Embedded system 8051 Microcontroller

ANKIT SHARMA
RESEARCH SCHOLAR
ME-VLSI DESIGN
PEC UNIVERSITY OF TECHNOLOGY
CHANDIGARH-160012
INDIA

OUTLINES
 INTRODUCTION TO COMPUTING
 Microprocessor
 Microcontroller 8051
 Real World Interfacing of 8051

DECIMAL AND BINARY NUMBER SYSTEMS
Human beings use base 10 (decimal) arithmetic
There are 10 distinct symbols, 0, 1, 2, …,9
 Computers use base 2 (binary) system
 There are only 0 and 1
These two binary digits are commonly referred to as bits

NUMBER SYSTEM
Decimal
Binary
Octal
Hexa Decimal

CONVERTING FROM DECIMAL TO BINARY
Divide the decimal number by 2 repeatedly
Keep track of the remainders
Continue this process until the quotient becomes zero
Write the remainders in reverse order to obtain the binary number

In general, we use an AC supply of 230V 50Hz, but this power has to
be changed into required form or voltage range for providing power
supply to different types of devices
Microcontrollers require a 5V DC supply, so AC 230V needs to be
converted into 5V DC using the step-down converter in their power
supply circuit.

4 BASIC STEPS
Step Down the Voltage Level
Convert AC to DC
Smoothing the Ripples using Filter
Regulated IC

STEP DOWN THE VOLTAGE LEVEL
Step-down transformer consists of two windings, namely primary and
secondary windings.
Primary can be designed using a less-gauge wire with more number
of turns as it is used for carrying low-current high-voltage power.
Secondary winding using a high-gauge wire with less number of turns
as it is used for carrying high-current low-voltage power. Transformers
works on the principle of Faraday’s laws of electromagnetic induction.
Actually, mutual induction between two or more winding is
responsible for transformation action in an Electrical Transformer.

CONVERT AC TO DC
230V AC power is converted into 12V AC (12V RMS), but required
power is 5V DC
For this purpose, 12V AC power must be primarily converted into DC
power then it can be stepped down to the 5V DC.
 AC power can be converted into DC using Rectifier.
 There are different types of rectifiers, such as half-wave rectifier, full-
wave rectifier and bridge rectifier.
 Due to advantages of bridge rectifier over half & full wave rectifier,
bridge rectifier is frequently used for converting AC to DC.

SMOOTHING THE RIPPLES USING FILTER
Output of diode bridge is a DC consisting of ripples also called
pulsating DC.
pulsating DC can be filtered using an inductor filter or a capacitor
filter or a resistor-capacitor-coupled filter for removing ripples.
 Consider a capacitor filter which is frequently used in most cases for
smoothing.

Capacitor is an energy storing element.
Capacitor stores energy while input increases from zero to a peak
value and, while supply voltage decreases from peak Value to zero,
capacitor starts discharging.
 This charging and discharging of the capacitor will make the
pulsating DC into pure DC.

REGULATING 12V DC INTO 5V DC USING VOLTAGE REGULATOR
 15V DC voltage can be stepped down to 5V DC voltage using a DC step-
down converter called as voltage reguIator IC7805.
 The first two digits ‘78’ of IC7805 voltage regulator represent positive
series voltage regulators and the last two digits ‘05’ represents the output
voltage of the voltage regulator

WHY DO WE NEED TO LEARN
MICROPROCESSORS / CONTROLLERS
The microprocessor is the core of computer systems
Now a days many communication, digital entertainment, portable
devices, are controlled by them.
A designer should know what types of components he needs, ways
to reduce production costs and product reliable.

DIFFERENT ASPECTS OF A
MICROPROCESSOR / CONTROLLER
Hardware :Interface to the real world
Software :Order how to deal with inputs

THE NECESSARY TOOLS FOR A
MICROPROCESSOR / CONTROLLER
CPU: Central Processing Unit
I/O: Input /Output
Bus: Address bus , Control Bus & Data bus
Memory: RAM & ROM
Timer
Interrupt
Serial Port
Parallel Port

MICROPROCESSOR:
Multipurpose
Re-Programmable
Digital Device
Semiconductor IC

MICROPROCESSOR (CONT.)
Works On:
Input data from outer world
Process it under control of stored
instructions /program in memory.
Provide desired result to the outer
world.

MICROCONTROLLER
Introduced in 1981 by Intel Corporation.
Microcontroller is a programmable digital processor with
necessary peripherals.
Both microcontrollers and microprocessors are complex
sequential digital circuits meant to carry out job according to
the program / instructions.

GENERAL PURPOSE MICROPROCESSORS
No RAM
 No ROM
 No I/O ports Many chips on mother’s board
A single chip

MICROCONTROLLER
CPU (microprocessor)
RAM
 ROM
 I/O ports
 Timer
 ADC and other peripherals

GENERAL PURPOSE MICROPROCESSORS
Must add RAM, ROM, I/O ports, and timers externally to make
them functional
Make the system bulkier and much more expensive
Have the advantage of versatility on the amount of RAM, ROM,
and I/O ports

DISADVANTAGES OF MICROPROCESSOR
The overall system cost is high
A large sized PCB is required for assembling all the components
Overall product design requires more time
Physical size of the product is big
 A discrete components are used, the system is not reliable

ADVANTAGES OF MICROCONTROLLER BASED SYSTEM
As the peripherals are integrated into a single chip, the overall
system cost is very less
The product is of small size compared to micro processor based
system
The system design now requires very little efforts
As the peripherals are integrated with a microprocessor the system is
more reliable
Though microcontroller may have on chip ROM,RAM and I/O ports,
additional ROM, RAM I/O ports may be interfaced externally if
required
On chip ROM provide a software security

MICROCONTROLLER
The fixed amount of on-chip ROM, RAM, and number of I/O ports
makes them ideal
In many applications, the space it takes, the power it consumes, and
the price per unit are much more critical considerations than the
computing power

8051 FAMILY HAS LARGEST NUMBER OF DIVERSIFIED
(MULTIPLE SOURCE) SUPPLIERS
Intel
Atmel
Philips/ Signetics
AMD
Infineon (formerly Siemens)
 Matra
Dallas Semiconductor/Maxim

EMBEDDED SYSTEMS
Embedded system means the processor is embedded into that
application.
An embedded product uses a microprocessor or microcontroller to
do one task and one task only There is only one application software
that is typically burned into ROM
A PC, in contrast with the embedded system, can be used for any
number of applications

It has RAM memory and an operating system that loads a variety of
applications into RAM and lets the CPU run them
PC contains or is connected to various embedded products
Each one peripheral has a microcontroller inside it that performs
only one task
In an embedded system, there is only one application software that
is typically burned into ROM.
Example：Printer, keyboard, Video game player

Home
 Appliances, intercom, telephones, security systems, garage door
openers, answering machines, fax machines, home computers,
TVs, cable TV tuner, VCR, camcorder, remote controls, video
games, cellular phones, musical instruments, sewing machines,
lighting control, paging, camera, pinball machines, toys, exercise
equipment
Office
 Telephones, computers, security systems, fax machines,
microwave, copier, laser printer, color printer, paging
Auto
Trip computer, engine control, air bag, ABS, instrumentation,
security system, transmission control, entertainment, climate
control, cellular

CHOOSING A MICROCONTROLLER
8-bit microcontrollers
Motorola’s 6811
Intel’s 8051
Zilog’s Z8
Atmel
Microchip’s PIC
 There are also 16-bit & 32-bit Uc made by various chip makers

CRITERIA FOR CHOOSING A MICROCONTROLLER
 Meeting the computing needs of the task at hand efficiently and cost
effectively
 Speed
 Packaging
 Power consumption
 The amount of RAM and ROM on chip
 The number of I/O pins and the timer on chip
 How easy to upgrade to higher performance or lower power
consumption
 Versions, software development tools, such as compilers, assemblers, simulator
and debuggers
 Cost per unit

COMPARISON OF THE 8051 FAMILY MEMBERS
 ROM type
 8031 no ROM
 80xx mask ROM
 87xx EPROM
 89xx Flash EEPROM
 89xx
 8951
 8952
 8953
 8955
 898252
 891051
 892051
Example (AT89C51,AT89LV51)
AT= ATMEL(Manufacture)
C = CMOS technology
LV= Low Power(3.0v)

INTEL INTRODUCED 8051, REFERRED AS MCS- 51, IN 1981
The 8051 is an 8-bit processor
The CPU can work on only 8 bits of data at a time
128 bytes of RAM
4K bytes of on-chip ROM
Two 16 bit timers
One serial port
Internal memory consists of on-chip ROM and on-chip data RAM
 Separate memory space for programs (code) and data

Four I/O ports, each 8 bits wide
6 interrupt sources
Full duplex UART
 On-chip clock oscillator
64K external code & data memory space.
 210 bit-addressable locations.
Operating frequency is 24MHz-33MHz
+5V Regulated DC power supply is required to operate .

MOST WIDELY USED REGISTERS
A (Accumulator)
 For all arithmetic and logic instructions
B, R0, R1, R2, R3, R4, R5, R6, R7
 DPTR (data pointer), and PC (program counter)
SP(stack pointer)
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

ACCUMULATOR
Accumulator, as its name suggests, is used as a general register to
accumulate the results of a large number of instructions.
For example, if you want to add the number 10 and 20, the
resulting 30 will be stored in Accumulator. Once you have a value
in Accumulator you may continue processing value or you may
store it in another register or in memory.

THE "R" REGISTERS
The "R" registers are a set of eight registers that are named R0, R1,
etc. up to and including R7.
Example:-
MOV A,R3 ;Move the value of R3 into the accumulator
ADD A,R4 ;Add the value of R4
MOV R5,A ;Store the resulting value temporarily in R5

THE "B" REGISTER
The "B" register is very similar to the Accumulator in the sense that it
may hold an 8-bit (1-byte) value.
The "B" register is only used by two 8051 instructions: MUL AB and
DIV AB. Thus, if you want to quickly and easily multiply or divide A
by another number, you may store other number in "B" and make
use of these two instructions.
Aside from the MUL and DIV instructions, "B" register is often used
as yet another temporary storage register much like a ninth "R"
register.

PROGRAM COUNTER(PC)
The Program Counter (PC) is a 2-byte address.
Indicate, where next instruction to execute is found in memory.
This means that it can access program addresses 0000 to FFFFH, a
total of 64K bytes of code
The first opcode is burned into ROM address 0000H, since this is
where the 8051 looks for the first instruction when it is booted. PC
incremented each time an instruction is executed.
We achieve this by the ORG statement in the source program
 PC is not always incremented by one. Some instructions need 2 or 3
bytes PC will be incremented by 2 or 3 in these cases.

DPTR
The Data Pointer (DPTR) is the 8051s only user-access 16-bit (2-byte)
register.
DPTR, is used to point to data. It is used by number of commands
which allow the 8051 to access external memory. When the 8051
accesses external memory it will access external memory at the
address indicated by DPTR.
While DPTR is most often used to point to data in external memory,
many programmers often take advantage of the fact that its the only
true 16-bit register available. It is often used to store 2-byte values
which have nothing to do with memory locations.

STACK IN 8051
Section of RAM used to store information
temporarily
Could be data or an address
CPU needs this storage area since there are
only limited number of registers
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 07h by default .
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

STACK POINTER(SP)
 The Stack Pointer is used to indicate where the next value to be
removed from the stack should be taken from.
When you push a value onto the stack, the 8051 first increments the
value of SP and then stores the value at the resulting memory
location.
When you pop a value off the stack, the 8051 returns value from the
memory location indicated by SP, and then decrements the value of
SP.

When the 8051 is initialized SP will be initialized to 07h.
 If you immediately push a value onto the stack, the value will be
stored in Internal RAM address 08h.
First the 8051 will increment the value of SP (from 07h to 08h) and
then will store the pushed value at that memory address (08h)
SP is modified directly by the 8051 by six instructions: PUSH, POP,
ACALL, LCALL, RET, and RETI.

PROGRAM STATUS WORD
Program status word (PSW) register, also referred to as flag
register, is an 8 bit register
Only 6 bits are used
These four are CY (carry), AC (auxiliary carry), P(parity), and OV
(overflow)
 They are called conditional flags, meaning that they indicate some
conditions that resulted after instruction was executed
The PSW3 and PSW4 are designed as RS0 and RS1, and are used to
change the bank
The two unused bits are user-definable

PROGRAM STATUS WORD
Bits of the PSW Register

INSTRUCTIONS
THAT AFFECT
FLAG BITS
Instruction CY OV AC
ADD X X X
ADDC X X X
SUBB X X X
MUL O X
DIV O X
DA X
RPC X
PLC X
SETB C 1
CLR C O
CPL C X
ANL C,bit X
ANL C, /bit X
ORL C, bit X
ORL C, /bit X
MOV c, bit X
CJNE X

EXAMPLE-ADD INSTRUCTION & PSW
Flag bits affected by ADD instruction are CY, P, AC, and OV
Show status of CY, AC and P flag after addition of 38H & 2FH in
following instructions.
 MOV A, #38H
 ADD A, #2FH ;after the addition A=67H, CY=0
Solution:
 38 (00111000) + 2F(00101111)= 67 (01100111)
 CY = 0 since there is no carry beyond the D7 bit
 AC = 1 since there is a carry from the D3 to the D4 bi
 P = 1 Accumulator has an odd number of 1s (it has five 1s)

PIN DESCRIPTION OF 8051
8051 family members (e.g, 8751, 89C51, 89C52, DS89C4x0)
Have 40 pins dedicated for various functions such as I/O, RD, WR,
address, data, and interrupts
Come in different packages, such as
DIP(dual in-line package)

QFP(quad flat package)
LLC(leadless chip carrier)
Some companies provide a 20-pin version of the 8051 with
reduced number of I/O ports for less demanding applications

VCC (40) & GND (20)
VCC（pin 40）
Vcc provides supply voltage to the chip.
The voltage source is +5V.
GND（pin 20）：ground

OSCILLATOR CIRCUIT:-
 The 8051 requires an external oscillator circuit.
 The oscillator circuit usually runs around 12MHz.
The crystal generates 12M pulses in one second.
The pulse is used to synchronize the system operation in a controlled
pace.
Used for synchronizing internal operations.
Pins XTAL1 & XTAL2 have been used.
Length of machine cycle depends on the frequency of crystal oscillator
connected to 8051.

XTAL1 AND XTAL2 (PIN 18 & PIN 19 )
8051 has on-chip oscillator but requires external clock to run it
 A quartz crystal oscillator is connected to inputs XTAL1 (pin19) and
XTAL2 (pin18)
Quartz crystal oscillator needs two capacitors of 30 pF value

RST（PIN 9）
RESET pin is an input and is active high (normally low)
Upon applying a high pulse to this pin, the microcontroller will
reset and terminate all activities
 This is often referred to as a power-on reset
Activating a power-on reset will cause all values in the registers to
be lost

RESET VALUE OF SOME 8051 REGISTERS:
Register Reset Value
PC 0000
DPTR 0000
ACC 00
PSW 00
SP 07
B 00
P0-P3 FF
RAM are all zero.

In order for the RESET input to be effective, it must have minimum
duration of 2 machine cycles
 In other words, the high pulse must be high for a minimum of 2
machine cycles before it is allowed to go low

𝐸𝐴 （PIN 31）
External access, is an input pin and must be connected to Vcc or GND
“𝐸𝐴 ” means active low
The 8051 family members all come with on-chip ROM to store
programs
EA pin is connected to Vcc
The 8031 and 8032 family members do no have on-chip ROM, so
code is stored on an external ROM and is fetched by 8031/32
EA pin must be connected to GND to indicate that the code is
stored externally

PSEN （PIN 29） & ALE （PIN 30）
Following two pins are used mainly in 8031-based systems
 𝑃𝑆𝐸𝑁 : Program store enable, is an output pin
This pin is connected to OE pin of the ROM
ALE :Address latch enable, is output pin and is active high
 𝑃𝑆𝐸𝑁 ＆ ALE are used for external ROM
The 8031 multiplexes address and data through port 0 to save pins
ALE pin is used for de multiplexing the address and data

I/O PORT PINS
The four 8-bit I/O ports P0, P1, P2 and P3 each uses 8 pins
All the ports upon RESET are configured as output, ready to be used
as input ports
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

PORT 0
Port 0 is also designated as AD0-AD7,
allowing it to be used for both address and
data
When connecting an 8051/31 to an
external memory, port 0 provides both
address and data
The 8051 multiplexes address and data
through port 0 pins
ALE indicates if P0 has address or data
When ALE=0, it provides data D0-D7
 When ALE=1, it has address A0-A7

PORT 0 (CONT’)
Port 0 is 8-bitbidirectional I/O port.
Port 0 pins can be used as high-impedance inputs.
P0 pin must be connected externally to 10K ohm pull-up resistor
P0 is an open drain, unlike P1, P2, and P3
Open drain is used for MOS chips in same way that open collector
is used for TTL chips
For output, pin latches are programmed to 0.
For input, pin latches are programmed to 1

PORT 1 & PORT 2
Port 1 & Port 2 is an 8-bit bidirectional I/0 port.
In 8051 with no external memory connection
Both P1 and P2 are used as simple I/O
In 8031/51external memory connections
Port 2 must be used along with P0 to provide the 16-bit address
for the external memory
P0 provides lower 8 bits via A0 – A7
 P2 is used for upper 8 bits of 16-bit address,
A8 – A15, and it cannot be used for I/O
For output, pin latches are programmed to 0.
For input, pin latches are programmed to 1.

PORT 3
Port 3 is an 8-bit bi-directional I/0 port.
 Port 3 does not need any pull-up resistors
 Port 3 has the additional function of providing some extremely
important signals

PORT 3 (CONT’)
P3 Bit Function Pin
P3.0 RxD 10
P3.1 TxD 11
P3.2 𝐼𝑁𝑇𝑂 12
P3.3 𝐼𝑁𝑇1 13
P3.4 T0 14
P3.5 T1 15
P3.6 𝑊𝑅 16
P3.7 𝑅𝐷 17
 RXD (P3.0): Serial input port
 TXD (P3.1): Serial output port
 INT0 (P3.2): External interrupt
 INT1 (P3.3): External interrupt
 T0 (P3.4): Timer 0 external input
 T1 (P3.5): Timer 1 external input
 WR (P3.6): External data memory write strobe
 RD (P3.7): External data memory read strobe

INTERRUPTS
An interrupt is an external or internal event that interrupts the
microcontroller to inform it that a device needs its service
A single microcontroller can serve several devices using interrupts
Whenever any device needs its service, the device notifies
microcontroller by sending it an interrupt signal
Upon receiving an interrupt signal, Microcontroller interrupts
whatever it is doing and serves device
Program associated with Interrupt is called interrupt service routine
(ISR) or interrupt handler

INTERRUPT SERVICE ROUTINE
For every interrupt, there must be an interrupt service routine (ISR),
or interrupt handler
When interrupt is invoked, microcontroller runs interrupt service
routine
For every interrupt, there is a fixed location in memory that holds the
address of its ISR
The group of memory locations set aside to hold the addresses of
ISRs is called Interrupt vector table

STEPS IN EXECUTING AN INTERRUPT
Upon activation of interrupt, microcontroller perform these steps
 It finishes the instruction it is executing and saves the address of the next
instruction (Program Counter) on the stack using PUSH operation.
 It also saves current status of all interrupts internally (i.e : not on stack)
 It jumps to a fixed location in memory, called interrupt vector table, that
holds address of the ISR

Microcontroller gets address of ISR from interrupt vector table and
jumps to it
It starts to execute the interrupt service subroutine until it reaches
the last instruction of subroutine which is RETI (return from interrupt)
Upon executing RETI instruction, microcontroller returns to the place
where it was interrupted
 First, it gets program counter (PC) address from stack by popping the
top two bytes of the stack into PC
 Then it starts to execute from that address

SIX INTERRUPTS IN 8051
Six interrupts are allocated as follows
 Reset – power-up reset
 Two interrupts are set aside for the timers:
 One for timer 0 and one for timer 1
 Two interrupts are set aside for hardware external interrupts
 P3.2 and P3.3 are for external hardware interrupts INT0 , & INT1
 Serial communication has a single interrupt that belongs to both
receive and transfer

INTERRUPT VECTOR TABLE
ORG 00h ;wake-up ROM reset
location
LJMP MAIN ;By-pass interrupt
vector table
ORG 30H
MAIN:
....
END

ENABLING AND DISABLING AN INTERRUPT
Upon reset, all interrupts are disabled (masked), means none will be
responded to by the microcontroller if they are activated
The interrupts must be enabled by software in order for the
microcontroller to respond to them
There is a register called IE (interrupt enable) that is responsible for
enabling (unmasking) and disabling (masking) the interrupts

IE (INTERRUPT ENABLE) REGISTER

ENABLING AND DISABLING AN INTERRUPT
To enable an interrupt, we take the following steps:
 Bit D7 of the IE register (EA) must be set to high to allow the rest of
register to take effect
 The value of EA
 If EA = 1, interrupts are enabled and will be responded to if their
corresponding bits in IE are high
 If EA = 0, no interrupt will be responded to, even if the associated bit in
the IE register is high

EXAMPLE
Show instructions to (a) enable serial interrupt, timer 0 Interrupt, and
external hardware interrupt 1 (ex1) (b) disable (Mask) timer 0 interrupt
(c) show how to disable all Interrupts with a single instruction.
Solution:
(A) MOV IE, #10010110B ; Timer 0, EX1 &
; enable serial
Another way
Setb IE.7 ;EA=1, global enable
Setb IE.4 ;enable serial interrupt
Setb IE.1 ;enable timer 0 interrupt
Setb IE.2 ;enable EX1
(B) CLR IE.1 ;mask (disable)
timer 0 interrupt only
(C) CLR IE.7 ;disable all
interrupts

The timer flag (TF) is raised when timer rolls over
If timer interrupt in IE register is enabled, whenever timer rolls over,
TF is raised, and microcontroller is interrupted in whatever it is
doing, and jumps to the interrupt vector table to service the ISR
In this way, microcontroller can do other until it is notified that the
timer has rolled over

EXTERNAL HARDWARE INTERRUPTS
The 8051 has two external hardware interrupts
 Pin 12 (P3.2) and pin 13 (P3.3) of the 8051, designated as INT0
and INT1, are used as external hardware interrupts
 The interrupt vector table locations 0003H and 0013H are set
aside for INT0 and INT1
There are two activation levels for the external hardware
interrupts
 Level trigged
 Edge trigged

LEVEL TRIGGERED INTERRUPT
In level-triggered mode, INT0 and INT1 pins are normally high
If a low-level signal is applied to them, it triggers the interrupt
Then the microcontroller stops whatever it is doing and jumps to the
interrupt vector table to service that interrupt
The low-level signal at the INT pin must be removed before execution
of the last instruction of ISR, RETI; otherwise, another interrupt will be
generated
This is called a level-triggered or level activated interrupt and is default
mode upon reset of the 8051

EXAMPLE
Assuming that INT1 pin is connected to a switch that is
normally high. Whenever it goes low, it should turn on an LED.
LED is normally off. When it is turned on it should stay on for a
fraction of a second. Till the time switch is pressed low, LED
should stay on.

1.Org 0000h ;Starting point
2.LJMP MAIN ;by-pass interrupt vector table ISR to turn on LED
 Org 0013h ;int1 ISR pin3.1=INT1
 Setb P1.3 ;turn on LED
 Mov r3,#255
 BACK: DJNZ R3,BACK ;keep LED on for a while
 Clr p1.3 ;turn off the LED
 RETI ;return from ISR MAIN program for initialization
3.Org 30h ;from 00 -> 30h to avoid interrupt rom location
4.MAIN: MOV IE,#10000100B ;enable external INT 1
5.HERE: SJMP HERE ;stay here until get interrupted
End

Pins P3.2 and P3.3 are used for normal I/O unless the INT0 and INT1
bits in the IE register are enabled
After hardware interrupts in the IE register are enabled, the controller
keeps check intn pin for a low-level signal once each machine cycle
According to one manufacturer’s data sheet, pin must be held in a low
state until the start of the execution of ISR . if intn pin is brought back
to a logic high before start of execution of ISR there will be no
interrupt
If INT pin is left at a logic low after RETI instruction of ISR, another
interrupt will be activated after one instruction is executed

SERIAL COMMUNICATION INTERRUPT
RI and TI Flags and Interrupts

SERIAL COMMUNICATION INTERRUPT
TI (transfer interrupt) is raised when last bit of the framed data,
stop bit, is transferred, indicating that SBUF register is ready to
transfer the next byte
RI (received interrupt) is raised when entire frame of data,
including stop bit, is received
 In other words, when the SBUF register has a byte, RI is raised to
indicate that received byte needs to be picked up before it is
lost (overrun) by new incoming serial data

In 8051, one interrupt set aside for serial communication
This interrupt is used for both send and receive data
Interrupt bit in IE register (IE.4) is enabled, when RI or TI is raised
 8051 gets interrupted and jumps to memory location 0023H
 To execute ISR we must examine TI and RI flags to see which one
caused the interrupt and respond accordingly

USE OF SERIAL COM IN 8051
The serial interrupt is used mainly for receiving data and is
never used for sending data serially
 This is like getting a telephone call in which we need a ring to be
notified
 If we need to make a phone call there are other ways to remind
ourselves and there is no need for ringing
 However in receiving the phone call, we must respond
immediately no matter what we are doing or we will miss the call

INTERRUPT PRIORITY
When 8051 is powered up, priorities are assigned according to the
following
Highest To Lowest Priority
External Interrupt 0 (INT0)
Timer Interrupt 0 (TF0)
External Interrupt 1 (INT1)
Timer Interrupt 1 (TF1)
Serial Communication (R1+T1)

INTERRUPT PRIORITY (CONT’)
We can alter the sequence of interrupt priority by assigning a
higher priority to any one of the interrupts by programming a
register called IP (interrupt priority)
To give a higher priority to any of interrupts, we make
corresponding bit in the IP register high
When two or more interrupt bits in IP register are set to high
 While these interrupts have a higher priority than others, they are
serviced according to the sequence

INTERRUPT PRIORITY REGISTER (BIT-ADDRESSABLE)
Priority bit=1 assigns high priority
Priority bit=0 assigns low priority

PROGRAMMING TIMERS
8051 has two timers/counters, they can be used either as
 Timers to generate a time delay or as
 Event counters to count events happening outside microcontroller
Both timer 0 and timer 1 are 16 bits wide since 8051 has an 8-bit
architecture, each 16-bits timer is accessed as two separate registers
of low byte and high byte

TIMER 0 & TIMER 1
Accessed as low byte and high byte
Low byte register is called TL0/TL1
High byte register is called TH0/TH1
Accessed like any other register
 Mov TL0, #4FH
 Mov R5, TH0

TMOD REGISTER
Both timers 0 and 1 use same register, called TMOD (timer
mode), to set various timer operation modes
TMOD is a 8-bit register,
 Lower 4 bits are for timer 0
 Upper 4 bits are for timer 1
 In each case,
 The lower 2 bits are used to set timer mode
 The upper 2 bits to specify operation

--- --- --- X X X X X x x x x x x x x

EXAMPLE
Indicate which mode and which timer are selected for each of the
following.
(a) MOV TMOD, #01H (b) MOV TMOD, #20H (c) MOV TMOD, #12H
Solution:
 We convert the value from hex to binary. we have:
 (a) TMOD = 00000001, mode 1 of timer 0 is selected.
 (b) TMOD = 00100000, mode 2 of timer 1 is selected.
 (c) TMOD = 00010010, mode 2 of timer 0, and mode 1 of timer 1 are
selected.

TMOD REGISTER-GATE
Timers of 8051 do starting and stopping by either software or hardware
control
 In using software to start and stop timer where GATE=0
 The start and stop of the timer are controlled by way of software by TR
(timer start) bits TR0 and TR1
The SETB instruction starts it, and it is stopped by CLR instruction
These instructions start and stop timers as long as GATE=0 in TMOD register
 The hardware way of starting and stopping timer by an external source is
achieved by making GATE=1 in the TMOD register

EXAMPLE
Find timer’s clock frequency and its period for various 8051-based
system, with crystal frequency 11.0592 MHz when C/T bit of TMOD is
0.
 1/12 × 11.0529 MHz = 921.6 MHz;
 T = 1/921.6 kHz = 1.085 us

EXAMPLE
Find value for TMOD if we want to program
timer 0 in mode 2, use 8051 XTAL for clock
source, and use instructions to start and stop
timer.
 TMOD = 0000 0010

MODE 1 PROGRAMMING
The following are the characteristics and operations of mode1:
1. It is a 16-bit timer;
 Allows value of 0000 to FFFFH to be loaded into timer’s register TL and
TH
2. After TH and TL are loaded with a 16-bit initial value, the timer must
be started
 This is done by SETB TR0 for timer 0 and SETB TR1 for timer 1
3. After the timer is started, it starts to count up
 It counts up until it reaches its limit of FFFFH

 When it rolls over from FFFFH to 0000, it sets high a flag bit called TF
(timer flag)
 Each timer has its own timer flag: TF0 for timer 0, & TF1 for timer 1
 This timer flag can be monitored
 When this timer flag is raised, one option would be to stop the timer
with instructions CLR TR0 or CLR TR1, for timer 0 and timer 1,
respectively
4. After the timer reaches its limit and rolls over, in order to repeat
the process
 TH and TL must be reloaded with the original value, and
 TF must be reloaded to 0

GENERATE A TIME DELAY
 Load the TMOD value register indicating which timer(timer 0 or timer 1 )is to be used and which
mode (0 or 1 ) is selected.
 Load registers TL and TH with initial count value
 Start the timer
 Keep monitoring the timer flag (TF) with the JNB TFx ,target instruction to see if it is raised
 Get out of the loop when TF become high
 Stop the timer
 Clear the TF flag for the next round
 Go back to step 2 to load TH and TL again

Timers can also be used as counters counting events happening
outside the 8051

C/T BIT IN TMOD REGISTER
The C/T bit in the TMOD registers decides the source of the clock
for the timer
 When C/T = 1, the timer is used as a counter and gets its ulses from
outside the 8051
 The counter counts up as pulses are fed from pins 14 and 15, these pins
are called T0 (timer 0 input) and T1 (timer 1 input)

MEMORY CAPACITY
Number of bits that a semiconductor memory chip can store is called
chip capacity
Units of Kbits (kilobits), Mbits (megabits), and so on
Distinguished from storage capacity of computer systems
Memory capacity of a memory IC chip is always given bits,
memory capacity of a computer system is given in bytes
16M memory chip – 16 megabits
 Computer comes with 16M memory – 16 megabytes

MEMORY ORGANIZATION
Memory chips are organized into number of locations within IC
 Each location can hold 1 bit, 4 bits, 8 bits, or even 16 bits,
depending on how it is designed internally
 Number of locations within a memory IC depends on the
address pins
Number of bits that each location can hold is always equal to
the number of data pins
 A memory chip contain 2 𝑋
location, where x is the number of address
pins
Each location contains y bits, where y is the number of data pins on
the chip
 The entire chip will contain 2 𝑋
.y bits

SPEED
One of the important feature of a memory chip is the speed at which
its data can be accessed
To access the data, the address is presented to the address pins, the
READ pin is activated, and after a certain amount of time has elapsed,
the data shows up at the data pins
Shorter this elapsed time, the better, and consequently, the more
expensive the memory chip
Speed of memory chip is referred to as its access time

SPEED (CONT’)
A given memory chip has 12 address pins and 4 data pins. Find: (a)
The organization, and (b) the capacity.
Solution:
(a) This memory chip has 4096 locations (212 = 4096), and each
location can hold 4 bits of data. This gives an organization
of 4096 × 4, often represented as 4K × 4.
(b) The capacity is equal to 16K bits since there is a total of
4K locations and each location can hold 4 bits of data.

Example
A 512K memory chip has 8 pins for data.
Find: (a) The organization
(b) The number of address pins for this memory chip.
Solution:
(a) A memory chip with 8 data pins means that each location within
the chip can hold 8 bits of data.
To find number of locations within this memory chip, divide the capacity
by number of data pins. 512K/8 = 64K; therefore, the organization for this
memory chip is 64K × 8
(b) The chip has 16 address lines since 216 = 64K

ROM (READ-ONLY MEMORY)
ROM is a type of memory that does not lose its contents when the
power is turned off
ROM is also called non volatile memory
 There are different types of read-only memory
 PROM
 EPROM
 EEPROM
 Flash EPROM
 Mask ROM

PROM (PROGRAMMABLE ROM)
PROM refers to the kind of ROM that the user can burn information
into
PROM is a user-programmable memory
For every bit of the PROM, there exists a fuse
 If the information burned into PROM is wrong, that PROM must be
discarded since its internal fuses are blown permanently
PROM is also referred to as OTP (one-time programmable)
Programming ROM, also called burning ROM, requires special
equipment called a ROM burner or ROM programmer

EPROM (ERASABLE PROGRAMMABLE ROM) (CONT’)
To program a UV-EPROM chip, following steps must be taken:
Its contents must be erased
To erase a chip, it is removed from its socket on the system
board and placed in EPROM erasure equipment to expose it to
UV radiation for 15-20 minutes
Program the chip
To program a UV-EPROM chip, place it in the ROM burner
To burn code or data into EPROM, ROM burner uses 12.5 volts,
Vpp in the UV-EPROM data sheet or higher, depending on the
EPROM type
 Place the chip back into its system board socket

EPROM (ERASABLE PROGRAMMABLE ROM) (CONT’)
There is an EPROM programmer (burner), and there is also separate
EPROM erasure equipment
 The major disadvantage of UV-EPROM, is that it cannot be
programmed while in the system board
Notice the pattern of the IC numbers
Ex. 27128-25 refers to UV-EPROM that has a capacity of 128K bits
and access time of 250 nanoseconds
27xx always refers to UV-EPROM chips

For ROM chip 27128, find the number of data and address pins.
Solution:
The 27128 has a capacity of 128K bits. It has 16K × 8 organization (all
ROMs have 8 data pins), which indicates that there are 8 pins for data,
and 14 pins for address (214 = 16K)

EEPROM (ELECTRICALLY ERASABLE PROGRAMMABLE ROM)
 EEPROM has several advantage over EPROM
 Its method of erasure is electrical and therefore instant, as opposed to
the 20- minute erasure time required for UVEPROM
 One can select which byte to be erased, in contrast to UV-EPROM, in
which the entire contents of ROM are erased
 One can program and erase its contents while it is still in the system
board
 EEPROM does not require an external erasure and programming
device
 The designer incorporate into the system board the circuitry to
program the EEPROM

FLASH MEMORY EPROM
Flash EPROM has become a popular user-programmable memory chip
 Process of erasure of the entire contents takes less than a second,
or might say in a flash
The erasure method is electrical
The major difference between EEPROM and flash memory is
Flash memory’s contents are erased, then the entire device is
erased
There are some flash memories are recently made so that the
erasure can be done block by block
One can erase a desired section or byte on EEPROM

FLASH MEMORY EPROM (CONT’)
It is believed that flash memory will replace part of the hard disk as a
mass storage medium
 The flash memory can be programmed while it is in its socket on
the system board
Widely used as a way to upgrade PC BIOS ROM
Flash memory is semiconductor memory with access time in
range of 100 ns compared with disk access time in the range of
tens of milliseconds
Flash memory’s program/erase cycles must become infinite, like
hard disks
Program/erase cycle refers to the number of times that a chip
can be erased and programmed before it becomes unusable

MASK ROM
Mask ROM refers to a kind of ROM in which the contents are
programmed by the IC manufacturer, not user programmable
 The terminology mask is used in IC fabrication
 Since the process is costly, mask ROM is used when the needed
volume is high and it is absolutely certain that the contents will not
change
 The main advantage of mask ROM is its cost, since it is significantly
cheaper than other kinds of ROM, but if an error in the data/code is
found, the entire batch must be thrown away

RAM (RANDOM ACCESS MEMORY)
RAM memory is called volatile memory since cutting off the power to
the IC will result in the loss of data
Sometimes RAM is also referred to as RAWM (read and write
memory), in contrast to ROM, which cannot be written to
There are three types of RAM
 Static RAM (SRAM)
NV-RAM (nonvolatile RAM)
Dynamic RAM (DRAM)

SRAM (STATIC RAM)
Storage cells in static RAM memory are made of flip-flops and
therefore do not require refreshing in order to keep their data
The problem with the use of flip-flops for storage cells is that each cell
require at least 6 transistors to build, and the cell holds only 1 bit of
data
 In recent years, the cells have been made of 4 transistors, which
still is too many
The use of 4-transistor cells plus the use of CMOS technology has
given birth to a high capacity SRAM, but its capacity is far below
DRAM

NV-RAM (NONVOLATILE RAM)
NV-RAM combines the best of RAM and ROM
The read and write ability of RAM, plus the nonvolatility of ROM
 NV-RAM chip internally is made of the following components
 It uses extremely power-efficient SRAM cells built out of CMOS
 It uses an internal lithium battery as a backup energy source
 It uses an intelligent control circuitry
The main job of this control circuitry is to monitor the Vcc pin
constantly to detect loss of the external power supply

DRAM (DYNAMIC RAM)
Dynamic RAM uses a capacitor to store each bit
 It cuts down number of transistors needed to build the cell
It requires constant refreshing due to leakage
The advantages and disadvantages of DRAM memory
 The major advantages are high density (capacity), cheaper cost per
bit, and lower power consumption per bit
The disadvantages is that
It must be refreshed periodically, due to the fact that the
capacitor cell loses its charge;
While it is being refreshed, the data cannot be accessed

PACKING ISSUE IN DRAM
In DRAM there is a problem of packing a large number of cells into a
single chip with normal number of pins assigned to addresses
 Using conventional method of data access, large number of pins
defeats the purpose of high density and small packaging
 For example, a 64K-bit chip (64K×1) must have 16 address lines
and 1 data line, requiring 16 pins to send in the address
The method used is to split the address in half and send in each
half of the address through the same pins, thereby requiring fewer
address pins

PACKING ISSUE IN DRAM (CONT’)
Internally, the DRAM structure is divided into a square of rows and
columns
The first half of the address is called the row and the second half is
called column
 The first half of the address is sent in through the address pins,
and by activating RAS (row address strobe), the internal latches
inside DRAM grab the first half of the address
After that, the second half of the address is sent in through the
same pins, and by activating CAS (column address strobe), the
internal latches inside DRAM latch the second half of the address

DRAM ORGANIZATION
In the discussion of ROM, we noted that all of them have 8 pins for
data
This is not the case for DRAM memory chips, which can have any of
the x1, x4, x8, x16 organizations

Discuss the number of pins set aside for address in each of the
following memory chips. (a) 16K×4 DRAM (b) 16K×4 SRAM
Solution :
Since 214 = 16K :
(a) For DRAM we have 7 pins (A0-A6) for the address pins and 2 pins for
RAS and CAS
(b) For SRAM we have 14 pins for address and no pins for RAS and CAS
since they are associated only with DRAM. In both cases we have 4 pins
for the data bus.

MEMORY ADDRESS DECODING
The CPU provides the address of the data desired, but it is the job of
the decoding circuitry to locate the selected memory block
Memory chips have one or more pins called CS (chip select), hich
must be activated for the memory’s contents to be accessed
Sometimes the chip select is also referred to as chip enable (CE)

MEMORY ADDRESS DECODING (CONT’)
In connecting a memory chip to CPU, note the following points
The data bus of the CPU is connected directly to the data pins of
the memory chip
Control signals RD (read) and WR (memory write) from the CPU are
connected to the OE (output enable) and WE (write enable) pins of
the memory chip

INTERNAL MEMORY OF 8051
 8051 implements a separate memory space for programs (code), data &
and external RAM.
 Both code and data may be internal, however, both expand using external
components to a maximum of 64K code memory and 64K data memory.
 Internal memory consists of on-chip ROM and on-chip data RAM.
 This is refereed to as a Harvard architecture.
 The early Mark I (1944) computer developed at Harvard was of this type
of architecture.
 Von Neumann at Princeton pointed out that it was not necessary to put
instructions and data in separate memories.
 Most machines have been Princeton architecture.

On-chip RAM contains a rich arrangement of general purpose storage,
bit addressable storage, register banks, and special function registers.
Registers and input/output ports are memory mapped and accessible
like any other memory location.
Stack resides in internal RAM, rather than in external RAM.
The internal data memory is accessed using an 8-bit address.

RAM MEMORY SPACE ALLOCATION IN 8051 UC

REGISTER BANKS IN 8051 MICROCONTROLLER

SPECIAL FUNCTION REGISTERS
ACC
B
PSW
SP
DPTR
IP
PMODE
PCON
TMODE
TCON etc.

SPECIAL FUNCTION REGISTERS
8051 has 21 special function registers (SFR) at the top of internal RAM
from address 80H to FFH.
Most of the addresses from 80H to FFH are not defined, except for 21
of them.
Some SFR’s are both bit-addressable and byte-addressable,
depending on the instruction accessing the register.
All 8051 CPU registers, I/O ports, timers and other architecture
components are accessible in 8051 C through SFRs

SFR REGISTERS AND THEIR ADDRESSES
The SFR (Special Function Register) can be accessed by their names
or by their addresses
MOV 0E0H,#55H ;is the same as
MOV A,#55h ;load 55H into A
MOV 0F0H,R0 ;is the same as
MOV B,R0 ;copy R0 into B
The SFR registers have addresses between 80H and FFH
Not all the address space of 80 to FF is used by SFR
The unused locations 80H to FFH are reserved and must not be
used by the 8051 programmer

SFR REGISTERS
AND THEIR
ADDRESSES
(CONT’)
F8 FF
F0 B F7
E8 EF
E0 Acc E7
D8 DF
D0 PSW D7
C8 CF
C0 C7
B8 IP BF
B0 P3 B7
A8 IE AF
A0 P2 A7
98 SCON SBUF 9F
90 P1 97
88 TCON TMOD TL0 TL1 TH0 TH1 8F
80 P0 SP DPL DPH PCON 87

SFR REGISTERS AND THEIR ADDRESSES

 Write code to send 55H to ports P1 and P2, using
 (a) their names (b) their addresses
 Solution :
 (a) MOV A,#55H ;A=55H
 MOV P1,A ;P1=55H
 MOV P2,A ;P2=55H
 (b) From Table 5-1, P1 address=80H; P2 address=A0H
 MOV A,#55H ;A=55H
 MOV 80H,A ;P1=55H
 MOV 0A0H,A ;P2=55H

B REGISTER
B register is used along with the accumulator for multiply and
divide operations.
 MUL AB: multiplies 8 bit unsigned values in A and B. and leaves the
16 bit result in A (low byte) and B (high byte).
 DIV AB: divided A by B, leaving the integer result in A and remainder
in B.
B register is bit-addressable.

PSW (PROGRAM STATUS WORD) / FLAG REGISTER

STACK POINTER
 Stack pointer (SP) is an 8-bit register at address
81H.
Register used to access stack is called SP (stack
pointer) register.
Stack pointer in 8051 is 8 bits wide, it can take
value 00 to FFH.
When 8051 powered up, SP register contains value
07.
It contains address of data item currently on top of
the stack.
Stack operations include pushing data on stack
and popping data off stack.

Pushing increments SP before writing data.
Popping from stack reads data and decrements SP.
8051 stack is kept in the internal RAM.
 Depending on the initial value of the SP, stack can have different
sizes
Example: MOV SP,#5FH
On 8051 this would limit the stack to 32 bytes since the
uppermost address of on chip RAM is 7FH.

DATA POINTER (DPTR)
Data pointer (DPTR): is used to access external data or code.
DPTR is a 16 bit register at addresses 82H (low byte) and 83H (high
byte).
The data pointer is used in operations regarding external RAM and
some instructions involving code memory.
Example: the following instructions write 55H into external RAM
location 1000H:
MOV A,#55H
MOV DPTR,#1000H
MOVX @DPTR,A

LCD INTERFACING
WITH
MICROCONTROLLER
16X2 LCD

The CPU can access data in various ways, which are called
addressing modes
 Immediate Addressing Mode
 Register Addressing Mode
 Direct Addressing Mode
 Register indirect Addressing Mode
 Indexed Addressing Mode
Addressing modes
Immediate
Addressing
Register
Addressing
Direct
Addressing
Register indirect
Addressing
Indexed
Addressing

1- IMMEDIATE ADDRESSING MODE
 The source operand is a constant
 The immediate data must be preceded by pound sign, “#”
 Can load information into any registers, including 16-bit DPTR
register
 DPTR can also be accessed as two 8-bit registers, the high byte
DPH and low byte DPL
 Operand comes immediately after op-code

EXAMPLES OF IMMEDIATE ADDRESSING MODE
MOV A,#25H // load 25H into A
MOV R4,#62 // load 62 into R4
MOV B,#40H // load 40H into B
MOV DPTR,#4521H // DPTR=4512H
MOV DPL,#21H // This is the same
MOV DPH,#45H // as above
MOV DPTR,#68975 //illegal!! Value > 65535 (FFFFH)

We can use EQU directive to
access immediate data
We can also use immediate
addressing mode to send
data to 8051 ports
Count EQU 30
... ...
MOV R4,#COUNT // R4=1EH
MOV P1,#55H // Sending
data to port

2- REGISTER ADDRESSING MODE
Register addressing mode involves the use of registers to hold the
data to be manipulated
 MOV A, R0 // copy the contents of R0 in to A.
MOV R1, A // copy the contents of A in to R1.
ADD A,R5 // add the content of R5 to content of A.
The movement of data between Rn registers is not allowed
MOV R4,R7 // Invalid Command
The source and destination registers must match in size
MOV DPTR,A // Error

3- DIRECT ADDRESSING MODE
In direct addressing mode, the data is in a RAM memory location
whose address is known, and this address is given as a part of the
instruction.
Contrast this with the immediate addressing mode in which the
operand itself is provided with the instruction.
MOV R0,40H ;save content of 40H memory location in R0
MOV 56H,A ;save content of A in 56H
 MOV PSW,R5 ; M(PSW)  R5

It is most often used direct addressing mode to access RAM
locations 30 – 7FH
The entire 128 bytes of RAM can be accessed
The register bank locations are accessed by the register names
MOV A,R4 ;which means copy R4 into A

PUSH/POP WITH DIRECT ADDRESSING MODE
 Only direct addressing mode is allowed for pushing or popping the stack
 PUSH A // Invalid
 Pushing the accumulator onto the stack must be coded as PUSH 0E0H
 Example
Show the code to push R5 and A onto the stack and then pop them back them
into R2 and B, where B = A and R2 = R5
Solution:
PUSH 05 ;push R5 onto stack
PUSH 0E0H ;push register A onto stack
POP 0F0H ;pop top of stack into B // now register B = register A
POP 02 // pop top of stack into R2 now R2=R6

4- REGISTER INDIRECT ADDRESSING MODE
In register indirect addressing mode, a register is used as a pointer to
the data.
If data is inside the CPU, only register R0 and R1 are used for this
purpose. R2-R7 cannot be used to hold the address of an operand
located in RAM when using this addressing mode.
When R0 and R1 are used as pointers , that is, when they hold the
address of RAM locations , they must be preceded by the “@” sign.
 MOV A,@R0 // move contents of RAM location whose address is held by R0 into A.
 MOV @R1,A // move contents of B into RAM whose address is held by R1

Example
Write a program to clear 16 RAM locations starting at RAM address
60H
Solution:
CLR A ;A=0
MOV R1,#60H ;load pointer. R1=60H
MOV R7,#16 ;load counter, R7=16
AGAIN: MOV @R1,A ;clear RAM R1 points to
INC R1 ;increment R1 pointer
DJNZ R7,AGAIN ;loop until counter=zero

5- INDEXED ADDRESSING MODE
Indexed addressing mode is used in accessing data elements of
look-up table entries located in program ROM space
Instruction used for this purpose is “@A+DPTR, @A+PC”.
 The 16-bit register DPTR and register “A” are used to form the data
element stored in on-chip ROM.
 In this instruction content of A are added to the 16-bit register
DPTR to form the 16-bit address of the needed data.

EXAMPLES
MOVC A,@A+DPTR ; A  ROM(A+DPTR)
MOVC A,@A+PC ; A ROM(A+PC)
JMP @A+DPTR ; PC  (A+DPTR)

Example
Write a program to copy the value 55H into RAM memory locations
40H to 41H using
(a) direct addressing mode
(b) register indirect addressing mode without a loop
(c) with a loop

Solution:
(a) MOV A,#55H //load A with value 55H
MOV 40H,A //copy A to RAM location 40H
MOV 41H.A //copy A to RAM location 41H
(b) MOV A,#55H //load A with value 55H
MOV R0,#40H //load the pointer. R0=40H
MOV @R0,A //copy A to RAM R0 points to
INC R0 //increment pointer. Now R0=41h
MOV @R0,A //copy A to RAM R0 points to

(c) MOV A,#55H //A=55H
MOV R0,#40H //load pointer.R0=40H,
MOV R2,#02 //load counter, R2=3
AGAIN: MOV @R0,A //copy 55 to RAM R0 points to
INC R0 //increment R0 pointer
DJNZ R2,AGAIN //loop until counter = zero

ASSEMBLING AND RUNNING AN 8051 PROGRAM
An Assembly language instruction consists of four fields:
[label : ] mnemonic [operands] [;comment]

INSTRUCTIONS THAT AFFECT FLAG SETTINGS
Instructions That Affect Flag Bits

INSTRUCTION SET OF 8051
Arithmetic Operation Group
Logical Operation Group
Data Transfer Group
Boolean Variable Manipulation Group
Program Branching Group

ARITHMETIC INSTRUCTIONS
These instructions are used to perform various mathematical
operations like addition, subtraction, multiplication, and division etc.

ADDITION WITHOUT CARRY
ADD A, Source //ADD the source operand to the accumulator
 ADD A, R1 // Add the content of register1 to Accumulator
 ADD A, Direct // ACCACC+ Data of memory location
 ADD A, #Data
 Add A, @Ri

ADDITION WITH CARRY
►ADDC A,Direct // Add data of memory to accumulator with carry
►ADDC A,Rn
►ADDC A,@Ri
►ADDC A,#2

EXAMPLES
MOV A,#25H ;load 25H into A
MOV R2,#34H ;load 34H into R2
ADD A,R2 ;add R2 to accumulator
;Executing the program above results in A = 59H

SUBSTRACTION WITH BORROW
►SUBB A,Direct
►SUBB A,Rn
►SUBB A,@Ri
►SUBB A,#Data

INCREMENT OPERATION
►INCA // Increment accumulator
►INCDirect // Increment data at memory location
►INCRn // Increment data of memory location
►INC@Ri // Increment data at address store by Ri

DECREMENT OPERATION
►DEC A // Decrement accumulator
►DEC Direct // Decrement data at memory location
►DEC Rn // Decrement data of memory location
►DEC @Ri // Decrement data at address store by Ri

ARITHMETIC OPERATION GROUP EXAMPLES
INCDPTR //Increment DPTR
DA A //Decimal adjust Accumulator(BCD)
MUL AB // Multiply A and B
DIV AB // Divide A by B

INSTRUCTION SET
Arithmetic Operation Group
Logical Operation Group
Data Transfer Group
Boolean Variable Manipulation Group
Program Branching Group

LOGICAL OPERATION GROUP
The logical instructions are instructions which are used for
performing some operations like AND, OR, NOT, X-OR and etc.

LOGICAL OPERATION-AND
►ANL A,Direct
►ANL A,Rn
►ANL A,@Ri
►ANL A,#Data
►ANL Direct,A
►ANL Direct,#Data

LOGICAL OPERATION-OR
ORL A, Direct
ORL A, Rn
ORL A, @Ri
ORL A, #Data
ORL Direct, A
ORL Direct, #Data

LOGICAL OPERATION-XOR
XRL A, Direct
XRL A, Rn
XRL A, @Ri
XRL A, #Data
XRL Direct, A
XRL Direct, #Data

LOGICAL OPERATION – ROTATE, SWAP
RR A // Rotate Accumulator right
If ACC=C3H (11000011), then the instruction results in ACC=E1H
(11100001) with the carry unaffected
RRC A //Rotate Accumulator right through carry
 If ACC=C3H (11000011), and the carry flag is 0, the instruction results in
ACC=61H (01100001) with the carry flag set
SWAP A //Swap nibbles with in the Accumulator
RL A //Rotate Accumulator left
RLC A //Rotate Accumulator left through carry

LOGICAL OPERATION -CLEAR & CLEAR
CLR A //Clear Accumulator
CPL A // Complement Accumulator

DATA TRANSFER GROUP
These instruction are used to transfer the data from source operand
to destination operand.
 All the store, move, load, exchange input and output instructions
belong to this group.

DATA TRANSFER-MOV
MOV destination, source ;copy source to destination
 MOV A, Direct
 MOV A, Rn
 MOV A, @Ri
 MOV A, #Data
 MOV Rn, Direct
 MOV Rn, @Ri
 MOV Rn, #Data

DATA TRANSFER-MOV
►MOV Direct, Direct
►MOV Direct, Rn
►MOV Direct, @Ri
►MOV Direct, #Data
►MOV Direct, A
►MOV @Ri, A
►MOV @Ri, #Data

DATA TRANSFER-MOV
MOV @Ri, Direct
MOV DPTR, #DATA16
MOVC A, @A+DPTR
MOVC A, @A+PC
MOVX A, @Ri
MOVX @Ri, A
MOVX @DPTR, A

EXAMPLES
MOV A,#55H // load value 55H into reg A
MOV R0,A //copy contents of A into R0 (A=R0=55H)
MOV R1,A //copy contents of A into R1 (A=R0=R1=55H)
MOV R2,A //copy contents of A into R2 (A=R0=R1=R2=55H)
MOV R3,#95H //load value 95H into R3 (R3=95H)
MOV A,R3 //copy contents of R3 into A (A=R3=95H)
MOVX A,@DPTR // Move external RAM to Accumulator

DATA TRANSFER-PUSH,POP
PUSH Direct // PUSH direct byte on to stack
POP Direct // POP direct byte from stack
XCH A,Rn //Exchange ACCUMULATOR with Register
XCH A,Direct// Exchange ACCUMULATOR with data at location
XCH A,@Ri
XCHD A,@Ri//Exchanges low-order nibble of Accumulator (bits 3
through 0), with that of internal RAM location indirectly addressed by
specified register.High-order nibbles (bits 7-4) of each register are not
affected

BOOLEAN VARIABLE-SET,CLEAR,COMPLEMENT
►CLRC //Clear the carry bit
►CLRbit // Clear the bit of any register
►SETB C // Set the carry bit=1
►SETB bit //Set the bit of any register
►CPLC //Complement the carry
►CPLbit //Complement the bit of any register

BIT ADDRESSABLE
EXAMPLE
SETB 07Fh ; Bit 7F 1
SETB 2F.7 ; same

BOOLEAN VARIABLE-AND,OR,MOV
►ANL C,bit
►ANL C,/bit
►ORL C,bit
►ORL C,/bit
►MOV C,bit
►MOV bit,C

INSTRUCTION SET
►Arithmetic Operation Group
►Logical Operation Group
►Data Transfer Group
►Boolean Variable Manipulation Group
►Program Branching Group

These instructions are used for jump, call instruction as well as
branch instruction.
 Program flow change in branch instruction if condition met
 Program flow always change in jump and call instruction

CONDITIONAL JUMP INSTRUCTIONS
JC // Jump if carry equal to one
JNC // Jump if carry equal to zero
JB // Jump if bit equal to one
JNB // Jump if bit equal to zero
JBC // Jump if bit equal to one and clear bit
JZ // Jump if A=Zero
JNZ // Jump if A not equal to zero
DJNZ // Decrement and Jump if not equal to zero.
CJNE A, P1,rel //compare and jump if not equal

CJNE A,Direct,rel//Compare and Jump if Not Equal
CJNE A,#Data,rel
CJNE Rn,#Data,rel
CJNE @Ri,#Data,rel
DJNZ Rn,rel //Decrement and jump if not zero
DJNZ Direct,rel
EXAMPLE
The Accumulator contains 34H. Register 7 contains 56H. The first
instruction in the sequence,
CJNE R7, # 60H, NOT_EQ ;
. . . . . . . . ;R7 = 60H.
NOT_EQ: JC REQ_LOW ;IF R7 < 60H.

►Unconditional Jump Instructions
The unconditional jump is a jump in which control is
transferred unconditionally to the target location
►In 8051 there two unconditional jumps. They are:
► SJMP addr16 // Short jump
► LJMP rel // Long jump

LJMP ADDR16
LJMP causes an unconditional branch to the indicated address, by
loading the high-order and low-order bytes of the PC (respectively)
with the second and third instruction bytes. The destination may
therefore be anywhere in the full 64K program memory address
space. No flags are affected.
LJMP
(PC) ← addr15-0

LCALL ADDR16
LCALL calls a subroutine located at indicated address.
Instruction adds three to program counter to generate address of next
instruction and then pushes the 16-bit result onto stack (low byte first),
incrementing the Stack Pointer by two.
 High-order and low-order bytes of the PC are then loaded,
respectively, with second and third bytes of LCALL instruction.
Program execution continues with instruction at this address. The
subroutine may therefore begin anywhere in the full 64K byte program
memory address space. No flags are affected.

LCALL OPERATION
LCALL
(PC) ← (PC) + 3
(SP) ← (SP) + 1
((SP)) ← (PC7-0)
(SP) ← (SP) + 1
((SP)) ← (PC15-8)
(PC) ← addr15-0

PROGRAM BRANCHING GROUP
►NOP
►Execution continues at the following instruction.
Operation
(PC) ← (PC) + 1

ACALL ADDR11
ACALL unconditionally calls a subroutine located at the indicated
address.
 The instruction increments the PC twice to obtain the address of the
following instruction, then pushes the 16-bit result onto the stack
(low-order byte first) and increments the Stack Pointer twice.
The destination address is obtained by successively concatenating
the five high-order bits of the incremented PC, opcode bits 7
through 5, and the second byte of the instruction.
 The subroutine called must therefore start within the same 2 K block
of the program memory as the first byte of the instruction following
ACALL. No flags are affected.

2-byte instructions where the 11-bit absolute address is specified as
the operand
Upper 5 bits of 16-bit PC address are not modified. The lower 11
bits are loaded from this instruction. So, the branch address must be
within the current 2K byte page of program memory (211 = 2048)
Example:
ACALL PORT_INIT ;PORT_INIT should be located within 2k bytes.
PORT_INIT: MOV P0, #0FH ;PORT_INIT subroutine

The label JMPADR is at program memory location 0123H. The
following instruction,
AJMP JMPADR
is at location 0345H and loads the PC with 0123H.
a10 a9 a8 0 0 0 0 1 a7 a6 a5 a4 a3 a2 a1 a0
Operation: AJMP
(PC) ← (PC) + 2
(PC10-0) ← page address

ASSEMBLER DIRECTIVE-DB
 DB directive is most widely used data directive in the assembler
 It is used to define the 8-bit data
 DB is used to define data in decimal, binary, hex,ASCII formats
 ORG 500H
 DATA1: DB 28 ;DECIMAL (1C in Hex)
 DATA2: DB 00110101B ;BINARY (35 in Hex)
 DATA3: DB 39H ;HEX
 ORG 510H
 DATA4: DB “2591” ;ASCII NUMBERS
 ORG 518H
 DATA6: DB “My name is Joe” ;ASCII CHARACTERS

ORG (ORIGIN)
The ORG directive is used to indicate beginning of the address
Number that comes after ORG can be either in hex and decimal
If the number is not followed by H, it is decimal & assembler will
convert it to hex

END
This indicates to the assembler the end of source (asm) file
The END directive is last line of an 8051 program
It’s mean that in the code anything after the END directive is
ignored by the assembler

EQU (EQUATE)
Used to define a constant without occupying memory location
EQU directive does not set aside storage for a data item but
associates a constant value with a data label
When label appears in program, its constant value will be
substituted for label
Assume that there is a constant used in many different places in
program, & programmer wants to change its value throughout

By the use of EQU, one can change it once and the
assembler will change all of its occurrences
 COUNT EQU 25
 ... ....
 MOV R3, #COUNT

STRUCTURE OF ASSEMBLY LANGUAGE
ORG 0H // start (origin) at 0
MOV R5,#25H // load 25H into R5
MOV R7,#34H // load 34H into R7
MOV A,#0 // load 0 into A
ADD A,R5 // add contents of R5 to A now A = A + R5
ADD A,R7 // add contents of R7 to A now A = A + R7
ADD A, #12H // add to A value 12H now A = A + 12H
HERE: SJMP HERE // stay in this loop
END // end of asm source file

Whenever the conventional ‘C’ language and its extensions are
used for programming embedded system, it is referred to as
Embedded C programming

WHY PROGRAM 8051 IN C
 Compilers produce hex files that is downloaded to ROM of microcontroller
 The size of hex file is the main concern
 Microcontrollers have limited on-chip ROM
 Code space for 8051 is limited to 64K bytes
 C programming is less time consuming, but has larger hex file size
 The reasons for writing programs in C
 It is easier and less time consuming to write in C than Assembly
 C is easier to modify and update
 You can use code available in function libraries

DATA TYPES
A good understanding of C data types for 8051 can help
programmers to create smaller hex files
 Unsigned char
 Signed char
 Unsigned int
 Signed int
 Sbit (single bit)
 Bit and sfr

UNSIGNED CHAR
The character data type is the most natural choice
8051 is an 8-bit microcontroller
Unsigned char is 8-bit data type in range of 0 – 255 (00 – FFH)
One of most widely used data types for the 8051
Counter value
ASCII characters
C compilers use the signed char as the default if we do not put the
keyword unsigned

UNSIGNED CHAR (CONT’)
 Write an 8051 C program to send values 00 – FF to
port P1.
 #include <reg51.h>
 void main(void)
 {
 unsigned char z;
 for (z=0;z<=255;z++)
 P1=z;
 }

DATA TYPES- UNSIGNED CHAR (CONT’)
Write an 8051 C program to send hex values for ASCII characters of 0, 1,
2, 3, 4, 5, A, B, C, and D to port P1.
 void main(void)
 {
 unsigned char mynum[]=“012345ABCD”;
 for (z=0;z<=10;z++)
 P1=mynum[z];
 }

UNSIGNED CHAR (CONT’)
Write an 8051 C program to toggle all the bits of P1 continuously.
#include <reg51.h>
void main(void)
{
for (;;)
{
p1=0x55;
p1=0xAA;
}

SIGNED CHAR
The signed char is an 8-bit data type
 Use the MSB D7 to represent – or +
Give us values from –128 to +127
We should stick with the unsigned char unless the data needs to be
represented as signed numbers

SIGNED CHAR (CONT’)
Write an 8051 C program to send values of –4 to +4 to port P1.
#include <reg51.h>
void main(void)
{
char mynum[]={+1,-1,+2,-2,+3,-3,+4,-4};
unsigned char z;
for (z=0;z<=8;z++)
P1=mynum[z];
}

UNSIGNED AND SIGNED INT
Unsigned int is a 16-bit data type
 Takes a value in the range of 0 to 65535 (0000 – FFFFH)
Define 16-bit variables such as memory addresses
Set counter values of more than 256
Since registers and memory accesses are in 8-bit chunks, the
misuse of int variables will result in a larger hex file
Signed int is a 16-bit data type
Use the MSB D15 to represent – or +
We have 15 bits for the magnitude of the number from –32768
to +32767

SINGLE BIT
 Write an 8051 C program to toggle bit D0 of the port P1 (P1.0) 50,000 times.
 sbit MYBIT=P1^0;
 void main(void)
 {
 unsigned int z;
 for (z=0;z<=50000;z++)
 {
 MYBIT=0;
 MYBIT=1;
 }
 }

BIT AND SFR
The bit data type allows access to single bits of bit-addressable
memory spaces 20 – 2FH
To access the byte-size SFR registers, we use the sfr data type
7FH
30H
2FH
20H
1FH
17H
10H
0FH
07H
08H
18H
00H
Register Bank 0
(Stack) Register Ban
Register Bank 2
Register Bank 3
Bit-Addressable RAM
Scratch pad RAM

TIME DELAY
 There are two way s to create a time delay in 8051 C
 Using the 8051 timer
 Using a simple for loop
 Be mindful of three factors that can affect the accuracy of the delay
 The 8051 design
 The number of machine cycle
 The number of clock periods per machine cycle
 The crystal frequency connected to the X1 – X2 input pins
 Compiler choice
 C compiler converts the C statements and functions to Assembly language
instructions
 Different compilers produce different code

 Write an 8051 C program to toggle bits of P1 continuously forever with some delay.
//Toggle P1 forever with some delay in between “on” and “off”
 void main(void)
 {
 unsigned int x;
 for (;;) //repeat forever
 {
 P1=0x55;
 for (x=0;x<40000;x++); //delay size
 //unknown
 P1=0xAA;
 for (x=0;x<40000;x++);
 }}

#include <reg51.h>
void MSDelay(unsigned int);
void main(void)
{
while (1) //repeat forever
{ p1=0x55;
MSDelay(250);
p1=0xAA;
MSDelay(250); }
}
void MSDelay(unsigned int itime)
{
unsigned int i,j;
for (i=0;i<itime;i++)
for (j=0;j<1275;j++);
}
Write 8051 program to toggle bits of P1 ports continuously with 250 ms.

LEDs are connected to bits P1 and P2. Write an 8051 C program that shows the
count from 0 to FFH (0000 0000 to 1111 1111 in binary) on the LEDs.
#include <reg51.h>
#define LED P2
void main(void)
{
P1=00; //clear P1
LED=0; //clear P2
for (;;) //repeat forever
{
P1++; //increment P1
LED++; //increment P2
}}

 Write an 8051 C program to monitor bit P1.5. If it is high, send 55H to P0;
otherwise, send AAH to P2.
 sbit mybit=P1^5;
 void main(void)
 {
 mybit=1; //make mybit an input
 while (1)
 {
 if (mybit==1)
 P0=0x55;
 else
 P2=0xAA;
 }}

 Write an 8051 C program to read the P1.0 and P1.1 bits and issue an ASCII character to P0
according to the following table.
P1.1 P1.0
0 0 send ‘0’ to P0
0 1 send ‘1’ to P0
1 0 send ‘2’ to P0
1 1 send ‘3’ to P0

 void main(void)
 {
 z=P1;
 z=z&0x3;
 switch (z)
 {
 case(0):
 {
 P0=‘0’;
 break;
 }
case(1):
{
P0=‘1’;
break;
}
case(2):
{
P0=‘2’;
break;
}
case(3):
{
P0=‘3’;
break;
}}}

After perfecting your project on programming side in KEIL, you'll
need to simulate it on PROTEUS to determine output of hardware
components and change it if need be.
This will completely ensure your project's success.

Place your components from library
Connect them accordingly
Load HEX file (if 8051 is involved)
Simulate the circuit

PLACING COMPONENTS
Click "Pick from library (P)" button
as shown in figure
Select any category
Select item from the list
Click OK

After selecting component, click anywhere in design area to select it
and then click again to place it

CONNECTING COMPONENTS
Place all the required components
Connect desired nodes by clicking at starting &ending points

LOAD HEX FILE
Double click the 8051 component to open its properties
Browse for the HEX file as shown and select it

SIMULATING THE CIRCUIT
Controls at left-bottom corner will help you simulate circuit in real
time

 #include<reg51.h>
 sbit LED = P1^0; // Pin P1.0 is named as LED
 void x(void); //Function declarations
 void delay(int a); //Function declarations
 int main(void)
 {
 x(); // Make all ports as output port
 while(1)
 {
 LED = 0; // Pin P1.0 Low
 delay(30000); // Half sec delay
 LED = 1; // Pin P1.0 High
 delay(30000); // Half sec delay
 }}
void delay(int a)
{
int i;
for(i=0;i<a;i++);
//null statement
}
void x(void)
{
P0 = 0x00;
P1 = 0x00;
P2 = 0x00;
P3 = 0x00;
}

INTERFACING LCD TO 8051
 LCD is finding widespread use replacing LEDs
 The declining prices of LCD
 The ability to display numbers, characters, and graphics
 Ease of programming for characters and graphics

Pin Symbol I/O Description
1 VSS -- Ground
2 VCC -- +5V power supply
3 VEE -- Power supply to control contrast
4 RS(Register select) I RS=0 to select command register,
RS=1 to select data register
5 R/W I R/W=0 for write,
R/W=1 for read
6 E I Enable
7 DB0 I/O The 8-bit data bus
15 BL+ -- Brightness
16 BL- -- Ground

RESISTER SELECT(RS) PIN
Control Register
 Used for instructing LCD on what to do
next. It's like talking to your LCD using
this register.
Data Register
 Used for displaying data on LCD.


No. Instruction Hex Decimal
1 Function Set: 8-bit, 1 Line, 5x7 Dots 0x30 48
2 Function Set: 8-bit, 2 Line, 5x7 Dots 0x38 56
5 Entry Mode 0x06 6
6
Display off Cursor off
(clearing display without clearing DDRAM content)
0x08 8
7 Display on Cursor on 0x0E 14
8 Display on Cursor off 0x0C 12
9 Display on Cursor blinking 0x0F 15
10 Shift entire display left 0x18 24
11 Shift entire display right 0x1C 30
12 Move cursor left by one character 0x10 16
13 Move cursor right by one character 0x14 20
14 Clear Display (also clear DDRAM content) 0x01 1

R/W (READ/WRITE) CONTROL PIN
Determine the flow of data.
Write Mode when you're sending something to LCD (data
or command)
Read Mode when you're reading from the LCD

 It is basically called 16x2 LCD
Sixteen columns & two rows thus it can display thirty-two (32)
characters at a time (sixteen characters in each row)
Connect potentiometers to Vee or BL controls to control the
contrast or Brightness yourself.

LCD PROGRAM CODE IN A STEPWISE FASHION

1- Unique name to each pin on LCD using #define directive so that
you can mention them easily

To differentiate between command and data as discussed above,
create two separate functions. As you can see, the only difference is
that of using RS pin according to requirement.

The delay function is general one and you can use it safely.

To send a command to LCD, use the lcdcmd function with
command as input.
To send data to LCD, use the lcddata function with the data as
input.
 LCD accepts only ASCII characters which means that every
character you send for display must be in ASCII

x = 'A'
lcddata ( x )
x = 65 //decimal value
lcddata ( x )
x = 0x41 //hexadecimal value
lcddata ( x )
lcddata ( 'A' )
where x can be of type char or int whether signed or unsigned.

KEIL-LCD INTERFACING
#include <REGX51.H> //header file
#define lcdport P2 //port 2 of microcontroller
#define rs P3_0 //port 3 pin 0 of microcontroller
#define rw P3_1 //port 3 pin 1 of microcontroller
#define e P3_2 //port 3 pin 2 of microcontroller
void lcdcmd (unsigned char); //function define for lcd command mode
void lcddata(unsigned char); //function define for lcd data mode
void delay (unsigned int); //function define for delay mode

void main()
{
//INITIALIZE LCD
lcdcmd(0x38); //for using 8-bit double row mode of LCD
lcdcmd(0x0E); //turn display ON for cursor blinking
lcdcmd(0x01); //clear screen
//PRINT A CHARACTER
lcddata(‘E');
lcddata(‘F');
lcddata('A');
while(1);
}
//*********Function to send command to LCD*******//
void lcdcmd(unsigned char value)
{
lcdport = value;
rw = 0; // write mode
rs = 0; //register select command register
e=1; //enable chip
delay(1); //delay
e=0; // disable chip
}

//**********Function to send data to LCD**********//
void lcddata(unsigned char value)
{
lcdport = value;
rw = 0; // write mode
rs = 1; //register select data mode
e=1; //enable lcd
delay(1);
e=0; //disable lcd
}
void delay(unsigned int msec) // Function to provide time delay in msec.
{
int i,j ;
for(i=0;i<msec;i++)
for(j=0;j<1275;j++);
}

PROTEUS- LCD INTERFACING SCHEMATIC

A Stepper Motor is a brushless, synchronous DC Motor.
 Performs many applications in field of robotics.
 The total rotation of motor is divided into steps.
 Angle of a single step is known as stepper angle of motor.
There are two types of stepper motors Unipolar and Bipolar.
Due to the ease of operation unipolar stepper motor is commonly
used by electronics hobbyists.
Stepper Motors can be easily interfaced with a microcontroller using
driver ICs such as L293D or ULN2003.

WORKING OF STEPPER MOTOR
 Stepper motors works on the principle of electromagnetic induction.
 Ordinary DC Brush motors rotate continuously when DC voltage is applied to
their terminals.
 While stepper motor need a sequence of digital pulses for one complete
rotation.
 Stepper motor contains a magnetic or soft iron rotor surrounded by
electromagnetic stators.
 Stator and rotor have poles which may be teethed depending on type of
stepper motor.
 Firstly one stator electromagnet is energised, this makes rotor teeth
magnetically attracted to electromagnet’s teeth.
 When rotor teeth gets align with first electromagnet, it gets misaligns with
next electromagnet. So when next electromagnet is turned on and first one is
turned off, rotor rotates to align with next one. This process is repeated to get
required rotation.

TYPES OF STEPPER MOTOR
Stepper Motors are classified in to three types depending upon its
construction.
Permanent Magnet Stepper
Variable Reluctance Stepper
Hybrid Stepper

PERMANENT MAGNET STEPPER
It has a permanent magnet in rotor and operates on repulsion and
attraction between permanent magnet rotor and stator electromagnets.
Stator and rotor poles of these types are not teethed. First a stator is
energized, it develops electromagnetic north and south poles. It rotates
rotor to align with magnetic field of stator.
Then other stators are energised in sequentially this rotates rotor to
align with new magnetic field.
Through this way we can rotate rotor through fixed steps.

VARIABLE RELUCTANCE STEPPER
These types of motors operates on principle that minimum
reluctance occurs with minimum gap and it has a non-magnetic
toothed soft iron rotor.
When a stator is energised, rotor rotates to have a minimum gap
between the stator and its teeth.
The rotor teeth is designed such that when it aligns with one stator,
they will get misaligned with then next stator.
Thus by energising stators sequentially we can rotate the rotor.

HYBRID STEPPER
These types of motors are a combination of Permanent Magnet and
Variable Reluctance techniques to achieve maximum power in a
small package size.
 It has a teethed magnetic rotor which can better guides magnetic
flux to preferred location in the air gap.
Usually electromagnets of stepper motor is energised using special
controlling circuits, such as microcontrollers.

Stepper Motors are classified into two, based on its winding
arrangement.
Unipolar Motors
Bipolar Motors

UNIPOLAR MOTORS
 A unipolar motor contains centre tapped windings.
 Usually centre connection of coils are tied together and used as power
connection.
 By using this arrangement a magnetic poles can be reversed without
reversing direction of current.
 Thus the commutation circuit can be made very simple.
 This ease of operation makes Unipolar Motor popular among
electronics hobbyists.

BIPOLAR MOTORS
Bipolar motors have no center tap connections.
Current through winding should be reversed to reverse magnetic
poles. So driving circuit is complicated.
We can solve this by using a H-bridge connection or by using ready
made chips such as L293D.
We can distinguish bipolar motors from unipolar motors by
measuring coil resistance.
In bipolar motors we can find two wires with equal resistance.

DRIVING OF STEPPER MOTOR – STEPPING MODES
Stepping Modes refers to sequence in which stator electromagnets
are energised to rotate stepper motor.
 There are three types of stepping modes.

WAVE DRIVE – ONE ON AT A TIME
Only one stator electromagnet is energised at a time.
 It has same number of steps as full step drive but torque is
significantly less.
 It is rarely used.

FULL DRIVE – TWO ON AT A TIME
Two stator electromagnets are energised at a time.
 It is the usual method used for driving and the motor will run at its
full torque in this mode of driving.

HALF DRIVE – ONE OR TWO ON AT A TIME
Alternatively one and two phases are energised.
 This mode is commonly used to increase angular resolution of
motor but the torque is less approximately 70% at its half step
position (when only a single phase is on)

DRIVING UNIPOLAR STEPPER MOTOR WITH 8051
Unipolar stepper motors can be used in three modes namely Wave
Drive, Full Drive and Half Drive mode.
Each drive have its own advantages and disadvantages, thus we
should choose required drive according to application and power
consumption.

WAVE DRIVE
One electromagnet is energized at a time. Generated torque will be
less when compared to full drive in which two electromagnets are
energized at a time but power consumption is reduced.
 It has same number of steps as in the full drive. This drive is
preferred when power consumption is more important than torque.
 It is rarely used.

FULL DRIVE
Two electromagnets are energized at a time, so the torque
generated will be larger when compared to Wave Drive.
 This drive is commonly used than others. Power consumption will
be higher than other modes.

HALF DRIVE
Alternatively one and two electromagnets are energized, so it is a
combination of Wave and Full drives.
 This mode is commonly used to increase angular resolution of
motor but the torque will be less, about 70% at its half step
position. We can see that the angular resolution doubles when
using Half Drive.

INTERFACING USING L293D
 Driving a bipolar stepper motor using 8051 microcontroller using L293D.
 24MHz crystal is connected to provide required clock for microcontroller.
 10μF capacitor and 10KΩ is used to provide Power On Reset (POR) for 8051
microcontroller.
 L293D is connected to pins P2.0, P2.1, P2.2, P2.3 of the microcontroller and
two pairs of L293D are enabled by tieing EN1, EN2 to 5V.
 Logic Voltage (5V) is connected to Vss pin and Motor Supply (12V) is
connected to the Vs pin of L293D.
 Center Tap of each windings of stepper motor is shorted and connected to the
motor supply.
 Now we can energize each winding of the motor by making corresponding pin
of L293D LOW.

INTERFACING BIPOLAR STEPPER MOTOR WITH 8051 USING L293D

INTERFACING UNIPOLAR STEPPER MOTOR WITH 8051USING L293D

INTERFACING UNIPOLAR STEPPER MOTOR WITH 8051 USING ULN2003

KEIL - C CODE FOR WAVE DRIVE
#include<reg51.h>
void delay(int);
void main()
{
do
{
P2=0x01; //0001
delay(1000);
P2=0x02; //0010
delay(1000);
P2=0x04; //0100
delay(1000);
P2=0x08; //1000
delay(1000);
}
while(1);
}
void delay( int k) //Delay function
{
int i,j;
for(i=0;i<k;i++)
{
for(j=0;j<100;j++)
{}
}
}

KEIL - C CODE FOR FULL DRIVE
#include<reg51.h>
void delay(int);
void main()
{
do
{
P2 = 0x03; //0011
delay(1000);
P2 = 0x06; //0110
delay(1000);
P2 = 0x0C; //1100
delay(1000);
P2 = 0x09; //1001
delay(1000);
}
while(1);
}
void delay(int k) //Delay function
{
int i,j;
for(i=0;i<k;i++)
{
for(j=0;j<100;j++)
{}
}
}

KEIL - C CODE FOR HALF DRIVE
#include<reg51.h>
void delay(int);
void main()
{
do
{
P2=0x01; //0001
delay(1000);
P2=0x03; //0011
delay(1000);
P2=0x02; //0010
delay(1000);
P2=0x06; //0110
delay(1000);
P2=0x04; //0100
delay(1000);
P2=0x0C; //1100
delay(1000);
P2=0x08; //1000
delay(1000);
P2=0x09; //1001
delay(1000);
} while(1);
}
void delay(int k) //Delay Function
{
int i,j;
for(i=0;i<k;i++)
{
for(j=0;j<100;j++)
{}
}}

INTERFACING BIPOLAR STEPPER MOTOR
Bipolar stepper motors have no center tap and having equal coil
resistances.
 It can be easily interfaced with a microcontroller using L293D DC
Motor Driver IC.

KEIL-CODE FOR BIPOLAR STEPPER MOTOR
#include<reg51.h>
void delay(int);
void main()
{
do
{
P2=0x01; //0001
delay(1000);
P2=0x04; //0100
delay(1000);
P2=0x02; //0010
delay(1000);
P2=0x08; //1000
delay(1000);
}while(1);
}
void delay(int k) //Delay function
{
int i,j;
for(i=0;i<k;i++)
{
for(j=0;j<100;j++)
{}
}
}

SEVEN SEGMENT DISPLAY INTERFACING

 Seven Segment Display which is most commonly known as SSD is
an output device which can be used to display information.
 SSDs are also composed of individual LEDs and work on the very same
principle. So basically they come in two forms
Common Anode Common Cathode

 FOR COMMON CATHODE
 Connect INPUT VOLTAGE to A
 Connect GROUND to COMMON
 FOR COMMON ANODE
 Connect INPUT VOLTAGE to COMMON
 Connect GROUND to A

PIN DESCRIPTION OF STANDARD SSD

SEVEN SEGMENT INTERFACING
 #include <reg51.H>
 #define input P1
 #define output P2
 unsigned char ssd[16] = { 0x7E, 0x30, 0x6D, 0x79, 0x33, 0x5B, 0x5F,0x70,0x7F, 0x7B, 0x77,
0x1F, 0x4E, 0x3D, 0x4F, 0x47 };
 void main()
 {
 //declare PORT1 as INPUT
 input = 0xFF;
 while(1)
 output = ssd[ input ];
 }

KEIL -SWITCH INTERFACE
#include <reg51.h>
#define button P1_0
#define led P1_1
void main()
{ button = 1;
led = 0;
while(1)
{
if(!button)
led = 1;
else
led = 0; }}

PROTEUS-SWITCH INTERFACE WITH 8051

Embedded system 8051 Microcontroller

Embedded system 8051 Microcontroller

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