Instruction Set of 8085 MCQ Quiz in தமிழ் - Objective Question with Answer for Instruction Set of 8085 - இலவச PDF ஐப் பதிவிறக்கவும்

Explanation:

Here two machine codes are being executed:

1. Opcode Fetch

2. Data to be uploaded in the accumulator

We can use the following instruction set:

MVI A, data

1 This is a 2-byte instruction.

2 Immediate addressing mode is used.

3 Total machine cycle = Opcode Fetch + Memory Read

4 Total machine cycle = 4T + 3T = 7T

Where T = T-states.

Calculation:

f = 2 MHz

\(T=\frac{1}{f}=\frac{1}{2}=0.5 \ μ s\)

Total machine cycle = 7T

Total machine cycle = 7 × 0.5 = 3.5 μs 

Concept:

For 8051 Microcontroller.

\(CY = \left\{ {\begin{array}{*{20}{c}} {0\;\left[ {When\;no\;carry\;from\;MSB\;i.e.\;{D_7}} \right]}\\ {1\;\left[ {When\;carry\;from\;MSB\;i.e.\;{D_7}} \right]} \end{array}} \right.\)

\(AC = \left\{ {\begin{array}{*{20}{c}} {0\;\left[ {When\;no\;carry\;from\;{D_3}\;to\;{D_4}} \right]}\\ {1\;\left[ {When\;carry\;from\;{D_3}\;to\;{D_4}} \right]} \end{array}} \right.\)

\(P = \left\{ {\begin{array}{*{20}{c}} {1\;\left[ {When\;odd\;no.\;of\;1's\;present} \right]}\\ {0\;\left[ {When\;even\;no.\;of\;1's \; present} \right]} \end{array}} \right.\)

Explanation:

Instruction 1: MOV A#9C
This instruction moves the immediate value 9C into the accumulator (A). The binary representation of 9C is 1001 1100.

After executing this instruction, the status of the flags remains unchanged, as the MOV instruction does not affect the flags.

Instruction 2: ADD A, #64H
This instruction adds the immediate value 64H to the accumulator (A). The binary representation of 64H is 0110 0100.

To perform the addition, we start from the least significant bit (LSB) and move towards the most significant bit (MSB), considering the carry from each bit addition.

Adding the LSB (0 + 0), we get:
Carry = 0
Result = 0

Adding the next bit (0 + 1), we get:
Carry = 0
Result = 1

Adding the next bit (1 + 0), we get:
Carry = 0
Result = 1

Adding the next bit (1 + 0), we get:
Carry = 0
Result = 1

Adding the next bit (1 + 1), we get:
Carry = 1
Result = 0

Adding the next bit (1 + 0), we get:
Carry = 1
Result = 1

Adding the next bit (0 + 1), we get:
Carry = 1
Result = 0

Adding the most significant bit (0 + 1), we get:
Carry = 1
Result = 1

After performing the addition, the carry (CY) flag is set to 1, indicating a carry-out from the MSB. The auxiliary carry (AC) flag is set to 1, as there is a carry from bit 3 to bit 4. The parity (P) flag is set to 1, as the result has an even number of set bits (1s).

Therefore, the correct option is:
4) CY = 1, AC = 1, P = 1

• INR is a 1-byte instruction as it requires 1-Byte, 3-Machine Cycles (Opcode Fetch, Memory Read, Memory Write) and 10 T-States.
• OUT is a 2-byte instruction as it requires 2-Bytes, 3-Machine Cycles (Opcode Fetch, Memory Read, I/O Write) and 10 T-States.
• STA is a 2-byte instruction requires 3-Bytes, 4-Machine Cycles (Opcode Fetch, Memory Read, Memory Read, Memory Write) and 13 T-States.

Stack pointer

A stack pointer is a small register that stores the memory address of the last data element added to the stack or, in some cases, the first available address in the stack.

PUSH Command

Push operation refers to inserting an element in the stack. Since there's only one position at which the new element can be inserted into the top of the stack, the new element is inserted at the top of the stack.

Explanation:

It is given that the stack pointer is pointing to the memory location 2000H. 

This means that 2000H is the memory location that is at the top of the stack.

After the execution of instruction PUSH D, a new element is inserted at the top of the stack. In this element, the data 1050H will be stored.

The memory address of the newly added element is given by 2000H + 0001H.

2000 H = 0010 0000 0000 0000

0001 H = 0000 0000 0000 0001

2000H + 0001H = 0001 1111 1111 1111

0001 1111 1111 1111 = 1FFEH

The stack pointer would be pointing at the memory address 1FFEH

The following instructions are run

MVI B, 00 H;         moves 00 H to register B ⇒ B = 00 H

MVIA, 1CH;           moves 1CH to register A ⇒ A = 1CH

DCR B;                    Decrements the contents of B (i.e. 00 H) by one. So B = FFH

DAA;                      DAA instruction changes the binary values of contents of the accumulator to BCD.

I. If the value of the lower four bits in the accumulator is greater than 9 or if the AC flag is set, the instruction adds 0110 to the lower four bits

II. If the value of the higher four bits in the accumulator is greater than 9 or if cy flag is set, the instruction adds 0110 to higher-order 4-bits.

A = 1CH = 0001 1100, the lower order bits are greater than 9, therefore DAA adds 0110 to lower bits to adjust the binary result to BCD.

\({\rm{A}}:1{\rm{\;CH}} = \begin{array}{*{20}{c}} {\underline {\begin{array}{*{20}{c}} {0001}&{1100}\\ {}&{0110} \end{array}} }\\ {\begin{array}{*{20}{c}} {0010}&{0010} \end{array}} \end{array}\)

So, A = 22 H

STA TEMP;       Stores the contents of the accumulator at memory location TEMP

∴ Contents of TEMP are 22 H

For the 8085A microprocessor, given instruction is LHLD 2100_H

This operation load registers L and H with the content in the memory at location 2100_H and 2101_H respectively

i.e. (H) ← M (2101_H) & (L) ← M (2100_H)

where L is lower address data and H is higher address data.

LDA 3000H → Fetch, Read, Read, Read (4 memory cycles)

LXI D, F0F1H → Fetch, Read, Read (3 memory cycles)

LDA 2000 H

⇒ 3 B instruction

⇒ OPFC + OPRC + OP2RC + MRC

⇒ 4 machine cycles

⇒ 4T + 3T + 3T + 3T

⇒ 13 T

LOOP: MVI C, 78 H; Moves 78 H to register C. So, C = 78 H

DCR C; Decreases the contents of C by 1.

JNZ LOOP; Jumps at LOOP if zero flag is not set.

HLT; Halts the execution of the program

The execution of the program remains in LOOP till the zero flag is set. The zero flag is set when the contents of C becomes zero. During each iteration of the loop, the contents of C are reduced by one. The initial contents of C are 78 H. The decimal equivalent of 78 H is 120. So, the loop is executed 120 times.

Concept:

Data transfer operations do not affect the status of flags.

Application:

MOV and MVI are data transfer instructions, they do not affect any flag.

SUB and CMP are arithmetic and Logical instructions that are executed in ALU and hence they do affect flags.