryat locations 2040H and 2041H

PC = 2042H

If LABEL2 is the label representingan instruction at location 2038H

Relative offset is2038H 2042H = -10 = F6H

Opcode of SJMP LABEL1 is 80 F6

Relative Addressing

L Ch Thng 31

Ex: ORG 0000H

AGAIN: SETB P1.0

NOP

NOP

CLR P1.0

NOP

SJMP AGAIN

END

a. Find the opcode of SJMP AGAIN instruction?

b. Find the duration of P1.0=1? P1.0=0?

c. What is the result of this program?

Your Turn!

L Ch Thng 32


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Used in AJMP, ACALL instruction


Absolute Addressing

Memory map showing 2K pages

L Ch Thng 33

Within any 2K page, the upper five addressbits are the same for the source anddestination addresses.The lower 11 bits of the destination aresupplied in the instruction

Used in LCALL and LJMP instruction

Use full 16-bit address


Ex: LCALL SUBPROGRAM1

LJMP LABEL2

Long Addressing

L Ch Thng 34


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Base address (PC or DPTR) + Offset (A) Effective Address

Used with JMP or MOVC

Ex: MOVC A, @A+DPTR

MOVC A, @A+PC

JMP @A+DPTR

Indexed Addressing

L Ch Thng 35

Ex: An even number from 0 to 6 is in the Accumulator. The following

sequence of instructions branches to one of four AJMP

instructions in a jump table starting at JMP_TBL.

MOV DPTR, # JMP_TBL

JMP @A + DPTR

JMP_TBL: AJMP LABEL0

AJMP LABEL1

AJMP LABEL2

AJMP LABEL3

If the Accumulator equals 04H when starting this sequence,execution jumps to label LABEL2. Because AJMP is a 2-byte

instruction, the jump instructions start at every other address.

Indexed Addressing

L Ch Thng 36


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Instruction Types Data transfer

Arithmetic

Logical

Boolean variable

Program branching

L Ch Thng 37

8051 Instruction Set SummaryLegend

L Ch Thng 38


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8051 Instruction Set SummaryData Transfer

L Ch Thng 39

8051 Instruction Set SummaryData Transfer

L Ch Thng 40


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Data TransferMOV dest, source ; dest source

MOV A,#72H ;A=72HMOV R4,#62H ;R4=62HMOV B,0F9H ;B=the content of address F9H of RAM

MOV DPTR,#7634HMOV DPL,#34HMOV DPH,#76H

MOV P1,A ;move (out) A to Port 1

Note 1:MOV A,#72H MOV A,72H

Note 2:MOV A,R3 MOV A,3 ;R3 of bank 0 is location 3

L Ch Thng 41

Data TransferEx: Write a program to move (write) 40H to location 30H in internal

RAM using 2 methods of addressing: direct addressing and indirect

addressing.

Method 1: Direct addressing

ORG 0000H

MOV 30H,#40H

ENDMethod 2: Indirect addressing

ORG 0000H

MOV R0,#30H

MOV @R0,#40H

END

L Ch Thng 42


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Creating a Loop (1)Ex: Write a program to write 40H to internal RAM from location

30H to location 36H.

ORG 0000H

MOV 30H,#40H

MOV 31H,#40H

MOV 32H,#40H

MOV 33H,#40H

MOV 34H,#40H

MOV 35H,#40H

MOV 36H,#40HEND

L Ch Thng 43

Ex: Write a program to write 40H to internal RAM

from location 30H to location 36H.

(source)

ORG 0000H

MOV R5,#7 ;Loop=7

MOV R1,#30H;Address=30H

Again: MOV @R1,#40HINC R1

DEC R5

CJNE R5,#0,Again

END

L Ch Thng 44

Addr 30H

(Addr) 40H

AddrAddr+1

Loop=0?N

Y

Loop 7

LoopLoop-1

Creating a Loop (2)


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(source)ORG 0000H

MOV R5,#7;Loop=7


Again: MOV @R1,#40H

INC R1

DJNZ R5,Again

END

L Ch Thng 45

Creating a Loop (3)

Addr 30H

(Addr) 40H

AddrAddr+1

Loop=0?N

Y

Loop 7

LoopLoop-1



(source)ORG 0000H


Again: MOV @R1,#40H

INC R1CJNE R1,#37H,Again

END

L Ch Thng 46

Creating a Loop (4)

Addr 30H

(Addr)

40H

AddrAddr+1

Addr=37H?N

Y


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Your Turn!

Write a program to read data from internal RAM from

location 20H to location 29H and output to Port 1

L Ch Thng 47

Your Turn!Write a program to read data from internal RAM from location 20H

to location 29H and output to Port 1

Method 1: using DJNZ

ORG 0000H

MOV R7,#10

MOV R0,#20H

Loop:MOV P1,@R0 ;Output to Port1INC R0

DJNZ R7,loop

END

L Ch Thng 48


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Your Turn!Ex: Write a program to read data from internal RAM from location

20H to location 29H and output to Port 1

Method 2: using CJNE

ORG 0000H

MOV R0,#20H

Loop:MOV P1,@R0 ;Output to Port1

INC R0

CJNE R0,#2AH,loop

END

L Ch Thng 49

Data Transfer More ExamplesEx: Write a program to clear location 31H in internal RAM using 2

methods of addressing: direct addressing and indirect addressing.

Hint: to clear means to reset the data to zero.


ORG 0000H

MOV 31H,#0


ORG 0000H

MOV R0,#31H

MOV @R0,#0

END

L Ch Thng 50


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Ex: Write a program to move (write) the content of A to location

32H in internal RAM using 2 methods of addressing: direct

addressing and indirect addressing.


ORG 0000H

MOV 32H,A

END

Method 2: Indirect addressing

ORG 0000H

MOV R0,#32H

MOV @R0,AEND

L Ch Thng 51

Data Transfer More Examples

Ex: Write a program to move the content of location 33H in internal

RAM to register A using 2 methods of addressing: direct addressing

and indirect addressing.


ORG 0000H

MOV A,33H


ORG 0000H

MOV R0,#33H

MOV A,@R0

END

L Ch Thng 52



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RAM to register A using 2 methods of addressing: direct addressing

and indirect addressing.


ORG 0000H

MOV A,33H

END

Method 2: Indirect addressing

ORG 0000H

MOV R0,#33H

MOV A,@R0END

L Ch Thng 53



RAM to location 35H in internal RAM using 2 methods of

addressing: direct addressing and indirect addressing.


ORG 0000H

MOV 35H,34H


ORG 0000H

MOV R0,#34H

MOV A,@R0

INC R0

MOV @R0,A

END

L Ch Thng 54



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Ex: Write a program to write 40H to location 0030H in external

RAM.

ORG 0000H

MOV A,#40H

MOV DPTR,#0030H

MOVX @DPTR,A

END

L Ch Thng 55


Ex: Write a program to clear location 0031H in externalRAM.

ORG 0000H

CLR A

MOV DPTR,#0031H

MOVX @DPTTR,A

END

L Ch Thng 56



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Ex: Write a program to move the content of location 0032H in

externalRAM to register A.

ORG 0000H

MOV DPTR,#0032H

MOVX A,@DPTTR

END

L Ch Thng 57


Ex: Write a program to move the content of register A to location

0033H in externalRAM.

ORG 0000H

MOV DPTR,#0033H

MOVX @DPTTR,A

END

L Ch Thng 58



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Ex: Write a program to move the content of location 0034H in

externalRAM to location 0035H in externalRAM.

ORG 0000H

MOV DPTR,#0034H

MOVX A,@DPTTR

INC DPTR

MOVX @DPTR,A

END

L Ch Thng 59


MOV A,#ENTRY_NUMBER

MOV DPTR,#TABLE

MOVC A,@A+DPTR

TABLE: DB data1, data2, data3,

L Ch Thng 60

Look-up Table


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Given a packed-BCD number in location 33H of internal RAM. Write

a program that calculate the square of high decade of this number,

and store the result in internal RAM at address 34H

ORG 0

MOV A,33H

SWAP A

ANL A,#0FH

MOV DPTR,#TABLE

MOVC A,@A+DPTR

TABLE:

DB 0,1,4,9,16,25,36,49,64,81END

L Ch Thng 61

Look-up Table

MOV A, #ENTRY_NUMBER

CALL LOOK_UP

LOOK_UP:

INC AMOVC A, @A+PC

RET

TABLE:

DB data1, data2, data3,

L Ch Thng 62

Look-up Table


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8051 Instruction Set SummaryArithmetic Operations

L Ch Thng 63


L Ch Thng 64


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L Ch Thng 65

8051 Instruction Set SummaryFlag Effect

L Ch Thng 66


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Arithmetic Operations (1)

ADD A,#34H ; Immediate addressing

ADD A,34H ;Direct addressing

ADD A,R1 ;Register addressing

ADD A,@R1 ;Indirect addressing

L Ch Thng 67


MOV A,#23H ; A = 23H, CY = 0, P = 1, OV = 0

ADD A,#22 ; A = 39H, CY = 0, P = 0, OV = 0

ADD A,#11100111B ; A = 20H, CY = 1, P = 1, OV = 0

ADDC A,#09H ; A = 2AH, CY = 0, P = 1, OV = 0

SUBB A,#00100111B ; A = 03H, CY = 0, P = 0, OV = 0

SUBB A,#3 ; A = 00H, CY = 0, P = 0, OV = 0

SUBB A,#0F2H ; A = 0EH, CY = 1, P = 1, OV = 0

L Ch Thng 68


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MOV A,#45 ; A = 2DH, CY = 1, P = 0, OV = 0

MOV B,#12H ; B = 12H

MUL AB ; A = 2AH, B = 03H

MOV B,#10 ; B = 10

DIV AB ; A = 04H, B = 02H

MOV A,#-100 ; A = 9CH, CY = 0, P = 0, OV = 0

ADD A,#-50 ; A = 6AH, CY = 1, P = 0, OV = 1

MOV A,#120 ; A = 78H, CY = 1, P = 0, OV = 1

ADD A,#30 ; A = 44H, CY = 0, P = 0, OV = 1

L Ch Thng 69

Your Turn!

Given 10 8-bit unsigned numbers in internal RAM at starting

address 30H. Write a program to calculate the sum of these

numbers and store the result in internal RAM at address 2FH.

Assume that the result is less than or equal 255.

L Ch Thng 70


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Your Turn!

Given 10 8-bit unsigned numbers in internal RAM at starting

address 30H. Write a program to calculate the sum of these

numbers and store the result in internal RAM at address 2FH.

Assume that the result is less than or equal 255.

ORG 0

MOV R0,#30H

CLR A

MOV R2,#10

loop: ADD A,@R0

INC R0

DJNZ R2,loopMOV 2FH,A

END

L Ch Thng 71

Rotate instructions operate only on A

RL A

MOV A,#0F0H ; A 11110000

RR A ; A 11100001

RR A

MOV A,#0F0H ; A 11110000

RR A ; A 01111000

Rotate

72L Ch Thng


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RRC A

MOV A, #0A9H ; A A9H

ADD A, #14H ; A BDH (10111101), C0

RRC A ; A 01011110, C1

RLC A

MOV A, #3CH ; A 3CH(00111100)

SETB C ; C 1

RLC A ; A 01111001, C1

C

C

Rotate through Carry

73L Ch Thng

Note that a shift left is the same asmultiplying by 2, shift right is divide by 2

MOV A, #3 ; A 00000011 (3)

CLR C ; C 0

RLC A ; A 00000110 (6)

RLC A ; A 00001100 (12)

RRC A ; A 00000110 (6)

Rotate and Multiplication/Division

74L Ch Thng


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SWAP A

MOV A, #72H ; A 72H

SWAP A ; A = 27H

Swap

75L Ch Thng

8051 Instruction Set SummaryLogic Operations

L Ch Thng 76


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Logic Operations (1)

L Ch Thng 77

MOV A,#46H

MOV R3,A

ANL A,#0FH

MOV R7,A

MOV A,R3

ANL A,#0F0H

SWAP A

MOV R6,A

R6 = .; R7 = .

Logic Operations (2)

L Ch Thng 78

MOV R2,#05H

MOV R3,#07H

MOV A,R2

SWAP A

ORL A,R3

A = .


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8051 Instruction Set SummaryBoolean Variable

L Ch Thng 79

Boolean Variable

L Ch Thng 80

ORG 0

LOOP: MOV C,P1.0

ANL C,P1.1

MOV P1.2,C

SJMP LOOP

END


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Your Turn!

L Ch Thng 81

Your Turn!

L Ch Thng 82

ORG 0

LOOP: MOV C,P1.4

CPL C

ANL C,P1.5

CPL C

ORL C,P1.6

MOV P1.7,C

SJMP LOOP

END


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8051 Instruction Set SummaryProgram Branching

L Ch Thng 83

8051 Instruction Set SummaryProgram Branching

L Ch Thng 84


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Program Branching

L Ch Thng 85

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 locationwithin the 2k block of program memory.

SJMP(short jump)In this 2-byte instruction. The relative address range of 00-FFH isdivided into forward and backward jumps, that is , within -128 to

+127 bytes of memory relative to the address of the current PC.

SJMP : 8-bit offset

LJMP : 11-bit address (2KB segment)

AJMP : 16-bit

Unconditional

jumps

Conditional Jumps

L Ch Thng 86

JZ Jump if A=0

JNZ Jump if A 0

DJNZ Decrement and jump if 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


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CALL Instruction

L Ch Thng 87

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 thesubroutine must be within 2K byte range.

Call is similar to a jump, but

Call pushes PC on stack before branching

ACALL ; stack PC

; PC address 11 bit

LCALL ; stack PC

; PC address 16 bit

Call and Return

88L Ch Thng


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Return is also similar to a jump, but

Return instruction pops PC from stack to getaddress to jump to

RET ; PC stack

Call and Return

89L Ch Thng

MAIN: ...

ACALL SUBLABEL

...

...

SUBLABEL: ...

...

RETthe subroutine

call to the subroutine

Subroutines

90L Ch Thng


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DJNZ a For loop

L Ch Thng 91

Use DJNZ to create a for loop control

Ex: a 10-time loop

MOV R7,#10

LOOP: (begin loop)

(end loop)

DJNZ R7,LOOP(continue)

DJNZ a For loop

L Ch Thng 92

Use DJNZ to create a 1000-time loop?


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DJNZ an example

L Ch Thng 93

Ex: Write a program to write 40H to internal

RAM from location 30H to location 36H.

(source)

ORG 0000H

MOV R5,#7;Loop=7


Again: MOV @R1,#40H

INC R1

DJNZ R5,Again

END

Addr 30H

(Addr) 40H

AddrAddr+1

Loop=0?N

Y

Loop 7

LoopLoop-1

Write a program toclear ACC, then add 3 to the

accumulator ten times.

Solution:

MOV A,#0

MOV R2,#10

AGAIN: ADD A,#03

DJNZ R2,AGAIN ;repeat until R2=0 (10 times)

MOV R5,A

L Ch Thng 94

DJNZ another example


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Write a program to copy a block of 10 bytes from RAMlocation starting at 37h to RAM location starting at 59h.

L Ch Thng 95

DJNZ Your Turn!

Write a program to copy a block of 10 bytes from RAMlocation starting at 37h to RAM location starting at 59h.

Solution:MOV R0,#37h ; source pointerMOV R1,#59h ; dest pointerMOV R2,#10 ; counter

L1: MOV A,@R0MOV @R1,AINC R0INC R1DJNZ R2,L1

L Ch Thng 96

DJNZ Your Turn!


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L Ch Thng 97

Blinky Program

P1.0 1

Delay

P1.0 0

Delay

schematic

ORG 0LOOP: SETB P1.0

ACALL DELAYCLR P1.0ACALL DELAYSJMP LOOP

DELAY: MOV R6,#200DL1: MOV R7,#250

DJNZ R7,$DJNZ R6,DL1RETEND

(source)

Waveform? Period? Frequency?

L Ch Thng 98

Blinky Program


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DELAY: MOV R6,#200DL1: MOV R7,#250 1MC x 1

DJNZ R7,$ 2 MC x 250DJNZ R6,DL1RET

END

L Ch Thng 99

Blinky Program

501 MC




DJNZ R7,$DJNZ R6,DL1 2 MC x 200RETEND

L Ch Thng 100

Blinky Program

501 MC x 200 100,601 MC


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DJNZ R7,$DJNZ R6,DL1RET 2 MC

END

L Ch Thng 101

Blinky Program

100,601 MC100,603 MC

ORG 0LOOP: SETB P1.0 1 MC

ACALL DELAY 100,063 + 2 MCCLR P1.0 1 MCACALL DELAY 100,063 + 2 MCSJMP LOOP 2 MC



tH = 100,063 + 2 + 1 = 100,066 MCtL = 100,063 + 2 + 2 + 1 = 100,068 MC

L Ch Thng 102

Blinky Program


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If using 12 MHz crystal, 1 MC = 1 s

tH = 100,066 MC = 100,066 s

tL = 100,068 MC = 100,068 s

T = tH + tL = 200,134 s

f = 1/T = 4.99 Hz

L Ch Thng 103

Blinky Program

tH tL

T





tH = tL tDELAY 200 x 250 x 2 MC = 100,000 MC = 100,000 s T 200,000 s f 5 Hz

L Ch Thng 104

Blinky Program - Estimating


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ORG 0LOOP: CPL P1.0

ACALL DELAYSJMP LOOP



L Ch Thng 105

Blinky Program Alternative Method

P1.0 NOT (P1.0)

Delay

Write a program that creates a 10-KHz square wave at pin P1.3.

Assume that crystal is 24 MHz.

L Ch Thng 106

10-kHz square wave


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Write a program that creates a 10-KHz square wave at pin P1.3.


ORG 0000H

lap: CPL P1.3

ACALL delay50

SJMP lap

delay50:

MOV R4,#25

DJNZ R4,$

RET

END

L Ch Thng 107

10-kHz square wave

Write a program that creates a 10-KHz square wave with duty cycle

30% at pin P1.3. Assume that crystal is 24 MHz.

L Ch Thng 108

10-kHz square wave, duty cycle 30%


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Write a program that creates a 10-KHz square wave with duty cycle


ORG 0000H

lap: SETB P1.3

ACALL delay30

CLR P1.3

ACALL delay70

SJMP lap

Delay30: MOV R4,#15

DJNZ R4,$

RET

Delay70: MOV R4,#35

DJNZ R4,$

RET

ENDL Ch Thng 109


Write a program that creates a 100-kHz square wave at pin P1.1.


L Ch Thng 110

100-kHz square wave


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Write a program that creates a 100-kHz square wave at pin P1.1.


ORG 0000H

lap: CPL P1.1

NOP

NOP

SJMP lap

END

L Ch Thng 111

100-kHz square wave

Write a program that creates a 100-kHz square wave with duty cycle


L Ch Thng 112



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Write a program that creates a 100-kHz square wave with duty cycle


ORG 0000H

lap: SETB P1.2

NOP

NOP

NOP

CLR P1.2

NOP

NOP

NOP

SJMP lap

END

L Ch Thng 113


CJNE A,#05H,Skip

(Statement 1)

Skip: (Continue)

L Ch Thng 114

CJNE Equal/Not Equal (1)

A = 05H?N

Y

Statement 1


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Ex: Write a program to write 40H to internal RAM from location

30H to location 36H.

(source)

ORG 0000H


Again: MOV @R1,#40H

INC R1

CJNE R1,#37H,Again

END

L Ch Thng 115

An Example

Addr 30H

(Addr) 40H

AddrAddr+1

Addr=37H?N

Y

CJNE A,#05H,Not_Eq

(Statement 1)

SJMP Next

Not_Eq: (Statement 2)

Next: (Continue)

L Ch Thng 116

CJNE Equal/Not Equal (2)

A = 05H?N

Y

Statement 1Statement 2


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CJNE A,#05H,Next

Next: JNC GT_Eq

SJMP Continue

GT_Eq: (Statement 1)

Continue: (Continue)

L Ch Thng 117

CJNE Greater Than or Equal/Less Than (1)

A 05H?N

Y

Statement 1

CJNE A,#05H,$+3

JNC GT_Eq

SJMP Continue


Continue: (Continue)

Ex: Examine the content of A, if 5 A 10 then output A to

Port 1; if not, output A to Port 2.

ORG 0

CJNE A,#5,$+3

JC PORT2

CJNE A,#11,$+3JNC PORT2

MOV P1,A

SJMP DONE

PORT2:MOV P2,A

DONE: NOP

END

118L Ch Thng

An Example


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CJNE A,#05H,$+3

JNC GT_Eq

(Statement 2)

SJMP Next


Next: (Continue)

L Ch Thng 119

CJNE Greater Than or Equal/Less Than (2)

A 05H?N

Y

Statement 1Statement 2

MOV DPTR,#JUMP_TABLE

MOV A,INDEX_NUMBER

RL A

JMP @A+DPTR

JUMP_TABLE: AJMP CASE0

AJMP CASE1AJMP CASE2

L Ch Thng

Jump Table

120


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Given a 100-byte unsigned number string in external RAM at

address starting from 0100H. Write a program that sends positive

numbers to Port 1 and negative numbers to Port 2.

Hint:

- A positive number has MSB = 0.

- A negative number has MSB = 1.

- Use JB / JNB instruction

L Ch Thng

Bit Testing

121

Given a 100-byte unsigned number string in external RAM at

address starting from 0100H. Write a program that sends positive

numbers to Port 1 and negative numbers to Port 2.

ORG 0000H

MOV DPTR,#0100H

MOV R4,#100

loop: MOVX A,@DPTRJNB ACC.7,positive

MOV P2,A

SJMP next

positive: MOV P1,A

next: INC DPTR

DJNZ R4,loop

END

L Ch Thng

Bit Testing

122


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Ex: Given a 20-byte string in internal RAM, starting at

address 40H. Write a program that output even numbers

to Port 2.

123L Ch Thng

Your Turn!

Ex: Given a 20-byte string in internal RAM, starting at

address 40H. Write a program that output even numbers

to Port 2.

ORG 0

MOV R4,#20 ; Number of loops

MOV R0,#40H ; Address pointer

LOOP:

MOV A,@R0 ; Read data from internal RAM to A

JB ACC.0,NEXT ; skip if odd number

MOV P2,A ; Output to Port 2 if even number

NEXT:

INC R0

DJNZ R4,LOOP

END

124L Ch Thng

Your Turn!


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Ex: Given a 20-byte string in external RAM, starting at

address 4000H. Write a program that output odd numbers

to Port 2.

125L Ch Thng

Your Turn!

Ex: Given a 20-byte string in external RAM, starting at

address 4000H. Write a program that output odd numbers

to Port 2.

ORG 0

MOV R4,#20 ; Number of loops

MOV DPTR,#4000H ; Address of external RAM

LOOP:

MOVX A,@DPTR ; Read data from external RAM to A

JNB ACC.0,NEXT ; skip if even number

MOV P2,A ; Output to Port 2 if odd number

NEXT:

INC DPTR

DJNZ R4,LOOP

END

126L Ch Thng

Your Turn!


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L ChThng

Bit Testing

LOOP: MOV C,P1.0

JNB P1.1,SKIP

CPL C

SKIP: MOV P1.2,C

SJMP LOOP

P1.0

P1.1

P1.2

127

L ChThng

Example Problem

A 4-bit DIP switch and a common-anode 7-segment LED are

connected to an 8051 as shown in the following figure. Write a

program that continually reads a 4-bit code from the DIP switch and

updates the LEDs to display the appropriate hexadecimal character.

For example, if the code 1100B is read, the hexadecimal character

C should appear, thus, segments a through g respectively should

be ON, OFF, OFF, ON, ON, ON, and OFF. Note that setting an 8051port pin to 1 turns the corresponding segment ON.

128


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129

References

L Ch Thng

I. Scott MacKenzie , The 8051 Microcontroller, 2ndEdition, Prentice-Hall, 1995

Kenneth J. Ayala, The 8051 Microcontroller:Architecture, Programming, and Applications, West

Publishing Company

[email protected] , Lecture notes

8051 chap3 instruction ver 02

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