The document discusses various addressing modes of the 8051 microcontroller including immediate, register, direct, register indirect, indexed, and bit addressing. It provides examples of instructions using each addressing mode and describes how they work. Special function registers are discussed along with their addresses. Examples are given for accessing on-chip ROM and using lookup tables. The use of assembler directives like EQU and BIT are also demonstrated.

1. 04/25/14 DPD 1
8051 Addressing Modes

2. 04/25/14 DPD 2
Introduction
• Immediate Addressing
• Register Addressing
• Direct Addressing
• Register Indirect Addressing
• Indexed Addressing

3. 04/25/14 DPD 3
Immediate Addressing
• MOV A, #25H
• MOV R4, #62
• MOV B, #40H
• MOV DTPR, #2550H
Same as
– MOV DPL, #50H
– MOV DPH, #25H

4. 04/25/14 DPD 4
Register Addressing
• MOV A, R0
• MOV R2, A
• ADD A, R5
• ADD A, R7
• MOV R6, A
• MOV R4, R7 – invalid
• MOV DTPR, A – error
• MOV R7, DPL – valid
• MOV R6, DPH – valid

5. 04/25/14 DPD 5
Direct Addressing
• MOV R0, 40H
• MOV 56H, A
• MOV R4, 7FH
• MOV R2, #5
• MOV A, 2
• MOV B, 2
• MOV 7, 2
• MOV A, 4
• MOV A, R4
• MOV A, 7
• MOV A, R7

6. 04/25/14 DPD 6
SFR registers and their
addressing
• SFR – special function
register
• A, B, PSW, DPTR can
be addresses by
– Names
– Addresses
• A – E0H
• B – F0H
• MOV 0E0H, #55H
• MOV A, #55H
• MOV 0F0H, #25H
• MOV B, #25H
• MOV P1, A
• MOV 90H, A

7. 04/25/14 DPD 7
SFR …contd.
• SFR have addresses
between 80H and FFH
• 00 to 7FH are addresses of
RAM memory inside 8051
• Not all address spaces of
80 to FF is used by the
SFR.
• Unused locations are
reserved and must not be
used by the 8051
programmer.
FFH
F0 – B register
E0 – Accumulator
D0 – PSW
B0 – Port 3
A0 – Port 2
90 – Port 1
83 – DPH
82 – DPL
81 – SP
80H 80 – Port 0

8. 04/25/14 DPD 8
Stack and direct addressing
• PUSH A – invalid
• PUSH 0E0H – valid
• PUSH 05
• PUSH 06
• PUSH 0E0H
• PUSH 0F0H
• POP 02
• POP 03

9. 04/25/14 DPD 9
Register Indirect Addressing
• A register is used as a
pointer to the data
• Only R0 and R1 are
used for this purpose
• Must be preceded by
‘@’ sign
• MOV A, @R0
– Move contents of
RAM location whose
address is held by R0
into A
• MOV @R1, B
– Move contents of B
into RAM location
whose address is held
by R1

10. 04/25/14 DPD 10
Advantage of register indirect
addressing• Make accessing data
dynamic rather than static
MOV A, #55H
MOV 40H, A
MOV 41H, A
MOV 42H, A
MOV 43H, A
MOV 44H, A
MOV A, #55H
MOV R0, #40H
MOV @R0, A
INC R0
MOV @R0, A
INC R0
MOV @R0, A
INC R0
MOV @R0, A
INC R0
MOV @R0, A
MOV A, #55H
MOV R0, #40H
MOV R2, #05
AGAIN: MOV @R0, A
INC R0
DJNZ R2, AGAIN
• Looping is not possible in
direct addressing

11. 04/25/14 DPD 11
Limitation of register indirect
addressing
• Only R0 and R1 can be used for this
purpose
• They 8 bit wide
• Use is limited to internal RAM (30H –
7FH)
• To access external RAM or on-chip ROM
we need 16-bit pointer
– DPTR is used in this case

12. 04/25/14 DPD 12
Indexed addressing mode and
On-chip ROM access
• Used in accessing data elements of LUT
entries located in the program ROM space
of the 8051
• MOVC A, @A+DPTR
– C means code
• Contents of A is added to the 16-bit register
DPTR to form the 16-bit address of the
needed data

13. 04/25/14 DPD 13
Example
• Assume that the word ‘USA’ is burned into
ROM location stating 200H, and that the
program is burned into location stating at 0.
• Analyze how the program works and state
where ‘USA’ is stored after this program is
run.

14. 04/25/14 DPD 14
Solution
ORG 0000H
MOV DPTR, #200H
CLR A
MOVC A, @A+DPTR
MOV R0, A
INC DPTR
CLR A
MOVC A, @A+DPTR
MOV R1, A
INC DPTR
CLR A
MOVC A, @A+DPTR
MOV R2, A
HERE: SJMP HERE
ORG 200H
MYDATA: DB “USA”
END

15. 04/25/14 DPD 15
Analysis
• ROM location 200H – 202H have the following
content
– 200 = (‘U’)
– 201 = (‘S’)
– 202 = (‘A’)
• Start with DPTR = 200H, and A = 0
• MOVC A, @A+DPTR → moves the content of
ROM location 200H (200H + 0 = 200H) to
register A
• Register A contains 55H, the ASCII value for ‘U’
• This is moved to R0
• Next DPTR is incremented to make DPTR = 201H

16. 04/25/14 DPD 16
Example
• Assuming that ROM space starting at 250H
contains ‘India’
• Write a program to transfer the bytes into
RAM locations stating at 40H

17. 04/25/14 DPD 17
Solution
ORG 0000
MOV DPTR, #MYDATA
MOV R0, #40H
MOV R2, #5
BACK: CLR A
MOVC A, @A+DPTR
MOV @R0, A
INC DPTR
INC R0
DJNZ R2, BACK
HERE: SJMP HERE
ORG 250H
MYDATA: DB “India”
END

18. 04/25/14 DPD 18
Look-up table and MOVC
ORG 0
MOV DPTR, #300H
MOV A, #0FFH
MOV P1, A
BACK: MOV A, P1
MOVC A, @A+DPTR
MOV P2, A
SJMP BACK
• Write a program to the x
value from P1 and send x2
to P2 continuously
ORG 300H
XSQR_TABLE:
DB 0, 1, 4, 9, 16, 25, 36,
49, 64, 81
END
300 = 00H 303 = 09H
301 = 01H 304 = 10H
302 = 04H 305 = 19H

19. 04/25/14 DPD 19
Accessing RAM locations 30 –
7FH as scratch pad
• Write a program to
toggle P1 a total of
200 times
• Use RAM location
32H to hold your
counter value instead
of registers R0 – R7
MOV P1, #55H
MOV 32H, #200
LOOP: CPL P1
ACALL DELAY
DJNZ 32H, LOOP

20. 04/25/14 DPD 20
Bit-addressable RAM
• 16 bytes are bit-
addressable
• 20H – 2FH
• Provides 16 x 8 = 128 bits
• Addressed as 0 to 127 or
00H to 7FH
• 20H – 2FH RAM location
– Bit addressable
– Byte addressable
SETB 42H ;set bit 42H to 1
CLR 67H ;clear bit 67
CLR 0FH ;clear bit 0F
SETB 05

21. 04/25/14 DPD 21
Single bit instruction
SETB bit bit =1
CLR bit bit = 0
CPL bit bit = NOT bit
JB bit, target Jump to target if bit = 1
JNB bit, target Jump to target if bit = 0
JBC bit, target Jump to target if bit = 1, clear bit

22. 04/25/14 DPD 22
I/O port bit addresses
SETB 86H ;set bit P0.6
CLR 87H ;clear bit P0.7

23. 04/25/14 DPD 23
Example
• Write a program to save Acc in R7 of Bank 2
CLR PSW.3
SETB PSW.4
MOV R7, A
• How to check overflow?
JB PSW.2, TARGET
• Check if RAM location 37H contains an even
value. If so, send it to P2, else make it even and
send it to P2.
MOV A, 37H
JNB ACC.0, YES
INC A
YES: MOV P2, A

24. 04/25/14 DPD 24
Bit Directive
LED BIT P1.7
HERE: CPL LED
LCALL DELAY
SJMP HERE
SW BIT P1.7
LED BIT P2.0
HERE: MOV C, SW
MOV LED, C
SJMP HERE

25. 04/25/14 DPD 25
Using EQU directive
• A switch is connected
to pin P1.7
• Write a program to
check that status of the
switch and make the
following decision
– If SW = 0, send “No”
to P2
– If SW = 1, send “Yes”
to P2
SW EQU P1.7
MYDATA EQU P2
HERE: MOV C, SW
JC OVER
MOV MYDATA, #’N’
MOV MYDATA, #’O’
SJMP HERE
OVER: MOV MYDATA, #’Y’
MOV MYDATA, #’Y’
MOV MYDATA, #’Y’
SJMP HERE
END