Digital Electronics Test 5

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Digital Electronics Test 5

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Weekly Quiz Competition

Question 1
1 / -0

In a memory map the lowest address of an 8 k byte RAM is 1000 H. What is the highest address?

A
8192 H

B
2FFF H

C
7FFF H

D
3000 H

Solution

Size of RAM = 8 k byte
= 2³ × 2¹⁰ byte
= 2¹³ byte = (2000) H
Given starting address = (1000) H
Highest address = starting address + size of RAM – 1
= (1000) H + (2000) H - (0001) H
= (2FFF) H
Question 2
1 / -0

A single instruction to clear the lower four bits of the accumulator in 8085 assembly language is:

A
XRI 0FH

B
ANI F0H

C
XRI F0H

D
ANI 0FH

Solution

Let the contents of A are 25 H
ANI F0 H; Contents of A are ANDed with F0H and result is stored in A.
\(\begin{array}{*{20}{c}}{A:}&{0010\;0101}\\{F0H:}&{\frac{{1111\;0000}}{{\underline {0010\;0000} }}}\end{array}\)
= 20 H
So, A = 20 H
The contents of lower four bits of accumulator are cleared.
Question 3
1 / -0

How many times will the following loop be executed?
XRA A
MVI C 14 H
Loop: DCR C
JNZ Loop

A
Once

B
14 times

C
20 times

D
Depends on the initial value of A

Solution

XRA A; A = 00 H
MVI C, 14 H; C = 14 H
Loop: DCR C; decrements the value of C
JNZ Loop
As the value of C is 14 H, the loop executed 20 times.
Question 4
1 / -0

A 3 × 8 decoder with two enable inputs is to be used to address 8 blocks of memory. The size of each memory block when addressed form a sixteen bit bus with two MSBs used to enable the decoder will be _________ kB

A
2-2

B
24-21

C
27-26

D
29-22

Solution

The question is asking the size of each memory block
Total size of address bus = 16 bits
2 bits are used for enabling decoder
Number of bits available for addressing = 14
Total size of address that can be accessed is : 2¹⁴ bytes= 16 Kilo Bytes
This 16 Kilo Bytes of memory is divided among 8 memory block
Size of each memory block is 16/8 = 2 kB
Question 5
1 / -0

The following 8085 instructions are executed sequentially.
LDA      C200 H
MOV E, A
LDA C201 H
MOV    C, A
MVI      D, 00H
LXI        H, 0000 H
LOOP:       DAD    D
                   DCR    C
                   JNZ     Loop
                   SHLD C300 H
RST 1
The code shown above is to generate

A
Multiplication of two numbers

B
Multiple additions of the given number of times

C
Comparing two numbers which are in C200 H and C201 H and the result is stored in C300 H

D
Addition of two numbers

Solution

Let [C200 H] = 01 H
      [C201 H] = 02 H
LDA C200 H → [A] = 01 H
MOV E, A → [E] = 01 H
LDA C201 H → [A] = 02 H
MOV C, A → [C] = 02 H
MVI D, 00 H → [D] = 00 H
LXI H, 0000 H   → [HL] = 0000 H
Loop: DAD D →    [HL] = 000 + 0001 = 0001 H
           DCR C   → [C] = 0.2 H – 1 = 01 H
           JNZ loop → Jump to loop
Loop: Dad D → [HL] = 0001 + 0001 = 0002 H
           DCR C    → [C] = 01 H – 1 = 00 H
           JNZ Loop → goes to next line.
           SHLD C300 H → [C300 H] = 02 H
[C301 H] = 00 H
The result is the multiplication of two numbers.
Question 6
1 / -0

The following is an assembly language program for 8085 microprocessors:
When this program halts; the accumulator contains
Address
Instruction Code
Mnemonic
1000 H
3E, 06
MVI A, 06 H
1002 H
C6, 70
ADI 70 H
1004 H
32, 07, 10
STA 1007 H
1007 H
AF
XRA A
1008 H
76
HLT

A
00 H

B
06 H

C
70 H

D
76 H

Solution

MVI A, 06 H; A = 06 H
ADI 70 H; A = 06 H + 70 H = 76 H
STA 1007 H; (1007) → A = 76 H
XRA A; A = 0H
HLT
The accumulator value = 0H
Note: XRA clears the accumulator irrespective of the contents of the accumulator.
So, in the given program the contents of the accumulator are zero and it doesn’t depend on the previous instructions.
Question 7
1 / -0

8-bit signed integers in 2’s complement form are read in to the accumulator of an 8085 microprocessor from an I/O port using the following assembly language program segment with symbolic addresses.
BEGIN: IN PORT
RAL
JNC BEGIN
RAR
END HLT
This program

A
halts upon reading a negative number

B
halts upon reading a positive number

C
halts upon reading a zero

D
never halts

Solution

Let assume the accumulator loaded with 01 H = (0000 0001)₂
RAL: (Rotate left through carry)
After executing RAL,
For CY = 0, A = 0000 0010 and CY = 0
For CY = 1, A = 0000 0011 and CY = 0
As the carry is zero, JNC will make the program go to BEGIN and this will continue for infinite times.
Let assume the accumulator loaded with a negative number.
Port address = (1000 0000)₂
After executing RAL,
For CY = 0, A = 0000 0000 and CY = 1
For CY = 1, A = 0000 0001 and CY = 1
As CY = 1, JNC will not execute and the next instruction RAR will execute and then the program will halt.
So, whenever the program reads a negative number, it will jump out of the loop and get halted.

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Address	Instruction Code	Mnemonic
1000 H	3E, 06	MVI A, 06 H
1002 H	C6, 70	ADI 70 H
1004 H	32, 07, 10	STA 1007 H
1007 H	AF	XRA A
1008 H	76	HLT

Digital Electronics Test 5

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Digital Electronics Test 5

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Rank

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SHARING IS CARING

Question 1

8192 H

2FFF H

7FFF H

3000 H

Solution

Question 2

XRI 0FH

ANI F0H

XRI F0H

ANI 0FH

Solution

Question 3

Once

14 times

20 times

Depends on the initial value of A

Solution

Question 4

2-2

24-21

27-26

29-22

Solution

Question 5

Multiplication of two numbers

Multiple additions of the given number of times

Comparing two numbers which are in C200 H and C201 H and the result is stored in C300 H

Addition of two numbers

Solution

Question 6

00 H

06 H

70 H

76 H

Solution

Question 7

halts upon reading a negative number

halts upon reading a positive number

halts upon reading a zero

never halts

Solution

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