Microcontroller 8051 1

By: Engr. Abraiz Khattak
Wednesday, November 2, 2016Engr. Abraiz Khattak, DEE, CUSIT

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

 CPU for Computers
 No RAM, ROM, I/O on CPU chip itself
 Example ： Intel’s x86, Motorola’s 680x0
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

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

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
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

 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
Embedded System

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
1. availability of software development tools
• assemblers, debuggers, C compilers, emulator,
simulator, technical support
1. wide availability and reliable sources of the
microcontrollers.
Three criteria in Choosing a Microcontroller

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

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
Comparison of the 8051 Family Members

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)


 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. 

 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 

 /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.

 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.

 Using a quartz crystal
oscillator
 We can observe the
frequency on the XTAL2 pin.
C2
30pF
C1
30pF
XTAL2
XTAL1
GND


 Using a TTL oscillator
 XTAL2 is unconnected.
N
C
EXTERNAL
OSCILLATOR
SIGNAL
XTAL2
XTAL1
GND



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 µs
(b) 16 MHz / 12 = 1.333 MHz;
machine cycle = 1 / 1.333 MHz = 0.75 µs

0000DPTR
0007SP
0000PSW
0000B
0000ACC
0000PC
Reset ValueRegister
RAM are all zero.


 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).


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

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

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]

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
Bit Addressable
Page 359,360

SUBB A,source ;A=A-source-CY
SETB C ;CY=1
SUBB A,R5 ;A=A-R5-1
ADC A,source ;A=A+source+CY
SETB C ;CY=1
ADC A,R5 ;A=A+R5+1

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.
 CALL

ANL - ORL - XRL
EXAMPLE:
MOV R5,#89H
ANL R5,#08H
RR – RL – RRC – RLC A
EXAMPLE:
RR A

ORG 0H
MOV R5,#25H
MOV R7,#34H
MOV A,#0
ADD A,R5
ADD A,#12H
HERE: SJMP HERE
END
EDITOR
PROGRAM
ASSEMBLER
PROGRAM
LINKER
PROGRAM
OH
PROGRAM
Myfile.asm
Myfile.obj
Other obj file
Myfile.lst
Myfile.abs
Myfile.hex

 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

 RAM memory space allocation in the 8051
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

 PSW Register
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

Instructions that Affect Flag Bits:
Note: X can be 0 or 1

Example:
MOV A,#38H
ADD A,#2FH
38 00111000
+2F +00101111
---- --------------
67 01100111
CY=0 AC=1 P=1
Example:
MOV A,#88H
ADD A,#93H
88 10001000
+93 +10010011
---- --------------
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

 Immediate
 Register
 Direct
 Register Indirect
 Indexed

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”

MOV Rn, A ;n=0,..,7
ADD A, Rn
MOV DPL, R6
MOV DPTR, A
MOV Rm, Rn

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
Bit Addressable
Page 359,360

 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
 jump

 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= 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.

 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.

 Example:
Write a program to get the x value from P1 and send x2
to P2, continuously .
Solution:
ORG 0
MOV DPTR, #TAB1
MOV A,#0FFH
MOV P1,A
L01:
MOV A,P1
MOVC A,@A+DPTR
MOV P2,A
SJMP L01
;----------------------------------------------------
ORG 300H
TAB1: DB 0,1,4,9,16,25,36,49,64,81
END

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 )

 MUL AB ;B|A = A*B
MOV A,#25H
MOV B,#65H
MUL AB ;25H*65H=0E99
;B=0EH, A=99H
 MUL AB ;A = A/B, B = A mod
B
MOV A,#25
MOV B,#10
MUL AB ;A=2, B=5

 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
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

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

 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

 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,#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

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.

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.

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.

 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.

 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)

Port 1 （ pins 1-8 ）

8051 IC
D Q
Clk Q
Vcc
Load(L1)
Read latch
Read pin
Write to latch
Internal CPU
bus
M1
P1.X
pinP1.X
TB1
TB2
P0.x

 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

Output Input
Tri-state control
(active high)
L H Low
Highimpedance
(open-circuit)
HH
L H


D Q
Clk Q
Vcc
Load(L1)
Read latch
Read pin
Write to latch
Internal CPU
bus
M1
P1.X
pinP1.X
8051 IC
2. output pin is
Vcc1. write a 1 to the pin
1
0 output 1
TB1
TB2

D Q
Clk Q
Vcc
Load(L1)
Read latch
Read pin
Write to latch
Internal CPU
bus
M1
P1.X
pinP1.X
8051 IC
2. output pin is
ground1. write a 0 to the pin
0
1 output 0
TB1
TB2

 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.

 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

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

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

 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.

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
 Following are instructions for reading external pins of ports:

 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.

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=12. 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

 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.

 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.

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

 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?

 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.

8051 IC
D Q
Clk Q
Read latch
Read pin
Write to latch
Internal CPU
bus
M1
P0.X
pinP1.X
TB1
TB2
P1.x

 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.

P0.0
P0.1
P0.2
P0.3
P0.4
P0.5
P0.6
P0.7
DS5000
8751
8951
Vcc
10 K
Port0

 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.

D
74LS373ALE
P0.0
P0.7
PSEN
A0
A7
D0
D7
P2.0
P2.7
A8
A15
OE
OC
EA
G
8051 ROM

D
74LS373ALE
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

D
74LS373ALE
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

 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.

 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.

 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
10 ）
External interrupt ： /INT0, /INT1 （ Chapter 11 ）
Timer/counter ： T0, T1 （ Chapter 9 ）
External memory accesses in 8031-based
system ： /WR, /RD （ Chapter 14 ）

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

Microcontroller 8051 1

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