1

LOGO

Oleh. Junartho Halomoan (juned_new@yahoo.com)

Mikroprosesor dan AntarmukaDasar Mikroprosesor Intel 8088

Dasar Sistem Komputer [1]

JUN - I

TTELKOM

2

Mikroprosesor

Memory I/O

Data Bus

Address Bus

Control Bus

Dasar Sistem Komputer [2]

Dasar Sistem Komputer [3]

JUN - I

TTELKOM

3

Dasar Sistem Komputer [4]

Arsitektur Bus ada 3 buah:

Address:

If I/O, a value between 0000H and FFFFH is issued. If memory, it depends on the architecture: 20-bits (8086/8088) 24-bits (80286/80386SX) 25-bits (80386SL/SLC/EX) 32-bits (80386DX/80486/Pentium) 36-bits (Pentium Pro/II/III)

Dasar Sistem Komputer [5]

Data:

8-bits (8088) 16-bits (8086/80286/80386SX/SL/SLC/EX) 32-bits (80386DX/80486/Pentium) 64-bits (Pentium/Pro/II/III)

Control:

Most systems have at least 4 control bus connections (active low). MRDC (Memory ReaD Control), MWRC, IORC

(I/O Read Control), IOWC (I/O Write Control), .

JUN - I

TTELKOM

4

Dasar Sistem Komputer [5]

Dasar Sistem Komputer [6]

JUN - I

TTELKOM

5

GND

AD14

AD13

AD12

AD11

AD10

AD9

AD8

AD7

AD6

AD5

AD4

AD3

AD2

AD1

AD0

NMI

INTR

CLK

GND

VCC

AD15

A16/S3

A17/S4

A18/S5

A19/S6

HOLD

HLDA

ALE

READY

RESET

BHE/S7

MN/MX

RD

WR

M/IO

DT/R

DEN

INTA

TEST

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

31

30

29

28

27

26

25

24

23

22

21

40

39

38

37

36

35

34

33

328086

Data

Bus

GND

A14

A13

A12

A11

A10

A9

A8

AD7

AD6

AD5

AD4

AD3

AD2

AD1

AD0

NMI

INTR

CLK

GND

VCC

A15

A16/S3

A17/S4

A18/S5

A19/S6

HOLD

HLDA

ALE

READY

RESET

SS0

MN/MX

RD

WR

IO/M

DT/R

DEN

INTA

TEST

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

31

30

29

28

27

26

25

24

23

22

21

40

39

38

37

36

35

34

33

328088

Konfigurasi Pin 8088 [1]

GND

AD14

AD13

AD12

AD11

AD10

AD9

AD8

AD7

AD6

AD5

AD4

AD3

AD2

AD1

AD0

NMI

INTR

CLK

GND

VCC

AD15

A16/S3

A17/S4

A18/S5

A19/S6

HOLD

HLDA

ALE

READY

RESET

BHE/S7

MN/MX

RD

WR

M/IO

DT/R

DEN

INTA

TEST

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

31

30

29

28

27

26

25

24

23

22

21

40

39

38

37

36

35

34

33

328086

INTR: Interrupt Request

-Aktif pada saat level high

-Masukan

-Di monitor pada cycle clock

terakhir setiap instruksi

NMI: Nonmaskable interrupt

-Positif Edge triggered signal

-Masukan

-Tidak dapat di-masked

(enable/disable) secara s/w

Konfigurasi Pin 8088 [2]

JUN - I

TTELKOM

6

GND

AD14

AD13

AD12

AD11

AD10

AD9

AD8

AD7

AD6

AD5

AD4

AD3

AD2

AD1

AD0

NMI

INTR

CLK

GND

VCC

AD15

A16/S3

A17/S4

A18/S5

A19/S6

HOLD

HLDA

ALE

READY

RESET

BHE/S7

MN/MX

RD

WR

M/IO

DT/R

DEN

INTA

TEST

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

31

30

29

28

27

26

25

24

23

22

21

40

39

38

37

36

35

34

33

328086

Clock

- Mikroprosesor membutuhkan clock

untuk sinkronisasi semua aktifitasnya

- Maksimum 10 MHz

- Clok generator : i8284

Konfigurasi Pin 8088 [3]

GND

AD14

AD13

AD12

AD11

AD10

AD9

AD8

AD7

AD6

AD5

AD4

AD3

AD2

AD1

AD0

NMI

INTR

CLK

GND

VCC

AD15

A16/S3

A17/S4

A18/S5

A19/S6

HOLD

HLDA

ALE

READY

RESET

BHE/S7

MN/MX

RD

WR

M/IO

DT/R

DEN

INTA

TEST

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

31

30

29

28

27

26

25

24

23

22

21

40

39

38

37

36

35

34

33

328086

Reset

•Menghentikan aktifitas mp saat itu

•masukan

•Aktif ‘high’

•Kondisi register di dalam

mikroprosesor setelah reset:

Contentsregister

FFFFHCS

0000HDS

0000HSS

0000HES

0000HIP

0000HFLAG

KosongQUEUE

Konfigurasi Pin 8088 [4]

JUN - I

TTELKOM

7

GND

AD14

AD13

AD12

AD11

AD10

AD9

AD8

AD7

AD6

AD5

AD4

AD3

AD2

AD1

AD0

NMI

INTR

CLK

GND

VCC

AD15

A16/S3

A17/S4

A18/S5

A19/S6

HOLD

HLDA

ALE

READY

RESET

BHE/S7

MN/MX

RD

WR

M/IO

DT/R

DEN

INTA

TEST

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

31

30

29

28

27

26

25

24

23

22

21

40

39

38

37

36

35

34

33

328086

READY:

•Masukan

•Aktif ‘high’

•Digunakan untuk memasukkan

waktu tunggu (wait state), untuk

akses memori atau I/O yang

lambat.

Konfigurasi Pin 8088 [5]

GND

AD14

AD13

AD12

AD11

AD10

AD9

AD8

AD7

AD6

AD5

AD4

AD3

AD2

AD1

AD0

NMI

INTR

CLK

GND

VCC

AD15

A16/S3

A17/S4

A18/S5

A19/S6

HOLD

HLDA

ALE

READY

RESET

BHE/S7

MN/MX

RD

WR

M/IO

DT/R

DEN

INTA

TEST

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

31

30

29

28

27

26

25

24

23

22

21

40

39

38

37

36

35

34

33

328086

Mode Min

Logic 1

Mode Max

Logic 0

0GT/ RQ

1GT/ RQLOCK

S2S1

0S

QS0QS1

* Pin-pin diatas digunakan untuk sinyal kontrol memori dan I/O

Konfigurasi Pin 8088 [6]

Fungsi pin 24 s/d 31 dari 8088 dan 8086 berubah sesuai mode yang

digunakan, max atau min

JUN - I

TTELKOM

8

Bus Timing [1]

Writing:

Dump address on address bus. Dump data on data bus. Issue a write (WR) and set M/IO to 1.

Bus Timing [2]

Reading:

Dump address on address bus. Issue a read (RD) and set M/IO to 1. Wait for memory access cycle.JU

N - ITTELK

OM

9

Bus Timing [3]

Bus Timing [4]

During T1:

The address is placed on the Address/Data bus. Control signals M/IO, ALE and DT/R specify memory or I/O, latch the

address onto the address bus and set the direction of data transfer on data bus.

During T2:

8086 issues the RD or WR signal, DEN, and, for a write, the data. DEN enables the memory or I/O device to receive the data for writes and the 8086 to receive the data for reads.

During T3:

This cycle is provided to allow memory to access data. READY is sampled at the end of T2. If low, T3 becomes a wait state. Otherwise, the data bus is sampled at

the end of T3. During T4:

All bus signals are deactivated, in preparation for next bus cycle. Data is sampled for reads, writes occur for writes.

JUN - I

TTELKOM

10

ESCSSSDS

4321

CONTROLSYSTEM

AH ALBH BLCH CLDH DL

SPBPSIDI

OPERANDFLAGS

C-BUS

IP

ALU

INSTRUCTIONSTREAM

BYTEQUEUE

A- BUS

BIU

EU

Diagram Internal intel 8088

Fungsi BIU (BUS Interface Unit)

Sebagai antarmuka (pengalamatan) dengan peripheral di luar mikroprosesor.

Bertanggung jawab terhadap semuaoperasi bus eksternal, seperti : Instruction fetch Operasi baca-tulis memori atau I/O Antrian instruksi dan perhitungan alamat (PA)

JUN - I

TTELKOM

11

Fungsi EU (Execution Unit)

Bertanggung jawab terhadap decoding danexecuting instruksi

Mengambil instruksi dari Queue Transfer data dari dan ke general purpose

RegistersCheck & update flag .Perhitungan alamat operand (EA)Memberi perintah ke BIU untuk operasi memori

atau I/O

Register 8088/8086 [1]

JUN - I

TTELKOM

12

Nama RegisterBitKategori

AX,BX,CX,DX16general

AH,AL,BH,BL,CH,CL,DH,DL8

SP,BP16pointer

SI,DI16index

CS,DS,SS,ES16segment

IP16instruction

FR16flag

Catatan : register general 16 bit tidak bisa digunakan bersamaan dengan

register general 8 bit, contoh : mov ax,bl tidak dilayani

Register 8088/8086 [2]

Register General Purpose [1]

EAX: Accumulator: Referenced as EAX, AX, AL or AH.

Used for mult, div, etc. Used to hold an offset.

EBX: Base Index:

Used to hold the offset of a data pointer. ECX: Count:

Used to hold the count for some instructions, REP and LOOP.

Used to hold the offset of a data pointer.

JUN - I

TTELKOM

13

Register General Purpose [2]

EDX: Data:

Used to hold a portion of the result for mult, of the operand for div.

Used to hold the offset of a data pointer. EBP: Base Pointer:

Holds the base pointer for memory data transfers. EDI: Destination Index:

Holds the base destination pointer for string instructions.

ESI: Source Index:

Holds the base source pointer for string instructions.

Register Special Purpose

EIP: Instruction Pointer:

Points to the next instruction in a code segment. 16-bits (IP) in real mode and 32-bits in protected

mode. ESP: Stack Pointer:

Used by the stack, call and return instructions. EFLAGS:

Store the state of various conditions in the microprocessor

JUN - I

TTELKOM

14

Register Flag [1]

5 flag bit yang paling kanan berubahsesudah instruksi operasi aritmatikadan operasi logika sedangkan instruksioperasi transfer data dan kontrol tidakmerubah register flag

Register Flag [2]

C (Carry):

Holds the carry out after addition or the borrow after subtraction.

Also indicates error conditions. P (Parity):

0 for odd number of bits and 1 for even. Obsolete feature of the 80x86.

A (Auxiliary Carry):

Highly specialized flag used by DAA and DAS instructions after BCD addition

or subtraction.

JUN - I

TTELKOM

15

Register Flag [3]

Z (Zero):

1 if the result of an arithmetic or logic instruction is 0. S (Sign):

1 if the sign of the result of an arith. or logic instruction is negative.

T (Trap):

Trap enable. The microprocessor interrupts the flow of instructions on conditions

indicated by the debug and control registers.

Register Flag [4]

I (Interrupt):

Controls the operation of the INTR (Interrupt request) pin. If 1, interrupts

are enabled. Set by STI and CLI instructions.

D (Direction):

Selects with increment or decrement mode for the DI and/or SI registers during

string instructions. If 1, registers are automatically decremented. Set by

STD and CLD instructions.

O (Overflow):

Set for addition and subtraction instructions.

JUN - I

TTELKOM

16

Register Segment [1]

CS (Code Segment):

In real mode, this specifies the start of a 64KB memory segment. In protected mode, it selects a descriptor. The code segment is limited to 64KB in the

8086-80286 and 4 GB in the 386 and above.

Register Segment [2]

DS (Data Segment):

Similar to the CS except this segment holds data. ES (Extra Segment):

Data segment used by some string instructions to hold destination data.

SS (Stack Segment):

Similar to the CS except this segment holds the stack. ESP and EBP hold offsets into this segment.

FS and GS: 80386 and up.

Allows two additional memory segments to be deÞned.

JUN - I

TTELKOM

17

Akses Memori

Sebelum 8088, uP mepunyai memorisekitar 64 kB

Digunakan untuk penyimpan : os, program aplikasi dan data

Peta Alamat Memori & I/O i8088

External Memory Address Space

Code segment

64 k byte

Data Segment

64 K Byte

Stack segment

64 k Byte

Extra Segment

64 k Byte

IP

SPBPSIDI

CSDSSSES

AH ALBHCHDH

BLCLDL

Input / output

Address space

00000

FFFFF

0000

FFFFLimit PC

9FFFF

JUN - I

TTELKOM

18

Peta Alamat I/O i8088/86

BIOS

Peta Memori [1]

8 bit

00000

FFFFF

Operating System

Program

AplikasiProgram

Data

00200

?

JUN - I

TTELKOM

19

Peta Memori [2] - App. 48 kB

Aplikasi : 48 kB

Dimisalkan pembagian besar memori aplikasiuntuk data dan program adalah sebagaiberikut:

• 10 kB <= data• 38 kB <= program

Besar memori untuk bagian data biasanyalebih kecil dibandingkan bagian program Tentukan alamat memori program dan data

dengan tanpa segmentasi atau menggunakansegmentasi

Peta Memori [3]- App. 48 kB

Tanpa segment:

Alamat Program Alamat awal program 200h Alamat akhir program 99FFhAlamat Data Alamat awal data 9A00h Alamat akhir data C1FFh

Contoh: Ambil data ke-100 acc

mov acc,[9A64]

Dengan segmen

Alamat Program Alamat awal program 0000h

di Code Segment Alamat akhir 97FFh di Code

SegmentAlamat Data Alamat awal data 0000h di

Data Segment Alamat akhir data 27FFh

Contoh : Ambil data ke-100 acc

mov acc,[0064]

JUN - I

TTELKOM

20

Peta Memori [4] - Segmentasi

Segmentasi

Membagi memori menjadi blok-blok 64 kB secara dinamis Dipisahkan blok memori menjadi blok fungsi

• 64 kB untuk program Code Segment• 64 kB untuk data Data Segment• 64 kB untuk stack Stack Segment• 64 kB untuk tambahan data (option) Extra

Segment

Statis vs Dinamis

BIOS

64 KB

64 KB

64 KB

64 KB

64 KB

64 KB

64 KB

64 KB

64 KB

64 KB

64 KB

64 KB

64 KB64 KB

64 KB64 KB

64 KB

64 KB

64 KB

64 KB

OS

BIOS

DS

SS

CS

ES

OS

CS

DS

SS

ES

Peta Memori [5]-statis/dinamis

JUN - I

TTELKOM

21

Peta Memori [6]-MultiSegment

Karena tersedia lebih dari 4 segment, maka bisa dibuat lebih dari 1 kelompok segment

multi tasking Pindah program (task) hanya mengubah

segmentSetiap program bisa dimulai dari alamat 0000

Program1

0000

Program2

0000

Program3

0000

Peta Memori [7]-Contoh Dinamis

00000

FFFFF

DS

10100 DS =10000100

DS

0700 20700 DS = 2000

DS

0064 2070F DS = XXXX

CS

CS

CSJUN - I

TTELKOM

22

Segmen untuk Program

Format : CS : IP

physical address A19-A0=2E5F3IP=95F3

CS=2500 0Adder

Geser kiri CS satu digit

Segmen untuk Data

Format : DS : EA (Effective Address)

physical address A19-A0=2E5F3EA=95F3

DS=2500 0Adder

Geser kiri DS satu digit

Perhitungan alamat fisik [1]

Segmen untuk Stack

Format : SS : SP

physical address A19-A0=2E5F3SP=95F3

SS=2500 0Adder

Geser kiri SS satu digit

Perhitungan alamat fisik [2]

JUN - I

TTELKOM

23

H/W Segment

Program

IP offset

CS segment

Data Offset sesuai mode

pengalamatannya

DS segment

Stack SP offset

SS segment

ES

CSSS

DS

432

1

CONTROL SYSTEM

AH ALBH BL

CH CL

DH DL

SP

BPSI

DI

OPERANDFLAGS

C-BUS

IP

ALU

INSTRUCTION

STREAM

BYTE

QUEUE

A- BUS

BIU

EU

Perhitungan alamat fisik [3]

H/W Alamat

Alamat program

IP offset CS segment Alamat Fisik:

IP + CS(digeser)

ES

CSSS

DS

4

32

1

CONTROL SYSTEM

AH ALBH BL

CH CL

DH DL

SP

BPSI

DI

OPERANDFLAGS

C-BUS

IP

ALU

INSTRUCTION

STREAM

BYTE

QUEUE

A- BUS

BIU

EU

Perhitungan alamat fisik [4]

JUN - I

TTELKOM

24

Perhitungan alamat fisik [5]

Contoh:

Jika CS=24F6H dan IP=634AH, Tentukan:

a) The logical address, b) The offset address c) The physical address, d) The lower range of the code segment e) The upper range of the code segment

Solusi

a) The logical address/ Alamat Logika ; 24F6:634A b) The offset address/ Alamat Offtset ; 634A c) The Physical address/ Alamat fisik ; 24F60+634A= 2B2AA d) The lower range of the code segment: 24F6:0000 =>

24F60+0000 =24F60 e) The upper range of the code segment: 24F6:FFFF =>

24F60+FFFF=34F5F

Perhitungan alamat fisik [6]

Ciri Alamat Segment:

Tidak bisa diletakkan di alamat awal selain xxxx0 Alamat segmen terakhir pasti F000 Contoh : CS= F100 tidak boleh

» Min CS:IP F100:0000 alamat fisikF1000

» Maks CS:IP F100:FFFF alamat fisik100FFF diluar batas maksimum FFFFF

JUN - I

TTELKOM

25

PUSH DXPUSH DIPUSH AXSTART

SS:1236

242424SS:1235

B6B6B6SS:1234

8585SS:1233

C2C2SS:1232

5FSS:1231

93SS:1230

Instruksi PUSH

Asumsi : SS=5FFF SP=1236, AX=24B6,DI=85C2,DX=5F93Perhatikan isi stack setiap dijalankan perintah sbb:

PUSH AX PUSH DI PUSH DX (gunakan prinsip little endian )

SP=1234 SP=1232 SP=1230

POP BXPOP DXPOP AXSTART

SS:1236

242424SS:1235

B6B6B6SS:1234

8585SS:1233

C2C2SS:1232

5FSS:1231

93SS:1230

Instruksi POP

Asumsi : SP=1230 Perhatikan isi stack setiap dijalankan perintah berikut : POP AX POP DX POP BX (gunakan prinsip LIFO)

AX=5F93 DX=85C2 BX=24B6

JUN - I

TTELKOM

26

Instruksi MOV [1]

MOV instruction:

MOV destination, source ; copy source operand to destination

Example: (8-bit )

MOV CL,55H

MOV DL,CL

MOV BH,DL

MOV AH,BH

Instruksi MOV [2]

Example: (16-bit)

MOV CX,468FH

MOV AX,CX

MOV BX,AX

MOV DX,BX

MOV DI,AX

MOV SI,DI

MOV DS,SI

MOV BP,DS

JUN - I

TTELKOM

27

Instruksi MOV [3]

Examples (diperiksa Instruksi yang ilegal):

MOV BX,14AFH

MOV SI,2345H

MOV DI,2233H

MOV CS,2A3FH

MOV DS,CS

MOV FR,BX

MOV DS,14AFH

Instruksi ADD [1]

ADD instruction

ADD destination, source ; add the source operand to destination

Example 8 bit:

MOV AL,24H

MOV DL,11H

ADD AL,DL

MOV CH,24H

MOV BL,11H

ADD CH,BL

MOV CH,24H

ADD CH,11H

JUN - I

TTELKOM

28

Instruksi ADD [2]

Example 16bit:

MOV AX,34EH

MOV DX,6A5H

ADD DX,AX

MOV CX,34EH

ADD CX,6A5H

Q&A uP 8088 [1]

Q : IO atau Memori yang diakses oleh uP pada saatpertama kali bootup (sesudah reset) ?

A : Memori

Q : Tahu dari mana yang diakses memori?

A : Register CS di isi FFFF

Q : Memangnya di peta I/O tidak ada segmen?

A : Tidak ada karena peta I/O hanya 64 kB

Q : Apa yang anda tahu tentang segmen? Ukuran? Jenis?

A : Ukuran segmen 64kB, Jenis 4 : Code, Data, Stack, Extra

Q : Alamat (IO / Memori) berapa yang diakses oleh uPpada saat pertama kali bootup (sesudah reset) ?

A : Alamat Memori FFFF0 yang diakses oleh uP pada saatpertama kali bootup (sesudah reset)

JUN - I

TTELKOM

29

Q&A uP 8088 [2]

Q : kenapa 8088 dibagi menjadi BIU dan EU?

A : sesuai fungsi, yaitu BIU menghitung alamat, EU menghitung data

Q : apa bedanya ALU di BIU dengan ALU di EU?

A : ALU di BIU menggabungkan 16 bit offset dengan16 bit segmen menjadi 20 bit alamat fisik, sedangkanALU di EU menghitung data 8/16 bit dengan data 8/16 bit menjadi 8/16/32 bit

LOGO

Thank You!Please study this subject at home

JUN - I

TTELKOM