microprocesser

Top Government and Private Engineering Colleges / Institutes

2009-09-15T22:42:00.000-07:00

Top Government and Private Engineering Colleges / Institutes
Rank
Name of Institute
City
1
Indian Institute of Technology - IIT
Kharagpur
2
Indian Institute of Technology - IIT
Bombay
3
Indian Institute of Technology - IIT
Kanpur
4
Indian Institute of Technology - IIT
Delhi
5
Indian Institute of Technology - IIT
Madras
6
Birla Institute of Technology & Science - BITS
Pilani
7
Indian Institute of Technology - IIT
Roorkee
8
Institute of Technology, Banaras Hindu University - IT-BHU
Varanasi
9
Indian Institute of Technology - IIT
Guwahati
10
College of Engineering, Anna University
Chennai
11
Indian School of Mines University - ISMU
Dhanbad
12
Jadavpur University, Faculty of Engineering
Calcutta
13
Birla Institute of Technology, Mesra
Ranchi
14
Delhi College of Engineering - DCE
Delhi
15
National Institute of Technology - NIT
Trichy
16
National Institute of Technology - NIT
Warangal
17
PEC University of Technology
Chandigarh
18
National Institute of Technology - NIT
Suratkal
19
PSG College of Technology
Coimbatore
20
Bengal Engineering & Science University, Shibpur
Howrah
21
Hapar Institute of Engineering & Technology
Patiala
22
Motilal Nehru National Institute of Technology
Allahabad
23
Netaji Subhas Institute of Technology - NSIT
Delhi
24
Visvesvaraya National Institute of Technology - VNIT
Nagpur
25
Indian Institute of Technology - IIT
Hyderabad
26
Govt College of Engineering
Pune
27
Indian Institute of Information Technology - IIIT
Allahabad
28
National Institute of Technology - NIT
Rourkela
29
Harcourt Butler Tech Institute
Kanpur
30
Malaviya National Institute Of Technology - MNIT
Jaipur
31
Sardar Vallabhbhai National Institute of Technology - SVNIT
Surat
32
National Institute of Technology - NIT
Calicut
33
Maulana Azad National Institute of Technology - MANIT
Bhopal
34
Veermata Jijabai Technological Institute - VJTI
Mumbai
35
College of Engineering
Visakhapatnam
36
Manipal Institute of Technology
Manipal
37
National Institute of Technology - NIT
Kurukshetra
38
Coimbatore Institute of Technology
Coimbatore
39
Vellore Institute of Technology - VIT
Vellore
40
National Institute of Technology - NIT
NIT
41
Jawaharlal Nehru Technological University - JNTU
Hyderabad
42
National Institute of Technology - NIT
Durgapur
43
R.V College Of Engineering - RVCE
Bangalore
44
ICT Mumbai University
Mumbai
45
Dhirubhai Ambani Institute of Information and Communication Technology - DAIICT
Gandhinagar
46
Maharashtra Institute of Technology - MIT
Pune
47
University Visvesvaraya College of Engineering - UVCE
Bangalore
48
SSN College of Engineering
Chennai
49
UCE, Osmania University
Hyderabad
50
Nirma University of Science & Technology
Ahmedabad
51
College of Engineering
Trivandrum
52
Sardar Patel Coll of Engineering
Mumbai
53
National Institute of Technology - NIT
Hamirpur
54
PES Institute of Technology
Bangalore
55
SRM Institute of Science & Technology
Chennai
56
BMS College of Engineering
Bangalore
57
Amrita Institute of Technology & Science
Coimbatore
58
Shanmugha Arts, Science, Technology & Research Academy - SASTRA
Thanjavur
59
National Institute of Engineering
Mysore
60
Bangalore Institute of Technology
Bangalore
61
Chaitanya Bharathi Institute of Technology
Hyderabad
62
MEPCO Schlenk Engineering College
Sivakasi
63
ICFAI Institute of Science & Technology
Hyderabad
64
Jawaharlal Nehru Technological University - JNTU
Kakinada
65
SJ College of Engineering
Mysore
66
National Institute of Technology - NIT
Jalandhar
67
MS Ramaiah Institute of Technology
Bangalore
68
Satyabhama Engineering College
Chennai
69
Karunya University
Coimbatore
70
Sri Ramdeobaba KN Engineering College
Nagpur
71
Kongu Engineering College
Erode
72
National Institute of Technology - NIT
Raipur
73
SGS Institute of Technology & Science
Indore
74
National Institute of Technology - NIT
Patna
75
Indian Institute of Information Technology - IIIT
Bangalore

T.N.Engineering College(A)

2009-09-15T06:19:00.000-07:00

T.N.Engineering College(B)

2009-09-15T06:18:00.000-07:00

B.S. Abdur Rahman University (B.S. Abdur Rahman Crescent Engineering College), Chennai
Bharath University (Bharath Institute of Higher Education and Research), Chennai
B.K.R. College of Engineering and Technology, Arakkonam
Balaji Institute of Engineering and Technology, Thiruporur
Bannari Amman Institute of Technology, Erode
Bethlahem Institute of Engineering, Karungal
Bhajarang Engineering College, Tiruvallur Taluk
Bharath Niketan Engineering College, Theni
Bharathidasan Engineering College, Tirupattur Taluk
Bharathidasan Institute of Engineering and Technology, Trichy
Bharathiyar Institute of Engineering for Women, Attur Taluk

T.N.College of Engineering (C)

2009-09-15T06:17:00.000-07:00

T.N.College of Engineering(D)

2009-09-15T06:16:00.000-07:00

D
Dhaanish Ahmed College of Engineering, Chennai
Dhanalakshmi College of Engineering, Chennai
Dr. MGR University (Dr. M.G.R. Educational and Research Insitute), Chennai
D.M.I. College of Engineering, Kancheepuram
D.M.I. Engineering College, Kanyakumari
Dhanalakshmi Srinivasan College of Engineering and Technology, Mamallapuram
Dhanalakshmi Srinivasan Engineering College, Perambalur
Dr. G.U. Pope College of Engineering, Thoothukudi
Dr. Mahalingam College of Engineering and Technology, Pollachi
Dr. N.G.P. Institute of Technology, Coimbatore
Dr. Nagarathinam,s College of Engineering, Namakkal
Dr. Nalini Institute of Engineering and Technology, Dharapuram
Dr. Navalar Nedunchezhiyan College of Engineering, Cuddalore
Dr. Paul's Engineering College, Vanur Taluk Dr. Sivanthi Aditanar College of Engineering, Tuticorin

T.N.College of Engineering (E)

2009-09-15T06:15:00.000-07:00

Easwari Engineering College, Chennai
E.S. College of Engineering and Technology, Villupuram
EASA College of Engineering and Technology, Coimbatore
Edayathangudi G.S. Pillay Engineering College, Nagapattinam
Einstein College of Engineering, Tirunelveli Taluk
Er. Perumal Manimekalai College of Engineering, Hosur
Erode Builder Educational Trust,s Group of Institutions, Kangeayam
Erode Sengunthar Engineering College, Erode
Excel College of Engineering and Technology, Namakkal
Excel College of Engineering for Women, Komarapalayam Excel Engineering College, Komarapalayam

T.N.Engineering College(F)

2009-09-15T06:14:00.000-07:00

Francis Xavier Engineering College, Palayamkottai

T.N.College of Engineering(G)

2009-09-15T06:13:00.002-07:00

T.N. College of Engineering (H)

2009-09-15T06:13:00.001-07:00

Hindustan University (Hindustan College of Engineering), Chennai
Hindustan Institute of Technology, Coimbatore
Hindusthan College of Engineering and Technology, Coimbatore
Hindusthan Institute of Technology, Coimbatore

T.N.College of Engineering(I)

2009-09-15T06:12:00.000-07:00

T.N.College of Engineering (J)

2009-09-15T06:11:00.000-07:00

T.N.Engineering College(K)

2009-09-15T06:10:00.000-07:00

T.N.Engineering College(L)

2009-09-15T06:09:00.000-07:00

Latha Mathavan Engineering College, Madurai
Lakshmi Chand Rajani College of Engineering and Technology, Thiruthani
Lord Jegannath College of Engineering and Technology, Kumarapuramthoppur Post
Lord Venkateshwara Engineering College, Kanchipuram Taluk
Loyola Institute of Technology, Sriperumbudur Taluk Loyola Institute of Technology and Science, Thovalai

T.N.College of Engineering(M)

2009-09-15T06:08:00.000-07:00

T.N.Engineering College(N)

2009-09-15T06:07:00.000-07:00

Nehru Institute of Engineering and Technology, Coimbatore
National Institute of Technology, Trichy
N.P.R. College of Engineering and Technology, Natham
Nandha Engineering College, Erode
Nandha Institute of Technology, Erode
Narasu's Sarathy Institute of Technology, Omalur Taluk
Narayanaguru College of Engineering, Kanyakumari
National College of Engineering, Nanguneri Taluk
National Engineering College, Kovilpatti
New Prince Shri Bhavani College of Engineering and Technology, Chennai
Noorul Islam University (Noorul Islam College of Engineering), Thuckalay Nyruthi Institute of Technology and Science, Coimbatore

T.N.Engineering College(O)

2009-09-15T06:06:00.000-07:00

Odaiyappa College of Engineering and Technology, Periyakulam Taluk Oxford Engineering College, Trichy

T.N.Engineering College(P)

2009-09-15T06:04:00.002-07:00

T.N Engineering College(R)

2009-09-15T06:04:00.001-07:00

R.M.D. Engineering College, Gummidipoondi Taluk
R.M.K. Engineering College, Gummidipoondi Taluk
R.V.S. College of Engineering and Technology, Dindugul
R.V.S. College of Engineering and Technology, Sulur
Raja Rajeswari Engineering College, Chennai
Rajiv Gandhi College of Engineering, Sriperumbudur
R.M.K. College of Engineering and Technology, Gummidipoondi Taluk
Rabidhranath Tagore College of Engineering for Women, Sankari Taluk
Raja College of Engineering and Technology, Madurai
Rajalakshmi Engineering College, Sriperumbudur Taluk
Rajalakshmi Institute of Technology, Sriperumbudur Taluk
Ranganathan Engineering College, Coimbatore
Ranipet Engineering College, Wallajah Taluk
Roever College of Engineering and Technology, Perambalur
Rover Engineering College, Perambalur
Rrase College of Engineering, Sriperumbudur

T.N. Engineering College(S)

2009-09-15T06:02:00.000-07:00

T.N, Engineering College(U)

2009-09-15T06:01:00.000-07:00

T.J. Institute of Technology, Chennai
Tagore Engineering College, Chennai
Thangavelu Engineering College, Chennai
Thanthai Periyar Govt. Institute of Technology, Vellore
Thiagarajar College of Engineering, Madurai
T.J.S. Engineering College, Gummudipoondi
Tagore Institute of Engineering and Technology, Attur Taluk
Tamil Nadu College of Engineering, Karumathampatti Post
Tamilnadu School of Architecture, Coimbatore
Tejaa Shakthi Institute of Technology for Women, Karumathampatti
The Kavery College of Engineering, Mettur
The Kavery Engineering College, Mettur
The New Royal College of Engineering and Technology, Thirukkalukundram Taluk
The Rajaas Engineering College, Vadakkangulam
Theni Kammavar Sangam College of Technology, Theni
Thirumalai Engineering College, Krishnapuram Post
Thiruvalluvar College of Engineering and Technology, Vandavasi
Trichy Engineering College, Konalai

T.N. Engineering College

2009-09-15T05:59:00.000-07:00

University College of Engineering, Tindivanam

University College of Engineering, Villupuram

Udaya School of Engineering, Kanyakumari

Ultra College of Engineering & Technology for Women, Madura

i
United Institute of Technology, Coimbatore

University College of Engineering, Arni

Unnamalai Institute of Technology, Thoothukudi

T.N.Engineering College ( v)

2009-09-15T05:54:00.000-07:00

Roll your NAME

2009-09-15T05:45:00.000-07:00

Clear command word

Step 3: Find display RAM command word.

Step 2: Find program clock command word

Step 1: Find keyboard/display command word

Statement: Write an assembly language program to your name from right to left

HARDWARE FOR ROLLING HELLO123

Fig. shows the interfacing of eight 7-segment digits to 8085 through 8279. As shown in the figure eight display lines (Bo-B3 and Ao-A3) are buffered with the help of transistor and used to drive display digits. These buffered lines are connected in parallel to all display digits. So, Sl and S2 lines are decoded and decoded lines are used for selection of one of the eight digits

SOFTWARE FOR ROLLING THE NAME - J.BINU

To roll the above namewe have to load 7-segment codes for characters within the message and it is necessary to configure 8279 in right entry mode

Source program:
LXI B, 6200B : Initialize lookup table pointer
MVI C, 08H : Initialize counter
MVI A, 10H : Initialize keyboard/display in right entry mode
OUT 8IH : Mode
MVI A, 3EH : Initialize prescaler count
OUT 8IH
MVI A, D0H : Clear Display
OUT 8IH
MVI A, 90H : Initialize 8279 in write display
OUT 81H : RAM mode
BACK : MOV A, M : Get the 7-segment code
OUT 80H : Write 7-segment code in display RAM
INX H : Increment lookup table pointer
DCR C : Decrement counter
JNZ BACK : if count = 0 stop, otherwise go to back
HLT : Stop program execution

Roll a message - 'HELL0123'

2009-09-15T05:30:00.000-07:00

step 3

step 2

clear command word

Step 1: Find keyboard/display command word.

Statement: Write an assembly language program to roll message 'HELL0123' from right to left

HARDWARE FOR ROLLING HELLO123

Interfacing of 4x4 matrix keyboard and 4 digit 7 segment display

2009-09-15T05:24:00.000-07:00

step 4

steo 3

step 2

step 1

Statement: Interface 4 x 4 matrix keyboard and 4 digit 7-segment display and write an tssembly language program to read keycode of the pressed key and display same key on :he 7 segment display.

HARDWARE FOR 4x4 MATRIX KEYBOARD & 4 DIGIT 7 SEGMENT DISPLAY INTERFACE

Fig. shows interfacing diagram. Here, 4 scan lines are sufficient to scan matrix keyboard and to select display digits. Hence decoded mode is used.

Source program:
MVI A, 00H : Initialize keyboard/display in encoded
OUT 81H : scan keyboard 2 key lockout mode
MVI A, 34H
OUT 81H : Initialize prescaler count
MVI A, 0BH : Load mask pattern to enable RST 7.5
SIM : mask other interrupts
EI : Enable Interrupt
HERE: JMP HERE : Wait for the interrupt
Interrupt service routine
MVI A, 40H : Initialize 8279 in read FIFO RAM mode
OUT 81H
IN 80H : Get keycode
MVI H, 62H : Initialize memory pointer to point
MOV L, A : 7-Segment code
MVI A, 80H : Initialize 8279 in write display RAM mode
OUT 81H
MOV A, M : Get the 7 segment code
OUT 80H : Write 7-segment code in display RAM
EI : Enable interrupt
RET : Return to main program

Interfacing of eight 7-segment digits

2009-09-15T01:45:00.000-07:00

Statement: Interface 8/7-segment digits (common cathode) to 8085 through 8279 and write an 8085 assembly language program to display 1 to 8 on the eight seven segment digits. External clock frequency is 3 MHz.

HARDWARE FOR EIGHT SEVEN SEGMENT DIGITS INTERFACE

Source program:
LXI B, 6200B : Initialize lookup table pointer
MVI C, 08H : Initialize counter
MVI A, 00H : Initialize keyboard/display
OUT 8IH : Mode
MVI A, 3EH : Initialize prescaler count
OUT 8IH
MVI A, 90H : Initial size 8279 in write Display
OUT 8IH : RAM-mode
BACK : MOV A, M : Get the 7-segment code
OUT 80H : Write 7-segment code in display RAM
INX H : Increment lookup table pointer
DCR C : Decrement counter
JNZ BACK : if count = 0 stop, otherwise go to back
HLT : Stop program execution

8 x 4 Matrix Keyboard Interface

2009-09-15T01:43:00.000-07:00

Statement: Interface an 8 x 4 matrix keyboard to 8085 through 8279.

ource program:
MVI A, 00H : Initialize keyboard/display in encoded
OUT 81H : scan keyboard 2 key lockout mode
MVI A, 34H
OUT 81H : Initialize prescaler count
MVI A, 0BH : Load mask pattern to enable RST 7.5
SIM : mask other interrupts
EI : Enable Interrupt
HERE: JMP HERE : Wait for the interrupt
Interrupt Subroutine:
MVI A, 40H : Initialize 8279 in read FIFO
OUT 81H : RAM mode
IN 80H : Read FIFO RAM (keycode)
EI : Enable Interrupt
RET : Return to main program

8 x 8 Keyboard Interface(With Interrupt signal)

2009-09-15T01:37:00.000-07:00

step 1

step 2

step 3

Statement: Interface an 8 x 8 matrix keyboard to 8085 through 8279 in 2-key lockout mode and write an assembly language program to read keycode of the pressed key. The external clock frequency is 2MHz. Use I/O mapped I/O technique.

HARDWARE FOR 8 x 8 MATRIX KEYBOARD INTERFACE(With Interrupt)

Fig. shows the interfacing of 8 x 8 matrix keyboard in interrupt driven keyboard mode. In the interrupt driven mode interrupt line from 8279 is connected to the one of the interrupt input of 8085 except INTR. Here, INT line from 8279 is connected to the interrupt RST 7.5 of 8085. Other signal connections are same as in the non interrupt mode

Note: In the interrupt driven keyboard, when key is pressed, key code is loaded into FIFO RAM and interrupt is generated. This interrupt signal is used to tell CPU that there is a keycode in the FIFO RAM. CPU then initiates read command with in the interrupt service routine to read key code from the FIFO RAM.

8 x 8 Keyboard Interface(Without Interrupt signal)

2009-09-15T01:32:00.000-07:00

Source program:
MVI A, 00H : Initialize keyboard/display
OUT 81H : in encoded scan keyboard-2 keylockout mode
MVI A, 34H
OUT 81H : Initialize prescaler count
BACK: IN 81H : Read FIFO status word
ANI 07H : Mask bit B3 to B7
JZ BACK : If 0, key is not pressed wait for key press else read FIFO RAM
MVI A, 40H : Initialize 8279 in read
OUT 81H : FI FO RAM mode
IN 80H : Read FIFO RAM (keycode)
HLT : Stop program execution

Seven Segment Display Interface (Eight Digits

2009-09-15T01:28:00.000-07:00

Statement: Interface an 8-digit 7 segment LED display using 8255 to the 8085 microprocessor system and write an 8085 assembly language routine to display message on the display.

HARDWARE FOR EIGHT DIGIT SEVEN SEGMENT DISPLAY INTERFACE

Fig. shows the multiplexed eight 7-segment display connected in the 8085 system using 8255. In this circuit port A and port B are used as simple latched output ports. Port A provides the segment data inputs to the display and port B provides a means of selecting a display position at a time for multiplexing the displays. A0-A7 lines are used to decode the addresses for 8255. For this circuit different addresses are:
PA = 00H PB = 01H
PC = 02H CR = 03H.
The register values are chosen in Fig. such that the segment current is 80 mA. This current is required to produce an average of 10 mA per segment as the displays are multiplexed. In this type of display system, only one of the eight display position is 'ON' at any given instant. Only one digit is selected at a time by giving low signal on the corresponding control line. Maximum anode current is 560 mA (7-segments x 80 mA = 560 mA), but the average anode current is 70 mA.

Source program:
SOFTWARE TO INITIALIZE 8255:
MVI A, 80H : Load control word in AL
OUT CR : Load control word in CR
SUBROUTINE TO DISPLAY MESSAGE ON MULTIPLEXED LED DISPLAY:
SET UP REGISTERS FOR DISPLAY:
MVI B, 08H : load count
MVI C, 7FH : load select pattern
LXI H, 6000B : starting address of message
DISPLAY MESSAGE:
DISP 1: MOV A, C : select digit
OUT PB
MOV A, M : get data
OUT PA : display data
CALL DELAY : wait for some time
DISP 1: MOV A, C
RRC
MOV C, A : adjust selection pattern
INX H
DCR B : Decrement count
JNZ DISP 1 : repeat 8 times
RET
Note: This "display message subroutine" must be called continuously to display the 7-segment coded message stored in the memory from address 6000H.
INTERFACING SCHEME

Delay subroutine:
Delay: LXI D, Count
Back: DCX D
MOV A, D
ORA E
JNZ Back
RET

Keyboard interface(64-key-matrix-keyboard)

2009-09-15T01:24:00.000-07:00

Statement: Interface a 64-key matrix keyboard to the 8085 microprocessor using 8255. Write an 8085 assembly language program to initialize 8255 and to read the key code

HARDWARE FOR MATRIX KEYBOARD INTERFACE
Fig. shows a matrix keyboard with 64 keys connected to the 8085 microprocessor using 8255. A matrix keyboard reduces the number of connections, thus the number of interfacing lines. In this example, the keyboard with 64 keys, is arranged in 8 x 8 (8 rows and 8 columns) matrix. This requires sixteen lines from the microprocessor to make all the connections instead of 64 lines if the keys are connected individually. The interfacing of matrix keyboard requires two ports: one input port and other output port. Rows are connected to the input port, port A and columns are connected to the output port, port B
SOFTWARE FOR MATRIX KEYBOARD INTERFACE

Source program
MVI A, 90H : Initialize Port A as input and
OUT CR : Port B as Output
START: MVI A, 00 : Make all scan lines zero
OUT PB
BACK: IN PA
CPI FF : Check for key release
JNZ BACK : If not, wait for key release
CALL DELAY : Wait for key debounce
BACK 1: IN PA
CPI FF : Check for key press
JZ BACK 1 : If not, wait for key press
CALL DELAY : Wait for key debounce
MVI L, 00H : Initialize key counter
MVI C, 08H
MVI B, FEH : Make one column low
NEXTCOL: MOV A, B
OUT PB
MVI D, 08H : Initialize row counter
IN PA : Read return line status
NEXTROW: RRC : Check for one row
JNC DISPLAY : If zero, goto display else continue
INR L : Increment key counter
DCR D : Decrement row counter
JNZ NEXTROW : Check for next row
MOV A, B
RLC : Select the next column
MOV B, A
DCR C : Decrement column count
JNZ NEXTCOL : Check for last column if not repeat
JMP START : Go to start
INTERFACING SCHEME

Delay subroutine:
Delay: LXI D, Count
Back: DCX D
MOV A, D
ORA E
JNZ Back
RET
FLOWCHART

Source Program
Delay routine

Stepper Motor Control

2009-09-15T01:17:00.000-07:00

Statement: Interface a Stepper Motor to the 8085 microprocessor system and write an 8085 assembly language program to control the Stepper Motor.

HARDWARE FOR STEPPER MOTOR CONTROL
motor is a digital motor. It can be driven by digital signal. Fig. shows the typical 2 phase motor rated 12V /0.67 A/ph interfaced with the 8085 microprocessor system using 8255. Motor shown in the circuit has two phases, with center-tap winding. The center taps of these windings are connected to the 12V supply. Due to this, motor can be excited by grounding four terminals of the two windings. Motor can be rotated in steps by giving proper excitation sequence to these windings. The lower nibble of port A of the 8255 is used to generate excitation signals in the proper sequence. These excitation signals are buffered using driver transistors. The transistors are selected such that they can source rated current for the windings. Motor is rotated by 1.80 per excitation
SOFTWARE FOR STEPPER MOTOR CONTROL

As port A is used as an output port, control word for 8255 is 80H.
Stepper Motor Control Program:
6000H Excite code DB 03H, 06H,
09H, OCH : This is the code sequence for clockwise rotation
Subroutine to rotate a stepper motor clockwise by 360° - Set the counts:
MVI C, 32H : Set repetition count to 50ıο
START: MVI B, 04H : Counts excitation sequence
LXI H, 6000H : Initialize pointer
BACK1: MOV A, M : Get the Excite code
OUT PORTA : Send Excite code
CALL DELAY : Wait
INX H : Increment pointer
DCR B : Repeat 4 times
JNZ BACK l
DCR C
JNZ START : Repeat 50 times
RET
INTERFACING SCHEME

Delay subroutine:
Delay: LXI D, Count
Back: DCX D
MOV A, D
ORA E
JNZ Back
RET

FLOWCHARTS

Source Program
Delay routine
Stepper motor subroutine

Traffic Light Control

2009-09-15T01:07:00.000-07:00

HARDWARE FOR TRAFFIC LIGHT CONTROL

Statement: Design a microprocessor system to control traffic lights. The traffic light arrangement is as shown in Fig. The traffic should be controlled in the following manner.
1) Allow traffic from W to E and E to W transition for 20 seconds. 2) Give transition period of 5 seconds (Yellow bulbs ON) 3) Allow traffic from N to 5 and 5 to N for 20 seconds 4) Give transition period of 5 seconds (Yellow bulbs ON) 5) Repeat the process.

Fig. shows the interfacing diagram to control 12 electric bulbs. Port A is used to control lights on N-S road and Port B is used to control lights on W-E road. Actual pin connections are listed in Table 1 below.

The electric bulbs are controlled by relays. The 8255 pins are used to control relay on-off action with the help of relay driver circuits. The driver circuit includes 12 transistors to drive 12 relays. Fig. also shows the interfacing of 8255 to the system

Source program:
MVI A, 80H : Initialize 8255, port A and port B
OUT 83H (CR) : in output mode
START: MVI A, 09H
OUT 80H (PA) : Send data on PA to glow R1 and R2
MVI A, 24H
OUT 81H (PB) : Send data on PB to glow G3 and G4
MVI C, 28H : Load multiplier count (40ıο) for delay
CALL DELAY : Call delay subroutine
MVI A, 12H
OUT (81H) PA : Send data on Port A to glow Y1 and Y2
OUT (81H) PB : Send data on port B to glow Y3 and Y4
MVI C, 0AH : Load multiplier count (10ıο) for delay
CALL: DELAY : Call delay subroutine
MVI A, 24H
OUT (80H) PA : Send data on port A to glow G1 and G2
MVI A, 09H
OUT (81H) PB : Send data on port B to glow R3 and R4
MVI C, 28H : Load multiplier count (40ıο) for delay
CALL DELAY : Call delay subroutine
MVI A, 12H
OUT PA : Send data on port A to glow Y1 and Y2
OUT PB : Send data on port B to glow Y3 and Y4
MVI C, 0AH : Load multiplier count (10ıο) for delay
CALL DELAY : Call delay subroutine
JMP START

Delay Subroutine:
DELAY: LXI D, Count : Load count to give 0.5 sec delay
BACK: DCX D : Decrement counter
MOV A, D
ORA E : Check whether count is 0
JNZ BACK : If not zero, repeat
DCR C : Check if multiplier zero, otherwise repeat
JNZ DELAY
RET : Return to main program

Flashing of LEDs

2009-09-15T01:05:00.000-07:00

Statement: Design a system (both Software and Hardware) that will cause 4 LEDs to flash 10 times when a push button switch is pressed. Use 8255. Assume persistence of vision to be 0.1 seconds.

Source program:
LXI SP, 2000 H : Initialize stack pointer
MVI A, 90H
OUT CR : Initialize 8255
BACK: IN PA : [Read status
ANI 01 : of push
JNZ BACK : button]
MVI B, 0AH : Initialize counter
AGAIN: MVI A, 00H : Load data to light LEDs
OUT PC : Send data on port C
CALL Delay : Call. Delay of 0.1 sec
MVI A, FFH : Load data to switch off LEDs
OUT PC : Send data on port C
CALL Delay : Call Delay of 0.1 sec
DCR B : Decrement count
JNZ AGAIN : If not zero repeat
JMP BACK : Jump back to read status
INTERFACING SCHEME

Delay subroutine:
Delay: LXI D, Count
Back: DCX D
MOV A, D
ORA E
JNZ Back
RET

Blink port C bit 0 of 8255

2009-09-15T01:00:00.000-07:00

Statement: Write a program to blink Port C bit 0 of the 8255. Assume address of control word register of 8255 as 83H. Use Bit Set/Reset mode.

Source program:
BACK: MVI A, 0lH : Load bit pattern to make PCο high
OUT 83H : Send it to control word register
CALL DELAY : Call Delay subroutine
MVI A, 00H : Load bit pattern to make PCο Low
OUT 83H : Send it to control word register
CALL Delay : Call Delay subroutine
JMP BACK : Repeat
FLOWCHART

Delay subroutine:
Delay: LXI D, Count
Back: DCX D
MOV A, D
ORA E
JNZ Back
RET

Initialize 8255

2009-09-15T00:57:00.000-07:00

Statement:Write a program to initialize 8255 in the configuration given below

Sample 1:
Write a program to initialize 8255 in the configuration given below:
1. Port A: Simple input
2. Port B: Simple output
3. Port CL: Output
4. Port Cu: Input
Assume address of the control word register of 8255 as 83H.
Sample 2:
Write a program to initialize 8255 in the configuration given below:
1. Port A: Output with handshake
2. Port B: Input with handshake
3. Port CL: Output
4. Port Cu: Input
Assume address of the control word register of 8255 as 23H.

Receive message using 8251

2009-09-15T00:54:00.000-07:00

Statement: Write a assembly program to receive 25 bytes from an CRT terminal to 8085 for the following requirements.
i) Baud rate x 16
ii) Stop bits 2

Source program:
LXI H, 2300 H : Initialize memory pointer
MVI C, FFH : Initialize counter to accept 25 characters
MVI A, 00H
OUT FFH
OUT FFH : Dummy mode word
OUT FFH
MVI A, 40H : Reset command word
OUT FFH : Reset 8251 A
MVI A, CAH : Mode word initialization
OUT FFH
MVI A, 14 H : Command word initialization
OUT FFH
CHECK: IN FFH
ANI 02 H : Check RxRDY
JZ CHECK : Is RxRDY ? If not, check again
IN FEH : Get the character
MOV M, A : save the character
INX H : Increment memory pointer
DCR C : Decrement memory pointer
OUT FEH : Send character to the transmitter
JNZ CHECK : If not zero, accept next character
HLT : Stop program execution

Transmit message using 8251

2009-09-15T00:47:00.000-07:00

Statement: Write a assembly program to transmit a message from an 8085 to a CRT terminal for the following requirements and draw the interfacing diagram.
i) A message of 50 characters is stored as ASCII characters (without parity) in memory locations starting at 2200H.
ii) Baud rate x 16
iii) Stop bits 2

Solution Description:
CRT terminal uses normal RS 232C standard serial communication interface. Therefore, to transmit data to CRT it is necessary to have RS 232C interface at the sending end.
Fig. shows the interfacing of 8251 with RS 232C to 8085.
As shown in the Fig. three RS-232C signals (TxD, RxD are Ground) are used for serial communication between the CRT terminal and the 8085 system.
Line drivers and receivers are used to transfer logic levels from TTL logic to RS-232C logic.
For RS-232C the voltage level +3V to +15V is defined as logic 0 and voltage level from -3V to -15V is defined as logic 1.
The line driver, MC 1488, converts logic 1 of TIL to approximately -9V and logic a of TIL to approximately +9V. These levels at the receiving end are again converted by the line receiver, MC1489, into TTL compatible logic.

Source program:
LXI H, 2200H : Initialize memory pointer to pointer the message
MVI C, 32H : Initialize counter to send 50 characters
MVI A, 00H
OUT FFH
OUT FFH : Dummy mode word
OUT FFH
MVI A, 40H : Reset command word
OUT FFH : Reset 8251A
MVI A, CAH : Mode word initialization
OUT FFH
MVI A, 11H : Command word initialization
OUT FFH
CHECK: IN FFH
ANI 0lH : Check TxRDY
JZ CHECK : Is TxRDY I? if not, check again
MOV A, M : Get the character in accumulator
OUT FEH : Send character to the transmitter
INX H : Increment memory pointer
DCR C : Decrement counter
JNZ CHECK : if not zero, send next character
HLT : Stop program execution

Receive ASCII character through SID pin (8085)

2009-09-13T04:45:00.000-07:00

Statement: An ASCII character is being received on SID pin of 8085. Write a program in assembly language of 8085 to assemble this character and store it in memory. Write comment for each instruction.

Source program:
LXI SP, 27FFH
LXI H, 2000H : Memory pointer
RIM : Read SID
ANI 80H : Check D7 bit of Accumulator
CALL Delay : 1/2 bit time delay for stop bit
MVI B, 08H : Initialize bit counter
MVI D, 00H : Clear data register
UP1: ALL Delay : 1 bit time
RIM : Read SID line
ANI 80H : Mask bits B6 - Bo
ORA D : OR data bit with previous bits
RRC
MOV D, A : Store data bit at appropriate position
DCR B
JNZ UP1
RLC : Shift left to correct result
MOV M, A : Store result
RIM : Read stop bit
ANI 8OH
CZ error : If not stop bit call error
HLT : Terminate program.
Delay subroutine:
Delay: LXI D, Count
Back: DCX D
MOV A, D
ORA E
JNZ Back
RET

Generate square wave from SOD pin (8085)

2009-09-13T04:43:00.000-07:00

Statement: Write a program to output square wave of 1 kHz frequency on the SOD pinof 8085 for 5 seconds. Operating frequency of 8085 is 2 MHz.

Source program
LXI SP, 27FFH : Initialize stack pointer
LXI B, 1388H : Initialize counter with count 5000.
BACK: MVI A, COH
SIM : Send high on SOD pin
CALL DELAY : Wait for 0.5 msec
MVI A, 40H : Send low on SOD pin
CALL DELAY : wait for. 5 msec
DCX B : Decrement count by 1
MOV A, C
ORA B : Check if count = 0
JNZ BACK : If not, repeat
HLT : Stop program execution
Delay subroutine:
Delay: LXI D, Count
Back: DCX D
MOV A, D
ORA E
JNZ Back
RET

Output byte from SOD pin (8085)

2009-09-13T04:39:00.000-07:00

Statement: Write a program to output contents of B register LSB to MSB on the SOD pin.

Source program:
MVI C, 08H : Initialize count with 8
MOV A, B
BACK: RRC : Rotate B register contents right
MOV B, A : Save contents of register B
JNC SKIP : If no carry skip
MVI A, COH
SIM : If carry, send high on SOD
JMP NEXT
SKIP: MVI A, 40H
SIM : If no carry, send low on SOD.
NEXT: CALL DELAY : Wait for specific time
DCR C : Decrement count by 1
JNZ BACK : if count = 0 Stop, if not repeat
HLT : Stop program execution

Delay subroutine:
Delay: LXI D, Count
Back: DCX D
MOV A, D
ORA E
JNZ Back
RET

HEX to binary conversion(8085)

2009-09-13T04:37:00.000-07:00

Statement: Convert an 8 bit hex no to its binary form & store in memory.

Source Program:

LXI H, 4150 : Initialize memory pointer
MVI B, 08 : count for 8-bit
MVI A, 54
LOOP : RRC
JC LOOP1
MVI M, 00 : store zero it no carry
JMP COMMON
LOOP2: MVI M, 01 : store one if there is a carry
COMMON: INX H
DCR B : check for carry
JNZ LOOP
HLT : Terminate the program

HEX to Decimal conversion (8085)

2009-09-13T04:36:00.000-07:00

Statement: Convert the HEX number in memory to its equivalent decimal number

Source Program:
LXI H, 4150 ; Point to data
LXI B, 0000 ; Initialize hundreds= 0, Tens=0
MOV A, M ; Get hex data to A
LOOP: SUI 64
JC LOOP 1
INR B ; hundreds= hundreds+1
JMP LOOP
LOOP 1: ADI 64 ; if subtracted extra, add it clear carry flag
LOOP 2: SUI 0A
JC LOOP 3
INR C ; Tens=tens+1
JMP LOOP 2
LOOP 3: ADI 0A ; If subtracted extra, add it again
INX H ; A = Units
MOV M, B ; store hundreds
MOV B, A ; Combine Tens in C &
MOV A, C ; Units in A to form a
RLC ; Single 8-bit number
RLC
RLC
RLC
ADD B
INX H
MOV M, A ; Store tens & Units
HLT

ASCII to Decimal Conversion (8085)

2009-09-13T04:33:00.000-07:00

Statement: convert the ASCII number in memory to its equivalent decimal number

Source Program:
LXI H, 4150 : Point to data
MOV A, M : Get operand
SUI 30 : convert to decimal
CPI 0A : Check whether it is valid decimal number
JC LOOP : yes, store result
MVI A, FF : No, make result=FF
LOOP: INX H
MOV M, A
HLT : (A) = (4151)
.

Note: The ASCII Code (American Standard Code for Information Interchange) is commonly used for communication. It is a seven bit code. In this code number 0 through 9 are represented as 30 through 39 respectively and letters A through Z are represented as 41H through 5AH. Therefore, by subtracting 30H we can convert an ASCII number into its decimal equivalent.
google_protectAndRun("ads_core.google_render_ad", google_handleError, google_render_ad);

Find the ASCII character (8085)

2009-09-13T04:29:00.000-07:00

Statement: Write an assembly language program to convert the contents of the five memory locations starting from 2000H into an ASCII character. Place the result in another five memory locations starting from 2200H.

Sample Problem
(2000H) = 1
(2001H) = 2
(2002H) = 9
(2003H) = A
(2004H) = B
Result:(2200H) = 31
(2201H) = 32
(2202H) = 39
(2203H) = 41
(2204H) = 42
Source program:
LXI SP, 27FFH : Initialize stack pointer
LXI H, 2000H : Source memory pointer
LXI D, 2200H : Destination memory pointer
MVI C, O5H : Initialize the counter
BACK: MOV A, M : Get the number
CALL ASCII : Call subroutine ASCII
STAX D : Store result
INX H : Increment source memory pointer
INX D : Increment destination memory pointer
DCR C : Decrement count by 1
CJNZ : if not zero, repeat
HLT : Stop program execution subroutine ASCII
ASCII: CPI, OAH : Check if number is OAR
JNC NEXT : If yes go to next otherwise continue
ADI 30H
JMP LAST
NEXT: ADI 37H
LAST: RET : Return to main program
Subroutine:
Subroutine 'ASCII' converts a hexadecimal digit to ASCII.The digit is passed using accumulator and the result is stored in accumulator.Stack starts From 27FEH to 27FDH.
FLOWCHART

google_protectAndRun("render_ads.js::google_render_ad", google_handleError, google_render_ad);
google_protectAndRun("ads_core.google_render_ad", google_handleError, google_render_ad);
google_protectAndRun("ads_core.google_render_ad", google_handleError, google_render_ad);

Note: The ASCII Code (American Standard Code for Information Interchange) is commonly used for communication. In such cases we need to convert binary number to its ASCII equivalent. It is a seven bit code. In this code number 0 through 9 are represented as 30 through 39 respectively and letters A through Z are represented as 41H through 5AH. Therefore, by adding 30H we can convert number into its ASCII equivalent and by adding 37H we can convert letter to its ASCII equivalent.

Find the 7-segment codes for given numbers(8085)

2009-09-13T04:27:00.000-07:00

Statement: Find the 7-segment codes for given 5 numbers from memory location 6000H and store the result from memory location 7000H.

Sample Problem: (6000) H = 8AH
Source Program
LXI H, 6200H : Initialize lookup table pointer
LXI D, 6000H : Initialize source memory pointer
LXI B, 7000H : Initialize destination memory pointer
BACK: LDAX D : Get the number
MOV L, A : A point to the 7-segment code
MOV A, M : Get the 7-segment code
STAX B : Store the result at destination memory location
INX D : Increment source memory pointer
INX B : Increment destination memory pointer
MOV A, C
CPI O5H : Check for last number
JNZ BACK : If not repeat
HLT : End of program

Binary to BCD conversion(8085)

2009-09-13T04:24:00.000-07:00

Statement: Write a main program and a conversion subroutine to convert the binary number stored at 6000H into its equivalent BCD number. Store the result from memory location 6100H.

Sample Problem: (6000) H = 8AH
1.8AH ? 64H (Decimal 100) :. Divide by 64H (Decimal 100)
8AH/64H ? Quotient = 1, Remainder = 26H
26H < 2 =" 1" quotient =" 3," remainder =" O8H" 1 =" 3" 0 =" O8H">

2-Digit BCD to binary conversion(8085)

2009-09-13T04:22:00.000-07:00

Statement: Convert a 2-digit BCD number stored at memory address 2200H into its binary equivalent number and store the result in a memory location 2300H.

Sample Problem
(2200H) = 67H
(2300H) = 6 x OAH + 7 = 3CH + 7 = 43H
Source Program:
LDA 2200H : Get the BCD number
MOV B, A : Save it
ANI OFH : Mask most significant four bits
MOV C, A : Save unpacked BCDI in C register
MOV A, B : Get BCD again
ANI FOH : Mask least significant four bits
RRC : Convert most significant four bits into unpacked BCD2
RRC
RRC
RRC
MOV B, A : Save unpacked BCD2 in B register
XRA A : Clear accumulator (sum = 0)
MVI D, 0AH : Set D as a multiplier of 10
Sum: ADD D : Add 10 until (B) = 0
DCR B : Decrement BCD2 by one
JNZ SUM : Is multiplication complete? i if not, go back and add again
ADD C : Add BCD1
STA 2300H : Store the result
HLT : Terminate program execution

Debug the delay routine (8085)

2009-09-13T04:20:00.000-07:00

Statement:The delay routine given below is in infinite loop, identify the error and correct the program.

Delay routine with error:
DELAY : LXI H, N
L1 : DCX H
JNZ L1
Sol.: 1) The fault in the above program is at instruction JNZ L1. This condition always evaluates to be true hence loops keep on executing and hence infinite loop.
2) Reason for infinite looping: - The instruction DCX H decrease the HL pair count one by one but it does not affect the zero flag. So when count reaches to OOOOH in HL pair zero flag is not affected and JNZ L1 evaluates to be true and loop continues. Now HL again decrements below OOOOH and HL becomes FFFFH and thus execution continues.
3) The modification in the program is as follows:

DELAY : LXI H, N :Load 16 bit count
L1 : DCX H : Decrement count
MOV A, L
ORA H : logically OR Hand L
JNZ L1 : If result is not 0 repeat

Generate and display the contents of decimal counter (8085)

2009-09-13T04:17:00.000-07:00

Statement:Write assembly language program to with proper comments for the following: To display decimal decrementing counter (99 to 00) at port 05 H with delay of half seconds between .each count. Write as well the delay routine giving delay of half seconds. Operating frequency of microprocessor is 3.072 MHz. Neglect delay for the main program.

Source Program:
MVI C, 99H : Initialize counter
BACK: MOV A, C
ANI OF : Mask higher nibble
CPI OF
JNZ SKIP
MOV A, C
SUI 06 : Subtract 6 to adjust decimal count
MOV D, A
SKIP: MOV A, C
OUT 05 : send count on output port
CALL Delay : Wait for 0.5 seconds
DCR C : decrement count
MOV A, C
CPI FF
JNZ BACK : If not zero, repeat
HLT : Stop execution
Delay subroutine:
Delay: LXI D, Count
Back: DCX D : 6 T-states
MOV A, D : 4 T-states
ORA E : 4 T-states
JNZ Back : 10 T-states
RET

Generate and display BCD down counter (8085)

2009-09-13T04:10:00.000-07:00

Statement:Write a program for displaying BCD down counter. Counter should count numbers from 99 to 00 and it should increment after every 1 sec. Assume operating frequency of 8085 equal to 3MHz. Display routine is available

Source Program 1:

LXI SP, 27FFH : Initialize stack pointer
MVI C, 99H : Initialize counter = 99
BACK:CALL Display : Call display subroutine
CALL Delay : Call delay subroutine
ADI 99H : See Addition below
DAA : Adjust for decimal
CPI 99H : Compare with last count
JNZ BACK :If no, repeat
HLT

Source Program2:
LXI SP, 27FFH : Initialize stack pointer
MVI C, 99H : Initialize counter = 99
BACK: CALL Display : Call display subroutine
CALL Delay : Call delay subroutine
MOV A, C : Get count
ANI 0FH : Check for lower nibble
JNZ SKIP : If it is not 0FH go to skip
MOV A,C : Else get the count
SBI ,06 : Subtract 06
MOV C,A : Store the count
DCR C : Decrement count
CPI 99H : Check it for last count
JNZ BACK : If not, repeat
HLT : Stop

Generate and display BCD up counter with frequency 1Hz (8085)

2009-09-13T04:05:00.000-07:00

Statement:Write a program for displaying BCD up counter. Counter should count numbers from 00 to 99H and it should increment after every 1 sec. Assume operating frequency of 8085 equal to 3MHz. Display routine is available.

Source Program:

LXI SP, 27FFH : Initialize stack pointer
MVI C, OOH : Initialize counter
BACK: CALL Display : Call display subroutine
CALL Delay : Call delay subroutine
MOV A, C
ADI A, 0 1 : Increment counter
DAA : Adjust it for decimal
MOV C,A : Store count
CPI ,00 : Check count is > 99
JNZ BACK : If not, repeat
HLT : Stop
Delay Subroutine:
Delay:MVI B, Multiplier-count : Initialize multiplier count
BACK 1:LXI D, Initialize Count
BACK: DCX D : Decrement count
MOV A, E
ORA D : Logically OR D and E
JNZ BACK : If result is not a, repeat
DCR B : Decrement multiplier count
JNZ BACK 1 : If not zero, repeat
RET : Return to main program.

Generate and display binary up counter (8085)

2009-09-13T03:57:00.000-07:00

Statement:Write a program for displaying binary up counter. Counter should count numbers from 00 to FFH and it should increment after every 0.5 sec.
Assume operating frequency of 8085 equal to 2MHz. Display routine is available.

Source Program:

LXI SP, 27FFH : Initialize stack pointer
MVI C, OOH : Initialize counter
BACK: CALL Display : Call display subroutine
CALL Delay : Call delay subroutine
INR C : Increment counter
MOV A, C
CPI OOH : Check counter is > FFH
JNZ BACK : If not, repeat
HLT : Stop
Delay Subroutine:
Delay: LXI B, count : Initialize count
BACK: DCX D : Decrement count
MOV A, E
ORA D : Logically OR D and E
JNZ BACK : If result is not 0 repeat
RET : Return to main program

Multiply two 2-digit BCD numbers (8085)

2009-09-13T03:56:00.000-07:00

Statement: Write an assembly language program to multiply 2 BCD numbers

Source Program:

MVI C, Multiplier : Load BCD multiplier
MVI B, 00 : Initialize counter
LXI H, 0000H : Result = 0000
MVI E, multiplicand : Load multiplicand
MVI D, 00H : Extend to 16-bits
BACK: DAD D : Result Result + Multiplicand
MOV A, L : Get the lower byte of the result
ADI, 00H
DAA : Adjust the lower byte of result to BCD.
MOV L, A : Store the lower byte of result
MOV A, H : Get the higher byte of the result
ACI, 00H
DAA : Adjust the higher byte of the result to BCD
MOV H, A : Store the higher byte of result.
MOV A, B : [Increment
ADI 01H : counter
DAA : adjust it to BCD and
MOV B,A : store it]
CMP C : Compare if count = multiplier
JNZ BACK : if not equal repeat
HLT : Stop

Subtraction of two BCD numbers (8085)

2009-09-13T03:55:00.000-07:00

Statement: Subtract the BCD number stored in E register from the number stored in the D register

Source Program:
MVI A,99H
SUB E : Find the 99's complement of subtrahend
INR A : Find 100's complement of subtrahend
ADD D : Add minuend to 100's complement of subtrahend
DAA : Adjust for BCD
HLT : Terminate program execution

Add two 4-digit BCD numbers (8085)

2009-09-13T03:53:00.000-07:00

Statement: Add two 4 digit BCD numbers in HL and DE register pairs and store result in memory locations, 2300H and 2301H. Ignore carry after 16 bit.

Sample Problem:
(HL) =3629
(DE) =4738
Step 1 : 29 + 38 = 61 and auxiliary carry flag = 1
:.add 06
61 + 06 = 67
Step 2 : 36 + 47 + 0 (carry of LSB) = 7D
Lower nibble of addition is greater than 9, so add 6.
7D + 06 = 83
Result = 8367
Source program
MOV A, L : Get lower 2 digits of no. 1
ADD E : Add two lower digits
DAA : Adjust result to valid BCD
STA 2300H : Store partial result
MOV A, H : Get most significant 2 digits of number
ADC D : Add two most significant digits
DAA : Adjust result to valid BCD
STA 2301H : Store partial result
HLT : Terminate program execution.

Split a HEX data into two nibbles and store it (8085)

2009-09-13T03:51:00.000-07:00

Statement:Write a simple program to Split a HEX data into two nibbles and store it in memory

Source Program:
LXI H, 4200H : Set pointer data for array
MOV B,M : Get the data in B-reg
MOV A,B : Copy the data to A-reg
ANI OFH : Mask the upper nibble
INX H : Increment address as 4201
MOV M,A : Store the lower nibble in memory
MOV A,B : Get the data in A-reg
ANI FOH : Bring the upper nibble to lower nibble position
RRC
RRC
RRC
RRC
INX H
MOV M,A : Store the upper nibble in memory
HLT : Terminate program execution

Find the Square Root of a given number (8085)

2009-09-13T03:47:00.000-07:00

Statement:Write a program to find the Square Root of an 8 bit binary number. The binary number is stored in memory location 4200H and store the square root in 4201H.

Source Program:
LDA 4200H : Get the given data(Y) in A register
MOV B,A : Save the data in B register
MVI C,02H : Call the divisor(02H) in C register
CALL DIV : Call division subroutine to get initial value(X) in D-reg
REP: MOV E,D : Save the initial value in E-reg
MOV A,B : Get the dividend(Y) in A-reg
MOV C,D : Get the divisor(X) in C-reg
CALL DIV : Call division subroutine to get initial value(Y/X) in D-reg
MOV A, D : Move Y/X in A-reg
ADD E : Get the((Y/X) + X) in A-reg
MVI C, 02H : Get the divisor(02H) in C-reg
CALL DIV : Call division subroutine to get ((Y/X) + X)/2 in D-reg.This is XNEW
MOV A, E : Get Xin A-reg
CMP D : Compare X and XNEW
JNZ REP : If XNEW is not equal to X, then repeat
STA 4201H : Save the square root in memory
HLT : Terminate program execution
Subroutine:
DIV: MVI D, 00H : Clear D-reg for Quotient
NEXT:SUB C : Subtact the divisor from dividend
INR D : Increment the quotient
CMP C : Repeat subtraction until the
JNC NEXT : divisor is less than dividend
RET : Return to main program

Find the factorial of a number(8085)

2009-09-13T03:44:00.000-07:00

Statement: Program to calculate the factorial of a number between 0 to 8

Source program
LXI SP, 27FFH ; Initialize stack pointer
LDA 2200H ; Get the number
CPI 02H ; Check if number is greater than 1
JC LAST
MVI D, 00H ; Load number as a result
MOV E, A
DCR A
MOV C,A ; Load counter one less than number
CALL FACTO ; Call subroutine FACTO
XCHG ; Get the result in HL
SHLD 2201H ; Store result in the memory
JMP END
LAST: LXI H, 000lH ; Store result = 01
END: SHLD 2201H
HLT
Subroutine Program:
FACTO:LXI H, 0000H
MOV B, C ; Load counter
BACK: DAD D
DCR B
JNZ BACK ; Multiply by successive addition
XCHG ; Store result in DE
DCR C ; Decrement counter
CNZ FACTO ; Call subroutine FACTO
RET ; Return to main program

Data transfer from one memory block to other memory block(8085)

2009-09-13T03:43:00.000-07:00

Statement: Transfer ten bytes of data from one memory to another memory block. Source memory block starts from memory location 2200H where as destination memory block starts from memory location 2300H

Source Program:

LXI H, 4150 : Initialize memory pointer
MVI B, 08 : count for 8-bit
MVI A, 54
LOOP : RRC
JC LOOP1
MVI M, 00 : store zero it no carry
JMP COMMON
LOOP2: MVI M, 01 : store one if there is a carry
COMMON: INX H
DCR B : check for carry
JNZ LOOP
HLT : Terminate the program

Arrange in DESCENDING Order (8085)

2009-09-13T03:41:00.000-07:00

Statement: Arrange an array of 8 bit unsigned no in descending order

Source Program:

START:MVI B, 00 ; Flag = 0
LXI H, 4150 ; Count = length of array
MOV C, M
DCR C ; No. of pair = count -1
INX H ; Point to start of array
LOOP:MOV A, M ; Get kth element
INX H
CMP M ; Compare to (K+1) th element
JNC LOOP 1 ; No interchange if kth >= (k+1) th
MOV D, M ; Interchange if out of order
MOV M, A ;
DCR H
MOV M, D
INX H
MVI B, 01H ; Flag=1
LOOP 1:DCR C ; count down
JNZ LOOP ;
DCR B ; is flag = 1?
JZ START ; do another sort, if yes
HLT ; If flag = 0, step execution

Generate a delay of 0.4 seconds(8085)

2009-09-13T03:40:00.000-07:00

Statement:Write a program to generate a delay of 0.4 sec if the crystal frequency is 5 MHz

Calculation: In 8085, the operating frequency is half of the crystal frequency,
ie.Operating frequency = 5/2 = 2.5 MHz
Time for one T -state =
Number of T-states required =
= 1 x 106
Source Program:
LXI B, count : 16 - bit count
BACK: DCX B : Decrement count
MOV A, C
ORA B : Logically OR Band C
JNZ BACK : If result is not zero repeat

Program to generate fibonacci number(8085)

2009-09-13T03:39:00.000-07:00

Statement:Write an assembly language program to generate fibonacci number

Source Program:
MVI D, COUNT : Initialize counter
MVI B, 00 : Initialize variable to store previous number
MVI C, 01 : Initialize variable to store current number
MOV A, B :[Add two numbers]
BACK: ADD C :[Add two numbers]
MOV B, C : Current number is now previous number
MOV C, A : Save result as a new current number
DCR D : Decrement count
JNZ BACK : if count 0 go to BACK
HLT : Stop.

Program to test RAM (8085)

2009-09-13T03:36:00.000-07:00

Statement:To test RAM by writing '1' and reading it back and later writing '0' (zero) and reading it back. RAM addresses to be checked are 40FFH to 40FFH. In case of any error, it is indicated by writing 01H at port 10H

Source Program:
LXI H, 4000H : Initialize memory pointer
BACK: MVI M, FFH : Writing '1' into RAM
MOV A, M : Reading data from RAM
CPI FFH : Check for ERROR
JNZ ERROR : If yes go to ERROR
INX H : Increment memory pointer
MOV A, H
CPI SOH : Check for last check
JNZ BACK : If not last, repeat
LXI H, 4000H : Initialize memory pointer
BACKl: MVI M, OOH : Writing '0' into RAM
MOV A, M : Reading data from RAM
CPI OOH : Check for ERROR
INX H : Increment memory pointer
MOV A, H
CPI SOH : Check for last check
JNZ BACKl : If not last, repeat
HLT : Stop Execution

Sub routine to perform the task of DAA(8085)

2009-09-12T23:34:00.000-07:00

Statement:Assume the DAA instruction is not present. Write a sub routine which will perform the same task as DAA.

Sample Problem:
Execution of DAA instruction:
1. If the value of the low order four bits (03-00) in the accumulator is greater than 9 or if auxiliary carry flag is set, the instruction adds 6 '(06) to the low-order four bits.
2. If the value of the high-order four bits (07-04) in the accumulator is greater than 9 or if carry flag is set, the instruction adds 6(06) to the high-order four bits.

Source Program:
LXI SP, 27FFH : Initialize stack pointer
MOV E, A : Store the contents of accumulator
ANI 0FH : Mask upper nibble
CPI 0A H : Check if number is greater than 9
JC SKIP : if no go to skip
MOV A, E : Get the number
ADI 06H : Add 6 in the number
JMP SECOND : Go for second check
SKIP: PUSH PSW : Store accumulator and flag contents in stack
POP B : Get the contents of accumulator in B register and flag register contents in C register
MOV A, C : Get flag register contents in accumulator
ANI 10H : Check for bit 4
JZ SECOND : if zero, go for second check
MOV A, E : Get the number
ADI 06 : Add 6 in the number
SECOND: MOV E, A : Store the contents of accumulator
ANI FOH : Mask lower nibble
RRC
RRC
RRC
RRC : Rotate number 4 bit right
CPI 0AH : Check if number is greater than 9
JC SKIPl : if no go to skip 1
MOV A, E : Get the number
ADI 60 H : Add 60 H in the number
JMP LAST : Go to last
SKIP1: JNC LAST : if carry flag = 0 go to last
MOV A, E : Get the number
ADI 60 H : Add 60 H in the number
LAST: HLT

Divide 16-bit number with 8-bit number using shifting technique(8085)

2009-09-12T23:32:00.000-07:00

Statement:Divide the 16-bit unsigned number in memory locations 2200H and 2201H (most significant bits in 2201H) by the B-bit unsigned number in memory location 2300H store the quotient in memory location 2400H and remainder in 2401H.

Assumption: The most significant bits of both the divisor and dividend are zero.
Source program
MVI E, 00 : Quotient = 0
LHLD 2200H : Get dividend
LDA 2300 : Get divisor
MOV B, A : Store divisor
MVI C, 08 : Count = 8
NEXT: DAD H : Dividend = Dividend x 2
MOV A, E
RLC
MOV E, A : Quotient = Quotient x 2
MOV A, H
SUB B : Is most significant byte of Dividend > divisor
JC SKIP : No, go to Next step
MOV H, A : Yes, subtract divisor
INR E : and Quotient = Quotient + 1
SKIP:DCR C : Count = Count - 1
JNZ NEXT : Is count =0 repeat
MOV A, E
STA 2401H : Store Quotient
Mov A, H
STA 2410H : Store remainder
HLT : End of program.

Multiply two eight bit numbers with shift and add method(8085)

2009-09-12T23:29:00.000-07:00

Statement:Multiply the 8-bit unsigned number in memory location 2200H by the 8-bit unsigned number in memory location 2201H. Store the 8 least significant bits of the result in memory location 2300H and the 8 most significant bits in memory location 2301H.

Sample problem:

(2200) = 1100 (0CH)
(2201) = 0101 (05H)
Multiplicand = 1100 (1210)
Multiplier = 0101 (510)
Result = 12 x 5 = (6010)

Source program

LXI H, 2200 : Initialize the memory pointer
MOV E, M : Get multiplicand
MVI D, 00H : Extend to 16-bits
INX H : Increment memory pointer
MOV A, M : Get multiplier
LXI H, 0000 : Product = 0
MVI B, 08H : Initialize counter with count 8
MULT: DAD H : Product = product x 2
RAL
JNC SKIP : Is carry from multiplier 1 ?
DAD D : Yes, Product =Product + Multiplicand
SKIP: DCR B : Is counter = zero
JNZ MULT : no, repeat
SHLD 2300H : Store the result
HLT : End of program

Deleting string in a given array of characters(8085)

2009-09-12T23:27:00.000-07:00

Statement:Write an 8085 assembly language program to delete a string of 4 characters from the tenth location in the given array of 50 characters.
Solution: Shift bytes from location 14 till the end of array upwards by 4 characters i.e. from location 10 onwards.
Source Program:
LXI H, 2l0DH :Initialize source memory pointer at the 14thlocation of the array.
LXI D, 2l09H : Initialize destn memory pointer at the 10th location of the array.
MOV A, M : Get the character
STAX D : Store character at new location
INX D : Increment destination pointer
INX H : Increment source pointer
MOV A, L : [check whether desired
CPI 32H bytes are shifted or not]
JNZ REPE : if not repeat the process
HLT : stop

Inserting string in a given array of characters (8085)

2009-09-12T23:24:00.000-07:00

Statement:Write an 8085 assembly language program to insert a string of four characters from the tenth location in the given array of 50 characters.
Solution:
Step 1: Move bytes from location 10 till the end of array by four bytes downwards.
Step 2: Insert four bytes at locations 10, 11, 12 and 13.
Source Program:
LXI H, 2l31H : Initialize pointer at the last location of array.
LXI D, 2l35H : Initialize another pointer to point the last location of array after insertion.
AGAIN: MOV A, M : Get the character
STAX D : Store at the new location
DCX D : Decrement destination pointer
DCX H : Decrement source pointer
MOV A, L : [check whether desired
CPI 05H bytes are shifted or not]
JNZ AGAIN : if not repeat the process
INX H : adjust the memory pointer
LXI D, 2200H : Initialize the memory pointer to point the string to be inserted
REPE: LDAX D : Get the character
MOV M, A : Store it in the array
INX D : Increment source pointer
INX H : Increment destination pointer
MOV A, E : [Check whether the 4 bytes
CPI 04 are inserted]
JNZ REPE : if not repeat the process
HLT : stop

Find the number of negative, zero and positive numbers(8085)

2009-09-12T23:22:00.000-07:00

Statement: A list of 50 numbers is stored in memory, starting at 6000H. Find number of negative, zero and positive numbers from this list and store these results in memory locations 7000H, 7001H, and 7002H respectively.

Source Program:
LXI H, 6000H : Initialize memory pointer
MVI C, 00H : Initialize number counter
MVI B, 00H : Initialize negative number counter
MVI E, 00H : Initialize zero number counter
BEGIN:MOV A, M : Get the number
CPI 00H : If number = 0
JZ ZERONUM : Goto zeronum
ANI 80H : If MSB of number = 1i.e. if
JNZ NEGNUM number is negative goto NEGNUM
INR D : otherwise increment positive number counter
JMP LAST
ZERONUM:INR E : Increment zero number counter
JMP LAST
NEGNUM:INR B : Increment negative number counter
LAST:INX H : Increment memory pointer
INR C : Increment number counter
MOV A, C
CPI 32H : If number counter = 5010 then
JNZ BEGIN : Store otherwise check next number
LXI H, 7000 : Initialize memory pointer.
MOV M, B : Store negative number.
INX H
MOV M, E : Store zero number.
INX H
MOV M, D : Store positive number.
HLT : Terminate execution

Add parity bit to 7-bit ASCII characters(8085)

2009-09-12T23:19:00.000-07:00

Statement: Add even parity to a string of 7-bit ASCII characters. The length of the string is in memory location 2040H and the string itself begins in memory location 2041H. Place even parity in the most significant bit of each character

Source Program:
LXI H, 2040H
MOV C ,M : Counter for character
REPEAT:INX H : Memory pointer to character
MOV A,M : Character in accumulator
ORA A : ORing with itself to check parity.
JPO PAREVEN : If odd parity place
ORI 80H even parity in D7 (80).
PAREVEN:MOV M , A : Store converted even parity character.
DCR C : Decrement counter.
JNZ REPEAT : If not zero go for next character.
HLT : Terminate program execution

Transfer contents to overlapping memory blocks (8085)

2009-09-12T23:17:00.000-07:00

Statement: Write assembly language program with proper comments for the following:
A block of data consisting of 256 bytes is stored in memory starting at 3000H. This block is to be shifted (relocated) in memory from 3050H onwards. Do not shift the block or part of the block anywhere else in the memory
Source Program:
Two blocks (3000 - 30FF and 3050 - 314F) are overlapping. Therefore it is necessary to transfer last byte first and first byte last.
MVI C, FFH : Initialize counter
LX I H, 30FFH : Initialize source memory pointer 3l4FH
LXI D, 314FH : Initialize destination memory pointer
BACK: MOV A, M : Get byte from source memory block
STAX D : Store byte in the destination memory block
DCX H : Decrement source memory pointer
DCX : Decrement destination memory pointer
DCR C : Decrement counter
JNZ BACK : If counter 0 repeat
HLT : Stop execution

Separate even numbers from given numbers(8085)

2009-09-12T23:15:00.000-07:00

Statement: Write an assembly language program to separate even numbers from the given list of 50 numbers and store them in the another list starting from 2300H. Assume starting address of 50 number list is 2200H.

Source Program:
LXI H, 2200H : Initialize memory pointer l
LXI D, 2300H : Initialize memory pointer2
MVI C, 32H : Initialize counter
BACK:MOV A, M : Get the number
ANI 0lH : Check for even number
JNZ SKIP : If ODD, don't store
MOV A, M : Get the number
STAX D : Store the number in result list
INX D : Increment pointer 2
SKIP: INX H : Increment pointer l
DCR C : Decrement counter
JNZ BACK : If not zero, repeat
HLT : Stop

Add each element of array with the elements of another array (8085)

2009-09-12T23:14:00.000-07:00

Statement: Add 2 arrays having ten 8-bit numbers each and generate a third array of result. It is necessary to add the first element of array 1 with the first element of array-2 and so on. The starting addresses of array l, array2 and array3 are 2200H, 2300H and 2400H, respectively

Source Program:
LXI H, 2200H : Initialize memory pointer 1
LXI B, 2300H : Initialize memory pointer 2
LXI D, 2400H : Initialize result pointer
BACK: LDAX B : Get the number from array 2
ADD M : Add it with number in array 1
STAX D : Store the addition in array 3
INX H : Increment pointer 1
INX B : Increment pointer2
INX D : Increment result pointer
MOV A, L
CPI 0AH : Check pointer 1 for last number
JNZ BACK : If not, repeat
HLT : Stop

Add two decimal numbers of 6 digit each(8085)

2009-09-12T23:12:00.000-07:00

Statement: Two decimal numbers six digits each, are stored in BCD package form. Each number occupies a sequence of byte in the memory. The starting address of first number is 6000H Write an assembly language program that adds these two numbers and stores the sum in the same format starting from memory location 6200H.

Source Program:
LXI H, 6000H : Initialize pointer l to first number
LXI D, 6l00H : Initialize pointer2 to second number
LXI B, 6200H : Initialize pointer3 to result
STC
CMC : Carry = 0
BACK: LDAX D : Get the digit
ADD M : Add two digits
DAA : Adjust for decimal
STAX.B : Store the result
INX H : Increment pointer 1
INX D : Increment pointer2
INX B : Increment result pointer
MOV A, L
CPI 06H : Check for last digit
JNZ BACK : If not last digit repeat
HLT : Terminate program execution

Search a byte in a given number (8085)

2009-09-12T23:09:00.000-07:00

Statement: Search the given byte in the list of 50 numbers stored in the consecutive memory locations and store the address of memory location in the memory locations 2200H and 2201H. Assume byte is in the C register and starting address of the list is 2000H. If byte is not found store 00 at 2200H and 2201H.

Source program:
LX I H, 2000H : Initialize memory pointer 52H
MVI B, 52H : Initialize counter
BACK: MOV A, M : Get the number
CMP C : Compare with the given byte
JZ LAST : Go last if match occurs
INX H : Increment memory pointer
DCR B : Decrement counter
JNZ B : I f not zero, repeat
LXI H, 0000H
SHLD 2200H
JMP END : Store 00 at 2200H and 2201H
LAST: SHLD 2200H : Store memory address
END: HLT : Stop

Find the square of given number (8085)

2009-09-12T23:07:00.000-07:00

Statement:Find the square of the given numbers from memory location 6100H and store the result from memory location 7000H.

Source Program:
LXI H, 6200H : Initialize lookup table pointer
LXI D, 6100H : Initialize source memory pointer
LXI B, 7000H : Initialize destination memory pointer
BACK: LDAX D : Get the number
MOV L, A : A point to the square
MOV A, M : Get the square
STAX B : Store the result at destination memory location
INX D : Increment source memory pointer
INX B : Increment destination memory pointer
MOV A, C
CPI 05H : Check for last number
JNZ BACK : If not repeat
HLT : Terminate program execution

Calculate the sum of series of odd numbers(8085)

2009-09-12T23:05:00.000-07:00

Statement:Calculate the sum of series of odd numbers from the list of numbers. The length of the list is in memory location 2200H and the series itself begins from memory location 2201H. Assume the sum to be 16-bit. Store the sum at memory locations 2300H and 2301H

Sample problem:
2200H = 4H
2201H= 9AH
2202H= 52H
2203H= 89H
2204H= 3FH
Result = 89H + 3FH = C8H
2300H= H Lower byte
2301H = H Higher byte
Source program
LDA 2200H
MOV C, A : Initialize counter
LXI H, 2201H : Initialize pointer
MVI E, 00 : Sum low = 0
MOV D, E : Sum high = 0
BACK: MOV A, M : Get the number
ANI 0lH : Mask Bit 1 to Bit7
JZ SKIP : Don't add if number is even
MOV A, E : Get the lower byte of sum
ADD M : Sum = sum + data
MOV E, A : Store result in E register
JNC SKIP
INR D : Add carry to MSB of SUM
SKIP: INX H : Increment pointer

Calculate the sum of series of even numbers(8085)

2009-09-12T23:03:00.001-07:00

Statement:Calculate the sum of series of even numbers from the list of numbers. The length of the list is in memory location 2200H and the series itself begins from memory location 2201H. Assume the sum to be 8 bit number so you can ignore carries and store the sum at memory location 2210H

Sample problem:
2200H= 4H
2201H= 20H
2202H= l5H
2203H= l3H
2204H= 22H
Result 22l0H= 20 + 22 = 42H
= 42H
Source program:
LDA 2200H
MOV C, A : Initialize counter
MVI B, 00H : sum = 0
LXI H, 2201H : Initialize pointer
BACK: MOV A, M : Get the number
ANI 0lH : Mask Bit l to Bit7
JNZ SKIP : Don't add if number is ODD
MOV A, B : Get the sum
ADD M : SUM = SUM + data
MOV B, A : Store result in B register
SKIP: INX H : increment pointer
DCR C : Decrement counter
JNZ BACK : if counter 0 repeat
STA 2210H : store sum
HLT : Terminate program execution

Arrange in ascending order (8085)

2009-09-12T23:00:00.000-07:00

Statement:Write a program to sort given 10 numbers from memory location 2200H in the ascending order

Source program:
MVI B, 09 : Initialize counter
START : LXI H, 2200H: Initialize memory pointer
MVI C, 09H : Initialize counter 2
BACK: MOV A, M : Get the number
INX H : Increment memory pointer
CMP M : Compare number with next number
JC SKIP : If less, don't interchange
JZ SKIP : If equal, don't interchange
MOV D, M
MOV M, A
DCX H
MOV M, D
INX H : Interchange two numbers
SKIP:DCR C : Decrement counter 2
JNZ BACK : If not zero, repeat
DCR B : Decrement counter 1
JNZ START
HLT : Terminate program execution

Count number of one's in a number (8085)

2009-09-12T22:58:00.000-07:00

Statement:Write a program to count number of l's in the contents of D register and store the count in the B register.

Source program:
MVI B, 00H
MVI C, 08H
MOV A, D
BACK: RAR
JNC SKIP
INR B
SKIP: DCR C
JNZ BACK
HLT

Find the largest of given numbers(8085)

2009-09-12T22:54:00.000-07:00

Statement:Find the largest number in a block of data. The length of the block is in memory location 2200H and the block itself starts from memory location 2201H.
Store the maximum number in memory location 2300H. Assume that the numbers in the block are all 8 bit unsigned binary numbers

Sample problem
(2200H) = 04
(2201H) = 34H
(2202H) = A9H
(2203H) = 78H
(2204H) =56H
Result = (2202H) = A9H
Source program
LDA 2200H
MOV C, A : Initialize counter
XRA A : Maximum = Minimum possible value = 0
LXI H, 2201H : Initialize pointer
BACK: CMP M : Is number> maximum
JNC SKIP : Yes, replace maximum
MOV A, M
SKIP: INX H
DCR C
JNZ BACK
STA 2300H : Store maximum number
HLT : Terminate program execution

Find the negative numbers in a block of data(8085)

2009-09-12T22:52:00.000-07:00

Statement:Find the number of negative elements (most significant bit 1) in a block of data. The length of the block is in memory location 2200H and the block itself begins in memory location 2201H. Store the number of negative elements in memory location 2300H

Sample problem
(2200H) = 04H
(2201H) = 56H
(2202H) = A9H
(2203H) = 73H
(2204H) = 82H
Result = 02 since 2202H and 2204H contain numbers with a MSB of 1.
Source program
LDA 2200H
MOV C, A : Initialize count
MVI B, 00 : Negative number = 0
LXI H, 2201H : Initialize pointer
BACK: MOV A, M : Get the number
ANI 80H : Check for MSB
JZ SKIP : If MSB = 1
INR B : Increment negative number count
SKIP: INX H : Increment pointer
DCR C : Decrement count
JNZ BACK : If count 0 repeat
MOV A, B
STA 2300H : Store the result
HLT : Terminate program execution

Divide a 16 bit number by a 8-bit number(8085)

2009-09-12T22:50:00.000-07:00

Statement:Divide 16 bit number stored in memory locations 2200H and 2201H by the 8 bit number stored at memory location 2202H. Store the quotient in memory locations 2300H and 2301H and remainder in memory locations 2302H and 2303H

Sample problem
(2200H) = 60H
(2201H) = A0H
(2202H) = l2H
Result = A060H/12H = 8E8H Quotient and 10H remainder
(2300H) = E8H
(2301H) = 08H
(2302H= 10H
(2303H) 00H
Source program
LHLD 2200H : Get the dividend
LDA 2202H : Get the divisor
MOV C, A
LXI D, 0000H : Quotient = 0
BACK: MOV A, L
SUB C : Subtract divisor
MOV L, A : Save partial result
JNC SKIP : if CY 1 jump
DCR H : Subtract borrow of previous subtraction
SKIP: INX D : Increment quotient
MOV A, H
CPI, 00 : Check if dividend <>

Multiply two 8-bit numbers(8085)

2009-09-12T22:48:00.001-07:00

Statement: Multiply two 8-bit numbers stored in memory locations 2200H and 2201H by repetitive addition and store the result in memory locations 2300H and 2301H

Sample problem:
(2200H) = 03H
(2201H) = B2H
Result = B2H + B2H + B2H = 216H
= 216H
(2300H) = 16H
(2301H) = 02H
Source program
LDA 2200H
MOV E, A
MVI D, 00 : Get the first number in DE register pair
LDA 2201H
MOV C, A : Initialize counter
LX I H, 0000 H : Result = 0
BACK: DAD D : Result = result + first number
DCR C : Decrement count
JNZ BACK : If count 0 repeat
SHLD 2300H : Store result
HLT : Terminate program execution

Calculate the sum of series of numbers(8085)

2009-09-12T22:45:00.000-07:00

Statement: Calculate the sum of series of numbers. The length of the series is in memory location 4200H and the series begins from memory location 4201H.
a. Consider the sum to be 8 bit number. So, ignore carries. Store the sum at memory location 4300H.
b. Consider the sum to be 16 bit number. Store the sum at memory locations 4300H and 4301H.

a. Sample problem
4200H = 04H
4201H = 10H
4202H = 45H
4203H = 33H
4204H = 22H
Result = 10 +41 + 30 + 12 = H
4300H = H
Source program:
LDA 4200H
MOV C, A : Initialize counter
SUB A : sum = 0
LXI H, 420lH : Initialize pointer
BACK: ADD M : SUM = SUM + data
INX H : increment pointer
DCR C : Decrement counter
JNZ BACK : if counter 0 repeat
STA 4300H : Store sum
HLT : Terminate program execution

b. Sample problem
4200H = 04H
420lH = 9AH
4202H = 52H
4203H = 89H
4204H = 3EH
Result = 9AH + 52H + 89H + 3EH = H
4300H = B3H Lower byte
4301H = 0lH Higher byte
Source program:
LDA 4200H
MOV C, A : Initialize counter
LXI H, 4201H : Initialize pointer
SUB A :Sum low = 0
MOV B, A : Sum high = 0
BACK: ADD M : Sum = sum + data
JNC SKIP
INR B : Add carry to MSB of SUM
SKIP: INX H : Increment pointer
DCR C : Decrement counter
JNZ BACK : Check if counter 0 repeat
STA 4300H : Store lower byte
MOV A, B
STA 4301H : Store higher byte
HLT :Terminate program execution

Alter the contents of flag register in 8085

2009-09-12T22:38:00.000-07:00

Statement: Write a set of instructions to alter the contents of flag register in 8085.

PUSH PSW : Save flags on stack
POP H : Retrieve flags in 'L'
MOV A, L : Flags in accumulator
CMA : Complement accumulator
MOV L, A : Accumulator in 'L'
PUSH H : Save on stack
POP PSW : Back to flag register
HLT :Terminate program execution

Left Shifting of a 16-bit data(8085)

2009-09-12T22:35:00.000-07:00

Statement: Program to shift a 16-bit data 1 bit left. Assume data is in the HL register pair.

Source program:

DAD H : Adds HL data with HL data

Right shift bit of data(8085)

2009-09-12T22:31:00.000-07:00

Statement:Write a program to shift an eight bit data four bits right. Assume that data is in register C.

Source program:
MOV A, C
RAR
RAR
RAR
RAR
MOV C, A
HLT

Execution format of instructions(8085)

2009-09-12T22:28:00.000-07:00

Statement:Read the program given below and state the contents of all registers after the execution of each instruction in sequence

Main program:
4000H LXI SP, 27FFH
4003H LXI H, 2000H
4006H LXI B, 1020H
4009H CALL SUB
400CH HLT
Subroutine program:
4100H SUB: PUSH B
4101H PUSH H
4102H LXI B, 4080H
4105H LXI H, 4090H
4108H SHLD 2200H
4109H DAD B
410CH POP H
410DH POP B
410EH RET
Note:
The table given gives the instruction sequence and the contents of all registers and stack after execution of each instruction

Unpack a BCD number(8085)

2009-09-12T22:26:00.000-07:00

Statement: Two digit BCD number is stored in memory location 4200H. Unpack the BCD number and store the two digits in memory locations 4300H and 4301H such that memory location 4300H will have lower BCD digit.

Sample problem
(4200H) = 58
Result = (4300H) = 08 and
(4301H) = 05
Source program
LDA 4200H : Get the packed BCD number
ANI FOH : Mask lower nibble
RRC
RRC
RRC
RRC : Adjust higher BCD digit as a lower digit
STA 4301H : Store the partial result
LDA 4200H : .Get the original BCD number
ANI OFH : Mask higher nibble
STA 4201H : Store the result
HLT : Terminate program execution

Pack the unpacked BCD numbers(8085)

2009-09-12T22:24:00.000-07:00

Statement: Pack the two unpacked BCD numbers stored in memory locations 4200H and 4201H and store result in memory location 4300H. Assume the least significant digit is stored at 4200H.

Sample problem:
(4200H) = 04
(4201H) = 09
Result = (4300H) = 94
Source program
LDA 4201H : Get the Most significant BCD digit
RLC
RLC
RLC
RLC : Adjust the position of the second digit (09 is changed to 90)
ANI FOH : Make least significant BCD digit zero
MOV C, A : store the partial result
LDA 4200H : Get the lower BCD digit
ADD C : Add lower BCD digit
STA 4300H : Store the result
HLT : Terminate program execution
NOTE:
BCD NO.: The numbers "0 to 9" are called BCD (Binary Coded Decimal) numbers. A decimal number 29 can be converted into BCD number by splitting

Finding Two's complement of a number(8085)

2009-09-12T22:22:00.000-07:00

Statement: Find the 2's complement of the number stored at memory location 4200H and store the complemented number at memory location 4300H.
Sample problem:
(4200H) = 55H
Result = (4300H) = AAH + 1 = ABH
Source program:
LDA 4200H : Get the number
CMA : Complement the number
ADI, 01 H : Add one in the number
STA 4300H : Store the result
HLT : Terminate program execution

Finding one's complement of a number(8085)

2009-09-12T22:20:00.000-07:00

Statement: Find the l's complement of the number stored at memory location 4400H and store the complemented number at memory location 4300H

Sample problem:
(4400H) = 55H
Result = (4300B) = AAB
Source program:
LDA 4400B : Get the number
CMA : Complement number
STA 4300H : Store the result
HLT : Terminate program execution

Subtract two 16-bit numbers(8085)

2009-09-12T22:18:00.000-07:00

Statement: Subtract the 16-bit number in memory locations 4002H and 4003H from the 16-bit number in memory locations 4000H and 4001H. The most significant eight bits of the two numbers are in memory locations 4001H and 4003H. Store the result in memory locations 4004H and 4005H with the most significant byte in memory location 4005H.
Sample problem
(4000H) = 19H
(400IH) = 6AH
(4004H) = I5H (4003H) = 5CH
Result = 6A19H - 5C15H = OE04H
(4004H) = 04H
(4005H) = OEH
Source program:
LHLD 4000H : Get first 16-bit number in HL
XCHG : Save first 16-bit number in DE
LHLD 4002H : Get second 16-bit number in HL
MOV A, E : Get lower byte of the first number
SUB L : Subtract lower byte of the second number
MOV L, A : Store the result in L register
MOV A, D : Get higher byte of the first number
SBB H : Subtract higher byte of second number with borrow
MOV H, A : Store l6-bit result in memory locations 4004H and 4005H.
SHLD 4004H : Store l6-bit result in memory locations 4004H and 4005H.
HLT : Terminate program execution

Add contents of two memory locations(8085)

2009-09-12T22:15:00.000-07:00

Statement: Add the contents of memory locations 40001H and 4001H and place the result in the memory locations 4002Hand 4003H

Sample problem:
(4000H) = 7FH
(400lH) = 89H
Result = 7FH + 89H = lO8H
(4002H) = 08H
(4003H) = 0lH
Source program:
LXI H, 4000H :HL Points 4000H
MOV A, M :Get first operand
INX H :HL Points 4001H
ADD M :Add second operand
INX H :HL Points 4002H
MOV M, A :Store the lower byte of result at 4002H
MVIA, 00 :Initialize higher byte result with 00H
ADC A :Add carry in the high byte result
INX H :HL Points 4003H
MOV M, A :Store the higher byte of result at 4003H
HLT :Terminate program execution

Add two 16-bit numbers (8085)

2009-09-12T22:03:00.000-07:00

Statement: Add the 16-bit number in memory locations 4000H and 4001H to the 16-bit number in memory locations 4002H and 4003H. The most significant eight bits of the two numbers to be added are in memory locations 4001H and 4003H. Store the result in memory locations 4004H and 4005H with the most significant byte in memory location 4005H.

Program - 5.a: Add two 16-bit numbers - Source Program 1
Sample problem:
(4000H) = 15H
(4001H) = 1CH
(4002H) = B7H
(4003H) = 5AH
Result = 1C15 + 5AB7H = 76CCH
(4004H) = CCH
(4005H) = 76H
Source Program 1:
LHLD 4000H : Get first I6-bit number in HL
XCHG : Save first I6-bit number in DE
LHLD 4002H : Get second I6-bit number in HL
MOV A, E : Get lower byte of the first number
ADD L : Add lower byte of the second number
MOV L, A : Store result in L register
MOV A, D : Get higher byte of the first number
ADC H : Add higher byte of the second number with CARRY
MOV H, A : Store result in H register
SHLD 4004H : Store I6-bit result in memory locations 4004H and 4005H.
HLT : Terminate program execution

Program - 5b: Add two 16-bit numbers - Source Program 2
Source program 2:
LHLD 4000H : Get first I6-bit number
XCHG : Save first I6-bit number in DE
LHLD 4002H : Get second I6-bit number in HL
DAD D : Add DE and HL
SHLD 4004H : Store I6-bit result in memory locations 4004H and 4005H.
HLT : Terminate program execution

Subtract two 8-bit numbers (8085)

2009-09-12T21:59:00.000-07:00

Statement: Subtract the contents of memory location 4001H from the memory location 2000H and place the result in memory location 4002H.
Program - 4: Subtract two 8-bit numbers
Sample problem:
(4000H) = 51H
(4001H) = 19H
Result = 51H - 19H = 38H
Source program:
LXI H, 4000H : HL points 4000H
MOV A, M : Get first operand
INX H : HL points 4001H
SUB M : Subtract second operand
INX H : HL points 4002H
MOV M, A : Store result at 4002H.
HLT : Terminate program execution

Add two 8-bit numbers (8085)

2009-09-12T21:52:00.000-07:00

Sample problem
(4000H) = 14H
(4001H) = 89H
Result = 14H + 89H = 9DH
Source program
LXI H 4000H : HL points 4000H
MOV A, M : Get first operand
INX H : HL points 4001H
ADD M : Add second operand
INX H : HL points 4002H
MOV M, A : Store result at 4002H
HLT : Terminate program execution

Exchange the contents of memory locations(8085)

2009-09-12T21:51:00.000-07:00

Program 1:
LDA 2000H : Get the contents of memory location 2000H into accumulator
MOV B, A : Save the contents into B register
LDA 4000H : Get the contents of memory location 4000Hinto accumulator
STA 2000H : Store the contents of accumulator at address 2000H
MOV A, B : Get the saved contents back into A register
STA 4000H : Store the contents of accumulator at address 4000H
Program 2:
LXI H 2000H : Initialize HL register pair as a pointer to memory location 2000H.
LXI D 4000H : Initialize DE register pair as a pointer to memory location 4000H.
MOV B, M : Get the contents of memory location 2000H into B register.
LDAX D : Get the contents of memory location 4000H into A register.
MOV M, A : Store the contents of A register into memory location 2000H.
MOV A, B : Copy the contents of B register into accumulator.
STAX D : Store the contents of A register into memory location 4000H.
HLT : Terminate program execution.
In Program 1, direct addressing instructions are used, whereas in Program 2, indirect addressing instructions are used.