Unit - 4

Architecture & Programming of 8085

Q1) Explain 8085 microprocessor architecture.

A1)

Fig. 8085 microprocessor architecture

8085 consists of the following functional units −

Accumulator

The accumulator is an 8-bit register used to perform arithmetic, logical, I/O & LOAD/STORE operations. It is connected to internal data bus & ALU.

Arithmetic and logic unit

The arithmetic and logic unit performs arithmetic and logical operations like Addition, Subtraction, AND, OR, etc. on 8-bit data.

General purpose register

8085 consists of six general purpose registers in 8085 processor that is B, C, D, E, H & L. Each register holds 8-bit data. These registers work in pair to hold 16-bit data and their pairing combination is like B-C, D-E & H-L.

Program counter

It is a 16-bit register used to store the memory address location of the next instruction to be executed. Microprocessor increments the program whenever an instruction is being executed, so that the program counter points to the memory address of the next instruction that is going to be executed.

Stack pointer

It is 16-bit register which works like stack, which gets incremented/decremented by 2 during push & pop operations.

Temporary register

It is an 8-bit register, which holds the temporary data of arithmetic and logical operations.

Flag register

It is an 8-bit register which holds 0 or 1 depending upon the result stored in the accumulator.

These are:

• Sign (S)
• Zero (Z)
• Auxiliary Carry (AC)
• Parity (P)
• Carry (C)

Instruction registers and decoder

It is an 8-bit register. When an instruction is fetched from memory then it is stored in the Instruction register. Instruction decoder decodes the information present in the Instruction register.

Timing and control unit

It provides timing and control signal to the microprocessor to perform operations. Following are the timing and control signals, which control external and internal circuits −

• Control Signals: READY, RD’, WR’, ALE
• Status Signals: S0, S1, IO/M’
• DMA Signals: HOLD, HLDA
• RESET Signals: RESET IN, RESET OUT

Interrupt control

It controls the interrupts during the process. When the microprocessor is executing the main program and whenever an interrupt occurs, the microprocessor shifts the control from the main program to process the incoming request. After the request is completed, the control returns to the main program.

There are 5 interrupt signals in 8085 microprocessors:

• INTR
• RST 7.5
• RST 6.5
• RST 5.5
• TRAP.

Serial Input/output control

It controls the serial data communication by using the two instructions: SID (Serial input data) and SOD (Serial output data).

Address buffer and address-data buffer

The content stored in the stack pointer and program counter is loaded into the address buffer and address-data buffer to communicate with the CPU.

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

Address bus and data bus

Data bus carries the data to be stored. It is bidirectional whereas address bus carries the location to where it should be stored, and it is unidirectional. It is used to transfer the data & Address I/O devices.

Q2) Explain Addressing modes of 8085.

A2)

Addressing modes of 8085

The instructions are used to transfer the data from one register to another register, from the memory to the register, and from the register to the memory without any change in the content.

Addressing modes in 8085 is divided into 5 groups −

      Immediate addressing mode

Here, the 8/16-bit data is specified in the instruction itself as one of its operand. For example: MVI B, 40F means 40F is copied into register B.

      Register addressing mode

Here, the data is copied from one register to another.

For example: MOV B, A: means data in register A is copied to register B.

      Direct addressing mode

Here, the data is directly copied from the given address to the register.

 For example: LDB 5008H: means the data at address 5008H is copied to register B.

      Indirect addressing mode

Here, the data is transferred from one register to another by using the address pointed by the register.

For example: MOV B, K: means data is transferred from the memory address pointed by the register K to the register B.

      Implied addressing mode

Here, it doesn’t require any operand; the data is specified by the opcode itself.

For example: CMP.

Q3) Explain Instruction sets of 8086 with examples.

A3)

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 246instructions.
• Each instruction is represented by an 8-bit binary value.
• These 8-bits of binary value is called Op-Code or Instruction Byte.

It is classified into 5 groups:

Data Transfer Instruction

These instructions transfer data between registers or between memory and registers. The instruction is used to copy data from source to destination while copying the contents of source cannot be modified.

For example  

MOV B,C     

MVI B,20H

LXI H, 4020H

LDA  6000H

Arithmetic Instructions

• Addition
• Subtraction
• Increment
• Decrement

For example:

ADD C

ADC M

ADI 40H

DAD HL

SUB B

SUI 20H

INR C

Logical Instructions

The logical operations are:

• AND
• OR
• XOR
• Rotate
• Compare
• Complement

For example

CMP C

ANA B

ANI 40H

XRA M

ORA C

RLC

Branching Instructions

These instructions alter the normal sequential flow conditionally or unconditionally

For example

JMP 2000H

RET

RST 6

Control Instructions

It controls the operation of microprocessor

For example

NOP

EI

SIM

Q4) Explain Programming in 8085.

A4)

Let us suppose that 10 data are stored in memory location from 9000H to 9009H. These data are to be added and let us store the result in memory location 9100H.

Algorithm:

Start.

• Load into register pair HL from memory location 9000H.
• Load into register pair DE from memory location 9100H.
• Move 09 into register C as counter.
• Move the content of memory M into accumulator A.
• Increment register pair HL by 1.
• Add content of memory M with the contents of the accumulator A.
• Decrement value of register C by 1.
• If no zero is present, go to step 6 else go to step 10.
• Store the content of accumulator A into memory location pointed by register pair DE.
• Terminate the program.

Program Code:

LXI H, 9000H

LXI D, 9100H

MVI C, 09H

MVI B, 00H

MOV A, M

UP:

INX H

ADD M

JNC DOWN

INR B

DOWN:

DCR C

JNZ UP

STAX D

MOV A, B

STA 9101H

HLT

Q5) Write an assembly language program to add two 8-bit numbers stored at address 2050 and address 2051 in 8085 microprocessor. The starting address of the program is taken as 2000. 

A5)

1. Load the first number from memory location 2050 to accumulator. 
2. Move the content of accumulator to register H. 
3. Load the second number from memory location 2051 to accumulator. 
4. Then add the content of register H and accumulator using “ADD” instruction and storing result at 3050
5. The carry generated is recovered using “ADC” command and is stored at memory location 3051

Program – 

Q6) Explain Data Transfer Instruction and Arithmetic instruction in 8085.

A6)

Data Transfer Instruction

These instructions transfer data between registers or between memory and registers. The instruction is used to copy data from source to destination while copying the contents of source cannot be modified.

For example  

MOV B,C     

MVI B,20H

LXI H, 4020H

LDA  6000H

Arithmetic Instructions

• Addition
• Subtraction
• Increment
• Decrement

For example:

ADD C

ADC M

ADI 40H

DAD HL

SUB B

SUI 20H

INR C

Q7) Explain Branching Instruction and Control instruction in 8085.

A7)

Branching Instructions

These instructions alter the normal sequential flow conditionally or unconditionally

For example

JMP 2000H

RET

RST 6

Control Instructions

It controls the operation of microprocessor

For example

NOP

EI

SIM