DÁP ÁN CẤU TRÚC MÁY TÍNH.pdf

8/10/2019 DP N CU TRC MY TNH.pdf

1/69

n tp :4 (12 cu )Cu hi 1:Thno l chhov chvn bntrnCard iu khin mn hnh v mn hnh?Trong chvn

bn, byte thuc tnh trongVideo RAM c ngha g?Trn mn hnh v Card iu khin mn hnh ca my tnh c haichlm vic: Chvn bn v chho.

-hnh l mt k t, mt k tl mt tp hp cc im sng csp xp thnh cc hng v ct (Tutheo phn gii m mi ktc kch thc l 8*14 im sng hay 9*14 im sng.v .v .).

Trong Card iu khin mn hnh c mt mch gii m to k tc np sn trong mt vi mch ROM thng qua mt mch chtv dch a k tra mn hnh. Tutheo phn gii ca mnhnh m kch thc ca mn hnh c thl 80 ct * 25 = 1000 kthng hay 80 ct * 25 hng = 2000 k t. Nh vy c ththyrng i tng iu khin ca Card iu khin mn hnh l rt t.Chvn bn chtn ti trn hiu hnh DOS Chho:

Card iu khin mn hnh khi cn iu khin n tng imsng trn mn hnh nu mnhnh c phn gii l 800 im sang* 600 im sng = 48,000 im sang th khi i tng iukhin ca card mn hnh sl 48,000 im sang ln hn gp nhiuln khi n lm vic chvn bn.

Chhoc sdng trn hiu hnh Windows. mimt im iu khin trn mn hnh, cn phi c thng tin vimnh, hay k t, mu sc, sng v vtr ca n. chvnbn thng tin vmu sc v sng ca mt k ttrn mn hnhc cha trong 1 byte v byte c gi l byte thuctnh.Vng nhca Video RAM c quy nh bt u ta chA0000h n BFFFFh. chvn bn thng ngi ta sdng


2/69

vng nhVideo RAM bt u ta chB0000h hoc B8000htutheo tng loi card v mn hnh Nh vy mi mt vtr caca mt k ttrn mn hnh stng ng vi hai byte ca bnh

Video RAM. Ngi ta quynh tnh ttri qua phi, ttrn xungdi, k tvtr ct u tin, hng u tin trn mn hnh sc nh s0, tip theo l 1, 2, .v .v . Gisa chu tin cam Video Ram ssdng bt u ta chB800h, mn hnh cphn gii 80 ct * 25 hng th ni dung ca Video RAM tach snh sau:

a chlogic a ch vtl

Ni dung

B800:0000 B8000 M ASCII ca k thng 1ct 1

B800:0001 B8001 Thuc tnh ca k tti hng1 ct 1

B800:0002 B8002 M ASCII ca k thng 1ct 2

B800:0003 B8003 Thuc tnh ca k tti hng1 ct 2

B800:07CE B87CE M ASCII ca k thng 25ct 80

B800:07CF B87CF Thuc tnh ca k tti hng25 ct 80

Cu hi 2: Nu mi quan h gia phn gii mn hnh, sbits m ho mu v ln ca bnhVideo RAM? Ti saongi ta li lun lun a ra cc gii php sdng giao dinCard AGP. AGP 2x, AGP4x, 8x ri PCI Exp 16x cho Card iukhin mn hnh trn my tnh?


3/69

Mi quan hgia phn gii mn hnh, sbits m ho mu vln ca Video RAM. chho: C hai tham slin quann im nh l:

- Vtr ca im nh- Thuc tnh ca im nh bao gm mu v sng.

Khnng mu cng ln th bnhVideo cng phi ln. Smuhin thti mt thi im ca 1 im nh lun l 2n (n l sbitsm ho mu)

Chng hn nh Card VGA c phn gii 640 * 480 c khnnghin th 256 mu (m ho bi 8 bits mu i hi b nh l640*480*8 = 2,457,600 bits. Card SVGA c phn gii1024*768 vc m ho bi 24 bits mu cn dung lng bnhl 1024*768*24 = 18,874,368 bits = 2,359,296 byte tng ngvi dung lng bnh2,5 MB.Nh vy c thkt lun rng, phn gii mn hnh cng ln, sbits m ho mu ca mn hnh

cng ln th i hi bnhVideo RAM cngln.Ta thy rng mun cho hnh nh trn mn hnh trung thc v rnt th phn gii v smu hin thcaca Card mn hnh cngphi ln, iu c ngha lhin thc 1 trang mn hnh ctnh nng nhtrn th lng thng tin phi c truyn tCPUnBnhVideo thng qua Bus cng phi ln. V dkhis lnh 3D, mt bnhhocp nht vi tc 30 hnh/giy ,

phn giI mn hnh l 640 * 480, v trca im nh c biudin bi 2 bytes, cn 1 ln ckhung hnh, 2 ln c/ghi bmZ, 8 ln c ghi ktcu v 1 ln c cc thnh phn bxung mu.Khi tc trao i dliu l: 640 * 480 * 2 byte*30 hnh/s *(1+2+8+1) = 640*480*60*12 = 221,184,000 bytes/sec.Di thng


4/69

ca Bus PCI chl 133MB/s nh vy ta thyngay rng nu dngBus PCI skhng p ng c biton xl hnh nh ny . nng cao tc truyn dliu gia CPU v Card mn hnh ngi

ta a ra chunAGP, Bus AGP c tn slm vic 66Mhz nhanhgp itn slm vic ca Bus PCI l 33Mhz v c 4 chhtrtruyn l 1X, 2X, 4X, 8X. Tc truyn 1X tngng vI 266MB/s, 2X tng ng vi 512MB/s, 4X l1GBs v ch8X l2,1 GBs v PCI Express tng ngvi tc truyn l 4,4GBs.R rng l xl hnh nhvi tc nhanh hn cho cht lnghnh nh cao th tctruyn gia CPU n Card Video cng phi

cao, ll do m ngi ta lin tc a ra cc chun AGP 1X,2X, 4X, 8X v PCI Express.

Cu hi 3: Trong my tnh IBM PC, chun Centronic khc vichun LPT imno? V sao hin nay ngi ta sdng chunLPT m khng sdng chun Centronic. C thdng phngphp no kim tra sc mt ca cc cng LPT trn mytnh?

Ni ghp Centronic l chun giao din my in u tindnh chocc my tnh h80x86 ca Intel. Ghp niny c 36 chn cnh s t1-36 trong c nhiuchn t cho php cc ngtn hiu c t ringnhm trnh tc ng ca nhiu qua ngngun.36 chn ny c chia thnh cc nhm sau:

Cc ng dliu - chuyn dliu tmytnh n my in

Cc tn hiu trng thi my in Xc nh trng thi my in timt thi im bt k.

Cc ng t - cp t cho tng tn hiu tn hiu dliu, tnhiu iu khin v trng thi.


5/69

Tmy tnh AT dng BVXL 80286 tri ni ghp my in dnggiao din DB-25P (25 chn). Vc bn ging nh chun Cetronickhc nhau chchun DB25P khng sdng cc ng tnhiu

t vcho cc chn d liu v iu khin. L do l vsau nymy in dng ngun Switching c khnngchu c sthay iln ca phti m khng gy nhiu do khng cn sdng ccng t ring bit cho tng tn hiu.

Khi bt ngun my tnh, ROMBIOS skim tra s c mt camy in qua cc cng DB-25P . a chca cc cng myin ccha trong a cht0040:0008 n 0040:000F ca vng nh(8bytes, mi cng my in cn 2 bytes).

a chc s LPT C ng dliu

C ng tr ngthi

C ng i ukhin

00408-00409 LPT1 03BC H 003BD H 03BE H0040A-0040B LPT2 0378 H 0379 H 037A H0040C-0040D LPT3 0278 H 0279 H 027AH

V d: Kim tra trn my tnh cng my in no c dng:C> DEBUG

d 40:98 L8

0040:0008 78 03 00 00 00 00 00 00

Nh vy trn my tnh tn ti 1 cng DB- 25P c a chc sl0378 H, khng c cng my in no na tn ti trn my tnh


6/69

Cu hi 4: Khi no xy ra hin tng trn bn phm. phnbit k tE v e ngi ta dng gii php no?

Khi phm c n m bn phm sc truyn ni tip sang bng

mch m, ty c chuyn thnh tn hiu song song v cct vo 1 bm trc khi CPU thc hin cm ca phm vac n. Bm bn phm ny lu c m qut ca 20 phm.Nu CPU bn khng kp c, thngtin sc cha y trong bm v xy ra hin tng trn bn phm. Khi bm ny y nsngng nhn phm n v gI 1 byte c gi trl 0FFh n CPU.Chng trnh phc vngt bn phm Int 09h u tin skim tragi trca byte ny , nu l 0FFh ngha l trn bm v n spht m thanh ra loa thng bo cho ngI s dng bit. Nukhng ngt Int 09h skim tra phm Shift, Alt v Ctrl. Sau Int09h sghi m qut v m ASCII vo mt bm bn phm nadi

32 bytes c a cht0041Eh n 0043Dh, ngt Int 16h sthc

hin c bm bn phm ny . Nu nh ngt Int 16h khng kpc d liu trong bm ny cng xy ra hin tng trn bnphm

Ta bit rng mi mt phm trn bn phm chcduy nht mt mqut bn phm. phn bit ktE v e hay ni cch khc lphn bit hai kt c cng mt m qut bn phm ngi ta sdng byte trng thi bn phm, byte trng thi thi bn phm thnht c lu tra ch0040:0017 v byte trng thi thi bnphm thhai c lu tra ch0040:0018. Nh vy tuthucvo bit biu din trng thi Capslock hay bit RightShift, hocLeftShift c mc logic 1 m chng ta xc nh c m ASCIIca k ttng ng vi m qut bn phm


7/69

Cu hi 5: Giao din USB hin nay c nhng u im g nibt so vi cc giao din khc nh RS-232, LPT. Truyn sliutrong giao dinUSB sdng nguyn l no. Nguyn l c u

im g ni bt.?USB l mt giao thc truyn dliu tun tgia my tnh vi ccthit bngoi vi. My tnh khi c coi nh chBus. Nguyntc kt ni nyhon ton tng tnh cch trao i thng tintrong mng my tnh ngha l mi thit bngoivi sc mytnh gn cho 1 a chIP (Internet Protocol). Do kt ni USBc nhng u im ni tri hn vi giao din RS 232 hay LPT nhsau:

Dliu sc truyn trn USB theo hai ch:

Chcao tc (Full speed mode) vi tc 12Mbs.

Chchm (low speed mode) vi tc 1,5Mbs.

C tc cao hn rt nhiu ln so vi hai giao din RS 232 hay

LPTNgoi ra giao din USB cn c HH htrhot plug andplay Ngi sdng c thcm thm hoc tho ra mt thit bngoi vi m khng cn tt my tnh hay ci t li hthng. Thitbngoi vi cng khng cn bn mch phtrn my chv chcnng ny c phn mm hthng m nhim. Khi pht hin ramt thit bngoi vi mi, hiu hnh stng ci t cc phnmm iu khin thit b. y l nhng tnh nngm cc ngoi vi

ni vi my tnh qua giao din RS232 v LPT khng thc c.Vi nguyn tc kt ni theo kiu mng my tnh nn giao dinUSB cho php kt ni c ti a 127 thit bbng cch ni tipnhau hoc dng Hub trung tm. Trong khi nu dng giao din


8/69

RS-232 hay LPT my tnh chc khnng qun l ti a n 4ngoi vi

Ngoi ra qu trnh truyn s liu trong giao din USB s dng

nguyn l truyn dng sdng bkhuych i vi sai, nguyn lny cho php trit tiu c nhiu lon tc ng ln ngtruynv m bo truyn c khong cch xa

Cu hi 6:Trong my tnh hin nay . Ni CPU, bnh,hBusv ngoi vi lm vic cng 1 tn s?iu l ng hay sai?Ti sao? Thit bno trn hthng Bus ca my tnh m bo

cho cc bphn trn lm vic cc tn skhc nhau?Ni trn my tnh hin nay, CPU, bnhhBus v ngoi vi lmvic cng 1 tn sl khng ng v cc l do sau y:

Trn my XT ban u sdng BVXL 8086/88 CPU, RAM v I/Ocng c ni vo 1 Bus

Nm 1987 Compaq a ra tng chia ng hhthng thnh 2

hkhc nhau, CPU v RAM lmvic cng 1 tn sc lp vitn sca I/O tng ny c thc hin trn my tnh sdngtrn my tnh sdng BVXL 80286 v 80386. Kt ni gia 2 Busl 1 thit bgi l Bridge (Cu ni)

Tmy tnh sdng BVXL 80486 Intel sdng kin trc Clockdoubling trong BVXL khi tn s lm vic ca BVXL bngtn slm vic ca Bus nhn 2. cc BVXL tin tin tn s lmvic ca BVXL c thgp 10,12 ln tn slm vic ca Bus.Nhvy ta thy rng kin trc Clock double v Bridge l thit bmbo cho CPU, bnh, hBus v ngoi vi lm vic cc tn skhc nhau.


9/69

Cu hi 7: Thno l hthng BUS, trn my tnhc bao nhiuloi Bus nu chc nngca tng loi Bus ? rng caBusa ch, Bus sliu quy nh khnng cthno ca my tnh?

LoiBus no c dng phbin hin nay trn cc my tnhIBM v tng thch?

Hthng Bus trn my tnh l thit bc dng kt ni tt ccc thnh phn trn my tnh vi nhau.C 3 loi bus: Bus dliu(data bus), bus a ch (addressbus) v bus iu khin (controlbus).

Address Bus : nhm tn hiu cho php xc nh a chca nhhay thit bvo ra trong qu trnh c/ghi thng tin ln bnhhay cng vo/ra.

Data Bus : nhm ng truyn ti data thc sgia cc thitbhthng do a chtrn address bus xc nh. rng cadata bus (sng dy dn) xc nh data trong mi ln truyn lbao nhiu.

Control Bus : nhm ng truyn cho cc tn hiu iu khin nh: cc thao tc l c hay ghi, thao tc c thc hin trn bnhhay trn thit b ngoi vi, nhn dng chu kbus v khi no thhon thnh thao tcrng bus chnh l sng dy dn hpthnh

rng ca bus a ch: Trn mi ng dy chc thc 1 trong

2 trng thi 0 hoc 1 nn bus crng n th c thnhn bit c2n a ch. Nhvy rng ca Bus a ch quy nh s nhtia m hmy tnh c thqun l c.rng ca bus sliu:c thit ktheo nguyn tc l bi ca 8 (8,16,32,64 bit) nh thmi ln


10/69

truyn 1 byte/2 bytes/4 bytes ty theo my . BrngData bus cngln th data truyn cng nhanh. Loi Bus c dng phbin trnccmy tnh IBM hin nay l Bus PCI. Busny c rng Bus

a chl 32 bits vrng bus sliu l 64 bits.Bus PCI tng thch vi ISA, EISA vMCA. Nh s dngcu ni bus, chun h tr cho cc bus tc chmISA,EISA vMCA.Bus PCI cho php thit lp chtngthit lp cu hnh

Cu hi 8: Qun l ngoi vi bng ngt c uim g so vi qun

l ngoi vi bng phngphp hi vng? Gii thch ti sao trnBUSPCI cho php c mt vi thit bc cng achngt? Vd nh IRQ 11 c th c dngcho n 6 thit b lAudio,SerialBusController, Network Controller, Mass StorageController v IEEE 1394.

Qun l ngoi vi bng phng php hi vng, c sau 1 khongthi gian nht nh my tnh hi trng thi ca ngoi vi, nu bt

thng tin biu din trngthi ca ngoi vi thay i th ngha lngoi vi c yu cu phc v, khi my tnh schy chngtrnh phc vngoi vi . Nh vy phn ng ca my tnh viyu cu c phc vca ngoi vi l khng tc thi.

Qun l ngoi vi bng ngt: Khi mt ngoi vi c yu cu cphc v, n s pht tn hiu yu cu c phc v - IRQs.

Ngaylp tc my tnh sdng tt ccc cngvic ang thc hinchy chng trinhphc vngoi vi . Nh vy phn ng camy tnh vI yu cu c phc vca ngoi vi trong trng hpny l tc thi. y chnh l u im ni tri ca vic qun l


11/69

ngoi vi bng ngt so vi vic qun l ngoi vi bng phng phphi vng

Trn cc hBus hin i nh MCA, EISA, PCI sdng tn hiu

IRQ l tnhiu mc.Do c khnng chia scc tn hiu ngt IRQs. Nghal c thc hai hay nhiu ngoi vi cng sdng mt tn hiu ngtIRQ.

Khi IRQ c mc tch cc, chng trnhiu khin ca cc ngoivi skim tra v nhn dng xem ngoi vi no trn Bus c yu cu

ngt. V ddi y cho thytrn Bus PCI tn hiu ngt IRQ 11c thchung cho n 6 ngoi vi

Cu hi 9: Di thng ca Bus l g? Cc bin php tngdithng ca Bus my tnh? Cc loi Bus ISA, MCA, EISA,PCI, AGP, PCI Exp c phttrin theo tiu ch no? Cc loiBus va ktrn Bus no c htrPlug and Play.

Di thng ca Bus l nng sut truyn hay tc m Bus c thtruyn s liu t thit bch nthit b t. n vo ca dithng l MB/s vc xc nh bi cng thc sau:

Di thng = tn sl/v ca Bus * rng Bus dliu (byte)

Tcng thc trn ta thy gii php tng di thng ca Bus l:

Tng tn slm vic ca Bus

Tng rng ca Bus sliu

Ngoi ra vic thay i giao thc ca Bus cng gp phn lm tngdi thng.Cc loi Bus ISA, MCA, EISA, PCI, AGP, PCI Expc xy dng v pht trin theo tiu ch: Di thng ca cc h


12/69

Bus ra i sau bao gicng lnhn di thng ca cc hBus thhtrc nhm m bo tc truyn dliu gia thit bchnthit btngy cng cao nh cc ng dng ho, qu trnh

c v ghi thng tin cc bnhngoi nh a cng chun ATA,chun SATA.v.v .

Trong cc loi Bus ktrn, cc hBus MCA, EISA,PCI, AGP vPCI Exp c htrPlug and Play cc mc t thp n caotheo nguyn tc cc hBus ra i sau bao gicng c khnngtngnh dng cu hnh (Plug and Play) cao hn hBus thhtrc

Cu hi 10:Thit bc gi l Cu nibridge trn ccmytnh hin nay dng thc hin chc nng g?C my loi?Mi loi c dng cthvo nhngcng vic g?

Nm 1987 Compaq a ra tng chia nghhthng thnh 2hkhc nhau, CPU vRAM lm vic cng 1 tn sc lp vitnsca I/O tng ny c thc hin trnmy tnh sdng

trn my tnh sdngBVXL 80286 v 80386. Kt ni gia 2 Busl1 thit bgi l Bridge (Cu ni)

Hin nay MainBoard ca my tnh c iu khinbi rt nhiumchphn cng c cc chc nngkhc nhau. Tt cchng cng trong mtmch tch hp cc ln gi l chipset. Cu trc phbin nht ca chipset thng c 2 chip gi l northand south

Bridges.North bridge: iu khin truyn sliu gia CPU,RAM,cng AGP v cng PCI Express * 16x.South Bridge: Lin kt viNorth bridge v truyn sliu n tt ccc ngoi vi cn li


13/69

Cu hi 11:Slng ngoi vi ti a m my tnh c thqunlc bng cng COM l bao nhiu? Cng LPT l bao nhiu?

kim tra sc mt ca cccng , ngi ta dng phngphp no? CngUSB l bao nhiu? Gii thch l do ti saogiaodin qua cng USB li cho php qun l sngoivi n 127thit b?

S lng cng COM v cng LPT m mytnh c th qun lc ti a l 4 thit bcho mi loi. Khi khi ng my tnh, h

iu hnh skim tra sc mt ca cc cngCOM v LPT c trnmy , a chca cc cng COM v LPT sc ghi vo trongcc nh ta ch0040:0000 n 0040:000F h.Do kimtra sc mt ca cc cngCOM v cng LPT trn my tnh ta cthsdng lnh D ca chng trnh Debug kim tra vi culnh nh sau:D 0040:0000 l10

Khi my tnh shin thni dung cacc nht0040:0000

n 0040:0007 l a chc cc cng tCOM 1 n COM 4. Nidung cc nh t 0040:0008 n0040:000F l a chca cccng tLPT1 n LPT4. Nu ni dung ca cc nh l 00ngha l cng COM hay cng LPT tng ng khng tn ti trnmy tnh.Slng ngoi vi qua giao din USB m my tnh cth qun ti a n 127 thitb bng cch ni tip nhau hocdngHub trung tm.. L do giao din USB c thqun l c sngoi vi ln n nhvy l do giao din ny sdng kt ni kiumng my tnh, mi thit bngoi vikhi c ni vo giao dinUSB sc cp mt a chIP . Hiu hnh qua c thnhndng c thit b.


14/69

Cu hi 12: Th no l byte thuc tnh trong b nh VideoRAM. Cho bit card mn hnhca my tnh ang ch02:vnbn, phn gii 80 ct x 25 dng, achbnhVideo bt

u tB8000h. Xc nh a ch logic ca byte k tv bytethuc tnh ng vi vtr dng12 ct 20 trn mn hnh.

Byte thuc tnh trong bnhVideo RAM c sdng khi mnhnh lm vic chvn bn (text) cho php biu din mu scv sng ca ca k ttrn mn hnh. Mimt byte thuc tnhtrong Video RAM tngng vi 1 vtr ca k ttrn mn hnhV dVideo RAM c a chtB8000H v c ln 16KB. Cca ch chn cha cc k t v a ch l cha thuc tnh cak t

a chlogic a chvt l Ni dungB800:0000 B8000 k thng 1 ct 1B800:0001 B8001 Thuc tnh ca k t ti

hng 1 ct 1

B800:0002 B8002 k thng 1 ct 2B800:0003 B8003 Thuc tnh ca k t ti

hng 1 ct 2B800:07CE B87CE k thng 25 ct 80B800:07CF B87CF Thuc tnh ca k t ti

hng 25 ct 80

Xc nh a chlogic ca byte k tv byte thuc tnh ca ktti vtr dng 12 ct 20 khi mn hnh lm vic ch02.S thtca k ttrn mn hnh tng ng vi vtr cho trc chphn gii 80 ct v 25 hng vi thtc tnh thng 0 ct 0c xc nh nh sau:


15/69

Stht= vtr dng * 80 + vtr ct= 12 * 80 +20 = 980

Vtr ca nhtrong Video RAM cha m ASCII v thuc tnhca k tc xc nh nh sau:

Vtr ca nh= sthtk t*2 = 980 *2 = 1,960 = 7A8h

Vy a chca cc nhsau scha m ASCII v thuc tnhCa k tti vvtr dng 12 ct 20 trn mn hnh:

nhB800:07A8 cha m ASCII ca k tti vtr dng 12 ct20

nhB800:07A9 cha thuc tnh ca k tti vtr dng 12 ct20


16/69

n tp 3 (12 cu )

Cu hi 1:Thno l vo/ ra theo mode a chi gin tip vtrc

tip (Vi my tnh IBM-PC sdng BVXL ca Intel)? Khnggian vo/ra ca hai mode achny khc nhau nh thno?Khi s dngmt phn khng gian b nh lm khnggianvo/ra. Khi nhng tn hiu iu khin noc sdngiu khin qu trnh c/ghi sliu tcng vo/ra. iu gl rt quan trong khi thc hin chng trnh ghi/ c cng loiny?

Phn bit vo/ra theo mode a chtrc tip v gin tip: Vo ratheo mode a ch trc tip l a chcacng vo/ra c nmtrc tip ngaytrong lnh vo ra. V dkhi cn c sliu tcngc a ch0FFh vo thanh ghi AL.IN AL, 0FFh ; c ni dungcng c a ch0FFh vo thanh ghi Al.

Vo ra theo mode a chgin tip l a chcng vo/ra c

nm trong thanh ghi DX. V d: Khi cn c sliu tcng c achl 0378h ta cn thc hin nh sau:

MOV DX, 0378h ; np a chcng vo thanh ghi DX.

IN AL, DX ; c sliu tcng 378h vo thnh ghi AL

Khng gian a chca hai Mode a chtrc tip v gin tip:Khi thc hin vo/ra theo mode a ch trc tip, a ch cngc m ho bi 8 bits a chtA0 n A7 do n c khnnga chho c 28= 256 cng vo v 28= 256 cng ra.

Khi thc hin vo/ra theo mode a chgin tip, a chcngc m ho bi 16 bits a chtA0 n A15 do n c kh


17/69

nng a chho c 216= 65,536 cng vo v 216= 65,536 cngra.

Khi sdng 1 phn khng gian bnhlm khng gian vo ra:

Trong cu trc BVXL ca mt shng nh Motorolla khng chtrvo/ra bng cc lnh vo ra. Khi thit kkhng gianvo/ra cho my tnh sdng cc BVXL loi ny ngi ta phidng mt phn ca khng gian b nh lm khng gian vo ra.(Khng gian a chvo ra c quy nh bi ngi thit khthng).

Khi mun thc hin trao i sliu vi cc cng vo/ra ngita phi sdng cc lnh trao i s liu vi bnh. V d: khicn c sliu tcng c a ch35000h vo thanh ghi AL caBVXL.

MOV AX, 3000h; khi ng thanh ghi on dliu.

MOV DS, AX.

MOV AL,[500]; c sliu tcng 3500h vo thanh ghi AL Khi cc tn hiu iu khin qu trinh c/ghi cng vo/ra lcc tn hiu c/ghi bnhMEMR hay MEMW

+)Cng vo ra trong trng hp ny sc a ch ho bng20bits a chdo cn c nhnglu sau:

Mch gii m cng vao/ra l mch gii m 20bits a ch(gii m

tuyt i)Khi thc hin cc lnh c ghi bnh nht thit phi khi tothanh ghi on cha a chcng vo/ra.


18/69

Cu hi 2: Trnh by s ghp ni Data Bus vi cc cng vo? Nhng tn hiu iu khin no sc dng iu khin

qu trnh c sliu tcng vo (trng hp vo/ra c khnggian a chring). Trnh by lnh ASM thc hin lnh trao isliu 8 bits v 16 bits.

ni ghp Bus vi cc cng vo ngi ta sdng bm mtchiu l vi mch 74LS244 ghp ni cc cng vo vi Bus sliu nhstrn.

Khi mch gii m scho ra tn hiu chn chip kt hp vi tnhiu c cng ni vo cc chn iu khin ca vi mch 74LS244thc hin iu khin qu trnh c sliu.


19/69

Lnh trao i sliu gia Bus v cng vo: Trao i sliu 8 bits( AL l thanh ghi cha sliu): ivi cng c a chbi 8bits (mode a chtrctip):

In Al, port#i vi cng c a chbi 16 bits (mode a chgin tip):

Mov dx, port#

In Al, dx

Trao i s liu 16 bits ( AX l thanh ghi cha sliu): i vi

cng c a chbi 8 bits (mode achtrc tip):In Ax, port#

i vi cng c a chbi 16 bits (mode a chgin tip):

Mov dx, port#

In Ax, dx

Cu hi 3: Trnh by sghp ni Data Bus vi cc cng ra? Nhng tn hiu iu khin no sc dng iu khinqu trnh ghi s liu tcng ra (trng hp vo/ra c khnggian a chring). Trnh by lnh ASM thc hin lnh trao isliu 8 bits v 16 bits.

thit kcc cng ra ngi ta sdng vi mch cht 74LS 373ghp ni bus sliu vi cc cng ra nhstrn.

Khi mch gii m scho tn hiu chn chp CS kt hp vi tnhiu ghi cng IOW ni vo cc chn iu khin ca vi mch74LS373 thc hin iu khin qu trnh ghi sliu.Lnh traoi sliu gia Bus v cngra:


20/69

Trao i s liu 8 bits ( AL l thanh ghi cha s liu): i vicng c achbi 8 bits (mode a chtrc tip):

Out port#, Al

i vi cng c a chbi 16 bits(mode a chgin tip):

Mov dx, port#

Out dx., Al

Trao i s liu 16 bits (AX l thanh ghi cha s liu): i vicng c a chbi 8 bits (mode a chtrc tip):

Out port#, Axi vi cng c a chbi 16 bits (mode a chgin tip):

Mov dx, port#

Out dx, Ax


21/69

Cu hi 4: Bvi xl truy cp (c/ghi) cc cng vo/ra nhnhng lnh no? C cc Mode a chnhthno? Scngvo/ra c tha chho c ti a l bao nhiu, ti sao? Tn

hiu iu khin no phn bit CPU ang truy cp vo/ra hay bnh?

My tnh IBM PC sdng BVXL ca hng Intel thc hin truycp cc cng vo ra bng cc lnh vo ra trc tip. l lnh IN vlnh OUT.

Lnh IN v lnh OUT c hai mode ach:

Mode a chtrc tip: Khng gian vo/ra c a chho bi 8bits a chtA0A7, mode ny a chcngc nm trctip trong lnh. V d:

IN AL, 30h; c cng c a ch30h vo thnh ghi AL

OUT 28h, AL; Ghi s liu t thanh ghi AL ra cng c a ch28h.

Khng gian vo/ra sdng mode a ch trc tip vi 8bits achta c tha chho c 28= 256 cng vo v 28= 256 cngra.

Mode a chgin tip: Khng gian vo/rac a chho bi 16bits a ch t A0 A15, mode a ch ny, a ch cacngvo/ra c ghi vo thanh ghi DX. V d:c sliu tcng

c a ch3f8h vo thanh ghi AL.MOV DX, 03f8h; Np a chcng vo thanh ghi DX

IN AL, DX; c sliu tcng 3f8h vo thanh ghi AL.


22/69


23/69

phc vthit bvo/ra .

u im ca phngphp ny l thit kphn cng n gin cthdng cho cc hti thiu.

Nhc im ca phngphp ny l khng p ng c tc thivi yucu ca ngoi vi.

Qun l bng phng php ngt: Mi thit bvo/ra cgn vimt tn hiu ngt. khi ngoi vi yu cu c phc v, n sphttn hiu IRQ n mch iu khinngt v qua thit bny thngbo cho CPU bit yu cu c phc v. Khi CPU nhn c

thng bo ny n sct ton btrng thi hin thi (a chCS:IP,thanh ghi RF) vo trong ngn xp v vn cvo shiu ngt xc nh c a chca chng trnh con phc vngt. CPU sthc hin chng trnh con phcv ngt p ng yu cu cangoi vi tng ng. Saukhi chng trnh con phc v ngt ktthc, trng thica CPU trc sc hi phc v CPU li

tip tcqu trnh trc . u im ca phng php ny lpng tc thi yu cu ca ngoi vi. Trn my tnhIBM PC v tngthch hnh thc qun l ngoi vi bngngt c p dng

Vector ngt l mt vng nhc di 1024 bytes (1 KB) cquy nh nm a cht00000h003FFh dng cha a chca cc chng trnh con phc vngt. Vecter ngt gm c 256phn tmi phn tgm c 4 byte (tng ng 1 KB bnh)

c dng cha a chon v lch ca chng trnh conphc vngt (hai byte thp cha a ch lch v 2 byte caocha a chon)


24/69

Hiu hnh ca my tnh IBM quy nh c 256 ngt c nhst00hFFh do mi phn tca vector ngt scha a chca 1 chng trnh con phc vngt. a chu tin ca mi

phn tngt trong bng vector ngt c xc nh gp 4 ln shiu ngt

Vi ngt c shiu l 14h th a chon vlch ca chngtrnh con phc v ngtny s c cha trong cc nh bngvector ngt nhsau.a chu tin ca phn tngt bng 4lnshiu ngt: 14h x 4 = 50h.

Nhvy nhc a ch00050h v 00051h scha a chlch ca chng trnh con phc vngt 14h.

nh00052h v nh00053h cha a chon ca chngtrnh con phc vngt c shiu 14h

Cu hi 6:Thno l ngt cng v ngt mm. Hiuhnh camy tnh IBM cho php qun l tia c bao nhiu ngt cng

v ngt mm.Khi mt ngt c kch hot, CPU xc nha chca chng trnh con phc vngt nhthno? (cthchotrng hp ngtmm, ngt cng). Xc nh cc nhchathng tin cho php xc nh a chlogic cachng trnhcon phc vngt c shiu ngt l 18h.

Ngt cng: (Hardware Interrupt) cpht sinh bi cc mch camy tnh khip li 1 skin no . N c dngiu khin

cc thit b ngoi vi quantrng trn my tnh nhbn phm, a,my in .v .v . Cc ngt ny c vi mchiu khin ngt8259A qun l.My tnh IBM PC XT c 8 ngt cng l:

IRQ0 Timer


25/69

IRQ1 Keyboard

IRQ2 Dng trong my AT

IRQ3 Serial port 2

IRQ4 Serial port 1

IRQ5 Hardisk

IRQ6 Floppy disk

IRQ7 Parallel port 1

Ngt mm (Software Interrup) Dochng trnh yu cu thchin 1chng trnh con no c sn trongROM hay trong RAM.Ngt mm cthc hin bi lnh INT .Ngoi ra cn c ngt phtsinh ni bc sinh ra khi trong qu trinh thchin chng trnhxy ra cc tnh hungnhtrn sliu bnh, chng trnhthchn gp phi php chia cho 0 hocdo yu cu ca ngi lp trnhtrong qutrnh thit kh thng c tht ccngt by kim

tra.Hiu hnh ca my tnh IBM PC v tng thch cho php qunl c 256ngt, mi ngt c gn 1 shiu ngtc nh st00h n 0FFh.Khi mt ngt c kch hot HH sxc nha chca chng trnh con phc vngt nhsau:

Vi ngt mm: Shiu ngt c nm ngay trong lnhINT, khi

H

H ch

c

n l

y s

hi

u ng

t

nhn v

i4

xc

nh

c

vtr ca 4 nhtrong bng vectorr ngt ni cha a chonv lch cachng trnh con phc vngt.

Vi ngt cng: HH s xc nh s hiu ca ngt tng ngthng qua vic c ni dung ca thanh ghi ICW2 trn vi mch


26/69

qun l ngt sau HH ly shiu ngt nhn vi 4 xcnh c vtr ca 4 nhtrong bng vectorr ngt ni cha achon v lch ca chng trnh con phc vngt.

Xc nh v tr nhcha a chon v lchca chngtrnh con phc vngt tng ng vingt c shiu ngt l 18h:

Vtr u tin ca nhtrong bng vector ngtcha a chcachng trnh con phc vngt c shiu l 18h l: 18h x 4 = 60h.

Nhvy hai nhc a chl 00060h, 00061h trong bng vectorngt scha a chlch, hai nhc a ch00062h, 00063h

scha a chon ca ngt c shiu 18hCu hi 7: nh ngha vai tr ca ngn xp v thanh ghi contrngn xp - SP khi thc hin thao tc ct giv hi phc sliu vo trong Stack (ngn xp). Khi khi ng my tnh SPc thit lp gi trnhthno. Khi no th SP c gi trtngv khi no th c gi trgim i

Ngn xp l mt vng nh c di 64 KB (tng ng 1segment) c my tnh sdng ct tnh trng tm thi ca hthng gm trng thi ca cc thanh ghi RF, CS, IP khi thc hincc thao tc ngt v l ni ct giv hi phc thng tin cc thanhghi a nng ca BVXL khi thc hin cc lnh Push v Pop.

Thanh ghi con trngn xp (SP) l thanh ghi c di 16 bits cchc nng lun trn nh ca ngn xp.nh ca ngn xp lvtr ca nhm SP trn, ti sdin ra thao tc phc hiv ct gisliu khi my tnh thc hin cc thao tc ngt v cclnh PUSH v POP. . .


27/69

a chca nh ngn xp l: SS:SP . khikhi ng my tnh IBMPC v tngthch gi trca thanh ghi SP bao gicng c npgi tr ln nht FFFEhCu trc ca ngn xp l cu trc LIFO

(Vo sau/ratrc) do vy khi my tnh thc hin qu trnh ctgitrng thi v thc hin cc lnh ct giPUSH th gitr cangn xp c gim i v khi my tnh thchn qu trnh hiphc trng thi hay cc lnh POPth gi tr ca ngn xp ctng ln. Mt thao tc ctgihay hi phc thng tin khin cho gitrca thanh ghi SP thay i 2 n v. Tng ng vi vic mytnh ct gihay hi phc c 2 byte thng tin

Cu hi 8 V sao ni iu quan trng nhttrong thao tc ngttrn my tnh l xc nhc shiu ngt. Vi ngt cng vngtmm lm thno my tnh xc nh cshiu ngt.mrng cc ngoi vic iu khin bng ngt cng.Trn mytnh IBM v tng thch hin nay, ngi tadng giiphp no? Cch no cho php xcnh c a chlch va chon cachng trnh con phc vngt c s hiu ngtl nn. trn mt my tnh hin c

iu quan trng nht vi thao tc ngt l xc nh c shiungt v khi xc nh c shiu ngt HH sthc hin nhn shiu ngt vi 4 xc nh c vtr u tin ca nh trongbng vector ngt cha a chlch v on ca chng trnhcon phc vngt tng ng.

Vi ngt mm c thc hin bng lnh INT, shiu ngt cnm ngay trong lnh. V d lnh Int 21h tng ng vi vicyu cu HH hnh thc hin 1 ngt c shiu l 21h.


28/69

Vi ngt cng: HH s xc nh s hiu ca ngt tng ngthng qua vic c ni dung ca thanh ghi ICW2 trn vi mchqun l ngt sau HH ly shiu ngt nhn vi 4 xc

nh c vtr ca 4 nhtrong bng vectorr ngt ni cha achon v lch ca chng trnh con phc vngt.

mrng cc ngoi vi c iu khin bng ngt cng c thdng hai vi mch iu khin ngt 8259 lm vic chni tngnhtrn my tnh IBM PC AT v tng thch cho php qun ln 15 ngoi vi bng ngt cng. Ngoi ra trn cc hBus phttrin ca my tnh nhBus PCI hin nay sdng tn hiu kchhot ngt l tn hiu mc th mt tn hiu ngt c thcho phpng vi nhiu ngoi vi

Vi mt my tnh hin c, thng dng chng trnhDebug xc nh c a chon v a chlchca mt chng trnhcon phc vngt bt k(nn h)nhsau:

Ly shiu ngt (nn h) nhn vi 4 xc nh c nhu

tin trong bng vector ngt l ni cha a chlch v a chon ca chng trnh con phc vngt.

Khi xc nh c nhtrong bng vector ngt. Sdng lnhD trong Debug:

D 0000: (nn h*4) L4

Khi trn mn hnh shin ra gi trca 4 nhc achlnlt l 0000: nn h*4, 0000: nn h*4 +1, 0000: nn h*4 +2 v 0000:nn h*4 +3.

a chca chnmg trnh con phc vngt c s hiu nn h cxc nh nhsau:2 nh0000: nn h*4 v 0000: nn h*4 +1 cha


29/69

a chlch ca chng trnh con phc vngt c shiu ngtnn h. (IP)

2 nh0000: nn h*4 +2 v 0000: nn h*4 +3 cha a chlch

ca chng trnh con phc vngt. c shiu ngt nn h (CS)Cu hi 9: L do no ngi ta sdng phng php truycp s liu trc tip vo bnh (DMA)? Vi cu trc ca vimch 8237 lp trn my tnh IBM PC XT cho php truyn tia my knh DMA, mt knh DMA mt ln thc hin DMA chophp truyn ti a mt khi sliu c dung lng l bao nhiu?

Ti sao? Nu vai tr ca 4 knh DMA trn my IBM XT? Trnmy tnh IBM AT/PC sknh DMA c c tng ln khng?

Khi my tnh cn phi chuyn mt slng ln sliu gia thitbngoi vi n bnhhay tvng ny n vng khc ca bnh.Nu s dng CPU th u tin CPU nhn thng tin ri sau chuyn n ni nhn. Qu trnh gii m lnh cn cn thm ccthng tin bxung do cng vic ny srt tn nhiu thi gian.

V vy Intel thit kmt biu khin v truy cp trc tip bnh(DMA)l vi mch 8237 cho php c thbqua CPU thchin truyn sliu trc tip gia bnhv ngoi vi hay gia bnhvi b nh, qu trnh ny ch thc hin truyn s liu mkhng c qu trnh gii m lnh v thc hin lnh nh trongCPU nhvy m lm cho qu trnh ny snhanh ln rt nhiu

V d: nu truyn 1 byte sliu t1 cngngoi vi n bnhthmy tnh phithc hin cc thao tc sau


30/69

Qu trnh ny nu thc hin bng truy cp trc tip th chmt c4 chu kng h.

Vi cu trc vi mch 8237 lp trn my tnh IBM PC XT chophp my tnh c 4 knh DMA.

V gi trcc thanh ghi a chhin hnh v m thin hnh cavi mch 8237 c di 16 bits do mt knh DMA mi ln thchin DMA cho php chuyn ti a ni dung sliu l 216= 65,536byte tng ng vi 64 KB sliu4 knh DMA tng ng trnmy tnh IBM PC l:

Knh DMA 0: c dng lm ti DRAM.Trong cc my

AT/PC khng sdng.Knh DMA 1: Khng sdng, ngi ta dng knh ny cho mng.

Knh DMA 2: Dng cho biu khin a mm.

Knh DMA 3: Dng cho biu khin a cng.

Trn my tnh AT ngi ta bxung thm mt vimch 8237 nado tng s lng knh DMA lnn 7 knh. Khi vi mch

8237#2 ni thm voc dng lm vi mch chv knh 0 can cdng ni tng vi vi mch 8237 #1. Cc knh 0,1, 2, 3ca vi mch 8237#1 c a ra trn phn64 chn ca Bus ISAdng truyn sliu 8 bitsgia I/O v bnh.


31/69

Cc knh DMA 5DMA 7 xut pht tvi mch 8237#2 c btr phn 36 chn ca Bus ISA v c dng truyn sliu16 bits

Cu hi 10:Phn bit vo/ra song song v vo/ra ni tip.unhc im ca 2 phng php truyn ny . Chngc sdng trong cc trng hp no. Khi khing, my tnh IBMPC skim tra v ct giachca cc cng vo ra song songv ni tip vou.kim tra sc mt ca cc cng COMvcng LPT trong my tnh IBM PCs ngi ta dngphngphp no? DTE v DCE l g

Vo ra song song l mi mt ln thc hin lnhIN hoc OUT th8bits hay 16 bits s liu ngthi c a ra hay nhn vo ticng.

Vo ra ni tip l mi mt ln thc hin lnh IN hoc OUT th chduy nht 1 bit sliu c a ra hay nhn vo ti cng. Munchuyn . 1 byte sliu th my tnh cn n 8 ln thc hin lnh

IN v OUT. Thao tc ny phi sdng 1 thanh ghi dch chuyntn hiu song song thnh ni tip (lnh OUT) v chuyn tn hiuni tip thnh song song (lnh IN)

Truyn s liu song song c tc cao hn nhiuln so vitruynsliu ni tip nhng khi Buslm vic tn scao thngbcan nhiu lm nhhng ti cht lng truyn tin. Mt khc

nutruyn khong cch xa th gi thnh s rt cao dos lngng truyn ln

Xut pht ttnh cht ca truyn sliu song song nn ngi tathng sdng phng thc truyn sliu song song khi truynsliu trong ni bmy tnh v truyn sliu gia thit bchv


32/69

thit btvi cly ngn nhtruyn sliu gia my tnh vi ccngoi vi thng qua cng kt ni LPT Phng thc truyn sliuni tip thng c dng khi truyn sliu gia thit bchv

thit b tc khong cch xa, khi thng c sgip cacc thit btruyn tin nhModem, Rounter.v.v.

Trn my tnh IBM PC cho php qun l ti a n4 cng vo rani tip theo chun RS 232 (cngCOM) c gn shiu tCOM1 n COM 4 v4 cng vo ra song song theo chun LPT cgnshiu tLPT1 n LPT 4. Khi khi ng mytnh, HH skim tra sc mt ca cc cng vora ni tip v vo ra songsong v ct a chcachng vo vng nhc a chl:

a ch ca cc cng vo ra ni tip c ct gi ta ch:0000:0400 n 0000:0407

a ch ca cc cng vo ra song song c ct gi t a ch0000:0408 n 0000:040F

Do xc nh c scng vo ra song song v ni tip trnmy tnh hin hnh ngi ta thng dng lnh D ca chng

trnh gri Debug quan st ni dung ca vng nhc a cht0000:0400 n 0000:400F. 8 byte u tin scha a chcacc cng COM t1 n 4 v 8 byte tip theo scha a chcacc cng LPT t1 n 4. Khi gi trca nhbng 00 khi cng COM v LPT tng ng khng tn ti.

Khi truyn sliu i xa ngi ta phn bitthit btruyn s liuthnh thit bu cuiv thit btruyn thng.


33/69

Thit bu cuiDTE (Data Terminalequipment): l thit bschng thu vpht s liu v dnhmy tnh, my Fax,myin thoi.v .v .

Thit btruyn thngDCE (Data Communication eqipment) lthit bch thc hin khuych i v truyn s liu trn ngtruyn v dnhmodem, router, switch.v.v.

Cu hi 11:Thno l phng php truyn vi sai, phng phpny c nhng u im g v chun vo/ra no trn my tnh

hin nay sdng phng php truyn ny. Modem ni vi mytnh bng cng g? . Mc logic 0v mc logic 1ca chunny trn my tnh tng ng vi in p no? Ti sao? Cchchuyn i in p ny trn my tnh.?

Tn hiu sc truyn trn hai dy tnhiu l U+ v U- c avo hai chnD+ v D- tng ng ca bkhuych ivi sai. Vistrn tn hiu trn u ra Ura sl:


34/69

Vic sdng bkhuych i vi sai c uim vi hai tn hiu U+v U- c xonvi nhau cho php trit tiu c nhiutc ngln ng truyn. Nu c mtnhiu lon U nh tc ng lnngtruyn th tn hiu s tc ng ngthi ln chai dy

tn hiu U+v U-, khi:

R rng tn hiu nhiu tc ng lnng truyn b trit tiukhi truynqua bkhuych i vi sai. Phng thctruyn sdng

khuych i vi sai cs dng trn giao thc truyn USBtrnmy tnh hin nay

Chun RS 232 quy nh mc in p nhsau:

Mc logc 0: Tng ng vi mc in pt+3V n + 25V .


35/69

Mc logic 1: Tng ng vi mc in pt -3V n - 25V .

Trn my tnh quy nh mc logic 0 v 1 trn cng COM theochun RS 232 nhsau:

Mc logc 0: Tng ng vi mc in p +12V

Mc logic 1: Tng ng vi mc in p -12V

Cu hi 12: Trnh by khi nim v track, sector, cluster v

bng FAT trn a mm v a cng. Nu 1 file c kch thcchc 8 bytes thi, hiu hnh sphi cp khng gian a lbao nhiu lu trfile ny (githit 1 sector c kch thc l1024 bytes, 1 cluster bao gm 4 sectors).

Track l cc vng trn ng tm nm trn bmt a m trn ng

i ta c th

ghi thngtin ln

a, s

track trn m

t m

t

a

phthuc vo kiu a v dung lng a. Trnmi mt rnh(track) ngi ta chia thnhnhiu cung nhcha thng tin gilsector, ssector trn mt track cng phthuc vo kiu a vdung lng a, slng thng tin lu trn mt sector thng


36/69

cthl 512 byte, 1024 byte.v.v . phthuc vodung lng av phin bn ca hiu hnh

Cluster hay cn gi l 1 lin cung l mt n vnhnht m acp pht cho hiu hnh khi lu trthng tin, mt cluster c thgm 1, 2, 4.v.v sector tuthuc vo loi a v hiu hnh. HH

bao gi cng s dng 1 s nguyn ln cc clusster khi lu trthng tin vo a

Bng FAT (File alocation Table) cn gi l bng nh vtp tin,n cung cp cho hiu hnh ltrnh tm kimthng tin ca ccfilel (tp tin) trn . Chnh v vym c 2 bng FAT ging ht


37/69

nhau cho php dphngkhi c 1 bng bhng, HH sc thngtin trn bngcn li.

Bng FAT lun c nh vsau bn ghi khIng. Slng cc

cha thng tin trng bng FAT(hay cn gi l di ca bngFAT) chnh bng slng cluster m a c c. Theo sphttrin cahiu hnh ngi ta sdng bng FAT 12, FAT 16,hayFAT 32 ngha l s lng cc cha thng tin trong bng FATc m ho bi 12, 16 hay 32 bits. V di ca mi cngtng ng l 12, 16 hay 32 bits.

Ni dung thng tin ca mi trong bngFAT ni nn:Tnh trng ca lin cung tng ngtrng hp lin cung khngchathng tin.

Lin cung tip theo scha thng tin camt file trn a, trnghp lin cungchthng tin. V dthng tin ghi trncc camt bng FAT 12 (16) nhdi y


38/69

n tp 2Cu hi 1:Trn bn bnh, bnhROM BIOStrong my

tnh IBM PCXT nm vnga chno?Nu chc nng ca bnhROM BIOStrong my tnh IBM PCXT .Khi khi ng mytnh th thanh ghi onm lnh (CS) v con tr lnh (IP)ckhi to nhthno vng nhROM BIOS sc quychiu u tin.

+)Khi khi ng my tnh th thanh ghi on m lnh (CS) v con

trlnh (IP) ckhi to nhthno vng nhROMBIOSsc quy chiu u tin.Vng nh ROM BIOS c nm vng a ch cao nht trong khng gian nh ca my tnhIBM PC XT. a ch cao nht ca ROM BIOS l FFFFF Hcn a ch thp nht ph thuc vo dung lng ca vi mchnh ROM

+)Khi khi ng my tnh th thanh ghi on m lnh (CS) v con

trlnh (IP) ckhi to nh thno vng nhROM BIOSsc quy chiu u tin.

BnhROM BIOS c dng cha cc chng trnh iuhnh mc c sca my tnh v l mt thnh phn ca Hiuhnh.

+)Khi khi ng my tnh th thanh ghi onm lnh (CS) v con

trlnh (IP) c khi to nhthno vng nhROMBIOSs c quy chiu u tin.Khi khi ng my tnh s khitothanh ghi on m lnh CS=FFFFh v thanh ghi con tr lnhIP=0000h, b vi x l 8088 nhn thng tin do CS:IP trti lFFFF:0000, ng vi a chvt l FFFF0h


39/69

Cu hi 2: Khi my vi tnh IBM PC sdng loi vi mch nhlloi DRAM (RAM ng) th my tnh cn cthm thao tc g bo ton dliu trn bnhRAM trong qu trnh lm vic? L

do ti sao? So snh gia SRAM v DRAM.Khi my vi tnh IBM PC sdng loi vi mch nhl loi DRAM(RAM ng) th my tnh phi c thao tc lm ti thng xuyn

L do ca vic phi thc hin lm ti lkhc phc hin tngtphng in ca tin dn ti mt dliu

Vcu to: Cc nhSRAM c ch to didng mch lt

(flip-flop), cn DRAM s dng tin nh tng bit.SRAMkhng cn lm ti thng xuyn, do vyc thtruy cp bt klc no, DRAM khi ang lm ti th khng th c hoc ghic. Do mi bit ca SRAM l mt flip-flop v mi flip-flop c6 transistor nn SRAM phi c nhiu transistor hn so vi DRAMv do vy c dung lng nhthp hn.

DRAM c u im chnh l dung lng (mt ) cao hn, chi phtrn mi bit rhn v tiu thin nng trn mibit thp hn sovi SRAM. SRAM thng c sdng trong cc v tr khngi hi dung lng ln nhng cn tc cao.

Cu hi 3: Trnh by vbn bnhtrn mytnh IBM PCXT ( ln, khng gian a ch, cc vng chnh ca bn nh... ). .1024 byte vng nhthp nht trongbn nhc

sdng vo mc ch g?My tnh XT vi bvi xl 8088/86 c 20 chn achc khnng qun l1 Mb bnh, (220= 1048576 = 1 Mb) ng vikhnggian a cht00000-FFFFFH.


40/69


41/69

+)1024 byte vng nhthp nht trongbn nhc sdng

cho bng vctngt ca my tnh.Cu hi 4: Trnh by bn cht ca bnhmrng(bng hnhv). Loi bnhny c nhng im g chung v ring so vi bnhpht trin.

Loi bnhny c nhng im g chungv ring so vi bnhpht trin.B nh m rng hay cn gi b nhbnh trng

(Expanded Memory) lgii php vt qua ro cn 640KB cahiu hnh DOS bng ccchng trnh trao i v bng cchadliu vo ra bnhchnh vi tc cao.


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Chun bnhmrng LIM (EMS phin bn 3.0) xut sdng64 K bnhlin nhau to thnh mt ca svng nhtrn chiuti 8 Mb ca bnhmrng.Khng gian nh64 K lin nhau nycchia thnh 4 phn gi l cc khungtrang (page frame), mikhung trangtruy nhp 16 K. a ch16 K ca trangthnht phibt u tbin ca tng16 K v d: 0000, 4000, 8000, C000......

Do 64 K bnhphi lin nhau nn nhiuboard nhmrng sdng cc achE0000-EFFFF cho 4 khung trangca bnhmrng. 64 K byte nyng vai tr ca ca schiu vo mtbnhln hn.Kch thc ca bnhln hn ny ban u cc i l 8Mb v sdng chlu dliu. M lnh khng c lu


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bnhmrng EMS 3.0

im chung vi bnhpht trin: Cngl vng nhln hn giihn 1MB cabnhPC XT. Nhng khc vi bnhpht trin,

bnhmrng phidng th thut phn mm truy cp, cnbnhpht trin c thtruy cp nhslng ng a chlnhn 20

Cu hi 5: Trnh by bn cht ca bnhphttrin (bng hnhv). Loi bnhny c nhng im g chungv ring so vi bnhm rng. H iu hnh qun l b nh pht trinbng

nhng cng cnoy l bnhtrn 1Mb, c c l do cc bVXL t80286 trvsau cnhiu hn 20 chn a chso vi 8086/88

C hai thut ngmi cc my tnh t80286 trvsau l:vng nhcaov bnhpht trin. 80286 c 24 chna ch, chophp truy nhp 16 Mb bnh, bt u t000000 ti FFFFFFH.

32 chn a chca 80386/486/ Pentium cho php truy nhp tia 4 Gb bnh, ta ch00000000 -FFFFFFFFH.

chthc, tt ccc bvi xl trnc thtruy nhp ti a 1Mb. truynhp bnhvt 1000000H, bvi xl phi chuynsang chbo vvi hi hiu hnh phc tp hn nhOS/2,UNIX hay Windows NT.

Khng gian nhtrn 100000H (trn1Mb) c gi l bnhphttrin. Nhm chun ho vic truy nhp bnh trn 1 Mb, nm1988, cc hngLotus, Intel, Microsoft v AST xydng tiuchun bnhpht trinXMS (Extended Memory Specification)


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Vng nhcao (HMA):l vng nh65520 byte t1000000 H n10FFEFH c thtruy nhp khng cn n hiuhnh c chbo v. Nhc dnga chA20 ca cc CPU t80286 trv

sau.im chung vi bnhmrng: Cngl vng nh ln hngii hn 1MB cabn nhPC XT. Nhng khc vi bnhmrng , bnhpht trin cc l do schn a chca cc CPUt80286 trvsau ln hn 20.


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Trong cc my tnh, slng cc nhthng rt ln, nn cn cphngphp tchc v tm chn a chthchhp. Phng phptm chn gi lgii m a ch.

Bvi xl 8086/8088 c 20 ng ach, nhng hu ht cc vimch nhkhng c schn, v vy cn c 1 mch lm nhimvphi hp gia ccng a ch. Mch ny gi lmch giim a ch. Mch gii ma chslm cho bnhc mrngv hiu quhn.

Mch gii m a chphbin cc hmy 8086/88 l 74LS138.

Mch c ba ng vo A, B, C dng gii m lira.Nhvy cc bt cao trong bus a chdng xc nh vi mchnhcn truycp (qua bgii m a ch). Cn ccbit thp dngxc nh vtr ca nhtrong tng chip nh.

Cu hi 8: Trnh by s gii m bnhROM BIOStrongmy tnh bit rng my tnh sdng bnhROM l vi mch

2764(8* 8 Kbits)

Trong tng s 256K byte ROM, bngmch ch ca my tnhPC/XT chsdng phn a ch vt l tF4000H n FFFFFHdnh cho BIOS v ngnngBASIC. Trong my XT IBM sdngchip 74LS138 gii m a chROM. Nhhnh vsau:


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ng vi mi chip ROM 8Kx8 c 13 ng a chA0-A12 i t

bus hthng, qua bm 74LS244 to racc chn XA0-XA12,ri c a ticcc chn a chca ROM.Cc bit a chA13-A15 c a ti ccu vo A, B v C ca 74LS138. Tutheothp ca A, B v C, mt trongcc u ra Y0-Y7 sc chn,khi chn skch hot mt trong cc chip ROM.

V thc hin c, cc chn , v G1 phi cho php m(enable). G2A ckch hot bng tn hiu u ra ca cngNAND4 u vo A16-A19. Chn G2B c kch hot bi tn hiu iukhinMEMR qua mch khuch i m vcui cng chn G1c tn hiu kch hot


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Cu hi 9: Trnh by vphi ghp gia CPU v bnh(cctnhiu iu khin, a chv sliu).

Bnhc ni vi CPU thng qua ccBus ca may tnh: Bus

iu khin, Bus a ch, Bus dliu. (hnh v)Bus dliu l cc ng a ch/ dliu ADO-AD7, Bus a chl cc ngAD0-AD7 v A8-A19.

Bus iu khin gm cc tn hiu iu khin bnhhot ng nhsau:ALE (Address latch enable). Dng sn xung ca tn hiuALE mcht a chnh. a chny c mnn c thtin

hnh vic ghi/c cho n khi ALE tr li mc cao.WR/RD(Write/read). Khi a chc m, bVXL stc ng hoc votn hiu WR (ghi) hoc tn hiu RD (c) tin hnh qutrnhghi/c bnh. Cc tn hiu ba trng thi u cmc logicthp v khng th m c hai cng mtlcIO/M (Input-Output/Memmory) Tn hiu cho php chn hoc l cng vo/rahoc l bnh. Tn hiu ny l cn thit v cng 1 a chc th

c dung cho bnhhoc cho mt cng I/O.DEN (data enable).phn nh thi giancho BUS. Hu ht cc hthng ni vi busdliu u phi qua mch m. Tn hiu DEN l tn hiuba trngthi hiu lc mc thp dng m(enable) mch m ny.DT/R (Data transmit/Receive). Dng xc nh chiu chuynd liu. Mc cao khi bx l chuyn d liu v mc thp khinhn dliu


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Cu hi 10:Trnh by chi tit cc vng nhtrong bn bnh

may tnh IBM PC XT .


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a chho bnh1 Mb ca 80x86

L mt sbiu din cc vtr c sdng trong bnh,

dng phn bit cc vtr v nhim vca chng640 Kb u tin c b tr lin tc t a ch00000 cho n9FFFFH v c gi l bnhqui c. Trong :1 Kb u tinca bnhqui c, ta ch00000 n 003FFH c dnh chobng vc tngt

- 256 byte tip theo, vng nht00400 n 004FFH c dnh

cho vng dliu ca BIOS.- 256 byte tip theo na, vng nht00500 -005FFH dng ghicc tham sca DOS (vng dliu DOS).


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- Vng 00700H - 9FFFFH dnh cho cc yu cu ca DOS, ghicu hnh my.

Thut ngvng nhtrn (upper memory area) mun ni n di

a chtA0000 n FFFFF vi tng bnhl 384K, trong :- 128 K tA0000H n BFFFFFH cho video RAM.

- Ta chC0000H n FFFFFH vi tng s 256K dnh choROM

Cu hi 11:Slng cc tn hiu a chv sliu ca hBustrn my tnh c lin quan g n cu trc ca bnhtrn mytnh. Vi cc my tnh c nhiu hn 20 ng a ch, khnggian bnhtrn 1M c Hiu hnhqun l nhthno?

S lng cc tn hiu a ch (bus a ch) ca mt CPU quytnh ln cakhng gian nhca my tnh sdngCPU , tcl quyt nh slng nhln nht c thc trong bnhmytnh.

Slng cc tn hiu a chln th khng gian nhcng ln. Gisslng tn hiu a chl n ths nhc thl 2n.Slngcc tn hiu sliu (bus sliu) quyt nh tc truy cp bnhcaCPU (hiu theo cch l dung lng dliu c th truy cptrong 1 chu kc ghi).

Khng gian nhtrn 100000H (trn1Mb) c gi l bnhpht

trintruy nhp bnhvt 1000000H, bvi xl phi chuyn sangchbo vv i hi hiu hnh phc tp hn nhOS/2,UNIX hay Windows NT.


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B nhpht trin c qun l da trnchun XMS (ExtendedMemory Specification)

Cu hi 12: Cc kiu Ton hng bnh c sdng trong

ngn ngAssembler, vd?Trong ngn ngAssembler s dng cckiu ton hng b nhsau:

- Ton hng bnhtrc tip: cha chny , a chca dliu trong bnhdo mt ton hng c trong lnh xcnh trc tip.a chny l mt tn hay mt biu thc hng. C thkt hp

thmton tchs[ ], ton t+, - v ton t(:).- Ton hng bnhgin tip Ton hngbnhgin tip cho phpdng ccthanh ghi v di chn cc gitrtrong bnh.Chc 4 thanh ghiBX, BP, SI, DI c thdng vi tonhng bnhgin tip. di c th lmt hng hay mt ton hng bnhtrc tip. Mt ton hng c nhiu di, cc di sc

cng li thnhmt di chung xc nh khi dch.Trong c cc chsau:

+ Chthanh ghi gin tip: Trong cha chthanh ghi gintip, a chcabnhcha ton hng c thanh ghixc nh.Cc thanh ghi s dng cho mcch ny l SI, DI v BX. Vtrongtrng hp ny , cc thanh ghi ny xcnh lch, cnthanh ghi on l DS.

VD: MOV AL,[BX]

+ Cha chcs: cha chny , cc thanh ghi csBX, BP v cngvi di c sdng xc nh achoffsetca dliu. di c thl 1 hng hay 1 ton hng bnhtrc


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tip. Mt ton hng c thc nhiu di, cc di sccng li thnh 1 di chung xc nh. a chon ngmnhng vi BX l DS cn SS l vi BP.

VD:MOV CX,[BX]+10+ Cha chchs: Trong chnycc thanh ghi chsDI,SI cng vi di xc nh offset ca d liu. Thanhghi onngm nh l DS.

VD:MOV DX,[SI]+5

+ Cha chchscs: Tng hpchai cha chcsv chsc gi l ha chch scs, trong sdng cthanh ghi csv thanh ghi chs.

VD MOV CL,[BX][DI]+8


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n tp 1 ( 12 cu )

Cu hi 1:Trnh by scu trc ca hthng mytnh theo

nguyn l Von Neuman?Nu chc nng ca tng bphn?c biu din di dng g, nm trongbphn no?Mtqu trnh my tnh thc hin mtlnh?

Bxl trung tm: ng vai tr cho nht trong hmy tnh.y tin hnh ton bcc tnh ton, cc php tnh shc v isv bin i cc m tng ng. y chnh l u no ca ccmy tnh. Khi bt u hot ng, n nhn m lnh tbnh, saugii m cc lnh ny thnh cc dy xung iu khin iu

khin cc khi tng ng thc hin theo ng trnh tv ni dungc trong m lnh.

Bnh(memory): Bnhthhin vic lu trthng tin v traoi thng tin. cc hmy vi tnh, ngi ta sdng bnhbn


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dn bao gm ROM, RAM. Trong ROM c thcha cc chngtrnh iu khin hot ng ca ton hkhi bt my . Cn mt phnchng trnh iukhin h thng, chng trnh ng dng

thngc lu RAM.Bnhmy tnh thc cht l bnhtrong phnbit vi khinim bnhngoi m nhiu ti liuhin vn hay sdng chthit bt(Bng, at)

Khi phi ghp vo/ra (I/O) tokhnng giao tip my tnh vithgii bn ngoi. Cc thit bngoi vi nhmn hnh, bnphm, my

in, cc bbin iD/A, A/D.... u lin hvi hmy tnh qua bphn ny m cthcho tng thit bl cc cng.

BUS a ch: Khi ghi/c bnhbxl trungtm sa ra trnbus ny a chca cc nhclin quan. Lu l bus a chlbus mt chiuxut pht tCPU.

BUS dliu thng c t8, 16, 20, 24, 32 n 64 ng dy tu

thuc vo tng CPU cth. BUS dliu l loi 2 chiu. Cc phntc u ra ni thngvi bus dliu u phi c trang bura 3 trng thi bam cho bus hot ng c bnhthng.

BUS iu khin thng gm hng chc dy tn hiukhc nhau, vxt theo cnhm th l lai bus 2 chiu.

Thit bvo: bin i tn hiu tnhin hoccho bi con ngisang dng m my bvi xl thc hin. Thit bvo thngdng nhtc bn phm, con chut, my qut scaner , thit bphntch nhn dng ting ni, CD-ROM.....

Thit bra: bin i cc m bn trong mytnh sau khi xl con ngi c thhiu c hoc iu khin cc thit bkhc.


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Thit bra thng dng nht c mn hnh, my in, faxmodem, myv, loa hay cc thit biu khin khc

+)M chng trnh v dliu trongmy tnh c biu din di

dngcc bt nhphn, c cha trongbnhca my tnh.+)Khi my tnh thc hin mt lnh, trctin n phi xc nh vtr ca lnh trong bnhthng qua cp thanh ghiCS v IP , sau n s np (nhn) lnh vo, thc hin gii m lnh, thchinlnh v cui cng l ghi kt qutrli bnh.

Cu hi 2: Gismy tnh thc hin on chngtrnh bt u

ti a ch 125Ah:0100h. Gi tr caDS, CS, SS, ES u l125Ah.

Hy cho bit:

- Chng trnh ny c phn mrng l .COM hay . EXE

- M chng trnh trn nm trong bphn no camy tnh?Ti vtr c a chvt l no?

- Bphn no thc hin cc lnh? Thc hin lnlt hay ngthi?

- Khi thc hin cc lnh th nhng thanh ghi CS vIP thay ithno

- Chng trnh ny c phn mrng l.COM v im vo ca mlnh

c b

t

u t

a chCS:0100h.

M chng trnh trn nm trong bnhca ca my tnh. Bt ut nhca chvt l l 126Ah. (125A0+0100)Cc lnh trnc thc hin trong nv thc hin EU ca b vi x l.Cclnh c thc hin ln lt theo th tttrn xung


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Khi thc hin 1 lnh no th gi trca thanh ghi IP c tngln thm 2 trvo vtr lnh tip theo cachng trnh.

Khi chun b thc hinlnh lnh u tin th ccthanh ghi CS =125Ah, IP = 0100h, cn khi thc hin lnh thnht th chng cgi tr: CS = 125Ah, IP = 0102h.

Khi thc hin lnh thhai th chng c gi tr: CS = 125Ah, IP =0104h

Cu hi 3: Cc my tnh ca IBMPC hin nay hot ng theonguyn l no?

Theo nguyn l ny, my tnh gm nhng bphn g?

V s khi ca my tnh v nu chc nng ca tng bphn?

Nhc im chnh ca nguyn l ny l g?

Nguyn l hot ng ca my tnh IBM -PC l hin nay l nguynl do Von Neumann xut

Theo nguyn l ny, my tnh gm 5 bphn l:

Bxl trung tm - CPU

Thit bvo - Input

Thit bra - OutputBnhlm vic - RAM

Bnhvnh cuROM


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Hthng Bus gm c: Bus a ch, Bus sliu v Bus iu khinc sdng lin kt cc thnh phn trong my tnh vi Bxl trung tm.

Bus a ch: l Bus 1 chiu xut pht tCPU v c sdng

a chho cc nhv cc cng vo/raBus sliu: l Bus 2 chiu m bo trao i thng tin gia CPUvi bnhhay thit bvo/ra

Bus iu khin: C th xut pht tCPU hay tc ng t bnngoi ln CPU dng iu khin chiu ca Bus sliu, hay ccqu trnh lm vic khc ca my tnh.

Bnhmy tnh l hthng con dng lu trtm thi cc chthca chng trnh v dliu sc thc hin bi my tnh. Nthung c gi l RAM. Bnhc chia thnh cc , mi c a chring dliu c thnp.


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H thng vo/ra (I/O), y l h thng con cho php my tnhtng tc vi cc thit bkhc v lin lc vi thgii bn ngoi.N cng chu trch nhim lu trchng trnh bng cch iu

khin a cng.Hn ch: Qu trnh thc hin lnh l qu trnh tun t, chai qutrnh tng tc vi lnh hoc vi sliu, khng ththc hin cnglc.

Cu hi 4: Bxl trung tm (CPU) ca my tnh lin kt vicc bphn khc ca hthng my tnh (ngoi vi, bnh) bng

cc nhm tn hiu no?Nu chc nng, hng truyn ca cc nhm tn hiu ?

Khi CPU c lnh thc hin, th m lnh chuyn tu tiu, theo ng no?

Bxl trung tm (CPU) ca my tnh lin kt vi cc bphnkhc ca hthng my tnh (ngoi vi, bnh) thng qua cc nhm

tn hiu c gi l hthng bus.Trong my tnh c hai nhm bus l bus hthng, ni gia CPUvi bnhchnh (trong cc my hin i ngy nay bus hthngc ni gia CPU v vi mch tng hp) v bus vo ra ni ghpgia cc thit bngoi vi ti CPU.

Hthng bus bao gm 3 loi bus l bus a ch, bus dliu v bus

iu khin.BUS a ch: Phc vvic chn nhhoc thit bvo/ra thngcc cc mch gii m a ch. Khi ghi/c bnhhoc thit bvo/ra, bxl trung tm sa ln bus ny a chca cc thitblin quan. y l bus mt chiu v xut pht tCPU.


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BUS dliu: c dng chuyn dliu v thng c t8, 16,20, 24, 32 n 64 ng dy tuthuc vo tng CPU cth. BUSdliu l loi 2 chiu. Cc phn tc u ra ni thng vi bus d

liu u phi c trang bu ra 3 trng thi bo m cho bushot ng c bnh thng.

BUS iu khin: Htrcho vic trao i cc thng tin iu khinv trng thi nhphn bit thit bc CPU truy nhp l bnhhay thit bvo/ra, thao tc truy nhp l c hay vit v.v......Busiu khin thng gm hng chc dy tn hiu khc nhau, v xttheo cnhm th l loi bus 2 chiu.

Khi CPU c lnh thc hin, th m lnh chuyn tbnhvoCPU thng qua bus dliu.

Cu hi 5: Phn bit a chlogic v a chvt l ca mt ngnnh?

Nu phng php CPU xc nh a chvt l da trn a ch

logic? Cho v dminh ho?Mt a chvt l c thc t nht v nhiu nht bao nhiu achlogic?

a chlogic ca mt ngn nhl a chgm: /c segment + /coffset, c vit theo cch nhsau: Thanh ghi on:Thanh ghilch,Cn a chvt l ca mt ngn nhchnh l sthtcangn nhtrong khng gian bnhca my tnh, c bt utnh t0 (ngn nhthnht c a chvt l l 0)

CPU xc nh a chvt l da trn a chlogic theo cch: achon sc dch tri 1 khong 4 bit kt hp vi a choffsetginguyn sto thnh cch xc nh a ch20 bit.


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Nhvy, a chvt l c xc nhtheo cng thc nhsau:achvt l = Thanh ghi on x 16 + Thanh ghi lch

V d

a chon 12A1 ---- Dch tri 4 bit 1 2 A 1 0

a choffset: 0224 ----- Ginguyn 0 2 2 4

a chto thnh 20 bit 1 2 C 3 4

-Mt a chvt l c thc t nht 1 v nhiu nht 4096 a chlogic

Cu hi 6:Bvi x l 8086/88 c bus a ch v d liu baonhiu bit?N c thtruy cp ti a c bao nhiu nh?achcc nhc nh snh thno?Khi CPU thc hinlnh: MOV [DI], AL nh c truy cp c a ch l baonhiu?

Cho bit gi trcc thanh ghi bn trong CPU trc lc thchin lnh l:

AX = 0000; BX = 0000; CX = 0000; DX = 0000; SP = FFEE;

BP = 0000; SI = 0000; DI = 0000; DS = 127C; ES = 127C; SS= 127C; CS = 127C; IP = 0100

Bvi xl 8086 c bus a chl 20 bit, bus dliu 16 bit. Cn

bvi xl 8088 c bus a chl 20 bit, nhng bus dliu ch8bit.


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Bvi xl 8086/88 sdng 20 ng a chnn c tha chho c 220= 1048576 nh. Tng ng vi khng gian nh1MB.

a chcc nhc nh snhsau: nhthnht c a chvt l l 0

nhthhai c a chvt l l 1

Khi CPU thc hin lnh: MOV [DI], AL nhc truy cp ca chon ghi trn thanh ghi on DS, a choffset ghi trnDI.

Vy a chlogic l DS:DI = 127C:0000, v a chvt l ca nhc truy cp l 127C0h.

Cu hi 7:on bnhl g?a chon bnhcha trongcc thanh ghi no?

di ca cc thanh ghi on l bao nhiu bit v son b

nhc tha chho ti a l bao nhiu on?Hy cho bit ngn nhc a chvt l l 256A3h c thnmtrong on bnhno, a chlch bao nhiu?

on bnhl vng trong bnhcha 216byte = 64 KB nhlin tip nhau c xc nh bng mt a chon tng ng btu ta ch0 v a chon ln nht l FFFFFh.

- Bn trong mi on, cc nhc gn a chgi l a chOffset, a chc tnh tu on l cc s16 bit v ln nhtl FFFFh

a chon bnhcha trong cc thanh ghi on l CS, DS, SSv ES.


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Cc thanh ghi on c di l 16bit, vy son nhiu nht cthl 216on.

Vi ngn nhc a chvt l l 256A3h c thc nhiu a ch

logic khc nhau.V dnhmt vi a chnhsau:

-256A:0003 tng ng on bnhc a ch256Ah v a chlch 0003h

-2569:0013 tng ng on bnhc a ch2569h v a chlch 0013h

Cu hi 8: Gismy tnh chy 1 on chng trnh cphnmrng l .COM. Vi trng thi ban u cc thanhghi on vcon trchstrong CPU nhsau: CS = DS =SS = ES = 0700H;IP = 0100; AX=0000; BX=0000;

CX=0000; DX=0000; BP=0000;SP=FFFE; SI=0000;

DI=0000.Hi:Khi CPU nhn m lnh tbnh, n sdng nhngthanh ghig xc nh a chca lnh cn c vo?Tni dung ccthanh ghi , CPU xc nh a chvt l ca byte m lnh nhthno?

Khi CPU nhn m lnh tbnh, n sdng cc thanh ghi CSv IP

xc

nh

a ch

c

a l

nh c

n

c vo t

b

nh

.

Trong thanh ghi CS dng chti on bnhm, IP chaa choffset trong on bnh.

Tni dung cc thanh ghi , CPU xc nh a chvt l cabyte m lnh theo cng thc:


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a chvt l = Thanh ghi on x 16 + Thanh ghi lch

Vi on chng trnh trn nm trong bnhbt u ti a ch0700h:0100h, th khi nhn lnh u tin, gi trcc thanh ghi cha

a chca lnh s l: CS=0700h, IP=0100h.a ch vt l cabyte m lnh u tin l:0700h*10h+0100h = 07100h

Cu hi 9: Thno l xl ng ng trong BVXL 80x86 caIntel? Do c cu trc bn trong no m bvi xl 8086 c khnng thc hin chng trnh theo kiu ng ng?

Giscc lnh u c thi gian nhn lnh l 1 chu k, thc

hin mt 4 chu k. Xc nh thi gian thc hin 5 lnh i n i+4ca CPU 8086.

Khnng xl ng ng c c l do cu trc bn trong cabvi xl 8086 c 2 n vhot ng tng i c lp l nvthc hin EU v n vgiao tip BIU. Nhsau:

Trong thi gian EU thc hin lnh th BIU vn c thm nhn

nhim vnhn trc cc lnh, ct tm vo hng i lnh v ccthanh ghi dliu.

Nhvy m khi bus bbn trong chu kc th EU thc hin cclnh trc v khi bus bn trong chu kghi th EU li thc hinlnh khc

Qu trnh thc hin mt lnh c thc chia thnh nhiu giai

on (micro instructions): Chng hn c thchia thnh 5 giai onnhsau:

c lnh (IF: Instruction Fetch)

Gii m lnh (ID: Instruction Decode)


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Thi hnh lnh (EX: Execute)

Thm nhp bnhtrong hoc nhy(MEM: Memory access)

Lu trkt qu(RS: Result Storing).

Cu trc ng ng c thbu din nhhnh vdi:

Ta thy thc hin 5 lnh trn. Nukhng c cu trc ngngmy tnhsmt 25 chu kxung ng h. Vi cutrc ng ngmy tnh chmt 9 chu kxung ng h

Cu hi 10: ngha ca thanh ghi c trong b vi x l? Saukhikhi ng, cc cCF, PF, AF, ZF, SF, OF c gi trno?

Xc nh gi trcc csau khi bvi xl8086 thc hin lnhsau:

MOV AL,056H

ADD AL,0AAH

Bit rng trc gi trcc bit ctrn v thanh ghi AL bng0.

L thanh ghi 16 bit dng lu gi thng tin vcc trng thicng tc ca EU hoc kt quphp ton do ALU thc hin. Cncvo trng thi cc cm ngi lp trnh c thsdng cc


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lnh thch hp tip theo cho bvi xl.8086/8088 chsdng 9bit ctrong s16 bit ghi thng tin

Trong cc ct1:6 phn nh trng thi ca kt qusau mtthao tc no . Cc ccn li t7:9 l ciu khin v c lp

hoc xo bng cc lnh ring.Sau khi khi ng, cc cc gi trnhsau: CF=0, PF= 0, AF=0 , ZF=0 , SF=0 , OF =0.Thc hinphp tinh trn nhsau:

056h = 0101 0110 v 0AAh = 1010 1010


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SS:SP chn vng ngn xp

ES: SP, DI chn vng sliu

CPU dng biu thc sau y xc nh a chvt l

a chvt l = Thanh ghi on x 16 + Thanh ghi lch

Byte m lnh nm trong on bnhc a chl 1034h v lch l 0200h sdng cc thanh ghi CS=1034h v IP=0200h.

a chvt l ca byte m lnh trn l: 10340 + 0200 = 10540h,

Cu hi 12: Trnh by cch ghp ni CPU Intel8088

vi Bus hthng (Data bus, Address bus, Control Bus) chti a trong my tnh PC/XT?

Trong cc my tnh hin i, CPU phi ghp vi bnh vngoi vi nhvi mch g, cc bus ni vi bnhv ngoi vi ccng tc khng?

chti a trong my tnh PC/XT,CPU truyn tn hiu iukhin ti khiiu khin knh thng tin (chip 8288).Khi nynhn v x l tn hiu iukhin ri mi pht cc tn hiuiukhin th cp ti knh thng tin. Lcny ghp ni CPUIntel8088 vi Bus hthng (Data bus, Address bus, ControlBus)c dng nhhnh v:

Trong :

Chip 8288 c sdng iu khin bus. Khi bvi xl hotng mode cc i th mch iu khin bus 8288 c dng cung cp cc tn hiu iu khin cho chlm vic ny.


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Vi mch 74LS373 c dng cht v m cc ng dy achvi cc bus c dn knh. V thng tin a chc trn buschtrong khong thi gian ngn nn mch cht c iu khin

bi tn hiu ALE sc sdng gili thng tin a ch.74LS245 l vi mch thu pht bus hai chiu 8 bit c 3 trng thidng m v iu khin chiu chuyn ng ca dliu

Trong cc my tnh hin i, CPU phi ghp vi bnhv ngoivi thng qua vi mch tng hp hay cn gi l chipset, lc ny ccbus ni CPU vi bnhv ngoi vi khng cng mt tc .

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