PROGRAMMING IN 8085

DEVELOPMENT OF PROGRAM

A program is a sequence of instructions written to tell a computer to perform a specific

function. The instructions are selected from the instruction set of the microprocessor. The

first step in writing a program is to think very carefully about the problem that you want

the program to solve. In other words, ask yourself many times, “What do I really want

this program to do?” If you don’t do this, you may write a program that works great but

does not do what you need it to do. As you think about the problem, it is a good idea to

write down exactly what you want the program to do and the order in which you want the

program to do it. At this point, you do not write down the instructions, you just divide the

given problem in small steps in terms of the operations the 8085 can perform. This

sequence of operations used to solve a programming problem is often called the

algorithm. For more complex problems, however, we develop a more extensive outline

before translating the algorithm into instructions. The most common way of representing

the algorithm for a program is flowchart. The steps necessary to write the program can be

represented in a pictorial format, called a flowchart. Generally, a flowchart is used for

two purposes: to assist and clarify the thinking process and to communicate the

programmer’s thoughts and logic to others. The below figure shows some of the common

flowchart symbols:

Arrow indicates the direction of the program execution.

Rectangle represents a process or an operation.

Diamond represents a decision making block.

Oval indicates the beginning or the end of the operation.

Double-sided represents a predefined process such as a sub-routine.

Rectangle

Circle with an represents a continuation (an entry or an exit) to a

Arrow different page.

Fig. 1 Flow Charting Symbols

Writing and Executing a program in a Single Board Microcomputer

Writing a simple program of adding two hexadecimal numbers 23H and 84H and saving

the result in a register in assembly language is illustrated below:

Algorithm

1. Load the 1st number 23H in one register.

2. Load the 2nd

number 84H in another register.

3. Add the contents of the two registers.

4. Save the result in any register.

5. End the program.

Flow Chart

The above steps can be represented in a pictorial format with the help of flowchart.

Fig.2 Flow Chart for a Program

START

Load 23H in

Register A

Load 84H in Register B

Add Register

B to A

Save sum in

Register C

STOP

Assembly Language

The steps described in the flowchart are translated into an Assembly Language as

follows:

Opcode Operand

1.

2.

3.

4.

5.

MVI

MVI

ADD

MOV

HLT

A, 23H

B, 84H

B

C, A

Machine Language

Now, the above mnemonics should be converted into machine language. By looking up

the machine code for each instruction in the instruction set, we can translate the program

into machine language as follows:

Mnemonics Machine Code (Hex)

1.

2.

3.

4.

5.

MVI A, 23H

MVI

ADD

MOV

HLT

3E}

23}

INSTRUCTION CYCLE:

An instruction is a command given to the computer to perform a specified operation on the given data.

Sequence of instructions written for a computer to perform a particular task is called program.

Program & data are stored in the memory. The microprocessor fetches one instruction from the

memory at a time & executes it. It executes all the instructions of the program one by one to produce

the final result. The necessary steps that a microprocessor carries out to fetch an instruction &

necessary data from the memory & to execute it constitute an instruction cycle.

In other words, an instruction cycle is defined as the time required completing the execution of an

instruction.

An instruction cycle consists of a fetch cycle & an execute cycle. The time required to fetch an opcode

(fetch cycle) is a fixed slot of time while the time required to execute an instruction (execute cycle) is

variable which depends on the type of instruction to be executed.

FETCH OPERATION:

The first byte of an instruction is its opcode. An instruction may be more than one byte long. The other

bytes are data of operand address. The program counter (PC) keeps the memory address of the next

instruction to be executed. In the beginning of a fetch cycle the content of the program counter, which

is the address of the memory location where opcode is available, is sent to the memory. The memory

places the opcode on the data bus so as to transfer it to the microprocessor. The entire operation of

fetching an opcode takes three clock cycles.

EXECUTE OPERATION:

The opcode fetched from the memory goes to the instruction register (IR). From the instruction

register it goes to the decoder circuitry which decodes the instruction. After the instruction is decoded,

execution begins. If the operand is in general purpose registers execution is immediately performed.

The time taken for decoding and execution is one clock cycle. If an instruction contains data or

operand and address which are still in the memory, the microprocessor has to perform some read

operations to get the desired data. After receiving the data it performs execute operation. A read cycle

is similar to a fetch cycle. In case of a read cycle the quantity received from the memory are data or

operand address instead of an opcode. In some instructions write operation is performed. In write cycle

data are sent from the microprocessor to the memory or an output device. Thus we see that in some

cases an execute cycle may involve one or more read or write cycles or both.

MACHINE CYCLE:

Machine cycle is defined as the time required completing the operation of accessing either memory or

I/O. In the 8085, the machine cycle may consist of three to six T states.

T-State:

T-State is defined as one sub-division of the operation performed in one clock period. These sub-

divisions are internal states synchronized with the system clock.

PROGRAMMING TECHNIQUES:

Microprocessor is very fast and accurate in processing the data. It is more efficient than human beings

when it is required to perform the repeated tasks. To perform repetitions tasks programmer must use

different programming techniques such as looping, indexing etc.

Looping:

Looping is the programming technique which is used to tell the processor to repeat the task. A loop

can be constructed by using jump instructions. Continuous loops can be constructed by using

unconditional jump instructions. Unless you reset the system continuous loop does not stop repeating

the tasks. For example: modulo ten counter which will be discussed later.

Conditional loops can be constructed by using conditional jump instructions. The specified tasks will

be repeated only when the conditions are met. For example: the delay loop, which will be discussed in

the next section.

INDEXING:

Indexing is the programming technique in which objects will be sorted in a sequential manner. In this

data bytes are stores in memory locations sequentially & those data bytes are referred to by their

memory address.

PROGRAMS

1.Write an Assembly Language Program to add 2- 16 bit numbers:

LHLD 9501H

XCHG

LHLD 9503H

MVI C, 00

DAD D

JNC LOOP1

INR C

LOOP1 SHLD 9505H

MOV A, C

STA 9507H

HLT

2.Write an ALP to add n- 8 bit numbers:

MVI D, 00

MVI C, 05

LXI H, 8030

MOV A, M

DCR C

AHEAD INX H

ADD M

JNC LOOP-1

INR D

LOOP-1 DCR C

JNZ AHEAD

STA 8056

MOV A, D

STA 8051

HLT

3.Write an ALP to perform 32-bit addition:

LXI H, 8500

MOV C, M

INX H

LXI D, 8600

XRA A

LOOP-1 LDAX D

ADC M

MOV M, A

INX H

INX D

DCR C

JNZ LOOP-1

MVI A, 00

RAL

MOV M, A

RST 5

4.Write an ALP to subtract 2-8 bit numbers:

LXI H, 8501

MOV A, M

INX H

SUB M

JNC LOOP-1

INR C

LOOP-1 INX H

MOV M, A

MOV A, C

INX H

MOV M, C

RST5

5.Write an ALP to ADD 2-BCD numbers:

LXI H, 8A00

MVI D, 00

MOV A, M

INX H

ADD M

DAA

JNC LOOP-1

INR D

LOOP-1 STA 8A03

MOV A, D

STA 8A04

RST5

6.Write an ALP to SUBTRACT 2-16 BIT NUMBERS:

LHLD 8100

XCHG

LHLD 8102

MOV A, E

SUB L

STA 8104

JNC LOOP

DCR D

LOOP MOV A, D

SUB H

STA 8105

RST5

7.Write an ALP to multiply 2-8bit numbers:

LXI H, 8A00

MOV B, M

XRA A

MOV C, A

INX H

AHEAD ADD M

JNC LOOP-1

INR C

LOOP-1 DCR C

JNZ AHEAD

INX H

MOV M, A

INX H

MOV M, C

RST5

8.Write an ALP to perform division of 1-8 bit number by another 8-bit number:

LXI H, 8900H

MOV A, M

INX H

MOV B, M

MVI C, FF

LOOP INR C

SUB B

JNC LOOP

ADD B

STA 8902H

MOV A, C

STA 8903H

HLT

TIME DELAYS:

Counters are constructed using software instructions to keep track of the events. Since the

counting is performed at such high speed, only the last count can be observed. To observe all the

counts, there must be an appropriate time delay between counts.

Designing a delay is very simple. A register is loaded with a number, depending on the delay

required, and then the register is decremented until it reaches zero by setting up a loop with a

conditional JUMP instruction. The loop causes the delay, depending upon the clock period of the

system.

Single Register Delay:

A count is loaded in a register and the loop is executed until the count reaches zero. The set of

instructions necessary to set up a delay loop are:

LOOPMVI B, FF

DCR B

JNZ LOOP

To calculate the time delay we must consider the T-states required for each instruction, and for the

number of times the instruction are executed in the loop. The clock frequency of 8085 is 3MHZ.

Clock period T= 1/f = 1/3*106

= 0.33*10-6

Register B is loaded with FFH (25510) therefore the loop is repeated 255 times.

The time delay can be calculated as follows:

TL=(T*Loop T states*N10)

Where:

TL=Time Delay in loop

T= System Clock Period.

N10=Equivalent decimal number of Hexadecimal count loaded in the delay register.

DCR & JNZ forms a 0 loop with a total of 14 (4+10) T- states.

Therefore:

TL= (0.33*10-6

*14*255)

TL=1.1781 *10-3

Eg:

Write an ALP to display FF and 00 alternatively with a delay:

MVI A, FF

BACK PUSH PSW

STA 8FF1

CALL UPDDT

CALL DELAY

POP PSW

CMA

JMP BACK

HLT