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1. list the four categories of 8085 instructions that manipulate data.
Ans:
Data
transfer group – Example: MOV, MVI, LXI.
Arithmetic
group – Example: ADD, SUB, INR.
Logical
group – Example: ANA, XRA, CMP.
Branch
group – Example: JMP, JNZ, CALL.
Ans:
The
opcode is the instruction that is executed by the CPU and the operand is the
data or memory location used to execute that instruction.
In MOV HL, MOV is the operand and HL is the opcode.
3. Write the machine code of the instruction MOV HA if the opcode is (01)2. The register code for H is (100)2 and the register code for A is 1112 .
Ans: the machine code is 01 100 111= 67 H
Identify the opcodes and the operands.
a) MVI H,47 H
2647H
2byte instruction.
Opcode =MVI
Operands=47H
b) ADI F5H
C6F5H
2bytes
Opcode=ADI
Operands= A(implied),F5H
c) SUB C
91C
1 byte
Opcode=SUB
Operands=A(implied),C
5. Find the hex codes from the following instructions. Identify opcodes and operands, and show the order of the entering the codes in the memory Find the hex codes
a) STA 2050H
Hexcode= 322050H
Opcode=STA
Operands=2050H
b) JNZ 2070H
Hexcode=C22070H
Opcode= JNZ
Operands=2070H
6. Find the hex machine code for the following instructions from the instruction sets listed on the back cover, and identify the number of bytes in each instructions.
|
Instruction |
Hex code |
bytes |
address |
|
|
MVI B,4FH |
06 4F |
2 |
2000 |
|
|
MVI C,78H |
0E 78 |
2 |
2002 |
|
|
MOV A,C |
79 |
1 |
2004 |
|
|
ADD B |
80 |
1 |
2005 |
|
|
OUT 07H |
D3 07 |
2 |
2006 |
|
|
HLT |
76 |
1 |
2008 |
|
|
|
|
|
|
|
Instruction | Hex code | bytes | address |
|
MVI B,4FH | 06 4F | 2 | 2000 |
|
MVI C,78H | 0E 78 | 2 | 2002 |
|
MOV A,C | 79 | 1 | 2004 |
|
ADD B | 80 | 1 | 2005 |
|
OUT 07H | D3 07 | 2 | 2006 |
|
HLT | 76 | 1 | 2008 |
|
|
|
|
|
|
|
Instruction |
address |
Hex code |
|
MVI A,8FH |
2020 |
3E8F |
|
MVI B,68H |
2022 |
0668 |
|
SUB B |
2024 |
90 |
|
ANI 0FH |
2025 |
E60F |
|
STA 2070H |
2027 |
32 2070 |
|
HLT |
2030 |
76 |
|
Instruction |
Hex |
Address |
|
IN F2H |
DB F2 |
2000 |
|
CMA |
2F |
2002 |
|
ORA A |
B7 |
2003 |
|
JZ START |
CA0020 |
2004 |
10. Write the logical steps to add the following two numbers. Both the numbers should be saved for future use. Save the numbers as accumulator.
A2H and 18 H
=>
i) Load A2H in one register
ii) Load 18H in second register
iii) Copy A2H in accumulator
iv) Add the data of second register to the accumulator
v) End of the program.
11. Translate the program in 10 to assembly language.
MVI
B,A2H
MVI
C,18H
MOV
A,B
ADD C
HLT
12. Data byte 28 H is stored in register B and data byte 97 H is stored in the accumulator. Show the contents of B, C and the accumulator after the execution of the following instructions.
MOV
A,B
MOV
C,A
=>After
the execution of the program,
A=28
H
B=28H
C=28H
13. In q 6 explain the potential result of the program if the code 07 H of the out instruction is omitted.
=>If the 07 H is omitted, the processor considers the opcode of the next instruction 76H(HLT) as the address of the output port, and outputs the contents of the accumulator to the address 76 H, and continues to the next code which is indeterminate.
14. In Q 8 explain the possible outcome if the second byte 07FH of the instruction ANI OFH is omitted.
=>The processor takes the opcode 32 H of the next instruction as the second byte of the ANI instruction. Then it takes the next code 20H as the opcode of the next instruction and continues.
15. Given three sets of hex codes, identify the mnemonics.
=>
a) 3E = MVI A, F2H
32 = STA 2032H
76= HLT
c) 06 = MVI B, 82H
78=MOV A,B
32= STA 2050 H
FF= RST 7
d) 06 = MVI B, 4FH
0E =MVI C, 37 H
78 =MOV A,B
81 =ADD C
00 =NOP
32 =STA 2035H
76= HLT
16. Identify and explain the results of Q 15 a) and b)
=>
a) this program, result in storing F2H into 2032H
1) Load F2H in accumulator A
2) Store the data from A to memory location 2032 H
3) End of program
b) This program results in storing 82H into 2050 H
1 ) Load 82 H in register B
2) Copy the data in B to A
3) Store the data from A to memory location 2050 H
4) Execute the instruction at 0038 H memory location
17. In Q 15. C) what does the code 00 represent? Data, low order address or an opcode?
=>00H is the hex code of opcode NOP. There is no operation.
18. In Q 15 c), explain what does the program does, calculate the sum and identify the memory location where the sum is stored.
=>This
program gives the sum of 4FH and 37H and store it in 2035H.
The
sum is 87H and it is stored in 2035 H
, MICROPROCESSOR){kind=link}
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