4221-Microcontroller-8051 89c52 51-1.ppt

8051 is one of the first and most popular microcontrollers also known as MCS-51. Intel introduced it in the year 1981. Initially, it came out as an N-type metal-oxide-semiconductor (NMOS) based microcontroller, but later versions were based on complementary metal-oxide-semiconductor(CMOS) technology. These microcontrollers were named 80C51, where C in the name tells that it is based on CMOS technology. It is an 8-bit microcontroller which means the data bus is 8-bit. Therefore, it can process 8 bits at a time. It is used in a wide variety of embedded systems like robotics, remote controls, the automotive industry, telecom applications, power tools, etc. The 8051 microcontroller is an 8-bit microcontroller that was developed by Intel in 1980. It's used in many embedded systems, including robotics, transportation, and automotive.
Features
Architecture
The 8051 has a Harvard architecture and a Reduced Instruction Set Computing (RISC) architecture.
Memory
The 8051 has 4 KB of on-chip Read Only Memory (ROM) and 128 bytes of Random Access Memory (RAM).
Registers
The 8051 has an 8-bit accumulator, an 8-bit B register, and several other 8-bit and 16-bit registers.
Buses
The 8051 has two buses, one for data and one for programs.
Interrupts
The 8051 can be interrupted by internal or external events, which can suspend the normal program execution.
Applications
Moving message LCD display
Fuel theft alarm
Water level indicator
RF based multipurpose device
Microcontroller-based tachometer
Microcontroller-based tone ring player
Microcontroller-based thermometer
Programming
The 8051 microcontroller can be programmed using the embedded C language and Keil software. The 8051 microcontroller is an 8-bit microcontroller that was developed by Intel in 1980. It's used in many embedded systems, including robotics, transportation, and automotive.
Features
Architecture
The 8051 has a Harvard architecture and a Reduced Instruction Set Computing (RISC) architecture.
Memory
The 8051 has 4 KB of on-chip Read Only Memory (ROM) and 128 bytes of Random Access Memory (RAM).
Registers
The 8051 has an 8-bit accumulator, an 8-bit B register, and several other 8-bit and 16-bit registers.
Buses
The 8051 has two buses, one for data and one for programs.
Interrupts
The 8051 can be interrupted by internal or external events, which can suspend the normal program execution.
Applications
Moving message LCD display
Fuel theft alarm
Water level indicator
RF based multipurpose device
Microcontroller-based tachometer
Microcontroller-based tone ring player
Microcontroller-based thermometer
Programming
The 8051 microcontroller can be programmed using the embedded C language and Keil software. The 8051 microcontroller is an 8-bit microcontroller that was developed by Intel in 1980. It's used in many embedded systems, including robotics, transportation, and automotive.
Features
Architecture
The 8051 has a Harvard architecture and a Reduced Instruction Set Computing (RISC) architectu

1. Monday, March 10, 2025 Arun kumar.B

2. Contents:
Introduction
Block Diagram and Pin Description of the 8051
Registers
Some Simple Instructions
Structure of Assembly language and Running
an 8051 program
Memory mapping in 8051
8051 Flag bits and the PSW register
Addressing Modes
16-bit, BCD and Signed Arithmetic in 8051
Stack in the 8051
LOOP and JUMP Instructions
CALL Instructions
I/O Port Programming
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3. Introduction
•CPU for Computers
•No RAM, ROM, I/O on CPU chip itself
•Example：Intel’s x86, Motorola’s 680x0
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CPU
General-
Purpose
Micro-
processor
RAM ROM
I/O
Port
Timer
Serial
COM
Port
Data Bus
Address Bus
General-Purpose Microprocessor System
Many chips on mother’s board
General-purpose microprocessor

4. •A smaller computer
•On-chip RAM, ROM, I/O ports...
•Example：Motorola’s 6811, Intel’s 8051, Zilog’s Z8 and PIC 16X
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RAM ROM
I/O
Port
Timer
Serial
COM
Port
Microcontroller
CPU
A single chip
Microcontroller :

5. Microprocessor
•CPU is stand-alone, RAM,
ROM, I/O, timer are separate
•designer can decide on the
amount of ROM, RAM and
I/O ports.
•expansive
•versatility
•general-purpose
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Microcontroller
•CPU, RAM, ROM, I/O and
timer are all on a single chip
•fix amount of on-chip ROM,
RAM, I/O ports
•for applications in which cost,
power and space are critical
•single-purpose
Microprocessor vs. Microcontroller

6. •Embedded system means the processor is embedded into that
application.
•An embedded product uses a microprocessor or microcontroller to
do one task only.
•In an embedded system, there is only one application software that
is typically burned into ROM.
•Example：printer, keyboard, video game player
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Embedded System

7. 1.meeting the computing needs of the task efficiently and cost
effectively
•speed, the amount of ROM and RAM, the number of I/O ports
and timers, size, packaging, power consumption
•easy to upgrade
•cost per unit
2.availability of software development tools
•assemblers, debuggers, C compilers, emulator, simulator,
technical support
3.wide availability and reliable sources of the microcontrollers.
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Three criteria in Choosing a Microcontroller

8. Block Diagram
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CPU
On-chip
RAM
On-chip
ROM for
program
code
4 I/O Ports
Timer 0
Serial
PortOSC
Interrupt
Control
External interrupts
Timer 1
Timer/Counter
Bus
Control
TxD RxDP0 P1 P2 P3
Address/Data
Counter
Inputs

9. Feature 8051 8052 8031
ROM (program space in bytes) 4K 8K 0K
RAM (bytes) 128 256 128
Timers 2 3 2
I/O pins 32 32 32
Serial port 1 1 1
Interrupt sources 6 8 6
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Comparison of the 8051 Family Members

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11. Pin Description of the 8051
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PDIP/Cerdip
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
40
39
38
37
36
35
34
33
32
31
30
29
28
27
26
25
24
23
22
21
P1.0
P1.1
P1.2
P1.3
P1.4
P1.5
P1.6
P1.7
RST
(RXD)P3.0
(TXD)P3.1
(T0)P3.4
(T1)P3.5
XTAL2
XTAL1
GND
(INT0)P3.2
(INT1)P3.3
(RD)P3.7
(WR)P3.6
Vcc
P0.0(AD0
)P0.1(AD1)
P0.2(AD2
)P0.3(AD3)
P0.4(AD4)
P0.5(AD5)
P0.6(AD6)
P0.7(AD7)
EA/VPP
ALE/PROG
PSEN
P2.7(A15)
P2.6(A14
)P2.5(A13
)P2.4(A12
)P2.3(A11
)P2.2(A10)
P2.1(A9)
P2.0(A8)
8051
(8031)


12. Pins of 8051（1/4）
•Vcc（pin 40）：
–Vcc provides supply voltage to the chip.
–The voltage source is +5V.
•GND（pin 20）：ground
•XTAL1 and XTAL2（pins 19,18）：
–These 2 pins provide external clock.
–Way 1：using a quartz crystal oscillator 
–Way 2：using a TTL oscillator 
–Example 4-1 shows the relationship between XTAL and the
machine cycle. 
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13. Pins of 8051（2/4）
•RST（pin 9）：reset
–It is an input pin and is active high（normally low）.
•The high pulse must be high at least 2 machine cycles.
–It is a power-on reset.
•Upon applying a high pulse to RST, the microcontroller will
reset and all values in registers will be lost.
•Reset values of some 8051 registers 
–Way 1：Power-on reset circuit 
–Way 2：Power-on reset with debounce 
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14. Pins of 8051（3/4）
•/EA（pin 31）：external access
–There is no on-chip ROM in 8031 and 8032 .
–The /EA pin is connected to GND to indicate the code is stored
externally.
–/PSEN ＆ ALE are used for external ROM.
–For 8051, /EA pin is connected to Vcc.
–“/” means active low.
•/PSEN（pin 29）：program store enable
–This is an output pin and is connected to the OE pin of the ROM.
–See Chapter 14.
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15. Pins of 8051（4/4）
•ALE（pin 30）：address latch enable
–It is an output pin and is active high.
–8051 port 0 provides both address and data.
–The ALE pin is used for de-multiplexing the address and data by
connecting to the G pin of the 74LS373 latch.
•I/O port pins
–The four ports P0, P1, P2, and P3.
–Each port uses 8 pins.
–All I/O pins are bi-directional.
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16. Figure 4-2 (a). XTAL Connection to 8051
•Using a quartz crystal oscillator
•We can observe the frequency on
the XTAL2 pin.
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C2
30pF
C1
30pF
XTAL2
XTAL1
GND


17. Figure 4-2 (b). XTAL Connection to an External Clock Source
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•Using a TTL oscillator
•XTAL2 is unconnected.
N
C
EXTERNAL
OSCILLATOR
SIGNAL
XTAL2
XTAL1
GND


18. Example :
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
Find the machine cycle for
(a) XTAL = 11.0592 MHz
(b) XTAL = 16 MHz.
Solution:
(a) 11.0592 MHz / 12 = 921.6 kHz;
machine cycle = 1 / 921.6 kHz = 1.085 s
(b) 16 MHz / 12 = 1.333 MHz;
machine cycle = 1 / 1.333 MHz = 0.75 s

19. RESET Value of Some 8051 Registers:
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0000DPTR
0007SP
0000PSW
0000B
0000ACC
0000PC
Reset ValueRegister
RAM are all zero.


20. Figure 4-3 (a). Power-On RESET Circuit
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30 pF
30 pF
8.2 K
10 uF
+
Vcc
11.0592 MHz
EA/VPP
X1
X2
RST
31
19
18
9


21. Figure 4-3 (b). Power-On RESET with Debounce
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EA/VPP
X1
X2
RST
Vcc
10 uF
8.2 K
30 pF
9
31


22. Pins of I/O Port
•The 8051 has four I/O ports
–Port 0 （pins 32-39）：P0（P0.0～P0.7）
–Port 1（pins 1-8） ：P1（P1.0～P1.7）
–Port 2（pins 21-28）：P2（P2.0～P2.7）
–Port 3（pins 10-17）：P3（P3.0～P3.7）
–Each port has 8 pins.
•Named P0.X （X=0,1,...,7）, P1.X, P2.X, P3.X
•Ex：P0.0 is the bit 0（LSB）of P0
•Ex：P0.7 is the bit 7（MSB）of P0
•These 8 bits form a byte.
•Each port can be used as input or output (bi-direction).
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

23. Monday, March 10, 2025 Arunkumar.B
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

24. Some Simple Instructions
MOV dest,source ; dest = source
MOV A,#72H ;A=72H
MOV A, #’r’ ;A=‘r’ OR 72H
MOV R4,#62H ;R4=62H
MOV B,0F9H ;B=the content of F9’th byte of RAM
MOV DPTR,#7634H
MOV DPL,#34H
MOV DPH,#76H
MOV P1,A ;mov A to port 1
Note 1:
MOV A,#72H ≠ MOV A,72H
After instruction “MOV A,72H ” the content of 72’th byte of RAM will replace in Accumulator.
8086 8051
MOV AL,72H MOV A,#72H
MOV AL,’r’ MOV A,#’r’
MOV BX,72H
MOV AL,[BX] MOV A,72H
Note 2:
MOV A,R3 ≡ MOV A,3
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25. ADDA, Source ;A=A+SOURCE
ADDA,#6 ;A=A+6
ADDA,R6 ;A=A+R6
ADD A,6 ;A=A+[6] or A=A+R6
ADD A,0F3H ;A=A+[0F3H]
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26. SETB bit ; bit=1
CLR bit ; bit=0
SETB C ; CY=1
SETB P0.0 ;bit 0 from port 0 =1
SETB P3.7 ;bit 7 from port 3 =1
SETB ACC.2 ;bit 2 from ACCUMULATOR =1
SETB 05 ;set high D5 of RAM loc. 20h
Note:
CLR instruction is as same as SETB
i.e:
CLR C ;CY=0
But following instruction is only for CLR:
CLR A ;A=0
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Bit Addressable
Page 359,360

27. SUBBA,source;A=A-source-CY
SETBC ;CY=1
SUBBA,R5 ;A=A-R5-1
ADC A,source;A=A+source+CY
SETBC ;CY=1
ADC A,R5 ;A=A+R5+1
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28. DEC byte ;byte=byte-1
INC byte ;byte=byte+1
INC R7
DEC A
DEC 40H ; [40]=[40]-1
CPL A ;1’s complement
Example:
MOV A,#55H;A=01010101 B
L01: CPL A
MOV P1,A
ACALL DELAY
SJMP L01
NOP & RET & RETI
All are like 8086 instructions.
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 CALL

29. ANL - ORL - XRL
EXAMPLE:
MOV R5,#89H
ANL R5,#08H
RR – RL – RRC – RLC A
EXAMPLE:
RR A
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30. Structure of Assembly language and
Running an 8051 program
ORG 0H
MOV R5,#25H
MOV R7,#34H
MOV A,#0
ADD A,R5
ADD A,#12H
HERE: SJMPHERE
END
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EDITOR
PROGRAM
ASSEMBLER
PROGRAM
LINKER
PROGRAM
OH
PROGRAM
Myfile.asm
Myfile.obj
Other obj file
Myfile.lst
Myfile.abs
Myfile.hex

31. Memory mapping in 8051
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•ROM memory map in 8051 family
0000H
0FFFH
0000H
1FFFH
0000H
7FFFH
8751
AT89C51
8752
AT89C52
4k
DS5000-32
8k 32k
from Atmel Corporation
from Dallas Semiconductor

32. •RAM memory space allocation in the 8051
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7FH
30H
2FH
20H
1FH
17H
10H
0FH
07H
08H
18H
00H
Register Bank 0
(Stack )Register Bank 1
Register Bank 2
Register Bank 3
Bit-Addressable RAM
Scratch pad RAM

33. 8051 Flag bits and the PSW register
•PSW Register
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CY AC F0 RS1 OVRS0 P--
CYPSW.7Carry flag
ACPSW.6Auxiliary carry flag
--PSW.5Available to the user for general purpose
RS1PSW.4Register Bank selector bit 1
RS0PSW.3Register Bank selector bit 0
OVPSW.2Overflow flag
--PSW.1User define bit
PPSW.0Parity flag Set/Reset odd/even parity
RS1 RS0 Register Bank Address
0 0 0 00H-07H
0 1 1 08H-0FH
1 0 2 10H-17H
1 1 3 18H-1FH

34. Monday, March 10, 2025 Arunkumar.B
Instructions that Affect Flag Bits:
Note: X can be 0 or 1

35. Example:
MOV A,#38H
ADD A,#2FH
38 00111000
+2F +00101111
---- --------------
67 01100111
CY=0 AC=1 P=1
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Example:
MOV A,#88H
ADD A,#93H
88 10001000
+93 [phone]
---- --------------
11B 00011011
CY=1 AC=0 P=0
Example:
MOV A,#9CH
ADD A,#64H
9C 10011100
+64 +01100100
---- --------------
100 00000000
CY=1 AC=1 P=0

36. Addressing Modes
•Immediate
•Register
•Direct
•Register Indirect
•Indexed
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37. Immediate Addressing Mode
MOV A,#65H
MOV A,#’A’
MOV R6,#65H
MOV DPTR,#2343H
MOV P1,#65H
Example :
Num EQU 30
…
MOV R0,Num
MOV DPTR,#data1
…
ORG 100H
data1: db “IRAN”
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38. Register Addressing Mode
MOVRn, A ;n=0,..,7
ADD A, Rn
MOVDPL, R6
MOVDPTR, A
MOVRm, Rn
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39. 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, 40H
MOV 56H, A
MOV A, 4 ; ≡ MOV A, R4
MOV 6, 2 ; copy R2 to R6
; MOV R6,R2 is invalid !
SFR register and their address
MOV 0E0H, #66H ; ≡ MOV A,#66H
MOV 0F0H, R2 ; ≡ MOV B, R2
MOV 80H,A ; ≡ MOV P1,A
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Bit Addressable
Page 359,360

40. 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.
Example:
Write a program to copy a block of 10 bytes from RAM location sterting at 37h to RAM
location starting at 59h.
Solution:
MOV R0,37h ; source pointer
MOV R1,59h ; dest pointer
MOV R2,10 ; counter
L1: MOV A,@R0
MOV @R1,A
INC R0
INC R1
DJNZ R2,L1
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 jump

41. Indexed Addressing Mode And On-Chip
ROM Access
•This mode is widely used in accessing data elements
of look-up table entries located in the program (code)
space ROM at the 8051
MOVCA,@A+DPTR
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.
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42. •Example:
Assuming that ROM space starting at 250h contains “Hello.”, write a program to transfer the
bytes into RAM locations starting at 40h.
Solution:
ORG 0
MOV DPTR,#MYDATA
MOV R0,#40H
L1:CLR A
MOVC A,@A+DPTR
JZ L2
MOV @R0,A
INC DPTR
INC R0
SJMP L1
L2:SJMP L2
;-------------------------------------
ORG 250H
MYDATA: DB “Hello”,0
END
Notice the NULL character ,0, as end of string and how we use the JZ instruction to
detect that.
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43. •Example:
Write a program to get the x value from P1 and send x
2
to P2, continuously .
Solution:
ORG 0
MOV DPTR, #TAB1
MOV A,#0FFH
MOV P1,A
L01:
MOV A,P1
MOVCA,@A+DPTR
MOV P2,A
SJMP L01
;----------------------------------------------------
ORG 300H
TAB1: DB 0,1,4,9,16,25,36,49,64,81
END
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44. 16-bit, BCD and Signed
Arithmetic in 8051
Exercise:
 Write a program to add n 16-bit number. Get n
from port 1. And sent Sum to LCD
a) in hex
b) in decimal
 Write a program to subtract P1 from P0 and
send result to LCD
(Assume that “ACAL DISP” display A to LCD )
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45. MUL & DIV
•MULAB ;B|A = A*B
MOVA,#25H
MOVB,#65H
MULAB ;25H*65H=0E99
;B=0EH, A=99H
•MULAB ;A = A/B, B = A mod B
MOVA,#25
MOVB,#10
MULAB ;A=2, B=5
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46. 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.
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7FH
30H
2FH
20H
1FH
17H
10H
0FH
07H
08H
18H
00H
Register Bank 0
(Stack )Register Bank
1
Register Bank 2
Register Bank 3
Bit-Addressable RAM
Scratch pad RAM

47. Monday, March 10, 2025 Arunkumar.B
Example:
MOV R6,#25H
MOV R1,#12H
MOV R4,#0F3H
PUSH 6
PUSH 1
PUSH 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

48. LOOP and JUMP Instructions
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,AGAING ;repeat until R2=0 (10 times)
MOV R5,A
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49. •Other conditional jumps :
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,#dataJump 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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50. SJMP and LJMP:
LJMP(long jump)
•LJMP is an unconditional jump.
•It is a 3-byte instruction in which the first byte is the opcode,
and the second and third bytes represent the 16-bit address of
the target location.
•The 20byte target address allows a jump to any memory
location from 0000 to FFFFH.
SJMP(short jump)
•In this 2-byte instruction. The first byte is the opcode and the
second byte is the relative address of the target location. 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.
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51. CJNE , JNC
Exercise:
Write a program that compare R0,R1.
If R0>R1 then send 1 to port 2,
else if R0<R1 then send 0FFh to port 2,
else send 0 to port 2.
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52. CALL Instructions
An another control transfer instruction is the CALL
instruction, which is used to call a subroutine.
LCALL(long call)
•In this 3-byte instruction, the first byte is the opcode
an the second and third bytes are used for the address
of target subroutine.
•Therefore, LCALL can be used to call subroutines
located anywhere within the 64K byte address space
of the 8051.
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53. ACALL (absolute call)
 ACALL is 2-byte instruction in contrast to LCALL, which
is 13 bytes.
 Since ACALL is a 2-byte instruction, the target address of
the subroutine must be within 2K bytes address because only
11 bits of the 2 bytes are used for the address.
 There is no difference between ACALL and LCALL in
terms of saving the program counter on the stack or the
function of the RET instruction.
 The only difference is that the target address for LCALL
can be anywhere within the 64K byte address space of the
8051 while the target address of ACALL must be within a 2K-
byte range.
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54. I/O Port Programming
•Port 1 is denoted by P1.
–P1.0 ~ P1.7
•We use P1 as examples to show the operations on ports.
–P1 as an output port (i.e., write CPU data to the external pin)
–P1 as an input port (i.e., read pin data into CPU bus)
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
Port 1（pins 1-8）

55. A Pin of Port 1
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8051 IC
D Q
Clk Q
Vcc
Load(L1)
Read latch
Read pin
Write to latch
Internal CPU
bus
M1
P1.X
pin
P1.X
TB1
TB2
P0.x

56. Hardware Structure of I/O Pin
•Each pin of I/O ports
–Internal CPU bus：communicate with CPU
–A D latch store the value of this pin
•D latch is controlled by “Write to latch”
–Write to latch＝1：write data into the D latch
–2 Tri-state buffer：
•TB1: controlled by “Read pin”
–Read pin＝1：really read the data present at the pin
•TB2: controlled by “Read latch”
–Read latch＝1：read value from internal latch
–A transistor M1 gate
•Gate=0: open
•Gate=1: close
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57. Tri-state Buffer
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Output Input
Tri-state control
(active high)
L H Low
Highimpedance
(open-circuit)
HH
L H


58. Writing “1” to Output Pin P1.X
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D Q
Clk Q
Vcc
Load(L1)
Read latch
Read pin
Write to latch
Internal CPU
bus
M1
P1.X
pin
P1.X
8051 IC
2. output pin is
Vcc
1. write a 1 to the pin
1
0
output 1
TB1
TB2

59. Writing “0” to Output Pin P1.X
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D Q
Clk Q
Vcc
Load(L1)
Read latch
Read pin
Write to latch
Internal CPU
bus
M1
P1.X
pin
P1.X
8051 IC
2. output pin is
ground
1. write a 0 to the pin
0
1
output 0
TB1
TB2

60. Port 1 as Output（Write to a Port）
•Send data to Port 1：
MOV A,#55H
BACK: MOV P1,A
ACALL DELAY
CPL A
SJMP BACK
–Let P1 toggle.
–You can write to P1 directly.
Monday, March 10, 2025 Arunkumar.B

61. Reading Input v.s. Port Latch
•When reading ports, there are two possibilities：
–Read the status of the input pin. （from external pin value）
•MOV A, PX
•JNB P2.1, TARGET ; jump if P2.1 is not set
•JB P2.1, TARGET ; jump if P2.1 is set
•Figures C-11, C-12
–Read the internal latch of the output port.
•ANL P1, A ; P1 ← P1 AND A
•ORL P1, A ; P1 ← P1 OR A
•INC P1 ; increase P1
•Figure C-17
•Table C-6 Read-Modify-Write Instruction (or Table 8-5)
•See Section 8.3
Monday, March 10, 2025 Arunkumar.B

62. Reading “High” at Input Pin
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D Q
Clk Q
Vcc
Load(L1)
Read latch
Read pin
Write to latch
Internal CPU bus
M1
P1.X pin
P1.X
8051 IC
2. MOV A,P1
external pin=High
1.write a 1 to the pin MOV
P1,#0FFH
1
0
3. Read pin=1 Read latch=0
Write to latch=1
1
TB1
TB2

63. Reading “Low” at Input Pin
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D Q
Clk Q
Vcc
Load(L1)
Read latch
Read pin
Write to latch
Internal CPU bus
M1
P1.X pin
P1.X
8051 IC
2. MOV A,P1
external pin=Low1.write a 1 to the pin
MOV P1,#0FFH
1
0
3. Read pin=1 Read latch=0
Write to latch=1
0
TB1
TB2

64. Port 1 as Input（Read from Port）
•In order to make P1 an input, the port must be programmed by writing 1 to
all the bit.
MOV A,#0FFH ;A=11111111B
MOV P1,A ;make P1 an input port
BACK: MOV A,P1 ;get data from P0
MOV P2,A ;send data to P2
SJMP BACK
–To be an input port, P0, P1, P2 and P3 have similar methods.
Monday, March 10, 2025 Arunkumar.B

65. Instructions For Reading an Input Port
Mnemonics Examples Description
MOV A,PX MOV A,P2
Bring into A the data at P2
pins
JNB PX.Y,.. JNB P2.1,TARGET Jump if pin P2.1 is low
JB PX.Y,.. JB P1.3,TARGET Jump if pin P1.3 is high
MOV C,PX.Y MOV C,P2.4 Copy status of pin P2.4 to CY
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•Following are instructions for reading external pins of ports:

66. Reading Latch
•Exclusive-or the Port 1：
MOV P1,#55H ;P1=01010101
ORL P1,#0F0H ;P1=11110101
1. The read latch activates TB2 and bring the data from the Q latch into
CPU.
•Read P1.0=0
2. CPU performs an operation.
•This data is ORed with bit 1 of register A. Get 1.
3. The latch is modified.
•D latch of P1.0 has value 1.
4. The result is written to the external pin.
•External pin (pin 1: P1.0) has value 1.
Monday, March 10, 2025 Arunkumar.B

67. Reading the Latch
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D Q
Clk Q
Vcc
Load(L1)
Read latch
Read pin
Write to latch
Internal CPU bus
M1
P1.X pin
P1.X
8051 IC
4. P1.X=1
2. CPU compute P1.X OR 1
0
0
1. Read pin=0 Read latch=1 Write to
latch=0 (Assume P1.X=0 initially)
1
TB1
TB2
3. write result to latch Read
pin=0 Read latch=0
Write to latch=1
1
0

68. Read-modify-write Feature
•Read-modify-write Instructions
–Table C-6
•This features combines 3 actions in a single instruction：
1. CPU reads the latch of the port
2. CPU perform the operation
3. Modifying the latch
4. Writing to the pin
–Note that 8 pins of P1 work independently.
Monday, March 10, 2025 Arunkumar.B

69. Port 1 as Input（Read from latch）
•Exclusive-or the Port 1：
MOV P1,#55H ;P1=01010101
AGAIN:XOR P1,#0FFH ;complement
ACALL DELAY
SJMP AGAIN
–Note that the XOR of 55H and FFH gives AAH.
–XOR of AAH and FFH gives 55H.
–The instruction read the data in the latch (not from the pin).
–The instruction result will put into the latch and the pin.
Monday, March 10, 2025 Arunkumar.B

70. Read-Modify-Write Instructions
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ExampleMnemonics
SETB P1.4SETB PX.Y
CLR P1.3CLR PX.Y
MOV P1.2,CMOV PX.Y,C
DJNZ P1,TARGETDJNZ PX, TARGET
INC P1INC
CPL P1.2CPL
JBC P1.1, TARGETJBC PX.Y, TARGET
XRL P1,AXRL
ORL P1,AORL
ANL P1,AANL
DEC P1DEC

71. You are able to answer this Questions:
•How to write the data to a pin？
•How to read the data from the pin？
–Read the value present at the external pin.
•Why we need to set the pin first？
–Read the value come from the latch（not from the external
pin）.
•Why the instruction is called read-modify write?
Monday, March 10, 2025 Arunkumar.B

72. Other Pins
•P1, P2, and P3 have internal pull-up resisters.
–P1, P2, and P3 are not open drain.
•P0 has no internal pull-up resistors and does not connects to
Vcc inside the 8051.
–P0 is open drain.
–Compare the figures of P1.X and P0.X. 
•However, for a programmer, it is the same to program P0, P1,
P2 and P3.
•All the ports upon RESET are configured as output.
Monday, March 10, 2025 Arunkumar.B

73. A Pin of Port 0
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8051 IC
D Q
Clk Q
Read latch
Read pin
Write to latch
Internal CPU
bus
M1
P0.X
pin
P1.X
TB1
TB2
P1.x

74. Port 0（pins 32-39）
•P0 is an open drain.
–Open drain is a term used for MOS chips in the same way
that open collector is used for TTL chips. 
•When P0 is used for simple data I/O we must connect it to
external pull-up resistors.
–Each pin of P0 must be connected externally to a 10K ohm
pull-up resistor.
–With external pull-up resistors connected upon reset, port 0
is configured as an output port.
Monday, March 10, 2025 Arunkumar.B

75. Port 0 with Pull-Up Resistors
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P0.0
P0.1
P0.2
P0.3
P0.4
P0.5
P0.6
P0.7
DS5000
8751
8951
Vcc
10 K
P
o
r
t
0

76. Dual Role of Port 0
•When connecting an 8051/8031 to an external memory, the 8051
uses ports to send addresses and read instructions.
–8031 is capable of accessing 64K bytes of external memory.
–16-bit address：P0 provides both address A0-A7, P2 provides
address A8-A15.
–Also, P0 provides data lines D0-D7.
•When P0 is used for address/data multiplexing, it is connected to the
74LS373 to latch the address.
–There is no need for external pull-up resistors as shown in
Chapter 14.
Monday, March 10, 2025 Arunkumar.B

77. 74LS373
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D
74LS373
ALE
P0.0
P0.7
PSEN
A0
A7
D0
D7
P2.0
P2.7
A8
A15
OE
OC
EA
G
8051 ROM

78. Reading ROM (1/2)
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D
74LS373
ALE
P0.0
P0.7
PSEN
A0
A7
D0
D7
P2.0
P2.7
A8
A12
OE
OC
EA
G
8051 ROM
1. Send address to
ROM
2. 74373 latches the
address and send to
ROM
Address

79. Reading ROM (2/2)
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D
74LS373
ALE
P0.0
P0.7
PSEN
A0
A7
D0
D7
P2.0
P2.7
A8
A12
OE
OC
EA
G
8051 ROM
2. 74373 latches the
address and send to
ROM
Address
3. ROM send the
instruction back

80. ALE Pin
•The ALE pin is used for de-multiplexing the
address and data by connecting to the G pin of
the 74LS373 latch.
–When ALE=0, P0 provides data D0-D7.
–When ALE=1, P0 provides address A0-A7.
–The reason is to allow P0 to multiplex address and
data.
Monday, March 10, 2025 Arunkumar.B

81. Port 2（pins 21-28）
•Port 2 does not need any pull-up resistors
since it already has pull-up resistors internally.
•In an 8031-based system, P2 are used to
provide address A8-A15.
Monday, March 10, 2025 Arunkumar.B

82. Port 3（pins 10-17）
•Port 3 does not need any pull-up resistors since it already has
pull-up resistors internally.
•Although port 3 is configured as an output port upon reset, this
is not the way it is most commonly used.
•Port 3 has the additional function of providing signals.
–Serial communications signal：RxD, TxD（Chapter 1
0）
–External interrupt：/INT0, /INT1（Chapter 11）
–Timer/counter：T0, T1（Chapter 9）
–External memory accesses in 8031-based system：/WR,
/RD（Chapter 14）
Monday, March 10, 2025 Arunkumar.B

83. Port 3 Alternate Functions
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17RDP3.7
16WRP3.6
15T1P3.5
14T0P3.4
13INT1P3.3
12INT0P3.2
11TxDP3.1
10RxDP3.0
PinFunctionP3 Bit


84. INSTRUCTIONS:
Instruction are classified in to 5 categories.
1.Data transfer
2.Arithmetic
3.Logical
4.Boolean
5.Jump instructions.
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85. 1.Data transfer group of instructions.
•MOV
•MOVX
•MOVC
•PUSH and POP
•XCH
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86. Immediate and register addressing mode
1. MOV A, #A
2.MOV A, Reg
3.MOV Reg , A
4.MOV Reg, # n
5.MOV DPTR, # nn
6.MOV Reg,Reg not allowed
Monday, March 10, 2025 Arunkumar.B

87. Direct addressing mode.
1.MOV A, add
2.MOV add, n
3.MOV REG, add
4.MOV add, reg
5.MOV add , #n
6.MOV add1 , add2 (used to move from reg to
reg)

Monday, March 10, 2025 Arunkumar.B

88. •Indirect addressing mode
•Data access from external memory.
•MOVX A, @ RP MOVX to move from
•MOVX A@DPTR external RAM
•MOVX @ RP, A
•MOVX @ DPTR, A
•MOVC A, @A+DPTR MOVC to get data from
•MOVC A, @A+PC external ROM

Monday, March 10, 2025 Arunkumar.B

89. PUSH address
Pop address
DATA Exchange
1.XCH A, Rr all modes except immediate
2.XCH A, add may be used in exchange
3.XCH A, @RP must always involve A
4.XCH A, @RP
5.XCHD A,@RP exchanges lower nibbles
Monday, March 10, 2025 Arunkumar.B

90. •Arithmetic group of Instructions
Increment and decrement instructions
1. INC A
2. INC Rr
3.INC add
4.INC @ Rp
5.INC DPTR DEC DPTR not allowed
Monday, March 10, 2025 Arunkumar.B

91. •Add A, #n Addc A, #n
•Add A, Rr Addc A, Rr
•Add A, add Addc A, add
•Add A, @Rp Addc A, @Rp
Subtract with borrow
1.SUBB A, #n
2.SUBB A, Rr
3.SUBB A, add
4.SUBB A, @Rp




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92. Subtract with borrow
1.SUBB A, #n
2.SUBB A, Rr
3.SUBB A, add
4.SUBB A, @Rp
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93. Multiplication
MUL AB
BA (A)*(B)
lower byte
Higher byte
Flags affected
Cy is cleared, ov affected depending on the
result in 8 register.
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94. DIVISION:
DIV AB
Reg A unsigned
Reg B
After division, Integer - quotient A
Integer - remainder B
Cy is cleared, AC is unaffected and even
OV is affected – SET TO 0 – DIV BY 0.
Monday, March 10, 2025 Arunkumar.B

95. Logical group of Instructions.
AND
ANL A, #n
ANL A, Rr
ANL A, add
ANL A, @Rp
ANL addr, A
ANL add #n
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96. OR XOR
ORL A, #n XRL A, #n
ORL A, Rr XRL A, Rr
ORL A, add XRL A, add
ORL A, @Rp XRL A, @Rp
ORL addr, A XRL add, A
ORL add, #n XRL add, #n
Monday, March 10, 2025 Arunkumar.B

97. Rotate always with respect to ACC
RL A Rotate left acc
RLC A Rotate left acc with carry
RR A
RRC A
SWAP A Lower nibble and higher nibbles are
exchanged
CLR A clears A
CPL A Compliment
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98. Branch group of instructions
Jump and call
1.Jump unconditionally
2.Decrement byte and jump if not equal
3.Compare bytes and jump if not equal
4.Jump on bit conditions
5.Call a subroutine and return from subroutine
Monday, March 10, 2025 Arunkumar.B

99. Unconditional jump:
It can be of 3 ranges
-Relative range – SJMP radd – 2 byte instn
-Absolute range
AJMP radd jump to any where within a

2 byte instruction page
address is 11 bit
-long range
LJMP Ladd 3 byte
Monday, March 10, 2025 Arunkumar.B

100. JMP @ A + DPTR indirect jump
Jump to the address which is obtained by adding A+
DPTR.
Flags are not affected in any these instructions.
Byte Jumps
All byte jumps are relative to pc.
DJNZ Rn, radd
DJNZ add, radd decrement contents of
memory.
Decrement and then jump.
None of the flags are affected.( No zero flag)
Monday, March 10, 2025 Arunkumar.B

101. CJNE A , add , radd
CJNE A, #N , radd
CJNE Rn, #n , radd
CJNE @ Rp, #n , radd
BIT jumps
All bit jumps are relative to pc
JC add jump on carry
JNC add jump on no carry.
JB b, add b- address bit, jump if
addressed bit is set to 2
JNB b, add jump if address bit reset.
Monday, March 10, 2025 Arunkumar.B

102. JBC b, add jump to relative addr if
addressed bit is set to one and clear
the addressed bit
JZ radd wrt acc
JNZ radd
Monday, March 10, 2025 Arunkumar.B