MMAL Final Solve

Question 1Marks

a. Using shift instructions, write the codes for 8086 microprocessor.08

1. Put the value 1234H in AX and multiply the value by 8.

   MOV AX, 1234H
   MOV CL, 3
   SHL AX, CL

2. Divide the number 2345H by 8 and put the quotient in AX.

   MOV AX, 2345H
   MOV CL, 3
   SHR AX, CL

b. Write an assembly program to find the smallest number in the given array, $[A]=[23,21,10,12,14].$ 10

.MODEL SMALL
.STACK 100H
 
.DATA
    ARRAY DB 23, 21, 10, 12, 14
    LENGTH DW 5
    SMALLEST DB ?
 
.CODE
MAIN PROC
    MOV AX, @DATA
    MOV DS, AX
    
    LEA SI, ARRAY
    MOV CX, LENGTH
    MOV AL, [SI]
    MOV SMALLEST, AL
    INC SI
    DEC CX
 
FIND_MIN:
    CMP CX, 0
    JE DONE
    MOV AL, [SI]
    CMP AL, SMALLEST
    JGE SKIP
    MOV SMALLEST, AL
 
SKIP:
    INC SI
    DEC CX
    JMP FIND_MIN
 
DONE:
    MOV AH, 4CH
    INT 21H
    
MAIN ENDP
END MAIN

c. Give a logic instruction to do each of the following:02

1. Set the MSB and LSB of BL register, leaving other bits unchanged.

   OR BL, 10000001b              ; OR 1 -> Set

2. Clear the even numbered bits of register AX, leaving the other bits unchanged.

   AND AX, 1010101010101010b     ; AND 0 -> Clear

Question 2

a. Write a 8086 assembly code to solve the summation of the given series: 08

$$1^2+2^2+3^2+4^2+\dots .+10^2$$

MOV SI, 1                ; Counter for Series
MOV BX, 0               
MOV CX, 10               ; Defualt counter for LOOP
 
Continue:               
	MOV AX, SI
	MUL SI               ; AX = AX * SI
	ADD BX, AX           ; ADD BX with AX
	INC SI
	LOOP Continue        ; Continue LOOP untill CX=0
 
MOV AX, BX               ; Store the sum of Series in AX
HLT

b. Name the registers that are involved in the stack operation of 8086 microprocessor. Suppose, the initial value of SP is 0100:1120 and FR=05H. Complete the table below for each instructions of the assembly language.10

Instructions	SP	Value	AX	BX
MOV AX, 1234H	$1120H$	-	$1234H$	-
MOV BX, 2345H	$1120H$	-	$1234H$	$2345H$
PUSH AX	$111EH$	$1234H$	$1234H$	$2345H$
POP BX	$1120H$	-	$1234H$	$1234H$
PUSHF	$111EH$	$0005H$	$1234H$	$1234H$
POP BX	$1120H$	-	$1234H$	$0005H$
NOT BX	$1120H$	-	$1234H$	$FFFAH$

Tip

• When a word (16 bits) is pushed, the SP is decremented by 2. Conversely, when a word is popped, the SP is incremented by 2.
• PUSHF pushes the 16-bit FLAGS register onto the stack (decrementing the Stack Pointer by 2), and POPF pops a word from the stack into the FLAGS register

c. When the stack is completely filled the stack area, SP=0, if a data is pushed onto the stack, what would happen to SP?02

Ans: When the stack is completely filled and SP=0, if a data is pushed onto the stack, SP would wrap around to FFFFH. This is a stack overflow condition and causes serious problems. The data overwrites memory outside the intended stack area. It can corrupt code, data, or other critical memory regions.

Question 3

Question 4

a. Suppose, R1 contains 64H, CY=1. What will be the value of R1 and CY after each successive instruction is executed:10

1. RR R1
2. RRC R1
3. INC R1
4. SWAP R1
5. RLC R1

Ans:

1. RR R1
   The RR instruction rotates the bits of the register to the right. The Least Significant Bit (LSB) moves to the Most Significant Bit (MSB) position.
   
   After RR R1,
   R1 = $32H$
   CY= $1$ (Unchanged)
   .

2. RRC R1
   The RRC instruction rotates the bits to the right through the Carry Flag. The LSB moves into the Carry Flag, and the previous value of the Carry Flag moves into the MSB.
   After RRC R1,
   R1 = $99H$
   CY = $0$
   .

3. INC R1
   The INC instruction adds 1 to the current value of the register.
   R1 = $99H+1H=9AH$
   CY = $0$
   .

4. SWAP R1 The SWAP instruction swaps the lower $4bit$ with higher $4bit$ . R1 = $A9H$
   CY = $0$ (Doesn’t effect carry flag)
   .

5. RLC R1
   The RLC instruction rotates the bits to the left through the Carry Flag. The MSB moves into the Carry Flag, and the previous value of the Carry Flag moves into the LSB.
   After RLC R1,
   R1 = $52H$
   CY = $1$

b. Show that the maximum external ROM size of 8051 microcontroller is $60KB$ and $64KB$ in different cases.08

Ans: The 8051 microcontroller architecture supports external memory expansion through a 16-bit address bus, allowing it to address external program memory (ROM) and external data memory (RAM). The maximum addressable external ROM size varies depending on the hardware configuration and the status of the External Access (EA) pin.

1. Maximum External ROM = 64KB
   This maximum is achieved when the internal ROM is bypassed entirely. When the External Access (EA) pin is connected to ground ($EA=0$), the microcontroller ignores its internal memory and fetches all instructions from the external ROM starting at address $0000\text{H}$. A 16-bit address bus allows the processor to access $2^{16}$ unique memory locations.

Address bus width = 16 bits 
Maximum addressable locations = 2^16 = 65,536 locations 
Each location = 1 byte 
Maximum external ROM = 65,536 bytes = 64KB

2. Maximum External ROM = 60KB
   This constraint applies when the microcontroller utilizes its on-chip program memory alongside external expansion. Standard 8051 variants contain $4\text{ KB}$ of internal ROM located at addresses $0000\text{H}$ to $0FFF\text{H}$. When the External Access (EA) pin is held high ($EA=1$), the processor executes instructions from the $4\text{ KB}$ internal ROM first. The microcontroller only switches to the external bus for addresses $1000\text{H}$ and above, effectively leaving $60\text{ KB}$ of accessible external program space ($64\text{ KB}-4\text{ KB}=60\text{ KB}$).

Total addressable code space = 64KB
Internal ROM space (EA=1) = 4KB 
Available external ROM = 64KB - 4KB = 60KB

With $\overline{EA}=HIGH$ (Internal ROM enabled):
With $\overline{EA}=LOW$ (External ROM only):

c. Distinguish between Microprocessor and Microcontroller. 02

Ans:

Microprocessor	Microcontroller
CPU is stand-alone. RAM, ROM, I/O are separate.	CPU, RAM, ROM, I/O are on a single chip.
Generally large in size.	Small and compact
More expensive	Cheaper than Microprocessor
Higher power consumption	Lower power consumption

Question 5

a. Demonstrate how memory is interfaced in 8051 micrcontroller with diagram.10

Question 6

a. With proper diagrams, explain the special function registers (SFR) of a 8051 microcontroller.10

Ans: Special Function Register (SFR) sits on top of 128 byte RAM of 8051 starting from memory location $80H$ to $FFH$. Total size of SRF is also 128 bytes.

SFR contains registers like microprocessor.

1. Accumulator (A & B)
   Performs math and logical operations. Mainly works A, involves B for multiplication and division.

Register	Size	Location
A	8 bit	$EOH$
B	8 bit	$FOH$

2. Status Register (Flag)
   Size = 8 Bit Address = $DOH$

CY	AX	FO	RS1	RS0	OV	-	P
Carry	Auxilary	Ext. Ram	Bank	Bank	Overflow		Parity

RS1	RS0	Bank
0	0	$B_0$
0	1	$B_1$
1	0	$B_2$
1	1	$B_3$

3. Pointer Register
   To indicate CPU which Bank we are using.
   $DPH\to$ Higher Bank ($B_2,B_3$) $83H$
   $DPL\to$ Lower Bank ($B_0,B_1$) $82H$
4. Stack Pointer
   Address = $81H$
5. I/O Port Register
   8 Bit Each
   $P_0\to 80H$
   $P_1\to 90H$
   $P_2\to A0H$
   $P_3\to B0H$

b. Draw the block diagram of a 8051 Microcontroller.08

c. Distinguish between RAM and ROM of 8051 microcontroller. 02

ROM	RAM
Read only memory	Random access memory
Program memory	Data memory
4KB ROM	128 Byte RAM
$\overline{EA}$ pin works under ROM	Not functional in RAM
Non Volatile	Volatile