Microprocessors and Microcontrollers Document Summary

EE6502
Microprocessors and
Microcontrollers
Presented by
C.GOKUL,AP/EEE
Velalar College of Engg & Tech , Erode
DEPARTMENTS: EEE {semester 05}
Regulation : 2013
1

Microprocessor
• Microprocessor (µP) is the “brain” of a computer
that has been implemented on one
semiconductor chip.
• The word comes from the combination micro and
processor.
• Processor means a device that processes
whatever(binary numbers, 0’s and 1’s)
To process means to manipulate. It describes all
manipulation.
Micro - > extremely small
3

Definition of a Microprocessor.
The microprocessor is a
programmable device that takes in numbers,
performs on them arithmetic or logical
operations according to the program stored in
memory and then produces other numbers as
a result.
4

Microprocessor ?
A microprocessor is multi
programmable clock driven
register based semiconductor
device that is used to fetch ,
process & execute a data
within fraction of seconds.
5

Applications
• Calculators
• Accounting system
• Games machine
• Instrumentation
• Traffic light Control
• Multi user, multi-function environments
• Military applications
• Communication systems
6

DIFFERENT PROCESSORS AVAILABLE
Socket
Processor
Pinless
Processor
Slot
Processor
ProcessorSl
ot
8

Development of Intel Microprocessors
• 8086 - 1979
• 286 - 1982
• 386 - 1985
• 486 - 1989
• Pentium - 1993
• Pentium Pro - 1995
• Pentium MMX -1997
• Pentium II - 1997
• Pentium II Celeron - 1998
• Pentium II Zeon - 1998
• Pentium III - 1999
• Pentium III Zeon - 1999
• Pentium IV - 2000
• Pentium IV Zeon - 2001
9

GENERATION OF PROCESSORS
Processor Bits Speed
8080 8 2 MHz
8086 16 4.5 – 10
MHz
8088 16 4.5 – 10
MHz
80286 16 10 – 20
MHz
80386 32 20 – 40
MHz
80486 32 40 – 133
MHz
10

GENERATION OF PROCESSORS
Processor Bits Speed
Pentium 32 60 – 233
MHz
Pentium
Pro
32 150 – 200
MHz
Pentium II,
Celeron ,
Xeon
32 233 – 450
MHz
Pentium
III, Celeron
, Xeon
32 450 MHz –
1.4 GHz
Pentium IV,
Celeron ,
Xeon
32 1.3 GHz –
3.8 GHz
Itanium 64 800 MHz –
3.0 GHz
11

Intel 4004
 Introduced in 1971.
 It was the first microprocessor
by Intel.
 It was a 4-bit µP.
 Its clock speed was 740KHz.
 It had 2,300 transistors.
 It could execute around
60,000 instructions per
second.
12

Intel 4040
Introduced in 1971.
It was also 4-bit µP.
13

Intel 8008
It was first 8-bit µP.
Its clock speed was
500 KHz.
Could execute
50,000 instructions
per second.
15

Intel 8080
Its clock speed was
2 MHz.
It had 6,000
transistors.
16

Intel 8085 Introduced in 1976.
Its clock speed was 3 MHz.
Its data bus is 8-bit and
address bus is 16-bit.
It had 6,500 transistors.
Could execute 7,69,230
instructions per second.
It could access 64 KB of
memory.
It had 246 instructions.
17

INTEL 8086
 It was first 16-bit µP.
 Its clock speed is 4.77 MHz, 8 MHz
and 10 MHz, depending on the
version.
 Its data bus is 16-bit and address
bus is 20-bit.
 It had 29,000 transistors.
 Could execute 2.5 million
instructions per second.
 It could access 1 MB of memory.
 It had 22,000 instructions.
 It had Multiply and Divide
instructions.
19

INTEL 8088
 It was also 16-bit µP.
 It was created as a
cheaper version of
Intel’s 8086.
 It was a 16-bit processor
with an 8-bit external
bus.
20

INTEL 80186 & 80188
 They were 16-bit µPs.
 Clock speed was 6 MHz.
 80188 was a cheaper
version of 80186 with an
8-bit external data bus.
21

INTEL 80286
 It was 16-bit µP.
 Its clock speed was 8
MHz.
 Its data bus is 16-bit
and address bus is 24-
bit.
 It could address 16 MB
of memory.
 It had 1,34,000
transistors.
22

INTEL 80386
 It was first 32-bit µP.
bit.
 It could address 4 GB of
memory.
 It had 2,75,000
transistors.
 Its clock speed varied
from 16 MHz to 33 MHz
depending upon the
various versions.
24

INTEL 80486
 It had 1.2 million
transistors.
from 16 MHz to 100
MHz depending upon
the various versions.
 8 KB of cache memory
was introduced.
25

INTEL PENTIUM
 It was originally named
80586.
MHz.
bit.
26

INTEL PENTIUM PRO
 It had 21 million
transistors.
 Cache memory:
 8 KB for instructions.
 8 KB for data.
27

INTEL PENTIUM II
MHz to 500 MHz.
 Could execute 333
million instructions per
second.
28

INTEL PENTIUM II XEON
 It was designed for
servers.
MHz to 450 MHz.
29

INTEL PENTIUM III
from 500 MHz to 1.4
GHz.
 It had 9.5 million
transistors.
30

INTEL PENTIUM IV
 Its clock speed was from
1.3 GHz to 3.8 GHz.
 It had 42 million
transistors.
31

INTEL DUAL CORE
 It is 32-bit or 64-bit µP.
32

INTEL CORE I5 INTEL CORE I7
36

Basic Terms
• Bit: A digit of the binary number { 0 or 1 }
• Nibble: 4 bit Byte: 8 bit word: 16 bit
• Double word: 32 bit
• Data: binary number/code operated by an
instruction
• Address: Identification number for memory
locations
• Clock: square wave used to synchronize various
devices in µP
• Memory Capacity = 2^n ,
n->no. of address lines
37

BUS CONCEPT
• BUS: Group of conducting lines that carries data ,
address & control signals.
CLASSIFICATION OF BUSES:
1.DATA BUS: group of conducting lines that carries
data.
2. ADDRESS BUS: group of conducting lines that
carries address.
3.CONTROL BUS: group of conducting lines that
carries control signals {RD, WR etc}
CPU BUS: group of conducting lines that directly
connected to µP
SYSTEM BUS: group of conducting lines that carries
data , address & control signals in a µP system
38

TRISTATE LOGIC
3 logic levels are:
• High State (logic 1)
• Low state (logic 0)
• High Impedance state
High Impedance: output is not being driven to any defined logic level
by the output circuit.
39

Basic Microprocessors System
Input
Devices
Processing
Data into
Information
Output
Devices
Control
Unit
Secondary Storage Devices
Arithmetic-
Logic
Unit
Primary Storage
Unit
Central Processing Unit
Keyboard,
Mouse
etc
Monitor
Printer
Disks, Tapes, Optical Disks
40

UNIT 1 Syllabus
• Hardware Architecture, pinouts
• Functional Building Blocks of Processor
• Memory organization
• I/O ports and data transfer concepts
• Timing Diagram
• Interrupts.
42

8085
PIN DIAGRAM &
ARCHITECTURE
43

X1 & X2
Pin 1 and Pin 2 (Input)
45
 These are also called
Crystal Input Pins.
 8085 can generate clock
signals internally.
 To generate clock
signals internally, 8085
requires external inputs
from X1 and X2.

RESET IN and RESET OUT
Pin 36 (Input) and Pin 3 (Output)
46
 RESET IN:
◦ It is used to reset the
microprocessor.
◦ It is active low signal.
◦ When the signal on this
pin is low for at least 3
clocking cycles, it forces
the microprocessor to
reset itself.
Presented by C.GOKUL,AP/EEE Velalar College of Engg & Tech , Erode

47
 Resetting the
microprocessor means:
◦ Clearing the PC and IR.
◦ Disabling all interrupts
(exceptTRAP).
◦ Disabling the SOD pin.
◦ All the buses (data,
address, control) are tri-
stated.
◦ Gives HIGH output to
RESET OUT pin.

48
 RESET OUT:
◦ It is used to reset the peripheral
devices and other ICs on the
circuit.
◦ It is an output signal.
◦ It is an active high signal.
◦ The output on this pin goes high
whenever RESET IN is given low
signal.
◦ The output remains high as long
as RESET IN is kept low.

SID and SOD
49
 SID (Serial Input
Data):
o It takes 1 bit input from
serial port of 8085.
o Stores the bit at the 8th
position (MSB) of the
Accumulator.
o RIM (Read Interrupt
Mask) instruction is used
to transfer the bit.

SID and SOD
50
 SOD (Serial Output
Data):
o It takes 1 bit from
Accumulator to serial port
of 8085.
o Takes the bit from the 8th
position (MSB) of the
Accumulator.
o SIM (Set Interrupt Mask)
instruction is used to
transfer the bit.

Interrupt Pins
51
 Interrupt:
• It means interrupting the normal execution of the
microprocessor.
• When microprocessor receives interrupt signal, it discontinues
whatever it was executing.
• It starts executing new program indicated by the interrupt
signal.
• Interrupt signals are generated by external peripheral devices.
• After execution of the new program, microprocessor goes
back to the previous program.

Sequence of StepsWheneverThere is
an Interrupt
52
 Microprocessor completes execution of current
instruction of the program.
 PC contents are stored in stack.
 PC is loaded with address of the new program.
 After executing the new program, the
microprocessor returns back to the previous
program.
 It goes to the previous program by reading the top
value of stack.

Five Hardware Interrupts in 8085
53
 TRAP
 RST 7.5
 RST 6.5
 RST 5.5
 INTR

Classification of Interrupts
54
 Maskable and Non-Maskable
 Vectored and Non-Vectored
 EdgeTriggered and LevelTriggered
 Priority Based Interrupts

Maskable Interrupts
55
 Maskable interrupts are those interrupts
which can be enabled or disabled.
 Enabling and Disabling is done by
software instructions.

Maskable Interrupts
56
 List of Maskable Interrupts:
• RST 7.5
• RST 6.5
• RST 5.5
• INTR

Non-Maskable Interrupts
57
 The interrupts which are always in
enabled mode are called non-maskable
interrupts.
 These interrupts can never be disabled
by any software instruction.
 TRAP is a non-maskable interrupt.

Vectored Interrupts
58
 The interrupts which have fixed memory
location for transfer of control from
normal execution.
 Each vectored interrupt points to the
particular location in memory.

Vectored Interrupts
59
 List of vectored interrupts:
• RST 7.5
• RST 6.5
• RST 5.5
• TRAP

Vectored Interrupts
60
 The addresses to which program control
goes:
 Absolute address is calculated by
multiplying the RST value with 0008 H.
Name Vectored Address
RST 7.5 003C H (7.5 x 0008 H)
RST 6.5 0034 H (6.5 x 0008 H)
RST 5.5 002C H (5.5 x 0008 H)
TRAP 0024 H (4.5 x 0008 H)

Non-Vectored Interrupts
61
 The interrupts which don't have fixed
memory location for transfer of control
from normal execution.
 The address of the memory location is
sent along with the interrupt.
 INTR is a non-vectored interrupt.

EdgeTriggered Interrupts
62
 The interrupts which are triggered at
leading or trailing edge are called edge
triggered interrupts.
 RST 7.5 is an edge triggered interrupt.
 It is triggered during the leading
(positive) edge.

LevelTriggered Interrupts
63
 The interrupts which are triggered at high
or low level are called level triggered
interrupts.
 RST 6.5
 RST 5.5
 INTR
 TRAP is edge and level triggered interrupt.

Priority Based Interrupts
64
 Whenever there exists a simultaneous
request at two or more pins then the
pin with higher priority is selected by the
microprocessor.
 Priority is considered only when there
are simultaneous requests.

Priority Based Interrupts
65
 Priority of interrupts:
Interrupt Priority
TRAP 1
RST 7.5 2
RST 6.5 3
RST 5.5 4
INTR 5

TRAP
Pin 6 (Input)
66
 It is an non-maskable interrupt.
 It has the highest priority.
 It cannot be disabled.
 It is both edge and level
triggered.
 It means TRAP signal must go
from low to high.
 And must remain high for a
certain period of time.
 TRAP is usually used for power
failure and emergency shutoff.

RST 7.5
Pin 7 (Input)
67
 It is a maskable interrupt.
 It has the second highest
priority.
 It is positive edge triggered
only.
 The internal flip-flop is
triggered by the rising
edge.
 The flip-flop remains high
until it is cleared by RESET
IN.

RST 6.5
Pin 8 (Input)
68
 It has the third highest
priority.
 It is level triggered only.
 The pin has to be held high
for a specific period of
time.
 RST 6.5 can be enabled by
EI instruction.
 It can be disabled by DI
instruction.

RST 5.5
Pin 9 (Input)
69
 It is a maskable
interrupt.
 It has the fourth highest
priority.
 It is also level triggered.
 The pin has to be held
high for a specific period
of time.
 This interrupt is very
similar to RST 6.5.

INTR
Pin 10 (Input)
70
 It has the lowest priority.
 It is also level triggered.
 It is a general purpose
interrupt.
 By general purpose we
mean that it can be used to
vector microprocessor to
any specific subroutine
having any address.

INTA
Pin 11 (Output)
71
 It stands for interrupt
acknowledge.
 It is an out going signal.
 It is an active low signal.
 Low output on this pin
indicates that
microprocessor has
acknowledged the INTR
request.

Address and Data Pins
72
 Address Bus:
• The address bus is used to send address to
memory.
• It selects one of the many locations in
memory.
• Its size is 16-bit.

Address and Data Pins
73
 Data Bus:
• It is used to transfer data between
microprocessor and memory.
• Data bus is of 8-bit.

AD0 – AD7
Pin 19-12 (Bidirectional)
74
 These pins serve the dual
purpose of transmitting lower
order address and data byte.
 During 1st clock cycle, these pins
act as lower half of address.
 In remaining clock cycles, these
pins act as data bus.
 The separation of lower order
address and data is done by
address latch.

A8 – A15
Pin 21-28 (Unidirectional)
75
 These pins carry the
higher order of address
bus.
 The address is sent from
microprocessor to
memory.
 These 8 pins are switched
to high impedance state
during HOLD and RESET
mode.

ALE
Pin 30 (Output)
76
 It is used to enable Address
Latch.
 It indicates whether bus
functions as address bus or data
bus.
 If ALE = 1 then
◦ Bus functions as address bus.
 If ALE = 0 then
◦ Bus functions as data bus.

S0 and S1
Pin 29 (Output) and Pin 33 (Output)
77
 S0 and S1 are called Status
Pins.
 They tell the current
operation which is in progress
in 8085.
S0 S1 Operation
0 0 Halt
0 1 Write
1 0 Read
1 1 Opcode Fetch

IO/M
Pin 34 (Output)
78
 This pin tells whether I/O
or memory operation is
being performed.
 If IO/M = 1 then
◦ I/O operation is being
performed.
◦ Memory operation is being
performed.

IO/M
Pin 34 (Output)
79
 The operation being performed is indicated by
S0 and S1.
 If S0 = 0 and S1 = 1 then
◦ It indicatesWRITE operation.
◦ It indicates Memory operation.
 Combining these two we get MemoryWrite
Operation.

Table Showing IO/M, S0, S1 and
Corresponding Operations
80
Operations IO/M S0 S1
Opcode Fetch 0 1 1
Memory Read 0 1 0
MemoryWrite 0 0 1
I/O Read 1 1 0
I/OWrite 1 0 1
Interrupt Ack. 1 1 1
Halt High Impedance 0 0

RD
Pin 32 (Output)
81
 RD stands for Read.
 It is an active low signal.
 It is a control signal used
for Read operation either
from memory or from
Input device.
 A low signal indicates that
data on the data bus must
be placed either from
selected memory location
or from input device.

WR
Pin 31 (Output)
82
 WR stands for Write.
 It is also active low signal.
 It is a control signal used
for Write operation either
into memory or into
output device.
 A low signal indicates that
data on the data bus must
be written into selected
memory location or into
output device.

READY
Pin 35 (Input)
83
 This pin is used to
synchronize slower
peripheral devices with
fast microprocessor.
 A low value causes the
microprocessor to
enter into wait state.
 The microprocessor
remains in wait state
until the input at this pin
goes high.

HOLD
Pin 38 (Input)
84
 HOLD pin is used to request
the microprocessor for DMA
transfer.
 A high signal on this pin is a
request to microprocessor
to relinquish the hold on
buses.
 This request is sent by DMA
controller.
 Intel 8257 and Intel 8237 are
two DMA controllers.

HLDA
Pin 39 (Output)
85
 HLDA stands for Hold
Acknowledge.
 The microprocessor uses
this pin to acknowledge the
receipt of HOLD signal.
 When HLDA signal goes high,
address bus, data bus, RD,
WR, IO/M pins are tri-
stated.
 This means they are cut-off
from external environment.

HLDA
Pin 39 (Output)
86
 The control of these
buses goes to DMA
Controller.
 Control remains at
DMA Controller until
HOLD is held high.
 When HOLD goes low,
HLDA also goes low
and the microprocessor
takes control of the
buses.

VSS andVCC
Pin 20 (Input) and Pin 40 (Input)
87
 +5V power supply is
connected toVCC.
 Ground signal is
connected toVSS.

THE 8085 AND ITS BUSSES
The 8085 is an 8-bit general purpose
microprocessor that can address 64K Byte of
memory.
It has 40 pins and uses +5V for power. It can run
at a maximum frequency of 3 MHz.
-The pins on the chip can be grouped into 6
groups:
Address Bus.
Data Bus.
Control and Status Signals.
Power supply and frequency.
Externally Initiated Signals.
Serial I/O ports. 88

The Address and Data Busses
 The address bus has 8 signal lines A8 – A15 which are
unidirectional.
 The other 8 address bits are multiplexed (time shared)
with the 8 data bits.
 So, the bits AD0 – AD7 are bi-directional and serve
as A0 – A7 and D0 – D7 at the same time.
During the execution of the instruction, these
lines carry the address bits during the early part,
then during the late parts of the execution, they
carry the 8 data bits.
 In order to separate the address from the data, we
can use a latch to save the value before the function
of the bits changes. 89

Flag Register
CYPACZS
D0D1D2D3D4D5D6D7
The flags are affected by the arithmetic and logical
instruction
91

Accumulator
 It is an 8 bit register
 For any arithmetic and logical instruction one of the data
should be in this register
 It is used for storing the result of any arithmetic and
logical manipulations.
 It is also called as A register
 All the data which are sent to I/O devices are sent via
A register.
92

Temporary register
 It is used to hold the data during the
operation of arithmetic and logical operation
93

Sign Flag
 If the D7 bit of the accumulator is set then
this flag is set i.e 1 meaning that the result is
in negative.
 Ex. 7-8 = -1
94

Carry flag
 During the arithmetic operation if a carry occurs then this
flag is set.
 Ex. F1+1F= 101
Carry
95

Zero flag
During the arithmetic/ logical
operation if the result is zero then this
flag is set.
Ex. FF-FF = 00
96

Parity flag
 After the of the arithmetic and logical
operation if the result is even then this flag is
set.
 Ex. 0A-02 = 08
97

Auxiliary carry flag
 During BCD arithmetic operation when a carry is
generated by D3 bit and passed on to D4 bit then
this flag is set.
 Ex. 1F+11 = 0001 1111 +
0001 0001
= 0010 0000
98

Timing and control
It synchronizes all the operation with the
clock and generates the communication
between the microprocessor and
peripherals
99

Instruction Register and decoder
The instruction is loaded in the
instruction register
The decoder decodes them and establishes
the operation that has to be performed
100

Register array
The W and Z register are temporary
registers
Used to hold the 8 bit data during the
execution and it is used internally .
It is not used by the programmer.
101

Control and status signals
Machine Cycle IO/M S1 S0
Opcode fetch 0 1 1
Memory read 0 1 0
Memory write 0 0 1
I/O read 1 1 0
I/O write 1 0 1
102

Arithmetic and Logical unit
It is an 8 bit register
It is used for performing addition,
subtraction and logical operation.
AND, OR, NOT, XOR, CMP are
some of the logical operation.
103

Program Counter
It is a 16 bit register
It is used to point out the address of
the next instruction which is to be
executed
104Presented by C.GOKUL,AP/EEE Velalar College of Engg & Tech , Erode

Stack pointer
 It is a 16 bit register
 It points the starting address of the stack .
105

Register Array
 B, C, D, E, H and L are general purpose
register
 All are 8 bit register
 If the are combined as BC, DE and HL
they can store 16 bit data
106

8085 Communication with Memory
 Involves the following three steps
1. Identify the memory location (with address)
2. Generate Timing & Control signals
3. Data transfer takes place
108

Example: Memory Read Operation
1
2
3
109

8085 Interfacing with Memory chips
8085
Memory
Interface
Memory
Chip
Address
Data
Control
Address
Data
Control
110

8085
Memory
Interface
Memory
Chip
AD0-AD7
Control
A0 – A7
Data
74LS373
A8-A15 A8-A15
ALE
111

8085
Memory
Interface
Program
Memory
AD0-AD7
IO/M
A0 – A7
Data
74LS373
A8-A15 A8-A15
ALE
RD
RD
CS
112

I/O ports &
Data transfer
concepts
113

Interfacing I/O devices with 8085
8085
I/O
Interface
I/O
Devices
Memory
Interface
Memory
Devices
System Bus
114

Techniques for I/O Interfacing
 Memory-mapped I/O
 Peripheral-mapped I/O
115

Memory-mapped I/O
 8085 uses its 16-bit address bus to identify a
memory location
 Memory address space: 0000H to FFFFH
 8085 needs to identify I/O devices also
 I/O devices can be interfaced using
addresses from memory space
 8085 treats such an I/O device as a memory
location
 This is called Memory-mapped I/O
116

Peripheral-mapped I/O
 8085 has a separate 8-bit addressing scheme
for I/O devices
 I/O address space: 00H to FFH
 This is called Peripheral-mapped I/O or
I/O-mapped I/O
117

8085 Communication with I/O devices
1. Identify the I/O device (with address)
2. Generate Timing & Control signals
 8085 communicates with a I/O device only if
there is a Program Instruction to do so
118

1.Identify the I/O device (with address)
1. Memory-mapped I/O (16-bit address)
2. Peripheral-mapped I/O (8-bit address)
119

2.Generate Timing & Control Signals
 Memory-mapped I/O
 Reading Input: IO/M = 0, RD = 0
 Write to Output: IO/M = 0, WR = 0
 Peripheral-mapped I/O
 Reading Input: IO/M = 1, RD = 0
 Write to Output: IO/M = 1, WR = 0
120

8085 Communication with I/O devices
 Identify the I/O device (with address)
 Generate Timing & Control signals
 Data transfer takes place
 8085 communicates with a I/O device only if
there is a Program Instruction to do so
121

Peripheral I/O Instructions
 IN Instruction
 Inputs data from input device into the
accumulator
 It is a 2-byte instruction
 Format: IN 8-bit port address
 Example: IN 01H
122

 OUT Instruction
 Outputs the contents of accumulator to an
output device
 It is a 2-byte instruction
 Format: OUT 8-bit port address
 Example: OUT 02H
123

----------Example Program----------
 WAP to read a number from input port (port
address 01H) and display it on ASCII display
connected to output port (port address 02H)
IN 01H ;reads data value 03H (example)into
;accumulator, A = 03H
MVI B, 30H;loads register B with 30H
ADD B ;A = 33H, ASCII code for 3
OUT 02H ;display 3 on ASCII display
124

Memory-mapped I/O Instructions
 I/O devices are identified by 16-bit addresses
 8085 communicates with an I/O device as if it
were one of the memory locations
 Memory related instructions are used
 For e.g. LDA, STA
 LDA 8000H
 Loads A with data read from input device with
16-bit address 8000H
 STA 8001H
 Stores (Outputs) contents of A to output
device with 16-bit address 8001H 125

----------Example Program----------
 WAP to read a number from input port (port
address 8000H) and display it on ASCII
display connected to output port (port
address 8001H)
LDA 8000H;reads data value 03H (example)into
;accumulator, A = 03H
MVI B, 30H;loads register B with 30H
ADD B ;A = 33H, ASCII code for 3
STA 8001H;display 3 on ASCII display
126

Timing Diagram is a graphical representation. It
represents the execution time taken by each
instruction in a graphical format. The execution
time is represented in T-states.
Instruction Cycle:
The time required to execute an instruction .
Machine Cycle:
The time required to access the memory or
input/output devices .
T-State:
•The machine cycle and instruction cycle takes
multiple clock periods.
•A portion of an operation carried out in one
system clock period is called as T-state.
128

Timing diagrams
• The 8085 microprocessor has 7 basic machine
cycle. They are
1. Op-code Fetch cycle(4T or 6T).
2. Memory read cycle (3T)
3. Memory write cycle(3T)
4. I/O read cycle(3T)
5. I/O write cycle(3T)
6. Interrupt Acknowledge cycle(6T or 12T)
7. Bus idle cycle
130

1.Opcode fetch cycle(4T or 6T)
132

OPCODE FETCH
• The Opcode fetch cycle, fetches the instructions from memory
and delivers it to the instruction register of the microprocessor
• Opcode fetch machine cycle consists of 4 T-states.
T1 State:
During the T1 state, the contents of the program counter are
placed on the 16 bit address bus. The higher order 8 bits are
transferred to address bus (A8-A15) and lower order 8 bits are
transferred to multiplexed A/D (AD0-AD7) bus.
ALE (address latch enable) signal goes high. As soon as
ALE goes high, the memory latches the AD0-AD7 bus. At
the middle of the T state the ALE goes low
133

T2 State:
During the beginning of this state, the RD’ signal goes low
to enable memory. It is during this state, the selected memory
location is placed on D0-D7 of the Address/Data multiplexed
bus.
T3 State:
In the previous state the Opcode is placed in D0-D7 of the A/D
bus. In this state of the cycle, the Opcode of the A/D bus is
transferred to the instruction register of the microprocessor.
Now the RD’ goes high after this action and thus disables the
memory from A/D bus.
T4 State:
In this state the Opcode which was fetched from the memory
is decoded.
134

• These machine cycles have 3 T-states.
T1 state:
• The higher order address bus (A8-A15) and lower order address
and data multiplexed (AD0-AD7) bus. ALE goes high so that the
memory latches the (AD0-AD7) so that complete 16-bit address
are available.
The mp identifies the memory read machine cycle from the
status signals IO/M’=0, S1=1, S0=0. This condition indicates the
memory read cycle.
T2 state:
• Selected memory location is placed on the (D0-D7) of the A/D
multiplexed bus. RD’ goes LOW
T3 State:
• The data which was loaded on the previous state is transferred
to the microprocessor. In the middle of the T3 state RD’ goes
high and disables the memory read operation. The data which
was obtained from the memory is then decoded.
136

3. Memory write cycle (3T)
137

• These machine cycles have 3 T-states.
T1 state:
• The higher order address bus (A8-A15) and lower order address
and data multiplexed (AD0-AD7) bus. ALE goes high so that the
memory latches the (AD0-AD7) so that complete 16-bit address
are available.
The mp identifies the memory read machine cycle from the
status signals IO/M’=0, S1=0, S0=1. This condition indicates the
memory read cycle.
T2 state:
• Selected memory location is placed on the (D0-D7) of the A/D
multiplexed bus. WR’ goes LOW
T3 State:
• In the middle of the T3 state WR’ goes high and disables the
memory write operation. The data which was obtained from
the memory is then decoded.
138

STA instruction
ex: STA 526A
141

It require 4 m/c cycles
13 T states
1.opcode fetch(4T)
2.memory read(3T)
3.memory read(3T)
4.Memory write(3T)
142

Timing diagram for IN C0H
• Fetching the Opcode DBH from the memory
4125H.
• Read the port address C0H from 4126H.
• Read the content of port C0H and send it to
the accumulator.
• Let the content of port is 5EH.
144

It require 3 m/c cycles
10 T states
opcode fetch(4T)
memory read(3T)
I/O read(3T)
145

OUT
instruction
Machines Cycles(10T):
1.instruction fetch(4T)
2.memory read (3T)
3.IO write (3T)
147

Timing diagram for MVI B, 43h
• Fetching the Opcode 06H from the memory
2000H. (OF machine cycle)
• Read (move) the data 43H from memory
2001H. (memory read)
149

8085 Interrupts
8085 has five interrupt inputs
1. TRAP
2. RST7.5
3. RST 6.5
4. RST5.5
5. INTR
154

U7
8085
36
1
2
5
6
9
8
7
10
11
29
33
39
35
12
13
14
15
16
17
18
19
21
22
23
24
25
26
27
28
30
31
32
34
3
374
38
40
20
RST-IN
X1
X2
SID
TRAP
RST 5.5
RST 6.5
RST 7.5
INTR
INTA
S0
S1
HOLD
READY
AD0
AD1
AD2
AD3
AD4
AD5
AD6
AD7
A8
A9
A10
A11
A12
A13
A14
A15
ALE
WR
RD
IO/M
RST-OT
CLKOSOD
HLDA
VCC
VSS
Interrupt pins of 8085 155

Types of Interrupts
• Interrupts of 8085 can be classified as
– Maskable (RST 7.5, RST 6.5, RST 5.5, INTR)
– Non-maskable (TRAP)
• An interrupt is a request for attention/service
• 8085 may choose to service/not-service a
maskable interrupt
• 8085 cannot ignore a service request from a
non-maskable interrupt
156

Interrupt process
• 8085 is executing its main program
• an interrupt is generated by an external
device
• 8085 pauses execution of main program
• 8085 calls the Interrupt service routine
• 8085 executes the Interrupt service routine
• 8085 returns to execution of main program
(from where it was paused)
157

Example: Blinking LED Display with
Interrupt-based Display-Pattern change
8085
Input
Switches
LED
Display
RST 7.5
(Display-Pattern)
Interrupt Switch
Peripheral-mapped I/OInterrupt I/O
158

Interrupt Service Routine (ISR)
• It is a subroutine
• 8085 calls an ISR in response to an
interrupt request by an external device
• ISRs must be located in memory at pre-
determined addresses known as Interrupt
Vectors
159

Interrupt Vector Table of 8085
Interrupt Interrupt Vector
TRAP 0024H
RST 7.5 003CH
RST 6.5 0034H
RST 5.5 002CH
Please Note: INTR is a non-vectored interrupt
160

Using Vectored Interrupts of 8085
• By default, all the vectored interrupts (except
TRAP) of 8085 are disabled
• 8085 vectored interrupts are enabled with
two instructions: EI and SIM
• EI (Enable Interrupt): 1-byte instruction that
sets the Interrupt Enable flip-flop
– It is internal to the processor & can be set or reset
by using software instructions
161

Using Vectored Interrupts
Step-1
• Set Interrupt Enable flip-flop by using EI
instruction to enable the interrupt process
Step-2
• Use SIM (Set Interrupt Mask) instruction to
set mask for RST 7.5, 6.5 and 5.5
interrupts
162

SIM Instruction
• It is a 1-byte instruction
• Reads Accumulator contents
• Enables/Disables interrupts accordingly
• Used for three different functions
– Set mask for RST 7.5, 6.5, 5.5 interrupts
– Additional control for RST 7.5
– Implement serial I/O
163

Accumulator bit pattern for SIM
D7 D6 D5 D4 D3 D2 D1 D0
SOD SDE XXX R7.5 MSE M7.5 M6.5 M5.5
0 = Available, 1 = Masked
Mask Set Enable, 0 = bits 0-2 ignored
1 = mask is set
IF 1, RESET RST 7.5
If 1, bit 7 is output to
serial output data latch
Serial Output Data,
ignored if bit 6 = 0 164

8085 Interrupt process for
Vectored-Interrupts
1. Enables Interrupt process by writing the EI
instruction in the main program
2. Set interrupt mask using SIM instruction
3. 8085 monitors the status of all interrupt
lines during the execution of each
instruction
165

4. When 8085 detects an interrupt signal from
an external device
• It completes execution of current
instruction
• Disables the Interrupt Enable flip-flop
5. Executes a CALL to Interrupt Vector
location for that interrupt
• Before the CALL is made, 8085 stores
return address in main program on stack
Vectored-Interrupts (Cont.)
166

6. 8085 executes the ISR written at the
specified interrupt vector location
• ISR should include the EI instruction to
Enable Interrupt again
• At the end of ISR, RET instruction
transfers the program control back to the
main program
Vectored-Interrupts (Cont.)
167

PROGRAMMING OF 8085 PROCESSOR
UNIT
2
168
Presented by
C.GOKUL,AP/EEE

UNIT 2 Syllabus
• Instruction -format and addressing
modes
• Assembly language format – Data
transfer, data manipulation& control
instructions
• Programming: Loop structure with
counting & Indexing – Look up table -
Subroutine instructions - stack.
169

Addressing Modes of 8085
• Format of a typical Assembly language instruction is
given below-
[Label:] Mnemonic [Operands] [;comments]
HLT
MVI A, 20H
MOV M, A ;Copy A to memory location whose
address is stored in register pair HL
LOAD: LDA 2050H ;Load A with contents of memory
location with address 2050H
READ: IN 07H ;Read data from Input port with
address 07H
171

• The various formats of specifying operands
are called addressing modes
• Addressing modes of 8085
1. Register Addressing
2. Immediate Addressing
3. Memory Addressing
4. Input/Output Addressing
172

1. Register Addressing
• Operands are one of the internal registers of
8085
• Examples-
MOV A, B
ADD C
173

2. Immediate Addressing
• Value of the operand is given in the
instruction itself
• Example-
MVI A, 20H
LXI H, 2050H
ADI 30H
SUI 10H
174

3. Memory Addressing
• One of the operands is a memory location
• Depending on how address of memory
location is specified, memory addressing is of
two types
– Direct addressing
– Indirect addressing
175

3(a) Direct Addressing
• 16-bit Address of the memory location is
specified in the instruction directly
• Examples-
LDA 2050H ;load A with contents of memory
STA 3050H ;store A with contents of memory
176

3(b) Indirect Addressing
• A memory pointer register is used to store the
address of the memory location
• Example-
MOV M, A ;copy register A to memory location
whose address is stored in register
pair HL
30HA 20H
H
50H
L
30H2050H
177

4. Input/Output Addressing
• 8-bit address of the port is directly specified in
the instruction
• Examples-
IN 07H
OUT 21H
178

Instruction set
 An instruction is a binary pattern designed
inside a microprocessor to perform a specific
function.
 A group of instruction together called as
instruction set.
 Group of instruction set is called as a
program.
180

Classification of instruction set
 According to word size or byte size it is
classified into 3 types.
 1 - byte instruction
 2 - byte instruction and
 3 - byte instruction
181

1. One-byte Instructions
• Includes Opcode and Operand in the same byte
• Examples-
Opcode Operand Binary Code Hex Code
MOV C, A 0100 1111 4FH
ADD B 1000 0000 80H
HLT 0111 0110 76H
182

2. Two-byte Instructions
• First byte specifies Operation Code
• Second byte specifies Operand
• Examples-
MVI A, 32H 0011 1110
0011 0010
3EH
32H
MVI B, F2H 0000 0110
1111 0010
06H
F2H
183

3. Three-byte Instructions
• First byte specifies Operation Code
• Second & Third byte specifies Operand
• Examples-
LXI H, 2050H 0010 0001
0101 0000
0010 0000
21H
50H
20H
LDA 3070H 0011 1010
0111 0000
0011 0000
3AH
70H
30H 184

Instruction Set of 8085
 An instruction is a binary pattern designed inside a
microprocessor to perform a specific function.
 The entire group of instructions that a
microprocessor supports is called Instruction
Set.
 8085 has 246 instructions.
 Each instruction is represented by an 8-bit binary
value.
 These 8-bits of binary value is called Op-Code or
Instruction Byte.

Classification of Instruction Set
 DataTransfer Instruction
 Arithmetic Instructions
 Logical Instructions
 Branching Instructions
 Control Instructions
186

1.DataTransfer Instructions
 These instructions move data between
registers, or between memory and
registers.
 These instructions copy data from source
to destination(without changing the
original data ).
187

Opcode Operand
MOV Rd, Rs
M, Rs
Rd, M
 This instruction copies the contents of the source
register into the destination register. (contents of the
source register are not altered)
 If one of the operands is a memory location, its location
is specified by the contents of the HL registers.
 Example: MOV B, C or MOV B, M
MOV-Copy from source to destination
188

A 20 B 20
A F
B 30 C
D E
H 20 L 50
A 20 B
BEFORE EXECUTION AFTER EXECUTION
MOV B,A
A F
B 30 C
D E
H 20 L 50
A F
B C
D E
H 20 L 50
A F
B C 40
D E
H 20 L 50
MOV M,B
MOV C,M
40 40
30
189

Opcode Operand
MVI Rd, Data
M, Data
 The 8-bit data is stored in the destination register or
memory.
 If the operand is a memory location, its location is
specified by the contents of the H-L registers.
 Example: MVI B, 60H or MVI M, 40H
MVI-Move immediate 8-bit
190

A F
B C
D E
H L
A F
B 60 C
D E
H L
AFTER EXECUTIONBEFORE EXECUTION
MVI B,60H
40HL=2050
2051H
204FH 204F
2051H
MVI M,40H
HL=2050
191

LDA-Load accumulator
Opcode Operand
LDA 16-bit address
 The contents of a memory location, specified by a 16-bit
address in the operand, are copied to the accumulator.
 The contents of the source are not altered.
 Example: LDA 2000H
192

A
30
A 30
30
LDA
2000H
2000H 2000H
193

Opcode Operand
LDAX B/D Register Pair
 The contents of the designated register pair point to a memory
location.
 This instruction copies the contents of that memory location into
the accumulator.
 The contents of either the register pair or the memory location are
not altered.
 Example: LDAX D
LDAX-Load accumulator indirect
194

A F
B C
D 20 E 30
A 80 F
B C
D 20 E 30
80 80
LDAX D
2030H
2030H
195

Opcode Operand
LXI Reg. pair, 16-bit data
 This instruction loads 16-bit data in the register pair.
 Example: LXI H, 2030 H
LXI-Load register pair immediate

A F
B C
H L
A 80 F
B C
H 90 L 30
30
90
50
LXI H,
2030
2030H
9030H
2031H
M=50
197

Opcode Operand
LHLD 16-bit address
 This instruction copies the contents of memory location
pointed out by 16-bit address into register L.
 It copies the contents of next memory location into
register H.
 Example: LHLD 2030 H
LHLD-Load H and L registers direct
198

A F
B C
H L
A 80 F
B C
H 85 L 00
00
85
60
LHLD
2030
2030H 8500H
M=60
199

Opcode Operand
STA 16-bit address
 The contents of accumulator are copied into the memory
location specified by the operand.
 Example: STA 2000H
STA-Store accumulator direct
200

A 50 A
50
50
STA
2000H
2000H 2000H
201

Opcode Operand
STAX Reg. pair
 The contents of accumulator are copied into the memory
location specified by the contents of the register pair.
 Example: STAX B
STAX-Store accumulator indirect
202

B 85 C 00
A=1AH
STAX B
1A8500H
203

Opcode Operand
SHLD 16-bit address
 The contents of register L are stored into memory
location specified by the 16-bit address.
 The contents of register H are stored into the next
memory location.
 Example: SHLD 2550H
SHLD-Store H and L registers direct
204

D E
H 70 L 80
SHLD
8500
80
70
8500H
8501H
205

Opcode Operand
XCHG None
 The contents of register H are exchanged with the
contents of register D.
 The contents of register L are exchanged with the
contents of register E.
 Example: XCHG
XCHG-Exchange H and L with D and E
206

D 20 E 40
H 70 L 80
D 70 E 80
H 20 L 40
XCHG
207

Opcode Operand
SPHL None
 This instruction loads the contents of H-L pair into SP.
 Example: SPHL
SPHL-Copy H and L registers to the
stack pointer
208

H 25 L 00
SP
BEFORE EXECUTION
AFTER EXECUTION
SPHL
SP 2500
H 25 L 00
209

Opcode Operand
XTHL None
 The contents of L register are exchanged with the
location pointed out by the contents of the SP.
 The contents of H register are exchanged with the next
location (SP + 1).
 Example: XTHL
XTHL-Exchange H and L with top of
stack
210

H 30 L 40
SP 2700
BEFORE EXECUTION
50
60
H
60
L
50
SP 2700
40
30
AFTER EXECUTION
XTHL
2700H
2701H
2702H
2700H
2701H
2702H
L=SP
H=(SP+1)
211

Opcode Operand Description
PCHL None Load program counter with H-
L contents
 The contents of registers H and L are copied into the
program counter (PC).
 The contents of H are placed as the high-order byte and
the contents of L as the low-order byte.
 Example: PCHL

Opcode Operand
PUSH Reg. pair
 The contents of register pair are copied onto stack.
 SP is decremented and the contents of high-order
registers (B, D, H,A) are copied into stack.
 SP is again decremented and the contents of low-order
registers (C, E, L, Flags) are copied into stack.
 Example: PUSH B
PUSH-Push register pair onto stack
213

Opcode Operand
POP Reg. pair
 The contents of top of stack are copied into register pair.
 The contents of location pointed out by SP are copied to the
low-order register (C, E, L, Flags).
 SP is incremented and the contents of location are
copied to the high-order register (B, D, H,A).
 Example: POP H
POP- Pop stack to register pair
215

Opcode Operand
IN 8-bit port address
 The contents of I/O port are copied into accumulator.
 Example: IN 8C H
IN- Copy data to accumulator from a
port with 8-bit address
217

10 A
10 A 10
BEFORE EXECUTION
AFTER EXECUTION
IN 80H
PORT
80H
PORT
80H
218

Opcode Operand
OUT 8-bit port address
 The contents of accumulator are copied into the I/O
port.
 Example: OUT 78H
OUT- Copy data from accumulator to a
port with 8-bit address
219

10 A 40
40 A 40
BEFORE EXECUTION
AFTER EXECUTION
OUT 50H
PORT
50H
PORT
50H
220

2.Arithmetic Instructions
 These instructions perform the
operations like:
◦ Addition
◦ Subtract
◦ Increment
◦ Decrement

Addition
 Any 8-bit number, or the contents of register, or
the contents of memory location can be added to
the contents of accumulator.
 The result (sum) is stored in the accumulator.
 No two other 8-bit registers can be added
directly.
 Example: The contents of register B cannot be
added directly to the contents of register C.
222

ADD R
M
Add register or memory to
accumulator
 The contents of register or memory are added to the
contents of accumulator.
 The result is stored in accumulator.
 If the operand is memory location, its address is specified by
H-L pair.
 Example: ADD B or ADD M
ADD
223

B C 05
D E
H L
B C 05
D E
H L
B C
D E
H 20 L 50
B C
D E
H 20 L 50
AFTER EXECUTION
BEFORE EXECUTION
A 09A 04
ADD C
A=A+C
ADD M
A=A+M
10
10
2050 2050
A 04
A 14
04+05=09
04+10=14 224

ADC R
M
Add register or memory to
accumulator with carry
 The contents of register or memory and Carry Flag (CY) are added to the
contents of accumulator.
 If the operand is memory location, its address is specified by H-L pair.
 All flags are modified to reflect the result of the addition.
 Example: ADC B or ADC M
ADC
225

B C 05
D E
H L
A 50
B C 20
D E
H L
A 56
ADC C
A=A+C+CY
CY 01
CY 1
A 06 A 37
H 20 L 50
H 20 L 50
ADC M
A=A+M+CY
30 302050H 2050H
06+1+30=37
50+05+01=56
226

ADI 8-bit data Add immediate to accumulator
 The 8-bit data is added to the contents of accumulator.
 All flags are modified to reflect the result of the addition.
 Example: ADI 45 H
ADI
227

A 03
ADI 05H
A=A+DATA(8)
A 08
03+05=08
228

ACI 8-bit data Add immediate to
accumulator with carry
 The 8-bit data and the Carry Flag (CY) are added to
the contents of accumulator.
 All flags are modified to reflect the result of the
addition.
 Example: ACI 45 H
ACI
229

CY 1
A 05
ACI 20H
A=A+DATA
(8)+CY
A 26
05+20+1=26 230

DAD Reg. pair Add register pair to H-L pair
 The 16-bit contents of the register pair are added to
the contents of H-L pair.
 The result is stored in H-L pair.
 If the result is larger than 16 bits, then CY is set.
 No other flags are changed.
 Example: DAD B or DAD D
DAD
231

D 12 E 34
H 23 L 45
D 12 E 34
H 35 L 79
DAD D
DAD D HL=HL+DE
DAD B HL=HL+BC
1234
2345 +
-------
3579
232

Subtraction
 Any 8-bit number, or the contents of register, or
the contents of memory location can be
subtracted from the contents of accumulator.
 The result is stored in the accumulator.
 Subtraction is performed in 2’s complement form.
 If the result is negative, it is stored in 2’s
complement form.
 No two other 8-bit registers can be subtracted
directly.
233

SUB R
M
Subtract register or memory
from accumulator
 The contents of the register or memory location are
subtracted from the contents of the accumulator.
 If the operand is memory location, its address is specified by
H-L pair.
 All flags are modified to reflect the result of subtraction.
 Example: SUB B or SUB M
SUB
234

B C 04
D E
H L
B C 04
D E
H L
B C
D E
H 20 L 50
B C
D E
H 20 L 50
AFTER EXECUTION
BEFORE EXECUTION
A 05A 09
SUB C
A=A-C
SUB M
A=A-M
10
10
2050 2050
A 14
A 04
09-04=05
14-10=04 235

SBB R
M
Subtract register or memory
from accumulator with borrow
 The contents of the register or memory location and Borrow Flag (i.e. CY)
are subtracted from the contents of the accumulator.
 If the operand is memory location, its address is specified by H-L pair.
 Example: SBB B or SBB M
SBB
236

B C 05
D E
H L
A 08
B C 05
D E
H L
A 02
SBB C
A=A-C-CY
CY 01
CY 1
A 06 A 03
H 20 L 50
H 20 L 50
SBB M
A=A-M-CY
02 022050H 2050H
08-05-01=02
06-02-1=03
237

SUI 8-bit data Subtract immediate from
accumulator
 The 8-bit data is subtracted from the contents of the
accumulator.
 Example: SUI 05H
SUI
238

A 08
SUI 05H
A=A-DATA(8)
A 03
08-05=03
239

SBI 8-bit data Subtract immediate from
accumulator with borrow
 The 8-bit data and the Borrow Flag (i.e. CY) is subtracted
from the contents of the accumulator.
 Example: SBI 45 H
SBI
240

CY 1
A 25
SBI 20H
A=A-DATA
(8)-CY
A 04
25-20-01=04
241

Increment / Decrement
 The 8-bit contents of a register or a
memory location can be incremented or
decremented by 1.
 The 16-bit contents of a register pair can
be incremented or decremented by 1.
 Increment or decrement can be
performed on any register or a memory
location.

INR R
M
Increment register or
memory by 1
 The contents of register or memory location are incremented
by 1.
 The result is stored in the same place.
 If the operand is a memory location, its address is specified by
 Example: INR B or INR M
INR
243

B 10 C
D E
H L
A
B 11 C
D E
H L
A
H
20
L
50
H
20
L
5010 112050H 2050H
INR M
M=M+1
B 10 C
D E
H L
A
BEFORE EXECUTION
INR B
R=R+1
10+1=11
10+1=11 244

INX R Increment register pair by 1
 The contents of register pair are incremented by 1.
 Example: INX H or INX B or INX D
INX
245

B C
D E
H 10 L 20
B C
D E
H 10 L 21
SPSP
INX H
RP=RP+1
1020+1=1021
246

DCR R
M
Decrement register or
memory by 1
 The contents of register or memory location are decremented
by 1.
 If the operand is a memory location, its address is specified by
 Example: DCR B or DCR M
DCR
247

B C
D E 19
H L
A
AFTER EXECUTION
B C
D E 20
H L
A
BEFORE EXECUTION
DCR E
R=R-1
H
20
L
50
H
20
L
5021 202050H
DCR M
M=M-1
2050H
21-1=20
20-1=19
248

DCX R Decrement register pair by 1
 The contents of register pair are decremented by 1.
 Example: DCX H or DCX B or DCX D
DCX
249

B C
D E
H 10 L 21
B C
D E
H 10 L 20
SPSP
DCX H
RP=RP-1
250

3.Logical Instructions
 These instructions perform logical operations on
data stored in registers, memory and status flags.
 The logical operations are:
◦ AND
◦ OR
◦ XOR
◦ Rotate
◦ Compare
◦ Complement
251

AND, OR, XOR
 Any 8-bit data, or the contents of register,
or memory location can logically have
◦ AND operation
◦ OR operation
◦ XOR operation
with the contents of accumulator.
252

ANA R
M
Logical AND register or
memory with accumulator
 The contents of the accumulator are logically ANDed with the contents of
register or memory.
 The result is placed in the accumulator.
 If the operand is a memory location, its address is specified by the contents
of H-L pair.
 S, Z, P are modified to reflect the result of the operation.
 CY is reset and AC is set.
 Example: ANA B or ANA M.
253

B 10 C
D E
H L
A
B 0F C
D E
H L
A 0A
AFTER EXECUTION
ANA B
A=A and R
B 0F C
D E
H L
A AA
BEFORE EXECUTION
CY AC CY 0 AC 1
CY AC CY 0 AC 1
A 11A 55
H 20 L 50 H 20 L 50
B3 B3
2050H
ANA M
A=A and M
2050H
1010 1010=AAH
0000 1111=0FH
0000 1010=0AH
0101 0101=55H
1011 0011=B3H
0001 0001=11H
254

ANI 8-bit data Logical AND immediate with
accumulator
 The contents of the accumulator are logically ANDed with the
8-bit data.
 S, Z, P are modified to reflect the result.
 CY is reset,AC is set.
 Example: ANI 86H.
255

CY AC
A B3
CY
0
AC 1
A 33
ANI 3FH
A=A and DATA(8)
1011 0011=B3H
0011 1111=3FH
0011 0011=33H
256

ORA R
M
Logical OR register or memory with
accumulator
 The contents of the accumulator are logically ORed with the contents of the
register or memory.
 If the operand is a memory location, its address is specified by the contents of H-L
pair.
 CY and AC are reset.
 Example: ORA B or ORA M.
257

CY AC
ORA B
A=A or R
1010 1010=AAH
0001 0010=12H
1011 1010=BAH
B 12 C
D E
H L
A AA
B 12 C
D E
H L
A BA
CY 0 AC 0
258

CY AC
ORA M
A=A or M
0101 0101=55H
1011 0011=B3H
1111 0111=F7H
H 20 L 50
A 55 A F7
CY
0
AC 0
H 20 L 50
B3 B3
2050H 2050H
259

ORI 8-bit data Logical OR immediate with
accumulator
 The contents of the accumulator are logically ORed with the
8-bit data.
 Example: ORI 86H.
260

CY AC
A B3
CY
0
AC
0
A BB
ORI 08H
A=A or DATA(8)
1011 0011=B3H
0000 1000=08H
1011 1011=BBH
261

XRA R
M
Logical XOR register or
 The contents of the accumulator are XORed with the contents of
the register or memory.
 If the operand is a memory location, its address is specified by the
contents of H-L pair.
 S, Z, P are modified to reflect the result of the operation.
 Example: XRA B or XRA M.
262

B 10 C
D E
H L
A
B C 2D
D E
H L
A 87
AFTER EXECUTION
XRA C
A=A xor R
B C 2D
D E
H L
A AA
BEFORE EXECUTION
CY AC CY 0 AC 0
1010 1010=AAH
0010 1101=2DH
1000 0111=87H
263

H 20 L 50
A 55
AFTER EXECUTION
XRA M
A=A xor M
BEFORE EXECUTION
CY AC CY 0 AC 0
0101 0101=55H
1011 0011=B3H
1110 0110=E6H
H 20 L 50
A E6
B3 B3
2050H 2050H
264

XRI 8-bit data XOR immediate with
accumulator
 The contents of the accumulator are XORed with the 8-
bit data.
 Example: XRI 86H.
265

CY AC
A B3
CY
0
AC
0
A 8A
XRI 39H
A=A xor DATA(8)
1011 0011=B3H
0011 1001=39H
1000 1010=8AH
266

Compare
 Any 8-bit data, or the contents of register,
or memory location can be compares for:
◦ Equality
◦ Greater Than
◦ LessThan
with the contents of accumulator.
 The result is reflected in status flags.
267

CMP R
M
Compare register or
 The contents of the operand (register or memory) are
compared with the contents of the accumulator.
 Both contents are preserved .
268

B 10 C
D E
H L
A
B C
D 20 E
H L
A 10
AFTER EXECUTION
CMP D
A-R
B C
D 20 E
H L
A 10
BEFORE EXECUTION
CY Z CY 01 Z 0
CY Z CY 0
ZF 1
A B8A B8
H 20 L 50 H 20 L 50
B8 B8
2050H
CMP M
A-M
2050H
A>R: CY=0
A=R: ZF=1
A<R: CY=1
A>M: CY=0
A=M: ZF=1
A<M: CY=1
10<20:CY=01
B8=B8 :ZF=01
269

CPI 8-bit data Compare immediate with
accumulator
 The 8-bit data is compared with the contents of
accumulator.
 The values being compared remain unchanged.
270

CY Z
A BA
CY
0
AC
0
A BA
CPI 30H
A-DATA
A>DATA: CY=0
A=DATA: ZF=1
A<DATA: CY=1
BA>30 : CY=00 271

Rotate
 Each bit in the accumulator can be shifted
either left or right to the next position.
272

RLC None Rotate accumulator left
 Each binary bit of the accumulator is rotated left by one
position.
 Bit D7 is placed in the position of D0 as well as in the Carry
flag.
 CY is modified according to bit D7.
 S, Z, P,AC are not affected.
 Example: RLC.
273

B7 B6 B5 B4 B3 B2 B1 B0CY
B6 B5 B4 B3 B2 B1 B0 B7B7
AFTER EXECUTION
BEFORE EXECUTION
274

RRC None Rotate accumulator right
 Each binary bit of the accumulator is rotated right by one
position.
 Bit D0 is placed in the position of D7 as well as in the Carry
flag.
 Example: RRC.
275

B7 B6 B5 B4 B3 B2 B1 B0 CY
B0 B7 B6 B5 B4 B3 B2 B1 B0
AFTER EXECUTION
BEFORE EXECUTION
276

RAL None Rotate accumulator left
through carry
 Each binary bit of the accumulator is rotated left by one
position through the Carry flag.
 Bit D7 is placed in the Carry flag, and the Carry flag is placed
in the least significant position D0.
 Example: RAL.
277

B7 B6 B5 B4 B3 B2 B1 B0CY
B6 B5 B4 B3 B2 B1 B0 CYB7
AFTER EXECUTION
BEFORE EXECUTION
278

RAR None Rotate accumulator right
through carry
 Each binary bit of the accumulator is rotated right by one
position through the Carry flag.
 Bit D0 is placed in the Carry flag, and the Carry flag is placed
in the most significant position D7.
 Example: RAR.

B7 B6 B5 B4 B3 B2 B1 B0 CY
CY B7 B6 B5 B4 B3 B2 B1 B0
AFTER EXECUTION
BEFORE EXECUTION
280

Complement
 The contents of accumulator can be
complemented.
 Each 0 is replaced by 1 and each 1 is
replaced by 0.
281

CMA None Complement accumulator
 The contents of the accumulator are complemented.
 No flags are affected.
 Example: CMA. A=A’
A 00 A FF
282

CMC None Complement carry
 The Carry flag is complemented.
 No other flags are affected.
 Example: CMC => c=c’
BEFORE EXECUTION
AFTER EXECUTION
C 00 C FF
283

STC None Set carry
 The Carry flag is set to 1.
 No other flags are affected.
 Example: STC CF=1
S-set (1) C-clear (0)
284

4.Branching Instructions
 The branch group instructions
allows the microprocessor to change
the sequence of program either
conditionally or under certain test
conditions.The group includes,
 (1) Jump instructions,
 (2) Call and Return instructions,
 (3) Restart instructions,
285

JMP 16-bit
address
Jump unconditionally
 The program sequence is transferred to the memory
location specified by the 16-bit address given in the
operand.
 Example: JMP 2034 H.
286

Jx 16-bit
address
Jump conditionally
operand based on the specified flag of the PSW.
 Example: JZ 2034 H.
287

Jump Conditionally
Opcode Description Status Flags
JC Jump if Carry CY = 1
JNC Jump if No Carry CY = 0
JZ Jump if Zero Z = 1
JNZ Jump if No Zero Z = 0
JPE Jump if Parity Even P = 1
JPO Jump if Parity Odd P = 0
A-Above , B-Below , C-Carry , Z-Zero , P-Parity
288

CALL 16-bit
address
Call unconditionally
operand.
 Before the transfer, the address of the next instruction after
CALL (the contents of the program counter) is pushed onto
the stack.
 Example: CALL 2034 H.
289

Call Conditionally
CC Call if Carry CY = 1
CNC Call if No Carry CY = 0
CP Call if Positive S = 0
CM Call if Minus S = 1
CZ Call if Zero Z = 1
CNZ Call if No Zero Z = 0
CPE Call if Parity Even P = 1
CPO Call if Parity Odd P = 0
290

RET None Return unconditionally
 The program sequence is transferred from the subroutine
to the calling program.
 The two bytes from the top of the stack are copied into
the program counter, and program execution begins at
the new address.
 Example: RET.
291

Return Conditionally
RC Return if Carry CY = 1
RNC Return if No Carry CY = 0
RP Return if Positive S = 0
RM Return if Minus S = 1
RZ Return if Zero Z = 1
RNZ Return if No Zero Z = 0
RPE Return if Parity Even P = 1
RPO Return if Parity Odd P = 0
292

RST 0 – 7 Restart (Software Interrupts)
 The RST instruction jumps the control to one of eight
memory locations depending upon the number.
 These are used as software instructions in a program to
transfer program execution to one of the eight locations.
 Example: RST 1 or RST 2 ….
293

Instruction Code Vector Address
RST 0 0*8=0000H
RST 1 1*8=0008H
RST 2 2*8=0010H
RST 3 3*8=0018H
RST 4 4*8=0020H
RST 5 5*8=0028H
RST 6 6*8=0030H
RST 7 7*8=0038H
294

5. Control Instructions
 The control instructions control the
operation of microprocessor.
295

NOP None No operation
 No operation is performed.
 The instruction is fetched and decoded but no operation
is executed.
 Example: NOP
296

HLT None Halt
 The CPU finishes executing the current instruction and
halts any further execution.
 An interrupt or reset is necessary to exit from the halt
state.
 Example: HLT
297

DI None Disable interrupt
 The interrupt enable flip-flop is reset and all the
interrupts except the TRAP are disabled.
 Example: DI

EI None Enable interrupt
 The interrupt enable flip-flop is set and all interrupts
are enabled.
 This instruction is necessary to re-enable the
interrupts (except TRAP).
 Example: EI
299

RIM None Read Interrupt Mask
 This is a multipurpose instruction used to read the status
of interrupts 7.5, 6.5, 5.5 and read serial data input bit.
 The instruction loads eight bits in the accumulator with
the following interpretations.
 Example: RIM
300

SIM None Set Interrupt Mask
 This is a multipurpose instruction and used to implement
the 8085 interrupts 7.5, 6.5, 5.5, and serial data output.
 The instruction interprets the accumulator contents as
follows.
 Example: SIM
302

8085 Assembly
Language
Programming
304

Example Data Transfer (Copy)
Operations / Instructions
1. Load a 8-bit number 4F in
register B
2. Copy from Register B to
Register A
3. Load a 16-bit number
2050 in Register pair HL
4. Copy from Register B to
Memory Address 2050
5. Copy between Input /
Output Port and
Accumulator
MVI B, 4FH
MOV A,B
LXI H, 2050H
MOV M,B
OUT 01H
IN 07H
305

Example Arithmetic
1. Add a 8-bit number 32H to
Accumulator
2. Add contents of Register B to
Accumulator
3. Subtract a 8-bit number 32H
from Accumulator
4. Subtract contents of Register
C from Accumulator
5. Increment the contents of
Register D by 1
6. Decrement the contents of
Register E by 1
ADI 32H
ADD B
SUI 32H
SUB C
INR D
DCR E
306

Example Logical & Bit Manipulation
1. Logically AND Register H
with Accumulator
2. Logically OR Register L with
Accumulator
3. Logically XOR Register B
with Accumulator
4. Compare contents of
Register C with Accumulator
5. Complement Accumulator
6. Rotate Accumulator Left
ANA H
ORA L
XRA B
CMP C
CMA
RAL
307

Example Branching
1. Jump to a 16-bit Address
2080H if Carry flag is SET
2. Unconditional Jump
3. Call a subroutine with its 16-bit
Address
4. Return back from the Call
5. Call a subroutine with its 16-bit
Address if Carry flag is RESET
6. Return if Zero flag is SET
JC 2080H
JMP 2050H
CALL 3050H
RET
CNC 3050H
RZ
308

Writing a Assembly Language Program
• Steps to write a program
–Analyze the problem
–Develop program Logic
–Write an Algorithm
–Make a Flowchart
–Write program Instructions using
Assembly language of 8085
309

Program 8085 in Assembly language to add two 8-
bit numbers and store 8-bit result in register C.
1. Analyze the problem
– Addition of two 8-bit numbers to be done
2. Program Logic
– Add two numbers
– Store result in register C
– Example
10011001 (99H) A
+00111001 (39H) D
11010010 (D2H) C
310

1. Get two numbers
2. Add them
3. Store result
4. Stop
• Load 1st no. in register D
• Load 2nd no. in register E
3. Algorithm
Translation to 8085
operations
• Copy register D to A
• Add register E to A
• Copy A to register C
• Stop processing
311

4. Make a Flowchart
Start
Load Registers D, E
Copy D to A
Add A and E
Copy A to C
Stop
• Load 1st no. in register D
• Load 2nd no. in register E
• Stop processing
312

5. Assembly Language Program
1. Get two numbers
2. Add them
3. Store result
4. Stop
a) Load 1st no. in register D
b) Load 2nd no. in register E
a) Copy register D to A
b) Add register E to A
a) Copy A to register C
a) Stop processing
MVI D, 2H
MVI E, 3H
MOV A, D
ADD E
MOV C, A
HLT
313

Program 8085 in Assembly language to add two 8-
bit numbers. Result can be more than 8-bits.
1. Analyze the problem
– Result of addition of two 8-bit numbers can
be 9-bit
– Example
10011001 (99H) A
+10011001 (99H) B
100110010 (132H)
– The 9th bit in the result is called CARRY bit.
314

0
• How 8085 does it?
– Adds register A and B
– Stores 8-bit result in A
– SETS carry flag (CY) to indicate carry bit
10011001
10011001
A
B
+
99H
99H
10011001 A1
CY
00110010 99H32H
315

• Storing result in Register memory
10011001
A
32H1
CY
Register CRegister B
Step-1 Copy A to C
Step-2
a) Clear register B
b) Increment B by 1
316

2. Program Logic
1. Add two numbers
2. Copy 8-bit result in A to C
3. If CARRY is generated
– Handle it
4. Result is in register pair BC
317

1. Load two numbers
in registers D, E
2. Add them
3. Store 8 bit result in C
4. Check CARRY flag
5. If CARRY flag is SET
• Store CARRY in
register B
6. Stop
• Load registers D, E
3. Algorithm
Translation to 8085
operations
• Stop processing
• Use Conditional
Jump instructions
• Clear register B
• Increment B
318

4. Make a Flowchart
Start
Load Registers D, E
Copy D to A
Add A and E
Copy A to C
Stop
If
CARRY
NOT SET
Clear B
Increment B
False
True
319

5. Assembly Language Program
MVI D, 2H
MVI E, 3H
MOV A, D
ADD E
MOV C, A
HLT
• Load registers D, E
• Stop processing
• Use Conditional
Jump instructions
• Clear register B
• Increment B
JNC END
MVI B, 0H
INR B
END:
320

Ascending & Descending order
329

Smallest Number in an Array
332

Largest Number in an Array
334

335

Basics
Microprocessor &
Microcontroller
336

What is Microcontroller?
Micro Controller
337
Very Small A mechanism that controls
the operation of a machine

 CPU for Computers
 No RAM, ROM, I/O on CPU chip itself
 Example: Intel's x86, Motorola’s 680x0
338

 A smaller computer
 On-chip RAM, ROM, I/O ports...
 Example: Motorola’s 6811, Intel’s 8051, Zilog’s
Z8 and PIC
339

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
 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
 Not Expansive
 Single-purpose
341

 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 etc.
Office
 Telephones, computers, security systems, fax machines,
microwave, copier, laser printer, color printer, paging etc.
 Auto
 Trip computer, engine control, air bag, ABS, instrumentation,
security system, transmission control, entertainment, climate
control, cellular phone, keyless entry
342

344
Presented by
C.GOKUL,AP/EEE
Regulation : 2013

UNIT 3 Syllabus
• Architecture of 8051
• Special Function Registers(SFRs)
• I/O Pins Ports and Circuits {Pin Diagram}
• Instruction set
• Addressing modes
• Assembly language programming
345

 The 8051 is a subset of the 8052
 The 8031 is a ROM-less 8051
 Add external ROM to it
 You lose two ports, and leave only 2 ports for I/O
operations
346

 Intel introduced 8051, developed in the year
1981.
 The 8051 is an 8-bit controller.
 D0-D7 DATA LINES
 A0-A15 ADDRESS LINES
348

Interrupt
Control
8bit
CPU
4K
ROM
256 B
RAM
OSC
Bus
Control
4 I/O Ports
Serial
Port
Timer 1
Timer 0
General Block Diagram of 8051
TXD RXD
P0 P1 P2 P3
349
External Interrupts
Counter
Inputs

 8 bit CPU
 On-chip clock oscillator
 4K bytes of on-chip Program Memory-ROM
 128 bytes of on-chip Data RAM
 64KB Program Memory address space
 64KB Data Memory address space
 32 bidirectional I/0 lines (Port 0,1,2,3)
Port 0 { P0.0-P0.7 } – 8 pins
Port 1 { P1.0-P1.7 } – 8 pins
Port 2 { P2.0-P2.7 } – 8 pins
Port 3 { P3.0-P3.7 } – 8 pins
350

 Two 16-bit timer/counters(Timer 1,Timer 0)
 One serial port
UART(Universal Asynchronous Receiver Transmitter)
 6-source interrupt structure
1. External interrupt INT0
2. Timer interrupt T0
3. External interrupt INT1
4. Timer interrupt T1
5. Serial communication interrupt
6. Timer Interrupt T2
 4 Register Banks (Bank 0, Bank 1, Bank 2, Bank 3)
each bank has R0-R7 registers
351

Pin Description
of the 8051
352

EA/VPP
• EA, “external access’’
• EA = 0, 8051 microcontroller access from
external program memory (ROM) only.
• EA = 1, then it access internal and external
program memories (ROMS).
354

I/O Port Pins
• The four 8-bit I/O ports
Port 0 { P0.0-P0.7 } – 8 pins
Port 1 { P1.0-P1.7 } – 8 pins
Port 2 { P2.0-P2.7 } – 8 pins
Port 3 { P3.0-P3.7 } – 8 pins
355

Port 3
• Port 3 can be used as input or output.
• Port 3 has the additional function of
providing some extremely important
signals
356

Pin Description Summary
PIN TYPE NAME AND FUNCTION
Vss I Ground: 0 V reference.
Vcc I Power Supply + 5V.
P0.0 - P0.7
I/O Port 0: Port 0 is also the multiplexed low-order address and
data bus during accesses to external program and data
memory.
P1.0 - P1.7
I/O Port 1: Port 1 is an 8-bit bi-directional simple I/O port.
P2.0 - P2.7
I/O Port 2: Port 2 is an 8-bit bidirectional I/O. Port 2 emits the
high order address byte
P3.0 - P3.7
I/O Port 3: Port 3 is an 8 bit bidirectional I/O port. Port 3 also
serves special features as explained.
357

Pin Description Summary
PIN TYPE NAME AND FUNCTION
RST I Reset: resets the device.
ALE O Address Latch Enable:
When ALE=0, it provides data D0-D7
When ALE=1, it has address A0-A7
PSEN* O Program Store Enable:
For External Code Memory, PSEN = 0
For External Data Memory, PSEN = 1
EA*/VPP I External Access Enable/Programming Supply Voltage:
EA = 0, 8051 microcontroller access from external
program memory (ROM) only.
EA = 1, then it access internal and external program
memories (ROMS).
358

Architecture of
8051
microcontroller
359

Program Counter(PC) : The program
counter always points to the address of the
next instruction to be executed.
Stack Pointer Register (SP) : It is an 8-bit
register which stores the address of the
stack top.
ALU: perform arithmetic & logical operations
Flags : Carry(C),Auxiliary Carry(AC),
Overflow(O) & Parity(P)
362

 Timing & Control: Timing and control unit
synchronises all microcontroller operations
with clock & generates control signals.
 DPTR: (Data Pointer) - 16 bit
 DPH-Data Pointer High – 8 bit
 DPL-Data Pointer Low – 8 bit
DPTR Register is usually used for storing data and
intermediate results.
363

8051
Program Memory,
Data Memory
structure
364

8051 Memory Structure
External
EXT INT
128
SFR
External
Program Memory Data Memory
64K 64K
EA = 0 EA = 1
4K
60K
365

Special
Function
Registers [SFR]
366

• A Register (Accumulator)
• B Register
• Program Status Word (PSW) Register
• Data Pointer Register (DPTR)
– DPH (Data Pointer High) , DPL(Data Pointer Low)
• Stack Pointer (SP) Register
• P0, P1, P2, P3 - Input/output port Registers
• Timer T0 - TH0 & TL0
• Timer T1 – TH1 & TL1
• Timer Control (TCON) Register
• Serial Port Control (SCON) Register
• Serial Buffer Control (SBUF) Register
• IP Register (Interrupt Priority)
• IE Register (Interrupt Enable)
367

8051 Register Bank Structure
4 MEMORY BANKS
Bank 0
R0 R1 R2 R3 R4 R5 R6 R7Bank 3
R0 R1 R2 R3 R4 R5 R6 R7
368

Program Status Word [PSW]
C AC F0 RS1 RS0 OV F1 P
Register Bank Select
Carry
Auxiliary Carry
User Flag 0
Parity
User Flag 1
Overflow
369
00-Bank 0
01-Bank 1
10-Bank 2
11-Bank 3

Data Pointer Register (DPTR)
It consists of two separate registers:
DPH (Data Pointer High) &
DPL (Data Pointer Low).
370

Stack Pointer (SP) Register
371
P0, P1, P2, P3 – Input / Output Registers
8 bit
8 bit
8 bit
8 bit
8 bit

8051
Interrupts

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 by two
ways:
1. Interrupt
2. Polling
373

Interrupt
– Upon receiving an interrupt signal, the
microcontroller interrupts whatever it is doing
and serves the device.
– The program which is associated with the
interrupt is called the interrupt service routine
(ISR) .

Steps in Executing an Interrupt
1. It finishes the instruction it is executing and saves the address of
the next instruction (PC) on the stack.
2. It also saves the current status of all the interrupts internally (i.e:
not on the stack).
3. It jumps to a fixed location in memory, called the interrupt
vector table, that holds the address of the ISR.
4. The microcontroller gets the address of the ISR from the
interrupt vector table and jumps to it.
5. It starts to execute the interrupt service subroutine until it
reaches the last instruction of the subroutine which is RETI
(return from interrupt).
6. Upon executing the RETI instruction, the microcontroller returns
to the place where it was interrupted.
375

Steps in executing an interrupt
• Finish current instruction and saves the PC on stack.
• Jumps to a fixed location in memory depend on type
of interrupt
• Starts to execute the interrupt service routine until
RETI (return from interrupt)
• Upon executing the RETI the microcontroller returns
to the place where it was interrupted. Get pop PC
from stack

Interrupt Sources
• Original 8051 has 6 sources of interrupts
– Reset (RST)
– Timer 0 overflow (TF0)
– Timer 1 overflow (TF1)
– External Interrupt 0 (INT0)
– External Interrupt 1 (INT1)
– Serial Port events (RI+TI)
{Reception/Transmission of Serial Character}

8051 Interrupt related Registers
• The various registers associated with the use of
interrupts are:
– TCON - Edge and Type bits for External Interrupts 0/1
– SCON - RI and TI interrupt flags for RS232 {SERIAL
COMMUNICATION}
– IE - interrupt Enable
– IP - Interrupts priority
379

Enabling and Disabling an Interrupt
• The register called IE (interrupt enable) that is
responsible for enabling (unmasking) and disabling
(masking) the interrupts.
380

Interrupt Enable (IE) Register
• EA : Global enable/disable.
• --- : Reserved for additional interrupt hardware.
• ES : Enable Serial port interrupt.
• ET1 : Enable Timer 1 control bit.
• EX1 : Enable External 1 interrupt.
• ET0 : Enable Timer 0 control bit.
• EX0 : Enable External 0 interrupt.
MOV IE,#08h
or
SETB ET1
--
381

Interrupt Priority (IP) Register
PS PT1 PX1 PT0 PX0Reserved
Serial Port
Timer 1 Pin
INT 1 Pin Timer 0 Pin
INT 0 Pin
Priority bit=1 assigns high priority
Priority bit=0 assigns low priority
383

384

Comparison to
Programming
concepts with 8085.
385
Assembly language programs : 8085 & 8051
NOTE: Refer Unit 2 & Unit 5

UNIT-4
Peripheral
interfacing
386
Presented by
C.GOKUL,AP/EEE
Regulation : 2013

UNIT 4 Syllabus
 Introduction: Memory Interfacing & I/O interfacing
• 8255 PPI {Parallel communication interface}
• 8259 {Programmable Interrupt controller }
• 8253/8254 Timer – {Timer {or counter}}
• 8237/8257 {DMA controller}
• 8251 USART {Serial communication interface}
• 8279 {Keyboard /display controller}
• A/D and D/A Interface {ADC 0800/0809,DAC 0800}
[Interfacing with 8085 & 8051]

Introduction to
peripheral
interfacing
388

Data Transfers
 Synchronous ----- Usually occur when
peripherals are located within the same
computer as the CPU. Close proximity
allows all state bits change at same
time on a common clock.
 Asynchronous ----- Do not require that
the source and destination use the
same system clock.
389

MEMORY DEVICES I/O DEVICES
Presented by C.GOKUL,AP/EEE , Velalar College of Engg & Tech, Erode
390

 interface memory (RAM, ROM, EPROM'...)
or I/O devices to 8086 microprocessor.
Several memory chips or I/O devices can
connected to a microprocessor. An address
decoding circuit is used to select the
required I/O device or a memory chip.
391

IO mapped IO V/s Memory Mapped
IO
Memory Mapped IO
 IO is treated as memory.
 16-bit addressing.
 More Decoder Hardware.
 Can address 216=64k
locations.
 Less memory is available.
IO Mapped IO
 IO is treated IO.
 8- bit addressing.
 Less Decoder
Hardware.
 Can address 28=256
locations.
 Whole memory address
space is available.
392

Memory Mapped IO
• Memory Instructions are
used.
• Memory control signals
are used.
• Arithmetic and logic
operations can be
performed on data.
• Data transfer b/w register
and IO.
IO Mapped IO
• Special Instructions are
used like IN, OUT.
• Special control signals
are used.
• Arithmetic and logic
operations can not be
performed on data.
• Data transfer b/w
accumulator and IO.
393

Parallel communication
interface
INTEL 8255
Presented by C.GOKUL,AP/EEE , Velalar College of Engg & Tech, Erode
394

8255 PPI
• The 8255 chip is also called as Programmable
Peripheral Interface.
• The Intel’s 8255 is designed for use with Intel’s
8-bit, 16-bit and higher capability
microprocessors
• The 8255 is a 40 pin integrated circuit (IC),
designed to perform a variety of interface
functions in a computer environment.
• It is flexible and economical.
395

8255 PIO/PPI
 It has 24 input/output lines which may be
individually programmed.
 2 groups of I/O pins are named as
Group A (Port-A & Port C Upper)
Group B (Port-B & Port C Lower)
 3 ports(each port has 8 bit)
Port A lines are identified by symbols PA0-PA7
Port B lines are identified by symbols PB0-PB7
Port C lines are identified by PC0-PC7 , PC3-PC0
ie: PORT C UPPER(PC7-PC4) , PORT C LOWER(PC3-PC0)
398

D0 - D7: data input/output lines for the
device. All information read from and
written to the 8255 occurs via these 8 data
lines.
CS (Chip Select). If this line is a logical 0, the
microprocessor can read and write to the
8255.
RESET : The 8255 is placed into its reset
state if this input line is a logical 1
399

• RD : This is the input line driven by the
microprocessor and should be low to
indicate read operation to 8255.
• WR : This is an input line driven by the
microprocessor. A low on this line
indicates write operation.
• A1-A0 : These are the address input lines
and are driven by the microprocessor.
400

Control Logic
 CS signal is the master Chip Select
 A0 and A1 specify one of the two I/O Ports
CS A1 A0 Selected
0 0 0 Port A
0 0 1 Port B
0 1 0 Port C
0 1 1 Control
Register
1 X X 8255 is not
selected
401

Block Diagram of 8255
(Architecture)
It has a 40 pins of 4 parts.
1. Data bus buffer
2. Read/Write control logic
3. Group A and Group B controls
4. Port A, B and C
403

1. Data bus buffer
 This is a tristate bidirectional buffer used
to interface the 8255 to system data bus.
Data is transmitted or received by the
buffer on execution of input or output
instruction by the CPU.
404

2. Read/Write control logic
 This unit accepts control signals ( RD, WR ) and
also inputs from address bus and issues
commands to individual group of control blocks
( Group A, Group B).
 It has the following pins.
CS , RD , WR , RESET , A1 , A0
405

3. Group A and Group B controls
• These block receive control from the CPU
and issues commands to their respective
ports.
Group A - PA and PCU ( PC7 –PC4)
Group B – PB and PCL ( PC3 –PC0)
a) Port A: This has an 8 bit latched/buffered
O/P and 8 bit input latch. It can be
programmed in 3 modes – mode 0, mode 1,
mode 2.
406

b) Port B: It can be programmed in mode 0,
mode1
c) Port C : It can be programmed in mode 0
407

Modes of Operation of 8255
 Bit Set/Reset(BSR) Mode
 Set/Reset bits in Port C
 I/O Mode
 Mode 0 (Simple input/output)
 Mode 1 (Handshake mode)
 Mode 2 (Bidirectional Data Transfer)
409

B3 B2 B1
Bit/pin of port C
selected
0 0 0 PC0
0 0 1 PC1
0 1 0 PC2
0 1 1 PC3
1 0 0 PC4
1 0 1 PC5
1 1 0 PC6
1 1 1 PC7
Concerned only with the 8-bits of Port C.
Set or Reset by control word
Ports A and B are not affected
411

a) Mode 0 (Simple Input or Output):
• Ports A and B are used as Simple I/O
Ports
• Port C as two 4-bit ports
• Features
– Outputs are latched
– Inputs are not latched
– Ports do not have handshake or
interrupt capability
2. I/O MODE
412

b) Mode 1: (Input or Output with
Handshake)
• Handshake signals are exchanged
between MPU & Peripherals
• Features
– Ports A and B are used as Simple I/O Ports
– Each port uses 3 lines from Port C as
handshake signals
– Input & Output data are latched
– interrupt logic supported
414

c) Mode 2: Bidirectional Data Transfer
• Used primarily in applications such as data
transfer between two computers
• Features
– Ports A can be configured as the bidirectional
Port
– Port B in Mode 0 or Mode 1.
– Port A uses 5 Signals from Port C as handshake
signals for data transfer
– Remaining 3 Signals from Port C Used as –
Simple I/O or handshake for Port B
415

Find control word
(1) Port A: output with handshake
(2) Port B: input with handshake
(3) Port CL: output (4)Port CU: input
 Solution:
1 0 1 0 1 1 1 0 = AEH
416

 Port A: Output, Port B: Output,
 Port CU: Output, Port CL: Output
Solution:
1 0 0 0 0 0 0 0 = 80H
The control word register for the above ports of Intel
8255 is 80H.
417

 Port A: Input, Port B: Input,
 Port CU: Input, Port CL: Input
Solution:
1 0 0 1 1 0 1 1 = 9BH
The control word register for the above ports of intel
8255 is 9BH.
418

1. This IC is designed to simplify the implementation of the interrupt interface in the 8088
and 8086 based microcomputer systems.
2. This device is known as a ‘Programmable Interrupt Controller’ or PIC.
3. It is manufactured using the NMOS technology and It is available in 28-pin DIP.
4. The operation of the PIC is programmable under software control (Programmable)and it
can be configured for a wide variety of applications.
5. 8259A is treated as peripheral in a microcomputer system.
6. 8259A PIC adds eight vectored priority encoded interrupts to the microprocessor.
7. This controller can be expanded without additional hardware to accept up to 64
interrupt request inputs. This expansion required a master 8259A and eight 8259A
slaves.
8. Some of its programmable features are:
· The ability to accept level-triggered or edge-triggered inputs.
· The ability to be easily cascaded to expand from 8 to 64 interrupt-inputs.
· Its ability to be configured to implement a wide variety of priority schemes.
8259 Programmable Interrupt Controller (PIC)

ASSINGMENT OF SIGNALS FOR 8259:
1. D7- D0 is connected to microprocessor data bus D7-D0 (AD7-AD0).
2. IR7- IR0, Interrupt Request inputs are used to request an interrupt and to connect to a slave
in a system with multiple 8259As.
3. WR - the write input connects to write strobe signal of microprocessor.
4. RD - the read input connects to the IORC signal.
5. INT - the interrupt output connects to the INTR pin on the microprocessor from the master,
and is connected to a master IR pin on a slave.
6. INTA - the interrupt acknowledge is an input that connects to the INTA signal on the system.
In a system with a master and slaves, only the master INTA signal is connected.
7. A0 - this address input selects different command words within the 8259A.
8. CS - chip select enables the 8259A for programming and control.
9. SP/EN - Slave Program/Enable Buffer is a dual-function pin.

When the 8259A is in buffered mode, this pin is an
output that controls the data bus transceivers in a
large microprocessor-based system.
When the 8259A is not in buffered mode, this pin
programs the device as a master (1) or a slave (0).
CAS2-CAS0, the cascade lines are used as outputs from
the master to the slaves for cascading multiple
8259As in a system.

Microprocessors and Microcontrollers Document Summary

Microprocessors and Microcontrollers Document Summary

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