Instructions of 8085 Microprocessor

Question 1
The clock frequency of an 8085 microprocessor is 5 MHz. If the time required to execute an instruction is 1.4 \mu s, then the number of T-states needed for executing the instruction is
A
1
B
6
C
7
D
8
GATE EC 2017-SET-1   Microprocessors
Question 1 Explanation: 
Given than,
f_{\mathrm{CLK}}=5 \mathrm{MHz}
Execution time =1.4 \mu \mathrm{s}
Execution time =n(T-\text { state })
n= number of T-states required to execute the instruction
T- state (or) T_{\mathrm{CLK}}=\frac{1}{f_{\mathrm{CLK}}}=0.2 \mu \mathrm{s}
So, \quad n=\frac{1.4 \mu \mathrm{s}}{T_{\mathrm{CLK}}}=\frac{1.4}{0.2}=7
Question 2
In an 8085 microprocessor, the contents of the accumulator and the carry flag are A7 (in hex) and 0, respectively. If the instruction RLC is executed, then the contents of the accumulator (in hex) and the carry flag, respectively, will be
A
4E and 0
B
4E and 1
C
4F and 0
D
4F and 1
GATE EC 2016-SET-3   Microprocessors
Question 2 Explanation: 
RLC: Rotate accumulator left by 1 bit without carry


Before RLC operation:
\begin{aligned} A &=A 7 H=(10100111)_{2} \\ C Y &=0 \end{aligned}
After RLC operation:
\begin{aligned} A &=(01001111)_{2}=4 \mathrm{FH} \\ \mathrm{CY} &=1 \end{aligned}
Question 3
In an 8085 system, a PUSH operation requires more clock cycles than a POP operation. Which one of the following options is the correct reason for this?
A
For POP, the data transceivers remain in the same direction as for instruction fetch (memory to processor), whereas for PUSH their direction B174has to be reversed.
B
Memory write operations are slower than memory read operations in an 8085 based system.
C
The stack pointer needs to be pre-decremented before writing registers in a PUSH, whereas a POP operation uses the address already in the stack pointer.
D
Order of registers has to be interchanged for a PUSH operation, whereas POP uses their natural order.
GATE EC 2016-SET-1   Microprocessors
Question 3 Explanation: 
The stack pointer needs to be pre-decremented before wiriting data into stack. For reading data from stack, such pre-decrement or pre-increment operations are not needed, as already stack pointer indicates the address of stack top from where the read operation takes place. Hence, PUSH operation requires more clock cycles than POP operation.
Question 4
In an 8085 microprocessor, which one of the following instructions changes the content of the accumulator?
A
MOV B,M
B
PCHL
C
RNZ
D
SBI BEH
GATE EC 2015-SET-2   Microprocessors
Question 4 Explanation: 
The only instruction that changes the contents of accumulator is SBI BEH.
Question 5
An 8085 assembly language program is given below. Assume that the carry flag is initially unset. The content of the accumulator after the execution of the program is
A
8 CH
B
64 H
C
23 H
D
15 H
GATE EC 2011   Microprocessors
Question 5 Explanation: 
MVI \;A,07\Rightarrow A=07H=(0000111)_{2}


\begin{aligned} \text { After } \mathrm{RLC}, \mathrm{A}&=(00001110)_{2}=\mathrm{OEH} \\ \mathrm{MOV} \mathrm{B}, \mathrm{A} &\Rightarrow \mathrm{B} \leftarrow \mathrm{A} \Rightarrow \mathrm{B}=\mathrm{OEH} \\ \mathrm{RLC} \quad &\Rightarrow \mathrm{A}=(00011100)_{2}=1 \mathrm{CH} \\ \mathrm{RLC} \quad &\Rightarrow \mathrm{A}=(00111000)_{2}=38 \mathrm{H} \\ \mathrm{ADDB} &\Rightarrow \mathrm{A} \leftarrow \mathrm{A}+\mathrm{B} \\ &\Rightarrow \mathrm{A}=38 \mathrm{H}+\mathrm{OEH}=46 \mathrm{H} \\ &\qquad=(01000110)_{2} \\ \mathrm{RRC} \quad &\Rightarrow \mathrm{A}=(00100011)_{2}=23 \mathrm{H} \end{aligned}
Question 6
For the 8085 assembly language program given below, the content of the accumulator after the execution of the program is
A
00H
B
45H
C
67H
D
E7H
GATE EC 2010   Microprocessors
Question 6 Explanation: 
\mathrm{MVI} \quad \mathrm{A}, 45 \mathrm{H} \Rightarrow \mathrm{A}=45 \mathrm{H}=01000101
\mathrm{MOV} \mathrm{B}, \mathrm{A} \Rightarrow \mathrm{B}=45 \mathrm{H}
\mathrm{STC} \Rightarrow \mathrm{CY}=1
\mathrm{CMC} \Rightarrow \mathrm{CY}=0
RAR


\begin{aligned} A &=00100010 \\ XR A B & \Rightarrow A \leftarrow(00100010) \oplus(01000101)\\ A&=01100111=67 \mathrm{H} \end{aligned}
Therefore, the content of the accumulator after the execution of the program is 67H.
Question 7
An 8085 executes the following instructions
2710 LXI H, 30A0 H
2713 DAD H
2414 PCHL
All address and constants are in Hex. Let PC be the contents of the program counter and HL be the contents of the HL register pair just after executing PCHL. Which of the following statements is correct ?
A
PC = 2715H
HL = 30A0H
B
PC = 30A0H
HL = 2715H
C
PC = 6140H
HL = 6140H
D
PC = 6140H
HL = 2715H
GATE EC 2008   Microprocessors
Question 7 Explanation: 
\begin{aligned} \mathrm{LXIH}, 30 \mathrm{AOH} & \Rightarrow \mathrm{HL}=30 \mathrm{AOH} \\ \mathrm{DADH} & \Rightarrow \mathrm{HL}=6140 \mathrm{H}\\ &(i.e. 30 \mathrm{AOH}+3 \mathrm{OAOH})\\ \mathrm{PCH} & \Rightarrow \mathrm{PC}=6140 \mathrm{H}\\ \end{aligned}
Therefore, contents are
P C=6140 H
H L=6140 H
Question 8
Consider an 8085-microprocessor system.
The following program starts at location 0100H.
LXI SP, OOFF
LXI H, 0701
MVI A, 20H
SUB M

If in addition following code exists from 019H onwards.
ORI 40 H
ADD M
What will be the result in the accumulator after the last instruction is executed ?
A
40 H
B
20 H
C
60 H
D
42 H
GATE EC 2005   Microprocessors
Question 8 Explanation: 
\begin{array}{l} 0109 H \text { ORI } 40 \mathrm{H} \\ 010 \mathrm{BH} \mathrm{ADD} \mathrm{M} \\ \text { Intial: } \mathrm{A}=00 \mathrm{H} \\ 0109 \mathrm{H}: \text { ORI } 40 \mathrm{H} \Rightarrow \mathrm{A} \leftarrow \mathrm{A}(\mathrm{OR}) 40 \mathrm{H}=40 \mathrm{H} \\ 010 \mathrm{BH}: \mathrm{ADDM} \Rightarrow \mathrm{A} \leftarrow \mathrm{A}+\mathrm{M}=40 \mathrm{H}+20 \mathrm{H}=60 \mathrm{H} \\ \therefore \quad \mathrm{A}=60 \mathrm{H} \end{array}
Question 9
Consider an 8085-microprocessor system.
The following program starts at location 0100H.
LXI SP, OOFF
LXI H, 0701
MVI A, 20H
SUB M
The content of accumulator when the program counter reaches 0109 H is
A
20 H
B
02 H
C
00 H
D
FF H
GATE EC 2005   Microprocessors
Question 9 Explanation: 
\begin{aligned} 0100H&: LXISP,00FFH\\ 0103H&: LXIH, 0107H\\ 0106H&: MVI A, 20H\\ 0108H&: SUBM \Rightarrow A\leftarrow A- M \end{aligned}
M contains the data of memory location whose
address is in HL pair.
Contents of HL pair =0107 \mathrm{H}
Contents of location 0106 \mathrm{H}= opcode of MVIA, Data
Contents of location 0107 \mathrm{H}=20 \mathrm{H}
\therefore \quad A-M=20 H-20 H=00 H
Question 10
It is desired to multiply the numbers 0AH by 0BH and store the result in the accumulator. The numbers are available in registers B and C respectively. A part of the 8085 program for this purpose is given below :
MVI A, 00H
LOOP:
------
------
-----
HLT
END
The sequence of instructions to complete the program would be
A
JNZ LOOP, ADD B, DCR C
B
ADD B, JNZ LOOP, DCR C
C
DCR C, JNZ LOOP, ADD B
D
ADD B, DCR C, JNZ LOOP
GATE EC 2004   Microprocessors
Question 10 Explanation: 
ADD B, OCR C, JNZ LOOP

There are 10 questions to complete.
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