m.sc. semester iv paper iii; unit i microprocessor 8085

Saikumar Chirra @ ECC - PrayagrajBy

Sources:- 1. Microprocessor 8085 – Ramesh Gaonkar

2. A course on 8085 Microprocessor – Prentice Hall

3. Microprocessors & its applications – P Raja

4. Microprocessor and Microcontroller Architecture and Interfacing

(Wikipedia)

M.Sc. Semester – IV

Paper – III; Unit – I

Microprocessor 8085 & Programs

Content :

1) Introduction to Microprocessor 8085

2) Architecture of Microprocessor 8085

3) Instruction set and

4) Simple programs

1) Introduction to Microprocessor 8085

Microprocessor is a Central Processing Unit (CPU) etched on a single chip.

A single Integrated Circuit (IC) has all the functional components of a CPU namely

Arithmetic Logic Unit (ALU), Control Unit and registers.

The 8085 microprocessor is an 8-bit processor that includes on its chip most of the logic

circuitry for performing computing tasks and for communicating with peripherals.

Features

8 bit microprocessor(8085 microprocessor can read or write or perform arithmetic and logical operations on 8-bit data at time)

It has 8 data lines and 16 address lines hence capacity is 216 = 64 kB of memory

Cock frequency is 3 MHz

It requires +5V power supply.

It is a single chip NMOS device implemented with 6200 transistors.

It provides 74 instructions with five addressing modes.

It provides 5 hardware interrupt and 8 software interrupts.

1. Introduction to …. Continue

Processor System Architecture

The typical processor system

consists of:

▪ CPU (central processing unit)

▪ ALU (arithmetic-logic unit)

▪ Control Logic

▪ Registers, etc…

▪ Memory

▪ Input / Output interfaces

Interconnections between these

units:

▪ Address Bus

▪ Data Bus

▪ Control Bus


Bus and CPU

Bus: A shared group of wires used for

communicating signals among devices

• address bus: the device and the location within

the device that is being accessed

• data bus: the data value being communicated

• control bus: describes the action on the address

and data buses

CPU: Core of the processor, where instructions

are executed

• High-level language: a = b + c

• Assembly language: add r1 r2 r3

• Machine language: 0001001010111010101

Memory and I/O

Memory: Where instructions (programs) and data are stored

• Organized in arrays of locations (addresses), each storing

one byte (8 bits) in general

• A read operation to a particular location always returns the

last value stored in that location

I/O devices: Enable system to interact with the world

• Device interface (a.k.a. controller or adapter) hardware

connects actual device to bus

• The CPU views the I/O device registers just like

memory that can be accessed over the bus. However,

I/O registers are connected to external wires, device

control logic, etc.

• Reads may not return last value written

• Writes may have side effects


2) Architecture of Microprocessor 8085

I. 8085 ARCHITECTURE

II. FUNCTION OF EACH BLOCK

III. 8085 BUS ORGANIZATION AND STRUCTURE

I. 8085 ARCHITECTURE (OR) BLOCK DIAGRAM

2). Architecture… Continue…

9

I. Intel 8085 CPU Block Diagram2). Architecture… Continue…

ALU

Register Sectiona) Temporary Register

b) General Purpose Register

c) Special Purpose Registers

a) Temporary Registers

• In 8085 available temporary register are temporary data register and W and Z

registers. These registers are not available to the programmer, but 8085 uses t

hem internally to hold temporary data during execution of some instructions.

• W and Z are two 8-bit temporary registers, used to hold 8-bit

data/address during execution of some instructions.

2). Architecture… Continue…II. FUNCTION OF EACH BLOCK

b). General Purpose Registers

B,C,D,E,H &L are used as general purpose register.

Each 8 bit long .

Apart from the above function these registers can also be used to work in pairs to

hold 16-bit data.

They can work in pairs such as B-C, D-E and H-L to store 16-bit data.

The H-L pair works as a memory pointer.

A memory pointer holds the address of a particular memory location.

They can store 16-bit address as they work in pair.


c). Special Purpose Register i. Accumulator

ii. Status or Flag Register

iii. Instruction Register

iv. Program Counter

v. Stack Pointer

i. Accumulator

• The accumulator is an 8-bit register that is a part of arithmetic/logic unit (ALU).

• This register is used to store 8-bit data and to perform arithmetic and logical operatio

ns.

• The result of an operation is stored in the accumulator. The accumulator is also ident

ified as register A.

• It also works as a via register for I/O accesses i.e. it reads data from input device and

similarly can transfer data to output device.



Flag register is a group of flip flops used to give status of different operations result.

The flag register is connected to ALU.

Once an operation is performed by ALU the result is transferred on internal data bus and status

of result will be stored in flip flops.

They are called Zero (Z), Carry (CY), Sign (S), Parity (P), and Auxiliary Carry (AC) flags.

A flag register contains a number of flags either to indicate certain conditions arising

after arithmetic and logical operations or to control certain operations.


The flags which indicate conditions are known as condition flags. The flags which are used to control certain

operations are called control flags.

A flag is a flip-flop which is set or reset by the processor to indicate certain conditions or set/reset by the

programmer to control certain operations.

• Carry flag(CY): If an operation performed in ALU generates the carry from D7 to next stage then CY flag is s

et, else it is reset.

• Auxiliary carry(AC): If an operation performed in ALU generates the carry from lower nibble (D0 to D3) to

upper nibble (D4 to D7) AC flag is set, else it resets.

• Zero flag(z): If an operation performed in ALU results 0value of entire 8-bits then zero flag is set, else it resets

• Sign flag(s): If MSB bit =0 then the number is positive, else it is negative.

• Parity flag(p):If the result contains even no. of ones this flag is set and for odd no. of ones this flag is reset.



II. FUNCTION OF EACH BLOCK

iii. Instruction Register

Temporary store for the current instruction of a program. Latest instruction sent here

from memory prior to execution. Decoder then takes instruction and decodes or interprets

the instruction. Decoded instruction then passed to next stage.

Instruction register is 8-bit register just like every other register of

microprocessor. Consider an instruction. The instruction may be anything like adding two

data's, moving a data, copying a data etc. When such an instruction is fetched from memory,

it is directed to Instruction register. So the instruction registers are specifically to store the

instructions that are fetched from memory.

There is an Instruction decoder which decodes the information's present in the Instruction

register for further processing.


iv. Program Counter

This 16-bit register deals with sequencing the execution of

instructions.

This register is a memory pointer.

Memory locations have 16-bit addresses, and that is why this is a 16-bit

register.

The microprocessor uses this register to sequence the execution of the

instructions.

The function of the program counter is to point to the memory address

from which the next byte is to be fetched. When a byte is being fetched,

the program counter is incremented by one to point to the next memory

location


v. Stack Pointer

The stack pointer is also a 16-bit register used as a memory pointer. It points to a memorylocation in R/W memory, called the stack. The beginning of the stack is defined by loading 16-bit address in the stack pointer.

Stack pointer is also a 16-bit register which is used as a memory pointer.

A stack is nothing but the portion of RAM (Random access memory).

Stack pointer maintains the address of the last byte that is entered into stack.

Each time when the data is loaded into stack, Stack pointer gets decremented.

Conversely it is incremented when data is retrieved from stack


ALU

The ALU performs the actual numerical and logic operation such as add,subtract, AND, OR, etc. Uses data from memory and from Accumulator toperform arithmetic. Always stores result of operation in Accumulator.

It is not accessible by user.

The word length of ALU depends upon of an internal data bus.

IT is 8 bit. It is always controlled by timing and control circuits.

It provides status or result of flag register.


Arithmetic and Logical Section

i. Instruction Decoder and Machine cycle Encoder

ii. Address Buffer

iii. Address/Data Buffer

iv. Increment/Decrement Address Latch

v. Interrupt Control

vi. Serial I/O Control Group

vii. Timing And Control


Instruction Decoder and Machine cycle Encoder

It accepts an OPCODE from instruction register, decodes it and gives

the decoded information to control logic.

The information includes what operation is to be performed, who is

going to perform it, etc. It means it will understand the instruction in

this block.

The decoded information is given to the timing and control unit that

provides control signals


Address Buffer

The contents of the stack pointer and program counter are loaded into the address buffer and address-data buffer.

These buffers are then used to drive the external address bus and address-data bus.

As the memory and I/O chips are connected to these buses, the CPU can exchange desired data to thememory and I/O chips.

The address-data buffer is not only connected to the external data bus but also to the internal data buswhich consists of 8-bits.

The address data buffer can both send and receive data from internal data bus.

The contents of the stack pointer and program counter are loaded into the address buffer and address-data buffer. These buffers are then used to drive the external address bus and address-data bus. As thememory and I/O chips are connected to these buses, the CPU can exchange desired data to the memoryand I/O chips.

The address-data buffer is not only connected to the external data bus but also to the internal data buswhich consists of 8-bits. The address data buffer can both send and receive data from internal data bus.


Increment/Decrement Address Latch

The 8-bit contents of a register or a memory location can be

incremented or decremented by 1.

This 16-bit register is used to increment or decrement the content of

program counter and stack pointer register by 1.

Increment or decrement can be performed on any register or a memory

location.


Interrupt Control

Consider that a microprocessor is executing the main program. Now wheneverthe interrupt signal is enabled or requested the microprocessor shifts thecontrol from main program to process the incoming request and after thecompletion of request, the control goes back to the main program.

For example an Input/output device may send an interrupt signal to notify thatthe data is ready for input. The microprocessor temporarily stops theexecution of main program and transfers control to I/O device. Aftercollecting the input data the control is transferred back to main program.


Serial I/O Control Group

The data transferred on to data bus is parallel data, but under

some conditions it is advantageous to use serial data transfer at

that time this control group is brought into application.

The input and output of serial data can be carried out using 2

instructions in 8085:

1. SID-Serial Input Data

2. SOD-Serial Output Data


Timing And Control

Timing and control unit is a very important unit as it synchronizes the registers and flow of data through

various registers and other units.

It accepts data from instruction decoder and generates micro steps to perform it ,in addition it accepts clock

inputs for synchronizing operations.

This unit consists of an oscillator and controller sequencer which sends control signals needed for internal and

external control of data and other units.

1. Control Signals: READY, RD(bar), WR(bar), ALE

2. Status Signals: S0, S1, IO/M(bar)

3. DMA Signals: HOLD, HLDA

4. RESET Signals: RESET IN(bar), RESET OUT.


Bus Organization

What is a Bus?

It is a set of pins, wires or signals having common functions asbus.

What is system Bus?

A bus is a bundle of wires that are grouped together to serve asingle purpose in microprocessor, generally there are three setsof communication lines that are called buses.

They are address bus, the data bus and control bus.


III. 8085 BUS ORGANIZATION AND STRUCTURE

Bus Structure


System Bus :-wires connecting memory &

I/O to microprocessor

Address Bus

Unidirectional

Identifying peripheral or memory location

Data Bus

Bidirectional

Transferring data

Control Bus

• Synchronization signals

• Timing signals

• Control signal

System Bus :-wires connecting memory &

I/O to microprocessor

Address Bus

Unidirectional

Identifying peripheral or memory location

Data Bus

Bidirectional

Transferring data

Control Bus

• Synchronization signals

• Timing signals

• Control signal


Address Bus

The bus over which the microprocessor sends out the address of a

memory location or I/O location is called as the address bus.

In 8085 microprocessor, Address bus is of 16 bits. This means that

Microprocessor 8085 can transfer maximum 16 bit address which

means it can address 65,536 different memory locations.

This bus is multiplexed with 8 bit data bus. So, the most significant

bits (MSB) of address goes through Address bus and LSB goes

through multiplexed data bus.

In 8085 address bus is 16 bit A0 –A15


Data Bus

It is bi-directional as Microprocessor requires to send orreceiver data. The data bus also works as bus in 8 bits long.

A data bus simply carries data. Internal buses carryinformation within the processor, while external buses carrydata between the processor and the memory. Typically, thesame data bus is used for both read/write operations.

When it is a write operation, the processor will put the data onto the data bus. When it is the read operation, the memorycontroller will get the data from the specific memory blockand put it in to the data bus.

The data bus in8085 is of 8 parallel lines D0 – D7


Control Bus

The control bus is used for sending control signals to the memory and

I/O devices. The CPU sends control signal on the control bus to enable

the outputs of addressed memory devices or I/O port devices.

Some of the control bus signals are as follows:

1.Memory read

2.Memory write

3.I/O read

4.I/O write.


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The 8085 MP Instruction set has 74 operation codes that results in 246 instructions.

The set includes all the 8080A instructions plus two additional (SIM & RIM related to serial

I/O) instructions.

There are 5 different kinds of I-sets:-

1. Data transfer/copy I-set:

2. Arithmetic I-set:

(*ADD *SUB *INR (increment1) *DCR (decrement1)

3. Logic and Bit Manipulation I-set:

*AND (ANA/I) *OR (ORA/I) *X-OR (XRA/I) *Compare(CMP/CPI)

4. Branch I-set:

5. Stack, Input/output and Machine control I-set:

3) Instruction set of 8085

33

Instruction format

Instructions typically consist of

Opcode (Operation Code)

◼ defines the operation to be performed (e.g. opcode for ADD B is 80H)

(Source/Destination) Operand

◼ what’s being operated on (e.g. In instruction MOV A, B the registers A (Destination) & B (Source) are

operabds)

The 8085 Instruction set typically have 0, 1 or 2 operands and consists of one,

two and three byte instructions. The first byte is always opcode.

(Writing a program using IS)

OpCode (1 byte) OpCode(1 byte)

OpCode (1 byte) Opd2 (1 byte)Opd1 (1 byte)

Opd (1 byte)

3. Instruction set Conti…

One-byte Instructions

Includes Opcode and Operand

in the same byte

Opcode Operand Binary Code Hex Code

MOV C, A 0100 1111 4FH

ADD B 1000 0000 80H

HLT 0111 0110 76H


MVI A, 32H 0011 1110

0011 0010

3EH

32H

MVI B, F2H 0000 0110

1111 0010

06H

F2H


LXI H, 2050H 0010 0001

0101 0000

0010 0000

21H

50H

20H

LDA 3070H 0011 1010

0111 0000

0011 0000

3AH

70H

30H

Two-byte Instructions

First byte specifies Operation Code

Second byte specifies Operand

Examples-

Three-byte Instructions

First byte specifies Operation

Code

Second & Third byte specifies

Operand

Examples-

35

Example (AVR instruction)

Subtraction with carry

Syntax: sbc Rd, Rr

Operation: Rd ← Rd – Rr – C

Rd: Destination register. 0 d 31

Rr: Source register. 0 r 31, C: Carry

Instruction format

OpCode uses 6 bits (bit 10 to bit 15).

Two operands share the remaining 10 bits.

0 0 0 0 1 0 r d r r r rd d d d

015


36

Instruction lengths

The number of bits an instruction has

For some machines – instructions all have the same length

E.g. MIPS machines

For other machines – instructions can have different lengths

E.g. M68K machine


37

Instruction encoding

Operation Encoding

2n operations needs at least n bits

Operand Encoding

Depends on the addressing modes and access space.

◼ For example: An operand in direct register addressing mode requires at most 3 bits if the the number of registers it can be stored is 8.

With a fixed instruction length, more encoding of operations means less available bits for encoding operands

Tradeoffs should be concerned


38

Example 1

A machine has:

16 bit instructions

16 registers (i.e. 4-bit register addresses)

Instructions could be formatted like this:

Maximally 16 operations can be defined.

OpCode Operand1 Operand2 Operand3


39

4) Simple programs

Program:

It is a sequence of instructions written to do specific task.

It instructs the microprocessor to perform a specific function.

Microprocessor understands Machine Language only!

Microprocessor cannot understand a program written

in Assembly language

A program known as Assembler is used to convert a

Assembly language program to machine language

Assembly

Language

Program

Assembler

Program

Machine

Language

Code

Assembly language program

to add two numbers

MVI A, 2H ;Copy value 2H in register A

MVI B, 4H ;Copy value 4H in register B

ADD B ;A = A + B

Note:

Assembly language is specific to a given

processor

For e.g. assembly language of 8085 is

different than that of Motorola 6800

microprocessor

Assembly Language of 8085

It uses English like words to

convey the action/meaning

called as MNEMONICS

For e.g.

MOV to indicate data transfer

ADD to add two values

SUB to subtract two values

4. Simple programs Conti…..

Graphic symbols used in the flow charts


Steps in Assembly Language Program

Steps to write a program

Analyze the Problem

Sample Problem

Make a Flowchart

Write program Instructions using Assembly language of 8085


Memory

Address (Hex)

Label Instruction Comments

Opcode Operands

XX00 As per need MOV B, C Move data from C register to B register

XX01 MOV A, M Move data from memory location M to

A register

Ex:

Program Format:

1. Program:

Add two 8-bit numbers and store 8-bit result in register C.

Analyze the Problem◼ Addition of two 8-bit numbers to be done

Sample Problem◼ Add two numbers

◼ Store result in register C

◼ Example

10011001 (99H) A

+00111001 (39H) D

11010010 (D2H) C


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

• Copy register D to A

• Add register E to A

• Copy A to register C

• Stop processing


Program


Memory

Address: Opcode: Operands: Comments:

MVI D, 2H ; Load 1st no. in register D

MVI E, 3H ; Load 2nd no. in register E

MOV A, D ; Copy register D to A

ADD E ; Add register E to A

MOV C, A ; Copy A to register C

HLT ; Stop processing

2 Program:

Add two 8-bit numbers. Result can be more than 8-bits (Use Carry).

Sample Problem

◼ Example

10011001 (99H) A

+10011001 (99H) B

100110010 (132H)

◼ The 9th bit in the result is called CARRY bit.


Analyze the Problem

Result of addition of two 8-bit

numbers can be 9-bit

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


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

Make a Flowchart

Start

Load Registers D, E

Copy D to A

Add A and E

Copy A to CStop

Is

Carry = 1 Clear B

Increment B

No


Yes

Program


Memory Mnemonics (OR)


MVI D, 2H ; Load 1st no. in register D

MVI E, 3H ; Load 2nd no. in register E

MOV A, D ; Copy register D to A

ADD E ; Add register E to A

MOV C, A ; Copy A to register C

JNC END ; Use Conditional Jump instructions

MVI B, 0H ; Clear register B

INR B ; Increment B


3 Program:

Calculate the sum of series of numbers. The length of the series is in memory location 2300H and series itself begins from memory

location 2301H. The result of sum is stored in memory location 2500H and the carry is stored on location 2501H.

Sample Problem

◼ Example

2300H = 04H

2301H = 9AH

2302H = 52H

2303H = 89H

2304H = 3EH

Result = 9A + 52 + 89 + 3E = 1B3 (1 will be carry )

2500H = B3H 2501H = 01H


Analyze the Problem

Let data of 4 numbers in 2301H to 23004H

Result 2500H and carry in 2501H

Make a Flowchart

Start

Sum = Sum + Pointer

Carry = Carry + 1

Stop

No


Yes

Sum = 0

Carry = 0

Pointer = 2301

Count = (2300H)

Is

Carry = 1

Pointer = Pointer + 1

Counter = Counter - 1

No

Yes

Is

Count = 0

Load (2500H) = Sum

(2501H) = Carry

Program


Memory


LDA 2300H

MOV C, A ; Initialize counter

LXI H, 23001H ; Initialize pointer

SUB A ; Sum = 0

MOV B, A ; Initialize carry = 0

BACK : ADD M ; Sum (L) = Sum + data

JNC Loop ; Check Carry

INR B ; Add carry

Loop : INX H ; Increment Pointer

DCR C ; Decrement counter

JNZ Back ; Check if counter not 0 repeat

STA 2500H ; Store Sum

MOV A, B

STA 2501H ; Store Carry


4 Program: Write an assembly language program in 8085 microprocessor to find maximum and

minimum of 10 numbers.

Sample Problem

1. Maximum number is stored in B register and minimum in C register

2. Load counter in D register

3. Load starting element in Accumulator, B and C register

4. Compare Accumulator and B register

5. If carry flag is not set then transfer contents of Accumulator to B. Else, compare Accumulator with

C register, if carry flag is set transfer contents of Accumulator to C

6. Decrement D register

7. If D>0 take next element in Accumulator and go to point 4

8. If D=0, store B and C register in memory

9. End of program


Analyze the Problem

List of numbers from 2050H to 2059H and output at 2060H and 2061H.

ADDRESS LABEL INSTRUCTION COMMENT

2000H LXI H, 2050H Load starting address of list

2003H MOV B, M Store maximum

2004H MOV C, M Store minimum

2005H MVI D, 0AH Counter for 10 elements

2007H LOOP MOV A, M Retrieve list element in Accumulator

2008H CMP B Compare element with maximum number

2009H JC MIN Jump to MIN if not maximum

200CH MOV B, A Transfer contents of A to B as A > B

200DH MIN CMP C Compare element with minimum number

200EH JNC SKIP Jump to SKIP if not minimum

2011H MOV C, A Transfer contents of A to C if A < minimum

2012H SKIP INX H Increment memory

2013H DCR D Decrement counter

2014H JNZ LOOP Jump to LOOP if D > 0

2017H LXI H, 2060H Load address to store maximum

201AH MOV M, B Move maximum to 2060H

201BH INX H Increment memory

201CH MOV M, C Move minimum to 2061H

201DH HLT Halt

Draw the Flow

Chart ?

m.sc. semester iv paper iii; unit i microprocessor 8085

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