Lecture-9.2: Addressing Modes related Mathematics in 8086 Microprocessor - Part-2(MPU)

  Addressing Modes related Mathematics in 8086 Microprocessor 

Rules of Segmentation process follows some rules as follows:

The starting address of a segment should be such that it can be evenly divided by 16.

Minimum size of a segment can be 16 bytes and the maximum can be 64 kB.

Segment	Offset Registers	Function
CS	IP	Address of the next instruction
DS	BX, DI, SI	Address of data
SS	SP, BP	Address in the stack
ES	BX, DI, SI	Address of destination data  (for string operations)

 

Physical Address Calculation

The formula to calculate the physical address (also called the absolute address) from a segmented address involves multiplying the segment address by a constant and then adding the offset address:

·       Effective Physical Address = Starting address of Segment + Offset 

 

Problem – 1: Let the content of different registers in 8086 be as follows:

DS = 1000H, SS = 2000H, ES = 3000H, BP = 7000H, SI = 5000H, DI = 6000H,

BX = 4000H

i.                MOV AX [BX]

ii.              MOV BX [SI] 

iii.            MOV CX [BP] 

iv.            MOV AL [DI]   

v.              MOV BH, SS: [SI] 

vi.            MOV CX, ES: [DI] 

vii.          MOV AH, [BX + 10H] 

Find the memory address/addresses from where the 8086 accesses the data while executing the following instructions:

 

Solution:

The formula for calculating the Physical Address is:

Physical Address=(Segment×10H) + Offset [Note: ×10H is for converting 16-bit to 20-bit address]

1)    MOV AX, [BX]

Default Segment: DS

Offset: BX

P.A = (DS × 10H) + BX

       = (1000H × 10H) + 4000H

       = 10000H+4000H

       = 14000H

So, the physical address is 14000H, 14001H

2)     MOV BX, [SI]

Default Segment: DS

Offset: SI

P.A = (DS × 10H) + SI

      = (1000H × 10H) + 5000H

      = 10000H + 5000H

      = 15000H

So, the physical address is 15000H,15001H

3)    MOV CX, [BP]

Default Segment: SS

Offset: BP

P.A = (SS × 10H) + BP

      = (2000H × 10H) + 7000H

      = 20000H + 7000H

      = 27000H

So, the physical address is 27000H

4)     MOV AL, [DI]

Default Segment: DS

Offset: DI

P.A = (DS × 10H) + DI

      = (1000H × 10H) + 6000H

      = 10000H + 6000H

      = 16000H

So, the physical address is 16000H

5)    MOV BH, SS: [SI]

Explicit Segment: SS

Offset: SI

P.A = (SS × 10H) + SI

      = (2000H × 10H) + 5000H

      = 20000H + 5000H

      = 25000H

So, the physical address is 25000H

6)     MOV CX, ES: [DI]

Explicit Segment: ES

Offset: DI

P.A = (ES × 10H) + DI

      = (3000H × 10H) + 6000H

      = 30000H + 6000H

      = 36000H

So, the physical address is 36000H,36001H

7)     MOV AH, [BX + 10H]

Default Segment: DS

Offset: BX + 10H

P.A = (DS × 10H) + (BX + 10H)

      = (1000H × 10H) + (4000H + 10H)

      = 10000H + 4010H

      = 14010H

So, the physical address is 14010H

 

Problem – 2: Let the content of different registers in 8086 be as follows:

DS = 2000H, SS = 8000H, ES = 5000H, BP = 0200H, SI = 0030H, DI = 0300H,

BX = 0800H, Value (displacement) = 0400H

i.                MOV AX, [BX] [SI]

ii.              ADD AL, [BP + 20H]

iii.            MOV CX, Values [BP]

iv.            MOV ES: [1500H], 30H

v.              MOV BX, [SI + 10H]

vi.            MOV CX, [BX + DI]

vii.          MOV AL, ES: [DI + 100H]

 

Find the memory address/addresses from where the 8086 accesses the data    while executing the following instructions:

 

Solution:

The formula for calculating the Physical Address is:

Physical Address=(Segment×10H) + Offset  [Note: ×10H is for converting 16-bit to 20-bit address]

 

i. MOV AX, [BX] [SI]

Segment: DS (Default Segment)

Offset: BX + SI

P.A = (DS × 10H) + [BX + SI]

      = (2000H × 10H) + (0800H + 0030H)

      = 20000H + 0830H

      = 20830H

So, the physical address is 20830H

ii. ADD AL, [BP + 20H]

Segment: SS (Stack Segment, used by default when BP is involved)

Offset: BP + 20H

P.A = (SS × 10H) + [BP + 20H]

      = (8000H × 10H) + (0200H + 20H)

      = 80000H + 0220H

      = 80220H

So, the physical address is 80220H

iii. MOV CX, Values [BP]

Segment: SS (Stack Segment, used by default when BP is involved)

Offset: BP

P.A = (SS × 10H) + [BP + Values]

      = (8000H × 10H) + (0200H + 0400H)

      = 80000H + 0600H

      = 80600H

So, the physical address is 80600H

iv. MOV ES: [1500H], 30H

Segment: ES (Explicitly specified segment)

Offset: 1500H

P.A = (ES × 10H) + (1500H + 30H)

      = (5000H × 10H) + (1500H + 30H)

      = 50000H + 1530H

      = 51530H

So, the physical address is 51530H

v. MOV BX, [SI + 10H]

Segment: DS (Default Segment)

Offset: SI + 10H

P.A = (DS × 10H) + (SI + 10H)

      = (2000H × 10H) +(0030H + 10H)

      = 20000H + 0040H

      = 20040H

So, the physical address is 20040H, 20041H

vi. MOV CX, [BX + DI]

Segment: DS (Default Segment)

Offset: BX + DI

P.A = (DS × 10H) + (BX + DI)

      = (2000H × 10H) + (0800H + 0300H)

      = 20000H + 0B00H

      = 20B00H

So, the physical address is 20B00H, 20B01H

vii. MOV AL, ES: [DI + 100H]

Segment: ES (Explicitly specified segment)

Offset: DI + 100H

P.A = (ES × 10H) + (DI + 100H)

      = (5000H × 10H) + (0300H + 100H)

      = 50000H + 0400H

      = 50400H

So, the physical address is 50400H

 

Problem – 3: Let the content of different registers in 8086 be as follows:

DS = 1000H, SS = 2000H, ES = 3000H, BP = 7000H, SI = 5000H, DI = 6000H, BX = 4000H

      i.          mov AX, [BX]

    ii.          mov BX, [SI]

  iii.          mov CX, [BP]

  iv.          mov AL, [DI]

     v.          mov BH, SS: [SI]

  vi.          mov CX, ES: [DI]

vii.          mov AX, [BX + DI]

viii.          mov BX, [BP + DI + 5]

  ix.          mov AH, [BX + 10H]

     x.          mov EX, DS: [BP + 4]

  xi.          mov BX, [SI - 5]

xii.          mov AX, [BX + 10]

Find the memory address/addresses from where the 8086 accesses the data while executing the following instructions: