Diploma in Electrical Engineering Switchgear and Protection UNIT-2 Full Notes 📝 State Board Of Technical Education [SBTE] BIHAR

1. UNIT – 2
(BY: SANDEEP GANGWAR SIR)
INSTRUCTION SET AND EXECUTION IN 8085
An instruction is a command to the microprocessor to perform a given task on a specified data. Each
instruction has two parts: one is task to be performed, called the operation code (opcode), and the
second is the data to be operated on, called the operand. The operand (or data) can be specified in
various ways. It may include 8-bit (or 16-bit) data, an internal register, a memory location, or 8-bit (or
16-bit) address. In some instructions, the operand is implicit.
Classification based on functionality:
I. Data transfer operations: This group of instructions copies data from source to destination.
The content of the source is not altered.
II. Arithmetic operations: Instructions of this group perform operations like addition,
subtraction, increment & decrement. One of the data used in arithmetic operation is stored
in accumulator and the result is also stored in accumulator.
III. Logical operations: Logical operations include AND, OR, EXOR, COMPARE, ROTATE. The
operations like AND, OR and EXOR uses two operands, one is stored in accumulator and
other can be any register or memory location. The result is stored in accumulator. NOT
operation requires single operand, which is stored in accumulator.

2. IV. IV. Branching operations: Instructions in this group (JUMP, CALL, RETURN) can be used to
transfer program sequence from one memory location to another either conditionally or
unconditionally.
1. Jump Instructions – The jump instruction transfers the program sequence to the memory
address given in the operand based on the specified flag. Jump instructions are 2 types:
Unconditional Jump Instructions and Conditional Jump Instructions.
(a) Unconditional Jump Instructions: Transfers the program sequence to the described
memory address.
JMP address Jumps to the address Ex- JMP 2050
(b) Conditional Jump Instructions: Transfers the programs sequence to the described
memory address only if the condition in satisfied.
2. Call Instructions – The call instruction transfers the program sequence to the memory
address given in the operand. Before transferring, the address of the next instruction after
CALL is pushed onto the stack. Call instructions are 2 types: Unconditional Call Instructions
and Conditional Call Instructions.
(a) Unconditional Call Instructions: It transfers the program sequence to the memory
address given in the operand.
CALL address Unconditionally calls Ex - CALL 2050
(b) Unconditional Call Instructions: It transfers the program sequence to the memory
address given in the operand.
3. Return Instructions – The return instruction transfers the program sequence from the
subroutine to the calling program. Return instructions are 2 types: Unconditional Jump
Instructions and Conditional Jump Instructions.

3. (a) Unconditional Return Instruction: The program sequence is transferred unconditionally
from the subroutine to the calling program.
RET Return from the subroutine unconditionally Ex – RET
(b) Conditional Return Instruction: The program sequence is transferred unconditionally
from the subroutine to the calling program only is the condition is satisfied.
V. Machine control operations: Instruction in this group (HLT, EI, DI) control execution of other
instructions and control operations like interrupt, halt etc.
Classification based on length:
I. One-byte instructions: Instruction having one byte in machine code. (MOV B,C)
II. I. Two-byte instructions: Instruction having two byte in machine code. (MVI B,23H)
III. II. Three-byte instructions: Instruction having three byte in machine code. (LDA 1234H)

4. Addressing Modes
The process of specifying the data to be operated on by the instruction is called addressing. The various
formats for specifying operands are called addressing modes. The 8085 has the following five types of
addressing:
1. Immediate addressing
2. Direct addressing
3. Register direct addressing
4. Indirect addressing
5. Implicit addressing
Immediate Addressing:

5. In this mode, the operand given in the instruction - a byte or word – transfers to the destination register
or memory location.
Ex: MVI A, 9AH
The operand is a part of the instruction.
The operand is stored in the register mentioned in the instruction.
Direct addressing:
Memory direct addressing moves a byte or word between a memory location and register. The memory
location address is given in the instruction.
Ex: LDA 850FH
This instruction is used to load the content of memory address 850FH in the accumulator.
Register Direct Addressing:
Register direct addressing transfer a copy of a byte or word from source register to destination register.
Ex: MOV B, C
It copies the content of register C to register B.
Indirect Addressing:
Indirect addressing transfers a byte or word between a register and a memory location.
Ex: MOV A, M
Here the data is in the memory location pointed to by the contents of HL pair. The data is moved to the
accumulator.
Implicit Addressing:
In this addressing mode the data itself specifies the data to be operated upon.
Ex: CMA
The instruction complements the content of the accumulator. No specific data or operand is mentioned
in the instruction.

6. NOTE: Some special instructions of 8085
PUSH - Push Two bytes of Data onto the Stack
POP - Pop Two Bytes of Data off the Stack
XTHL - Exchange Top of Stack with H & L
SPHL - Move content of H & L to Stack Pointer
IN - Initiate Input Operation
OUT - Initiate Output Operation
EI - Enable Interrupt System
DI - Disable Interrupt System
HLT – Halt
NOP - No Operation
INSTRUCTION EXECUTION:
Each instruction in 8085 microprocessor consists of two part- operation code (opcode) and operand. The
opcode is a command such as ADD and the operand is an object to be operated on, such as a byte or the
content of a register.
Instruction Cycle: The time taken by the processor to complete the execution of an instruction. An
instruction cycle consists of one to six machine cycles.
Machine Cycle: The time required to complete one operation; accessing either the memory or I/O
device. A machine cycle consists of three to six T-states.
T-State: Time corresponding to one clock period. It is the basic unit to calculate execution of instructions
or programs in a processor.
To execute a program, 8085 performs various operations as:
Opcode fetch
Operand fetch
Memory read/write
I/O read/write
External communication functions are:

7. Memory read/write
I/O read/write
Interrupt request acknowledge
NOTE:
The 8085 microprocessor has 5 (seven) basic machine cycles. They are
Opcode fetch cycle (4T)
Memory read cycle (3 T)
Memory write cycle (3 T)
I/O read cycle (3 T)
I/O write cycle (3 T)
Interrupts in 8085
Interrupts are the signals generated by the external devices to request the microprocessor to perform a
task. There are 5 interrupt signals, i.e. TRAP, RST 7.5, RST 6.5, RST 5.5, and INTR.
Interrupt are classified into following groups based on their parameter –
• Vector interrupt − In this type of interrupt, the interrupt address is known to the processor. For
example: RST7.5, RST6.5, RST5.5, TRAP.
• Non-Vector interrupt − In this type of interrupt, the interrupt address is not known to the processor
so, the interrupt address needs to be sent externally by the device to perform interrupts. For example:
INTR.
• Maskable interrupt − In this type of interrupt, we can disable the interrupt by writing some
instructions into the program. For example: RST7.5, RST6.5, RST 5.5
• Non-Maskable interrupt − In this type of interrupt, we cannot disable the interrupt by writing some
instructions into the program. For example: TRAP.
• Software interrupt − In this type of interrupt, the programmer has to add the instructions into the
program to execute the interrupt. There are 8 software interrupts in 8085, i.e. RST0, RST1, RST2, RST3,
RST4, RST5, RST6, and RST7.
• Hardware interrupt − There are 5 interrupt pins in 8085 used as hardware interrupts, i.e. TRAP, RST7.5,
RST6.5, RST5.5, INTA.

8. Note –
INTA is not an interrupt; it is used by the microprocessor for sending acknowledgement.
TRAP has the highest priority, then RST7.5 and so on.

9. Programs:
Problem – 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.
Ans.
Explanation –
1. LDA 2050 moves the contents of 2050 memory location to the accumulator.
2. MOV H, A copies contents of Accumulator to register H to A
3. LDA 2051 moves the contents of 2051 memory location to the accumulator.
4. ADD H adds contents of A (Accumulator) and H register (F9). The result is stored in A
itself. For all arithmetic instructions A is by default an operand and A stores the
result as well
5. MOV L, A copies contents of A (34) to L
6. MVI A 00 moves immediate data (i.e., 00) to A

10. 7. ADC A adds contents of A(00), contents of register specified (i.e A) and carry (1). As
ADC is also an arithmetic operation, A is by default an operand and A stores the result
as well
8. MOV H, A copies contents of A (01) to H
9. SHLD 3050 moves the contents of L register (34) in 3050 memory location and
contents of H register (01) in 3051 memory location
10. HLT stops executing the program and halts any further execution
Problem – Multiply two 8 bit numbers stored at address 2050 and 2051. Result is stored
at address 3050 and 3051. Starting address of program is taken as 2000.
Ans.
Explanation – Registers used: A, H, L, C, D, E
1. LHLD 2050 loads content of 2051 in H and content of 2050 in L
2. XCHG exchanges contents of H with D and contents of L with E
3. MOV C, D copies content of D in C
4. MVI D 00 assigns 00 to D
5. LXI H 0000 assigns 00 to H and 00 to L

11. 6. DAD D adds HL and DE and assigns the result to HL
7. DCR C decrements C by 1
8. JNZ 200A jumps program counter to 200A if zero flag = 0
9. SHLD stores value of H at memory location 3051 and L at 3050
10. HLT stops executing the program and halts any further execution
Problem - Write an assembly language program to add two 16 bit numbers by using 8 bit
operation.
Ans.
Explanation –
1. LDA 2050 stores the value at 2050 in A (accumulator)
2. MOV B, A stores the value of A into B register
3. LDA 2052 stores the value at 2052 in A
4. ADD B add the contents of B and A and store in A
5. STA 3050 stores the result in memory location 3050
6. LDA 2051 stores the value at 2051 in A
7. MOV B, A stores the value of A into B register

12. 8. LDA 2053 stores the value at 2053 in A
9. ADC B add the contents of B, A and carry from the lower bit addition and store in A
10. STA 3051 stores the result in memory location 3051
11. HLT stops execution
Problem – Write an assembly language program in 8085 microprocessor to find sum of
digit of an 8 bit number.
Ans.
Explanation – Registers used A, B, C
1. LDA 2050 –loads the content of memory location 2050 in accumulator A
2. MOV B, A –moves the value of accumulator A in register B
3. ANI 0F –performs AND operation in value of accumulator A and 0F
4. MOV C, A –moves the value of accumulator A in register C
5. MOV A, B –moves the value of register B in accumulator A

13. 6. RLC –instruction rotate the value of accumulator A, left by 1 bit. Since it is performed 4
times therefore this will reverse the number i.e swaps the lower order nibble with
higher order nibble
7. Repeat step 3
8. ADD C –add the content of register of C in accumulator A
9. STA 3050 –stores value of A in 3050
10. HLT –stops executing the program and halts any further execution