Microprocessor 8085 complete

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Chapter 1
Introduction to Microprocessor

Introduction
ROM RAM
I/O
interface
I/O
devices
CPU
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Computer: A computer is a programmable machine that receives input, stores and
manipulates data/information, and provides output in a useful format.
Basic computer system consist of a CPU, memory and I/O unit.
Block diagram of a basic computer system
Address bus
Data bus Control
bus

Basic Concepts of Microprocessors
 Microcomputer:- It is a programmable machine. The two principal
characteristics of a computer are:
 Responds to a specific set of instructions in a well-defined manner.
 It can execute a prerecorded list of instructions (a program)
 Its main components are
 CPU
 Input & Output devices
 Memory
 Microprocessor:- It is a programmable VLSI chip which includes ALU, register
circuits & control circuits. Its main units are-
 ALU
 Registers
 Control Unit
 Microcontroller:- Silicon chip which includes microprocessor, memory & I/O in
a single package
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Block Diagram
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Microcomputer
Microprocessor Microcontroller

Definition of the Microprocessor
 Microprocessor is a Programmable, Clock driven, Register based,
Electronic device that reads instruction from a storage device, takes the
data from input unit and process the data according to the instructions
and provides the result to the output unit.
 Programmable- Perform Different set operation on the data depending on the
sequence of instructions supplied by the programmer.
 Clock Driven – Whole task is divided into basic operations, are divided into
precise system clock periods.
 Register Based – Storage element
 Electronic Device – fabricated on a chip
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Microprocessor Based System with bus
Architecture
 ALU:- Arithmetic and logical operations like add, subtraction, AND & OR.
 Register Array: - Store data during the execution of program.
 Control Unit: Provides necessary timing & control signal. It controls the flow of data
between microprocessor and peripherals.
*Microprocessor is one component of microcomputer.
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• Memory:
 Stores information such as instructions and data in binary format (0 and 1).
 Sub-system” of microprocessor-based system. sub-system includes the registers
inside the microprocessor .
 Read Only Memory (ROM): used to store programs that do not need
alterations.
Random Access Memory (RAM) (R/WM): used to store programs
that can read and altered like programs and data.
• Input/output: Communicates with the outside world.
• System Bus: Communication path between the microprocessor and peripherals.
 group of wires to carry bits.

How does a Microprocessor works
 To execute a program, the microprocessor “reads” each
instruction from memory, “interprets” it, then “executes or
perform” it.
 The right name for the cycle is
 Fetch
 Decode
 Execute
 This sequence is continued until all instructions are
performed.
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Machine Language
 To communicate with computer, instruction is given in binary language.
 MP has 8 bit data so 2^8 = 256 combinations. So difficult to write
programs in set of 0’s and 1’s .
for eg:
 0011 1100 – is an instruction that increments the number in accumulator by 1
 1000 0000– is an instruction that add the number in register B to the accumulator
content , and keep the result in A.
So it is very tedious and error inductive. For convenience, written in Hexadecimal
code. For example 0011 1100 is equivalent to 3C H
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8085 Assembly Language
 Even program is written in Hexa decimal .. It is difficult to understand.
 Program is written in mnemonic.
For E.g.: binary code 0011 1100 (3C H in hexadecimal) is represented by INR
A
 INR A –INR stands for Increment, A is accumulator… this symbol suggest the
operation of incrementing the accumulator by 1
Similarly 1000 0000 is equivalent ( 80 H) is represented as
 ADD B– ADD stands for addition and B represents content in register B. this symbol
suggest that addition of the number in register B to the accumulator content , and
keep the result in A.
***So MP has 246 such bit pattern amounting to 74 different instruction for
performing various operations
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Chapter 2
Microprocessor Architecture & Pin
diagram

Features of Microprocessor- 8085
 8085 is developed by INTEL
 8 bit microprocessor: can accept 8 bit data simultaneously
 Operates on single +5V D.C. supply.
 Designed using NMOS technology
 6200 transistor on single chip
 It provides on chip clock generator, hence it does not require external clock
generator.
 Operates on 3MHz clock frequency.
 8bit multiplexed address/data bus, which reduce the number of pins.
 16address lines, hence it can address 2^16 = 64 K bytes of memory
 It generates 8 bit I/O addresses, hence it can access 2^8 = 256 I/O ports.
 5 hardware interrupts i.e. TRAP, RST6.5, RST5.5, RST4.5, and INTR
 It provides DMA.
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Internal Architecture (functional block diagram)of 8085

8085 Architecture…….cont…
8085 architecture consists of following blocks:
1. Register Array
2. ALU & Logical Group
3. Instruction decoder and machine cycle encoder, Timing and control
circuitry
4. Interrupt control Group
5. Serial I/O control Group
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8085 Architecture …… cont….
1. Registers Array : 14 register out of which 12 are 8 bit capacity and 2 of 16 bit. Classify into 4 types
(a) General purpose register: (user accessible)
 B,C,D,E,H,L are 8 bit register.(can be used singly)
 Can also be used for 16-bit register pairs- BC, DE & HL.
 Used to store the intermediate data and result
 H & L can be used as a data pointer(holds memory address)
(b) Special Purpose Register[A, Instruction Register and Flag]
(b.1) Accumulator (A): (user accessible)
 8 bit register
 All the ALU operations are performed with reference to the contents of Accumulator.
 Result of an operation is stored in A.
 Store 8 bit data during I/O transfer
(b.2) Instruction Register: (user not accessible)
 When an instruction is fetched from memory, it is loaded in IR. Then transferred to the
decoder for decoding.
 It is not programmable and can not be accessed through any instruction.
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(b.3) Flag Register(F): (user accessible)
 8 bit Register
 Indicates the status of the ALU operation.
 ALU includes 5 flip flop, which are set or reset after an operation
according to data conditions of the result in the accumulator.
(Flag Register)
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Flag Register…… cont….
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Flag Significance
C or CY (Carry) CY is set when an arithmetic operation generates a carry
out, otherwise it is 0 (reset)
P (Parity) P= 1; if the result of an ALU operation has an even number
of 1’s in A;
P= 0; if number of 1 is odd.
AC (Auxiliary carry) Similar to CY,
AC= 1 if there is a carry from D3 to D4 Bit
AC= 0 if there is a no carry from D3 to D4 Bit
(not available for user)
Z(zero) Z = 1; if result in A is 00H
0 otherwise
S(Sign) S=1 if D7 bit of the A is 1(indicate the result is -ive)
S= 0 if D7 bit of the A is 0(indicate the result is +ive)

(c) Temporary Register[ W, Z, Temporary data register]
 Internally used by the MP(user not accessible)
(c.1) W and Z register:
 8 bit capacity
 Used to hold temporary addresses during the execution of some
instructions
(c.2) Temporary data register:
 8 bit capacity
 Used to hold temporary data during ALU operations.
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(d) Pointer Register or special purpose [SP, PC]
(d.1) Stack Pointer(SP)
 16 bit address which holds the address of the data present at the top of the stack
memory
 It is a reserved area of the memory in the RAM to store and retrieve the temporary
information.
 Also hold the content of PC when subroutines are used.
 When there is a subroutine call or on an interrupt. ie. pushing the return address on a
jump, and retrieving it after the operation is complete to come back to its original
location.
(d.3) Program Counter(PC)
 16 bit address used for the execution of program
 Contain the address of the next instruction to be executed after fetching the instruction
it is automatically incremented by 1.
 Not much use in programming, but as an indicator to user only.
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In addition to register MP contains some latches and buffer
 Increment and decrement address latch
 16 bit register
 Used to increment or decrement the content of PC and SP
 Address buffer
 8 bit unidirectional buffer
 Used to drive high order address bus(A8 to A15)
 When it is not used under such as reset, hold and halt etc this buffer is used
tristate high order address bus.
 Data/Address buffer
 8 bit bi-Directional buffer
 Used to drive the low order address (A0 to A7) and data (D0 to D7) bus.
 Under certain conditions such as reset, hold and halt etc this buffer is used
tristate low order address bus.
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(2) ALU & Logical Group: it consists ALU, Accumulator, Temporary
register and Flag Register.
(a) ALU
 Performs arithmetic and logical operations
 Stores result of arithmetic and logical operations in accumulator
(b) Accumulator
 General purpose register
 Stores one of the operand before any arithmetic and logical
operations and result of operation is again stored back in
Accumulator
 Store 8 bit data during I/O transfer
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(2) ALU & Logical Group…………………..cont…………………
(c) Temporary Register
 8 bit register
 During the arithmetic and logical operations one operand is available in A
and other operand is always transferred to temporary register
For Eg.: ADD B – content of B is transferred into temporary register
before actual addition
(d) Flag Register
 Five flag is connected to ALU
 After the ALU operation is performed the status of result will be stored in
five flags.
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(3) Instruction decoder and machine cycle encoder, Timing and
control circuitry
(a) Instruction decoder and machine cycle encoder :
 Decodes the op-code stored in the Instruction Register (IR) and
establishes the sequence of events to follow.
 Encodes it and transfer to the timing & control unit to perform the
execution of the instruction.
(b) Timing and control circuitry
 works as the brain of the CPU
 For proper sequence and synchronization of all the operations of MP,
this unit generates all the timing and control signals necessary for
communication between microprocessor and peripherals.

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(4) Interrupt Control group
 Interrupt:- Occurrence of an external disturbance
 After servicing the interrupt, 8085 resumes its normal working sequence
 Transfer the control to special routines
 Five interrupts: - TRAP, RST7.5, RST6.5, RST5.5, INTR
 In response to INTR, it generates INTA signal
(5) Serial I/O control Group
 Data transfer red on D0- D7 lines is parallel data
 But under some condition it is used serial data transfer
 Serial data is entered through SID(serial input data) input (received)
 Serial data is outputted on SOD(serial output data) input (send)

8085 Pin Diagram
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Pin Configuration Functional Pin diagram

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 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 and Data Bus.
 Status Signals.
 Control signal
 Interrupt signal
 Power supply and Clock signal
 Reset Signal
 DMA request Signal
 Serial I/O signal
 Externally Initiated Signals.
8085 Pin Description

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 Address Bus (Pin 21-28)
 16 bit address lines A0 to A15
 The address bus has 8 signal lines A8 – A15 which are unidirectional.
 The other 8 address lines A0 to A7 are multiplexed (time shared) with the 8
data bits.
 Data Bus (Pin 19-12)
 To save the number of pins lower order address pin are multiplexed with 8 bit
data bus (bidirectional)
 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 (T1 state), then during the late parts(T2 state) of the
execution, they carry the 8 data bits.
The Address and Data Busses

Status Signals
Status Pins – ALE, S1, S0
1. ALE(Address Latch Enable): (Pin 30)
 Used to demultiplexed the address and data bus
 +ive going pulse generated when a new operation is started by uP.
 ALE = 1 when the AD0 – AD7 lines have an address
 ALE = 0 When it is low it indicates that the contents are data.
 This signal can be used to enable a latch to save the address bits from the AD lines.
2. S1 and S0 (Status Signal): (Pin 33 and 29)
 Status signals to specify the kind of operation being performed .
 Usually un-used in small systems. S1 S0 Operation
0 0 HALT
0 1 WRITE
1 0 READ
1 1 FETCH

Control Signals
Control Pins – RD, WR, IO/M(active low)
1. RD: Read(Active low) (Pin 32)
 Read Memory or I/O device
 Indicated that data is to be read either from memory or I/P device and data bus is ready
for accepting data from the memory or I/O device.
2. WR: Write(Active low) (Pin 31)
 Write Memory or I/O device
 Indicated that data on the data bus are to be written into selected memory or I/P device.
3. IO/M: (Input Output/Memory-Active low) (Pin 34)
 Signal specifies that the read/write operation relates to whether memory or I/O device.
 When (IO/M=1) the address on the address bus is for I/O device
 When (IO/M=0) the address on the address bus is for memory
IO/M(active low) RD WR Control Signal Operation
0 0 1 MEMR M/M Read
0 1 0 MEMW M/M write
1 0 1 IOR I/O Read
1 1 0 IOW I/O Write

Control and status Signals
When S0, S1 is combined with IO/M(active low), we get status of
machine cycle
Z= Tristate, X = don’t care condition
IO/M S1 S0 OPERATION
Control
Signal
0 1 1 Opcode fetch RD = 0
0 1 0 Memory read RD= 0
0 0 1 Memory write WR = 0
1 1 0 I/O read RD = 0
1 0 1 I/O write WR = 0
1 1 0
Interrupt
Acknowledge
INTA = 0
Z 0 0 Halt
RD, WR = Z
and INTA =1
Z x x Hold
Z x x Reset

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 They are the signals initiated by an external device to request the
microprocessor to do a particular task or work.
 There are five hardware interrupts called, (Pin 6-11)
 On receipt of an interrupt, the microprocessor acknowledges the interrupt
by the active low INTA (Interrupt Acknowledge) signal.
Interrupts

Power supply and Clock Signal
Vcc (Pin 40) : single +5 volt power supply
Vss (Pin 20) : Ground
There are 3 important pins in this group.
 X0 and X1 :((Pin 1-2)
 Crystal or R/C network or LC network connections to set the
frequency of internal clock generator.
 The frequency is internally divided by two.
 Since the basic operating timing frequency is 3 MHz, a 6 MHz
crystal is connected to the X0 and X1 pins.
 CLK (output): (Pin 37)
 Clock Output is used as the system clock for peripheral and devices
interfaced with the microprocessor.

Reset Signals
 Reset In (input, active low) (Pin 36)
 This signal is used to reset the microprocessor.
 The program counter inside the microprocessor is set to
zero(0000H)
 The buses are tri-stated.
 Reset Out (Output, Active High) (Pin 3)
 It indicates MP is being reset.
 Used to reset all the connected devices when the
microprocessor is reset.

DMA Request Signals
 DMA:
 When 2 or more devices are connected to a common bus, to prevent the devices from
interfering with each other, the tristate gates are used to disconnect all devices except the one
that is communicating at a given instant .
 The CPU controls the data transfer operation between memory and I/O device.
 DMA operation is used for large volume data transfer between memory and an I/O device
directly.
 The CPU is disabled by tri-stating its buses and the transfer is effected directly by external
control circuits.
 HOLD (Pin 38)
 This signal indicates that another device is requesting the use of address and data bus.
 So it relinquish the use of buses as soon as the current machine cycle is completed.
 MP regains the bus after the removal of a HOLD signal
 HLDA (Pin 39)
 On receipt of HOLD signal, the MP acknowledges the request by sending
out HLDA signal and leaves out the control of the buses.
 After the HLDA signal the DMA controller starts the direct transfer of data.
 After the removal of HOLD request HLDA goes low.

Serial I/O Signals
These pins are used for serial data communication
 SID (input) Serial input data (Pin 4)
 It is a data line for serial input
 Used to accept serial data bit by bit from external device
 The data on this line is loaded into accumulator bit 7 whenever a
RIM instruction is executed.
 SOD (output) Serial output data (Pin 5)
 It is a data line for serial output
 Used to transmit serial data bit by bit to the external device
 The 7th bit of the accumulator is outputted on SOD line when SIM
instruction is executed.

Externally Initiated signal
 Ready (input) (Pin 35)
 Memory and I/O devices will have slower response compared to
microprocessors.
 Before completing the present job such a slow peripheral may not be
able to handle further data or control signal from CPU.
 The processor sets the READY signal after completing the present job
to access the data.
 It synchronize slower peripheral to the processor.
 The microprocessor enters into WAIT state while the READY pin is
disabled.

8085 Programming register and
programming model
 The register which are programmable and available for the use are six general
purpose register, A, F, PC, SP.
8085 programming model

Fetching & Exécution Cycles
 Fetching Cycles
The fetch cycle takes the instruction required from memory,
stores it in the instruction register, and
 Moves the program counter on one so that it points to the next
instruction.
 Execute cycle
 The actual actions which occur during the execute cycle of an
instruction.
 Depend on both the instruction itself and the addressing mode
specified to be used to access the data that may be required.
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Fetching an instruction
Step 1: Instruction pointer (program counter) hold the
address of the next instruction to be fetch.
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Step 2
40
Fetching an instruction….Cont….

Step 3
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Step 4
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Step 5
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Step 6
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Data flow from memory to MPU
Steps and data flow, when the
instruction code 01001111 (4FH
– MOV C, A) stored in the
location 2005H, is being fetch.
Fetch Cycle: To fetch the byte,
the MPU needs to identify the
memory location 2005 and
enable the data flow from
memory
Step 1: MPU places the 16 bit memory address from PC on the address bus
Step 2: Control unit send the signal RD to enable memory chip
Step 3: The byte from the memory location is placed on the data bus.
Step 4: The byte is placed on the instruction decoder of the MPU and task is carried
out according to the instruction.

Timing: Transfer of byte from memory to MPU
How a data byte is transfer from
memory to the MPU.
It shows the five different group
of signals with clock
Step 1: At T1 higher order memory
address 20H is placed on the A15 – A8
and the lower order memory address
05H is placed on the bus AD7-AD0,
and ALE signal high. IO/M goes
low(memory related signal).
Step 2: During T2 RD signal is sent
out. RD is active during two clock
periods.
Step 3 : During T3, Memory is enabled then instruction byte 4FH is placed on the data
bus and transferred to MPU. When RD goes high it causes the bus to go into high
impedance state.
Step 4: During T4, the machine code or byte is decoded by the instruction decoder and
content of A is copied into register.

Buses Structure
 Various I/O devices and memories are connected to CPU by a group of lines
called as bus.
8085 Bus structure

De-multiplexing AD7-AD0
 AD7– AD0 lines are serving a dual purpose and that they need to be
demultiplexed to get all the information.
 The high order bits(20 H) of the address remain on the bus for three clock
periods. However, the low order bits (05H) remain for only one clock
period and they would be lost if they are not saved externally. The low order
bits of the address disappear when they are needed most.
 To make sure we have the entire address for the full three clock cycles, we
will use an external latch to save the value of AD7– AD0 when it is carrying
the address bits. We use the ALE signal to enable this latch. ALE signal is
connected to the enable (G) pin of the latch.

De-multiplexing AD7-AD0
 Given that ALE operates
as a pulse during T1, ALE is
high the latch is
transparent; output
changes according to input.
So during T1 output of
latch is 05H.
 When ALE goes low, the
data byte 05H is latched
until the next ALE, the
output of the latch
represents the low order
address bus A7- A0 after
latching operation.

Generating Control Signals
Fig: Generate Read/write control signal
for memory and I/O
Signals are used both for memory and I/O
related operations. So four different
control signals are generated by combining
the signals RD, WR and IO/M.
MEMR = Reading from memory
MEMW = writing into memory
IOR = Reading from input port
IOW = writing to an output port
Fig: 8085 De-multiplexed address and data
bus with control signal

Instruction Decoding & Execution
Step 1: Places the content of data bus (4FH) in the IR and decode the instruction.
Step 2: Transfer the content of the accumulator(82H) to the temporary register in the
ALU.
Step 3: Transfer the content of temporary register to C register.
Assume A= 82 H
Instruction MOV
C,A(4FH) is fetched.
List of steps in decoding
& execution

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Chapter 3
Instruction Set and addressing
modes

8085 Instruction
Since the 8085 is an 8-bit device it can have up to 28 (256)
instructions.
 However, the 8085 only uses 246 combinations that represent a total of 74
instructions.
 Most of the instructions have more than one format.
These instructions can be grouped into five different groups
Data Transfer Operations
Arithmetic Operations
Logic Operations
Branch Operations
Machine Control Operations

Instruction Formats
Each instruction has two parts.
Opcode (operation code)- The first part is the task or operation to be
performed.
Operand - The second part is the data to be operated on. Data can be
given in various form.
 It can specify in various ways: may include 8 bit/16 bit data, an internal register,
memory location or 8 bit /16 bit address.
For Eg.: MVI A,02H
Opcode Operand

Instruction word Size
An instruction is assembled in binary form(0,1) known as machine code or opcode.
Due to different ways of specifying data or operand the machine code are not same for
all the instruction
The size of an instruction signifies how much memory space is required to load an
instruction in the memory.8085 instructions are of following sizes:
One-byte or one word Instructions: opcode and operand in 8 bits
only i.e. one byte. Operand(s) are internal register and are coded into the
instruction.
e.g. MOV, ADD, ANA, SUB, ORA etc.
MOV A,B Move the content of B in A
01 111 000 = 78H 01 for MOV operation, 111 binary code for
register A and last 000 binary code for register B
Task Op- code Operand Binary
Code
Hex
Code
Copy the contents of the accumulator
in the register C.
MOV C,A 0100 1111 4FH
Add the contents of register B to the
contents of the accumulator.
ADD B 1000 0000 80H
Invert (compliment) each bit in the
accumulator.
CMA 0010 1111 2FH

Instruction Format or Size….Cont…
 Two-byte instructions: first byte is opcode in 8 bits and second byte is
operand either 8 bit data or 8 bit address.
e.g. MVI, ADI, ANI, ORI, XRI etc.
MVIB, 05H Move 05 in register B
06,05 MVIB, 05H in the code for m
Task Op- code Operand Binary
Code
Hex
Code
Load an 8-bit data byte in
the accumulator.
MVI A, Data 0011 1110
DATA
3E (First Byte)
Data(Second
Byte)
Mnemonics Hex code
MVI A, 32H 3E 32H

Instruction Format or Size….Cont…
 Three-byte instructions: first byte is opcode in 8 bits and second and
third byte are operand either 16 bit data or 16 bit address.
Operand 1 = lower 8 bit data/address
Operand 2 = Higher 8 bit data/address
opcode + data byte + data byte
e.g. LXI, LDA, STA, LHLD, SHLD etc.
LXIH , 2500H load HL pair with 2500H
21,00,25 LXIH , 2500H in code form
Task Op-
code
Operand Binary
Code
Hex
Code
Hex
Code
Transfer the program
sequence to the memory
Location 2085H.
JMP 2085H 1100 0011
1000 0101
0010 0000
C3
85
20
First Byte
Second
Byte
Third Byte

Opcode Format
 Opcode contains the information regarding about operation, register
used, memory to be used etc.
 Fixed for each instruction.
Register
pair
Code
BC 00
DE 01
HL 10
A, F, SP 11
Register Code
B 000
C 001
D 010
E 011
H 100
L 101
M 110
A 111

Addressing Modes
The various formats for specifying operands/data are called the ADDRESSING MODES. 8085
instructions can be classified in following addressing modes
 Register Addressing mode
 Data is provided through the registers .e.g. MOV, ADD, SUB, ANA, ORA, XRA etc.
For Ex. MOV A, B move the content of register B to register A.
78H opcode of MOV A,B
 Immediate Addressing mode
 Data is present in the instruction. Load the immediate data to the destination provided. e.g. MVI, LXI,
ADI, SUI, ANI, ORI etc.
For Ex. MVI A, 05H move the 05H in register A.
3E, 05 opcode of MVI A, 05H
 Direct Addressing mode
 Instructions have their operands in memory and the 16-bit memory address is specified in the
instruction
 The address of the operand is given in the instruction. e.g. LDA, STA, LHLD, SHLD etc.
For Ex.
STA 7500 store the content of accumulator in the memory location 7500 H
32 , 00, 75 opcode of STA 7500
IN 01 Read data from port 01.
DB,01 opcode of IN 01

Addressing Modes…..Cont…
Indirect Addressing mode
In this address of the operand is specified by the register pair. The address
stored in the register pair points to memory location e.g. LDAX, STAX,
PUSH, POP etc.
LXI H, 7500 load HL pair with 7500H
MOV A, M move the content of memory location,
whose address is in HL pair to the accumulator
Implicit or Implied Addressing mode
The operand is not specified in the instruction, specified within the opcode
itself.
Operand is supposed to be present in accumulator. e.g. CMA, CMC, STC
etc.
RAL Rotate the content of accumulator towards left.

Instruction Set
 Since the 8085 is an 8-bit device it can have up to 28 (256)
instructions.
 However, the 8085 only uses 246 combinations that represent a
total of 74 instructions.
 Most of the instructions have more than one format.
These instructions can be grouped into five different groups:
 Data Transfer Operations
 Arithmetic Operations
 Logic Operations
 Branch Operations
 Machine Control Operations

Instruction Set…..cont….
(A) Data Transfer operation
 These operations simply COPY the data from the source to the
destination.
 They transfer:
 Data between registers.
 Data Byte to a register or memory location.
 Data between a memory location and a register.
 Data between an I/O Device and the accumulator.
 The data in the source is not changed.
 Data transfer instructions never affect the flag bits.
e.g. LDA, STA, MOV, LDAX, STAX, MVI, LXI etc.

Instruction Set…..cont….
(B) Arithmetic Operation
 These instruction perform addition, subtraction and compare operations.
 These operations are always performed with accumulator as one of the
operands.
 The status of the result can be verified by the contents of the flag register.
Addition: Any 8-bit number, or the contents of a register or the contents of a
memory location can be added to the contents of the accumulator and the
sum is stored in the accumulator. The instruction DAD is an exception; it
adds 16-bit data directly in register pairs. Ex ADD, ADI
Subtraction - Any 8-bit number, or the contents of a register, or the contents
of a memory location can be subtracted from the contents of the accumulator
and the results stored in the accumulator. Subtraction is done by 2’s
compliment method and set carry flag to indicate borrow. SUB. SBI
Increment/Decrement - The 8-bit contents of a register or a memory
location can be incremented or decrement by 1. Similarly, the 16-bit contents
of a register pair (such as BC) can be incremented or decrement by 1.
INR,DCR.

Instruction Set… Cont….
(C) Logical Operation
Perform 8-bit basic logical operations with the content of the
accumulator
Logical instructions also modify the flag bits.
Op-codes for logical instructions include ANA, ANI, ORA, ORI, XRA,
XRI, CMA, CMC, RAL, RLC, RAR, RRC, CMP, CPI etc.
AND, OR Exclusive-OR - Any 8-bit number, or the contents of a register, or of a
memory location can be logically AND, Or, or Exclusive-OR with the contents of the
accumulator. The results are stored in the accumulator.
Rotate- Each bit in the accumulator can be shifted either left or right to the next
position.
Compare- Any 8-bit number, or the contents of a register, or a memory location can be
compared for equality, greater than, or less than, with the contents of the accumulator.
Complement - The contents of the accumulator can be complemented.

(D) Branch Operation
These instructions are used to transfer the program control:
To jump from one memory location to any other memory location within a
program
 From one program to another program called as a subroutine.
 Alters the sequence of program execution either conditionally or
unconditionally
 Unconditional branch instructions- Transfer the program to the specified label or
address JMP unconditionally i.e. without satisfying any condition.
Unconditional Program control instructions are
 Call & RET
 Conditional branch instructions -Transfer the program to the specified label or
address when certain condition is satisfied.
 JNC, JC, JNZ, JZ, JP, JM, JPE, JPO
 CNC, CC, CNZ, CZ, CP, CM, CPE, CPO
 RNC, RC, RNZ, RZ, RP, RM, RPE, RPO

(E) Machine Control Operation
These instructions include special instructions such as I/O data
transfer, perform machine related operation
 HLT – To halt the CPU
 NOP – To perform no operation
 SIM – To set the masking of hardware interrupts and serial
output data
 RIM – To read the status of interrupt mask and serial input
data
 EI – Enable Interrupt
 DI – Disable Interrupt

How to Write the Assemble Program
Write the instructions to load two hexadecimal
number 32H and 48H in registers A and B
respectively. Add two numbers and display the sum at
the LED output port PORT 1.
Divide the program into small steps.
 Load the numbers in register
 Add the number
 Display the sum at the LED output port PORT 1.
The steps listed in the problem, the sequence can be
represented in the block diagram, called flow chart
67

Write the Assemble Program..cont..
68
Start
Load Hex
Numbers
1
Add Numbers
2
Display Sum
3
End
4
Block 1: MVI A, 32H Load A with 32 H
MVI B, 48H Load B with 48 H
Block 2: ADD B Add two bytes and save sum in A
Block 3: OUT 01H Display accumulator content at
port 01H
Block 4: HALT End

Write the Assemble Program..cont..
69
Mnemonics Hex Code
MVI A, 32H 3E 2 Byte Instruction
32
MVI B, 48H 06 2 Byte Instruction
48
ADD B 80 1 Byte Instruction
OUT 01H D3 2 Byte Instruction
01
HALT 76 1 Byte Instruction

Storing in Memory and converting from
Hex code to binary Code
To store the program in R/W memory, assume the
memory ranges from 2000 H to 20FFH. Now to enter
the Program
1. Reset the system by pushing the RESET key.
2. Enter the first memory address using the HEX keys, where the program
should be written.(2000H)
3. Enter each machine code by pushing Hex keys. For Ex, to enter the first
machine code, push 3E and store keys. Store key, store the program in
memory location 2000H and upgrade the memory address t0 2001.
4. Repeat step 3 until the last machine code 76H.
5. Reset the system.
70
8085
Mnemonics
Hex Code
Binary
Code
Flowchart
Manual
lookup
Monitor
Program
To Memory
for Storage
Check the keys and convert
Hex code into binary code

Program Stored in memory as
71
Mnemonics Hex Code Memory Content Memory Address
MVI A, 32H 3E 0011 1110 2000
32 0011 0010 2001
MVI B, 48H 06 0000 0110 2002
48 0100 1000 2003
ADD B 80 1000 0000 2004
OUT 01H D3 1101 0011 2005
01 0000 0001 2006
HALT 76 0111 1110 2007
Eight Machine code, require eight memory location. MPU can only understand
binary code; everything else(Mnemonics, Hex code, comments) for the convenience
of human being.

Executing the Program
To execute the program, we need to tell the MPU
where the program begins by entering the memory
address 2000H.
 Push the execute Key, so MPU loads 2000H in PC.
 PC is transferred from the monitor program to our program.
 MPU begin to read the one machine cycle at a time, when it
fetches the complete instruction, it execute that instruction.
 For Eg. When it fetches the machine code stored in 2000 and
2001H and execute the instruction MVI A, 32H , thus load 32H in
register A.
 Continue to execute the instruction until it fetches the HLT
instruction.
72

How Does a MPU differentiate between
data and instruction code
MPU is a sequential machine. When the system is turns on, it
begins the execution of the code in memory. Now the question
is that how does the MPU differentiate between code and
data when both are binary numbers.
 MPU interprets the first byte , it fetches as an opcode.
 When the MPU is reset its PC is reset to 0000H. fetches the first
code from the location 0000H but in our example the memory
location is 2000H.
 From the previous program it fetches the 3EH .when it decode that
code it know that it is a two byte instruction. so it assume that
second code 32H is data.
 But if we forget to enter 32H and enter the next code 06H, instead
it , so MPU load 06H in the accumulator and interpret the next
code 48H as an opcode. And continue the execution in sequence.
We may get a totally unexpected result.
73

Introduction
 It represents the execution time taken by each instruction in a graphical format.
 It is the graphical representation of initiation of read/write and transfer of data
operations under the control of 3-status signals IO / M , S1, and S0. All the operation
is performed with respect to CLK signal.
 The combination of these 3-status signals identify read or write operation and
remain valid for the duration of the cycle.
Machine cycle status and control signal

Clock Signal - The 8085 divides the clock frequency provided at x1 and
x2 inputs by 2 which is called operating frequency.
Rise time and fall time
Single Signal - Single signal status is represented by a line. It may have
status either logic 0 or logic 1 or tri-state
T-State
1 Clock cycle
Ideal Practical

Group of signals - Group of signals is also called a bus. Eg:
Address bus, data bus
Machine cycle showing clock periods -

Cycles and States
From the above discussion, we can define terms that will become handy later on:
 T- State:
 One subdivision of an operation performed in one clock period.
 Each T states is precisely equal to one clock period.
 An instruction’s execution length is usually measured in a number of T-states.
 Machine Cycle:
 The time required to complete one operation of accessing memory, I/O, or
acknowledging an external request.
 This cycle may consist of 3 to 6 T-states.
 Various machine cycle in 8085 is
 Opcode fetch
 Memory read/write
 Input read/write
 Intereupt acknowledge
 Halt/hold
 Reset
Instruction Cycle:
 The time required to complete the execution of an instruction.
 In the 8085, an instruction cycle may consist of 1 to 6 machine cycles.

Processor Cycle
 The function of the microprocessor is divided into two cycle of the instruction
 Fetch
 Execute
 Number of instructions are stored in the memory in sequence.
 In the normal process of operation, the microprocessor fetches (receives or reads) and
executes one instruction at a time in the sequence until it executes the halt (HLT)
instruction.
 Thus, an instruction cycle is defined as the time required to fetch and execute an
instruction.
Instruction Cycle (IC) = Fetch cycle (FC) + Execute Cycle (EC)
Processor Cycle

Machine cycle 2 Machine cycle 5
Instruction cycle
Machine cycle1
T – State 1 T – State 2 T – State 3 T – State 6

The 8085 microprocessor has 7 basic machine cycles.
They are
1. Opcode fetch cycle (4T)
2. Memory read cycle or operand fetch(3 T)
3. Memory write cycle (3 T)
4. I/O read cycle (3 T)
5. I/O write cycle (3 T)
6. Interrupt Acknowledge
7. Bus Idle cycle

Opcode Fetch Machine Cycle
The first step of executing any instruction is the Opcode fetch cycle.
 In this cycle, the microprocessor brings in the instruction’s Opcode
from memory.
 To differentiate this machine cycle from the very similar “memory
read” cycle, the control & status signals are set as follows:
 IO/M=0(memory operation),
 s1 and s0 are both 1. (opcode fetch)
 This machine cycle has four T-states.
 The 8085 uses the first 3 T-states to fetch the opcode.
 T4 is used to decode and execute it.
 It is also possible for an instruction to have 6 T-states in an opcode fetch
machine cycle.

Timing Diagram for Opcode Fetch Machine Cycle
Timing Diagram of Opcode Fetch

Timing: Transfer of byte from memory to MPU
How a data byte is transfer from
memory to the MPU.
It shows the five different group of
signals with clock
Step 1: At T1 higher order memory
address 20H is placed on the A15 – A8
and the lower order memory address
05H is placed on the bus AD7-AD0, and
ALE signal high. IO/M goes low(memory
related signal).
Step 2: During T2 RD signal is sent out.
RD is active during two clock periods.
Step 3 : During T3, Memory is enabled then instruction byte 4FH is placed on the data
bus and transferred to MPU. When RD goes high it causes the bus to go into high
impedance state.
Step 4: During T4, the machine code or byte is decoded by the instruction decoder and
content of A is copied into register.

Data Flow
form
Memory to
MP
85

 This cycle is executed by the processor to read a data byte from
memory or to fetch operand in a multi byte instruction. For ex. 2 or 3
byte instruction because in 1 byte instruction the machine code is an
opcode; so operation is always an opcode fetch
 The instructions which have more than one byte word size will use
the machine cycle after the opcode fetch machine cycle.
The memory read machine cycle is exactly the same as the
opcode fetch except:
 IO/M= 0(memory operation),
 s1 = 1 and s0 = 0. (memory read)
 WR = 1 & RD = 0
 It only has 3 T-states
 First cycle is opcode fetch cycle.
 So this cycle requires
4T(opcode) +3T(memory read )=7 T states to execute.
Memory Read or Operand Fetch Machine Cycle

Timing Diagram of Memory Read Machine Cycle

Timing for execution of the instruction MVI A, 32H
MVIA, 32H is 2 byte instruction, so hex code for MVI A is 3E.
8085 decode the opcode and finds out a second byte need to be read
High Impedance state.

89
The execution times of the memory read machine cycle and the instruction
cycle are calculated as
Clock Frequency f = 2MHz
T state = clock period (1/f) = 0.5µsec
Execution time for opcode fetch = 4T * 0.5 = 2µsec
Execution time for Memory Read = 3T * 0.5 = 1.5µsec
Execution time for Instruction = 7T * 0.5 = 3.5µsec

Data Flow
form
Memory to
MP
90

The memory write machine cycle is executed by the processor
to write a data byte in a memory location.
The memory write machine cycle is exactly the same as the
memory read except:
 IO/M= 0(memory operation),
 s1 = 0 and s0 = 1. (memory read
 WR = 0 & RD = 1
First cycle is opcode fetch cycle.
So this cycle requires
4T(opcode) +3T(memory write )=7 T states to execute.
Memory Write Machine Cycle

Timing Diagram of Memory write Machine Cycle

Data Flow
form MPU
to Memory
93

Microprocessor executes this cycle to read content of I/O port
or to read 8 bit data present on an input ports through data bus.
The I/O read machine cycle is exactly the same as the memory
read except:
 IO/M= 1(I/O operation),
 s0 = 0 and s1 = 1. ( read)
 WR = 1 & RD = 0
This instruction reads the data from an input device and places
the data byte in the accumulator.
It accept the data from input device by using I/O read
signal(IOR).
Input/ Output Read Machine Cycle

Timing Diagram of I/O read Machine Cycle

Data Flow
form input
device to MP
96

Microprocessor executes this cycle to write a data into an I/O
port or to write 8 bit data present on an output ports.
The I/O write machine cycle is exactly the same as the memory
write except:
 s1 = 1 and s0 = 0. ( write)
 WR = 0 & RD = 1
This instruction places the content of the accumulator on the
data bus.
It transmit the data to output device by using I/O write
signal(IOW).
Input/output write Machine Cycle

Timing Diagram of I/O write Machine Cycle

Data Flow
form MPU to
output
device
99

Interrupt is hardware call of subroutine.
MPU executes this cycle to acknowledge INTR request
MPU determines the starting address of the interrupt service
routine by using interrupt information.
RSTn and CALL address is used as interrupt information.
Length of this cycle is 6T or 3T states.
So two interrupt acknowledge cycle
(A) Interrupt acknowledge of RSTn
(B) Interrupt acknowledge of CALL address
Interrupt Acknowledge Cycle

The RSTn instruction is one byte.
It is similar to opcode fetch cycle with
 INTA signal is activated instead of RD signal.
 s0 = 1 and s1 = 1.
Interrupt Acknowledge of RSTn

Timing Diagram of Restart instruction

The CALL instruction is Three byte. So MPU executes 3
interrupt acknowledge cycle.
The length of the first cycle is 6 T states while second and third
cycle is 3T states.
 PC is not incremented. IR is selected as destination register
during 1st interrupt acknowledge cycle and in 2nd and 3rd cycle Z
and W are selected as destination register
It is similar to opcode fetch cycle with
 s0 = 1 and s1 = 1.
Interrupt Acknowledge for CALL

Timing Diagram of CALL instruction

The MPU executes this cycle for internal operation. Few
situations where the machine cycle are neither read nor write.
 For Execution of DAD instruction: -
 this instruction add the content of a specified register pair to the
content of HL pair.
After opcode fetch, DAD requires extra 6 T states to add 16 bit
content .
This extra T states are divided into two machine cycle do not
involve any memory or I/O operation.
 During Internal Opcode generation(For TRAP and RST) –
 In response to TRAP interrupt, 8085 enters into a bus idle cycle
during which it invokes restart instruction, stores the content of
PC onto the stack and places 0024H on the PC.
Bus Idle Cycle

Timing Diagram of DAD instruction

When speed of memory system and I/O system are not
compatible with MPU timings, means longer time for read/
write data.
So MPU has to confirm whether a peripheral is to ready transfer
data op not
If READY pin s high the peripheral is ready. Otherwise it is
continue as long as READY is low.
During the wait state the content of the address bus, the data
bus and control bus are held constant.
The wait state gives an I/O port an extra clock time to output
valid data on the data bus.
WAIT State

Read Machine cycle with and without wait state

Introduction to 8085 based micro-computer

Microprocessor 8085 complete

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