Giáo Trình Kỹ Thuật Số -Nguyễn Trung Tập

Gio Trnh K Thut S

Bi:Nguyn Trung Tp

Gio Trnh K Thut S

Bi:Nguyn Trung Tp

Phin bn trc tuyn:< http://voer.edu.vn/content/col10236/1.1/ >

Hoc lieu Mo Vietnam - Vietnam Open Educational Resources

Ti liu ny v s bin tp ni dung c bn quyn thuc v Nguyn Trung Tp. Ti liu ny tun th giy php

Creative Commons Attribution 3.0 (http://creativecommons.org/licenses/by/3.0/).

Ti liu c hiu nh bi: August 9, 2010

Ngy to PDF: August 9, 2010

bit thng tin v ng gp cho cc module c trong ti liu ny, xem tr. 292.

Ni dung

1 Li ni u-k thut s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 Nguyn l ca vic VIT s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

3 CC H THNG S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54 Bin i qua li gia cc h thng s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Cc php tnh trong h nh phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 M ha . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Bi tp chng 1-k thut s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

8 HM LOGIC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 cc dng chun ca hm logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

10 RT GN HM LOGIC 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

11 RT GN HM LOGIC 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

12 RT GN HM LOGIC 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

13 BI TP-chng 2-kts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

14 CNG LOGIC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6915 CNG LOGIC C BN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7116 THNG S K THUT CA IC S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7917 H TTL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8718 HO MOS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9119 GIAO TIP GIA CC H IC S . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

20 BI TP CHNG 3-KTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

21 MCH T HP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

22 MCH GII M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

23 MCH A HP V MCH GII A HP . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 123

24 MCH SO SNH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

25 MCH KIM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

26 BI TP CHNG 4-KTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

27 MCH TUN T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 14128 MCH GHI DCH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

29 MCH M 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 157

30 MCH M 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169

31 MCH M 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

32 BI TP CHNG 5-KTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189

33 MCH LM TON . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 19134 Php tr s nh phn dng s b 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19535 Php tr s nh phn dng s b 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19736 Php ton vi s c du . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19937 Mch cng nh phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20338 Cng hai s nh phn nhiu bit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 20539 Mch tr nh phn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 21340 Mch nhn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 21741 Mch chia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 225

iv

42 B NH BN DN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23543 i cng v vn hnh ca b nh . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . 23744 Cc loi b nh bn dn 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24145 Cc loi b nh bn dn 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24946 Cc loi b nh bn dn 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 257

47 M RNG B NH . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263

48 BI TP CHNG 7-KTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269

49 BIN I AD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27150 Bin i tng t 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27951 Bin i tng t 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28552 Ti liu tham kho-k thut s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291Attributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .292

Chng 1

Li ni u-k thut s1

1.1 Li ni u

Gio trnh c bin son nhm cung cp cho sinh vin in t - Vin thng & T ng ha s kin thcc bn ca mt mn hc c coi l nn tng ca chuyn ngnh.

Ni dung gm tm chng- Chng 1 v 2 n tp mt s kin thc c bn v h thng s v hm logic m SV c th hc i

s Boole.- Chng 3 hc v Cng logic, phn t c bn ca cc mch s- Chng 4, 5 v 6 i vo cc loi mch s c th, bao gm Mch t hp, Mch tun t v Mch lm

ton. y l 3 chng nng ct ca mn hc.- Chng 7 s hc v B nh bn dn, SV s tm hiu y cu to v vn hnh ca cc loi b nh

bn dn , b nh chnh ca my tnh.- Cui cng, chng 8 s bn v loi mch gip cho con ngi giao tip vi my, l cc mch Bin i

tng t sang s v ngc li. hc tt mn hc SV cn c mt kin thc c bn v linh kin in t, gm Diod, Transistor BJT v

FET, phn vn hnh ch ngng v dn. Nu hc i s Boole nhng hc k trc th s tip thus d dng, tuy nhin, ni dung n tp chng 1 v 2 cng SV c th hc tip cc chng sau mtcch khng kh khn lm.

C th ni tt c cc mn hc c lin quan n k thut u t nhiu cn kin thc v K thut s nntrong iu kin cn kh khn khi phi c sch ngoi ng, hy vng y l mt ti liu khng th thiu trongt sch ca mt sinh vin chuyn ngnh in t-Vin thng & T ng ha.

Tc gi rt hy vng cung cp cho sinh vin mt ni dung phong ph trong mt gio trnh trang nhnhng chc khng th trnh khi thiu st. Rt mong c s gp ca c gi.

Cui cng tc gi xin thnh tht cm n Thc s Phm vn Tn c v ng gp nhiu kin qubu gio trnh c th hon thnh.

Cn th, thng 8 nm 2003Ngi vitNguyn trung Lp

1This content is available online at .

1

2 CHNG 1. LI NI U-K THUT S

Chng 2

Nguyn l ca vic VIT s1

2.1 Nguyn l ca vic VIT s

Mt s c vit bng cch t k nhau cc k hiu, c chn trong mt tp hp xc nh. Mi k hiutrong mt s c gi l s m (s hng, digit).

Th d, trong h thng thp phn (c s 10) tp hp ny gm 10 k hiu rt quen thuc, l cc cons t 0 n 9:

S10 = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}Khi mt s gm nhiu s m c vit, gi tr ca cc s m ty thuc v tr ca n trong s . Gi tr

ny c gi l trng s ca s m.Th d s 1998 trong h thp phn c gi tr xc nh bi trin khai theo a thc ca 10:199810 = 1x103 + 9x102 +9x101 + 9x100 = 1000 + 900 + 90 + 8Trong trin khai, s m ca a thc ch v tr ca mt k hiu trong mt s vi qui c v tr ca hng

n v l 0, cc v tr lin tip v pha tri l 1, 2, 3, ... . Nu c phn l, v tr u tin sau du phy l -1,cc v tr lin tip v pha phi l -2, -3, ... .

Ta thy, s 9 u tin (sau s 1) c trng s l 900 trong khi s 9 th hai ch l 90.C th nhn xt l vi 2 k hiu ging nhau trong h 10, k hiu ng trc c trng s gp 10 ln k

hiu ng ngay sau n. iu ny hon ton ng cho cc h khc, th d, i vi h nh phn ( c s 2) tht l ny l 2.

Tng qut, mt h thng s c gi l h b s gm b k hiu trong mt tp hp:Sb = {S0, S1, S2, . . ., Sb-1}Mt s N c vit:N = (anan-1an-2. . .ai . . .a0 , a-1a-2 . . .a-m)b vi ai SbS c gi tr:N = an bn + an-1bn-1 +an-2bn-2 + . . .+ aibi +. . . + a0b0 + a-1 b-1 + a-2 b-2 +. . .+ a-mb-m.=

Figure 2.1


3

4 CHNG 2. NGUYN L CA VIC VIT S

aibi chnh l trng s ca mt k hiu trong Sb v tr th i.

Chng 3

CC H THNG S1

3.1 CC H THNG S

3.1.1 H c s 10 (thp phn, Decimal system)

H thp phn l h thng s rt quen thuc, gm 10 s m nh ni trn.Di y l vi v d s thp phn:N = 199810 = 1x103 + 9x102 + 9x101 + 8x100 = 1x1000 + 9x100 + 9x10 + 8x1N = 3,1410 = 3x100 + 1x10-1 +4x10-2= 3x1 + 1x1/10 + 4x1/100

3.1.2 H c s 2 (nh phn, Binary system)

H nh phn gm hai s m trong tp hpS2 = {0, 1}.Mi s m trong mt s nh phn c gi l mt bit (vit tt ca binary digit).S N trong h nh phn:N = (anan-1an-2. . .ai . . .a0 , a-1a-2 . . .a-m)2 (vi ai S2)C gi tr l:N = an 2n + an-12n-1 +. . .+ ai2i +. . . + a020 + a-1 2-1 + a-2 2-2 + . . .+ a-m2-m

an l bit c trng s ln nht, c gi l bit MSB (Most significant bit) v a-m l bit c trng snh nht, gi l bit LSB (Least significant bit).

Th d: N = 1010,12 = 1x23 + 0x22 + 1x21 + 0x20 + 1x2-1 = 10,510

3.1.3 H c s 8 (bt phn ,Octal system)

H bt phn gm tm s trong tp hpS8 = {0,1, 2, 3, 4, 5, 6, 7}.S N trong h bt phn:N = (anan-1an-2. . .ai . . .a0 , a-1a-2 . . .a-m)8 (vi ai S8)C gi tr l:N = an 8n + an-18n-1 +an-28n-2 +. . + ai8i . . .+a080 + a-1 8-1 + a-2 8-2 +. . .+ a-m8-m

Th d: N = 1307,18 = 1x83 + 3x82 + 0x81 + 7x80 + 1x8-1 = 711,125101This content is available online at .

5

6 CHNG 3. CC H THNG S

3.1.4 H c s 16 (thp lc phn, Hexadecimal system)

H thp lc phn c dng rt thun tin con ngi giao tip vi my tnh, h ny gm mi su strong tp hp

S16 ={0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F }(A tng ng vi 1010 , B =1110 ,. . . . . . , F=1510) .S N trong h thp lc phn:N = (anan-1an-2. . .ai . . .a0 , a-1a-2 . . .a-m)16 (vi ai S16)C gi tr l:N = an 16n + an-116n-1 +an-216n-2 +. . + ai16i . . .+a0160+ a-1 16-1 + a-2 16-2 +. . .+ a-m16-m

Ngi ta thng dng ch H (hay h) sau con s ch s thp lc phn.Th d: N = 20EA,8H = 20EA,816 = 2x163 + 0x162 + 14x161 + 10x160 + 8x16-1

= 4330,510

Chng 4

Bin i qua li gia cc h thng s1

4.1 Bin i qua li gia cc h thng s

Khi c nhiu h thng s, vic xc nh gi tr tng ng ca mt s trong h ny so vi h kia l cnthit. Phn sau y cho php ta bin i qua li gia cc s trong bt c h no sang bt c h khc trongcc h c gii thiu.

4.1.1 i mt s t h b sang h 10

i mt s t h b sang h 10 ta trin khai trc tip a thc ca bMt s N trong h b:N = (anan-1an-2. . .ai . . .a0 , a-1a-2 . . .a-m)b vi ai SbC gi tr tng ng trong h 10 l:N = an bn + an-1bn-1 +. . .+ aibi +. . . + a0b0+ a-1 b-1 + a-2 b-2 +. . .+ a-mb-m.Th d:* i s 10110,112 sang h 1010110,112 = 1x24 + 0 + 1x22 + 1x2 + 0 + 1x2-1 + 1x2[U+2011]2= 22,7510* i s 4BE,ADH sang h 104BE,ADH=4x162+11x161+14x160+10x16-1+13x16-2 = 1214,67510

4.1.2 i mt s t h 10 sang h b

y l bi ton tm mt dy k hiu cho s N vit trong h b.Tng qut, mt s N cho h 10, vit sang h b c dng:N = (anan-1 . . .a0 , a-1a-2 . . .a-m)b = (anan-1 . . .a0)b + (0,a-1a-2 . . .a-m)bTrong (anan-1 . . .a0)b = PE(N) l phn nguyn ca Nv (0,a-1a-2 . . .a-m)b = PF(N) l phn l ca NPhn nguyn v phn l c bin i theo hai cch khc nhau:Phn nguyn:Gi tr ca phn nguyn xc nh nh trin khai:PE(N) = anbn +an-1bn-1 + . . .+ a1b 1+ a0b0

Hay c th vit liPE(N) = (anbn-1 +an-1bn-2 + . . .+ a1)b + a0Vi cch vit ny ta thy nu chia PE(N) cho b, ta c thng s l PE(N) = (anbn-1 +an-1bn-2

+ . . .+ a1) v s d l a0.


7

8 CHNG 4. BIN I QUA LI GIA CC H THNG S

Vy s d ca ln chia th nht ny chnh l s m c trng s nh nht (a0) ca phn nguyn.Lp li bi ton chia PE(N) cho b:PE(N) = anbn-1 +an-1bn-2 + . . .+ a1= (anbn-2 +an-1bn-3 + . . .+ a2)b+ a1Ta c s d th hai, chnh l s m c trng s ln hn k tip (a1) v thng s l PE(N)=

anbn-2 +an-1bn-3 + . . .+ a2.Tip tc bi ton chia thng s c c vi b, cho n khi c s d ca php chia cui cng,

chnh l s m c trng s ln nht (an)Phn l:Gi tr ca phn l xc nh bi:PF(N) = a-1 b-1 + a-2 b-2 +. . .+ a-mb-m

Hay vit liPF(N) = b-1 (a-1 + a-2 b-1 +. . .+ a-mb-m+1 )Nhn PF(N) vi b, ta c : bPF(N) = a-1 + (a-2 b-1 +. . .+ a-mb-m+1 ) = a-1+ PF(N).Vy ln nhn th nht ny ta c phn nguyn ca php nhn, chnh l s m c trng s ln

nht ca phn l (a-1) (s a-1 ny c th vn l s 0).PF(N) l phn l xut hin trong php nhn.Tip tc nhn PF(N) vi b, ta tm c a-2 v phn l PF(N).Lp li bi ton nhn phn l vi b cho n khi kt qu c phn l bng khng, ta s tm c dy s

(a-1a-2 . . .a-m).Ch : Phn l ca s N khi i sang h b c th gm v s s hng (do kt qu ca php nhn lun

khc 0), iu ny c ngha l ta khng tm c mt s trong h b c gi tr ng bng phn l ca s thpphn, vy ty theo yu cu v chnh xc khi chuyn i m ngi ta ly mt s s hng nht nh.

Th d:* i 25,310 sang h nh phnPhn nguyn: 25 : 2 = 12 d 1 a0 = 112 : 2 = 6 d 0 a1 = 06 : 2 = 3 d 0 a2 = 03 : 2 = 1 d 1 a3 = 1thng s cui cng l 1 cng chnh l bit a4: a4 = 1Vy PE(N) = 11001Phn l: 0,3 * 2 = 0,6 a-1 = 00,6 * 2 = 1,2 a -2 = 10,2 * 2 = 0,4 a-3 = 00,4 * 2 = 0,8 a-4 = 00,8 * 2 = 1,6 a-5 = 1 . . .Nhn thy kt qu ca cc bi ton nhn lun khc khng, do phn l ca ln nhn cui cng l 0,6,

lp li kt qu ca ln nhn th nht, nh vy bi ton khng th kt thc vi kt qu ng bng 0,3 cah 10.

Gi s bi ton yu cu ly 5 s l th ta c th dng y vPF(N) = 0,01001.Kt qu cui cng l:25,310 = 11001,010012* i 1376,8510 sang h thp lc phnPhn nguyn: 1376 : 16 = 86 s d = 0 a0 = 086 : 16 = 5 s d = 6 a1 = 6 & a2 = 5137610 = 560HPhn l: 0,85 * 16 = 13,6 a-1 = 1310=DH0,6 * 16 = 9,6 a -2 = 90,6 * 16 = 9,6 a-3 = 9Nu ch cn ly 3 s l: 0,8510= 0,D99H

9V kt qu cui cng:1376,8510 = 560,D99H

4.1.3 i mt s t h b sang h bk v ngc li

T cch trin khai a thc ca s N trong h b, ta c th nhm thnh tng k s hng t du phy v haipha v t thnh tha s chung

N = anbn +. . . +a5b5 +a4b4 +a3b3 +a2b2 +a1b1 +a0b0 +a-1 b-1 +a-2 b-2 +a-3 b-3. . .+a-mb-m

d hiu, chng ta ly th d k = 3, N c vit li bng cch nhm tng 3 s hng, k t du phyv 2 pha

N = ...+ (a5b2 +a4b1 + a3b0)b3 + (a2b2 + a1b1 + a0b0 )b0+ (a-1 b2 + a-2 b1 + a-3b0)b-3 +...Phn cha trong mi du ngoc lun lun nh hn b3 , vy s ny to nn mt s trong h b3 v lc

c biu din bi k hiu tng ng trong h ny.Tht vy, s N c dng:N = ...+A2B2+A1B1+A0B0 + A-1B-1 +...Trong :B=b3 (B0=b0; B1=b3; B2=b6, B-1=b-3 ....)A2= a8b2 +a7b1 + a6b0 = b3(a8b-1 +a7b-2 + a6b-3) < B=b3

A1= a5b2 +a4b1 + a3b0 = b3(a5b-1 +a4b-2 + a3b-3) < B=b3

A0= a2b2 + a1b1 + a0b0 = b3(a2b-1 +a1b-2 + a0b-3) < B=b3

Cc s Ai lun lun nh hn B=b3 nh vy n chnh l mt phn t ca tp hp s to nn h B=b3

Ta c kt qu bin i tng t cho cc h s k khc.Tm li, i mt s t h b sang h bk, t du phy i v hai pha, ta nhm tng k s hng, gi tr

ca mi k s hng ny (tnh theo h b) chnh l s trong h bk .Th d:* i s N = 10111110101 , 011012 sang h 8 = 23

T du phy, nhm tng 3 s hng v hai pha (nu cn, thm s 0 vo nhm u v cui 3 shng m khng lm thay i gi tr ca s N):

N = 010 111 110 101 , 011 0102Ghi gi tr tng ng ca cc s 3 bit, ta c s N trong h 8N = 2 7 6 5 , 3 2 8* i s N trn sang h 16 = 24

Cng nh trn nhng nhm tng 4 s hngN = 0101 1111 0101 , 0110 10002N = 5 F 5 , 6 8 16T kt qu ca php i s t h b sang h bk, ta c th suy ra cch bin i ngc mt cch d dng:

Thay mi s hng ca s trong h bk bng mt s gm k s hng trong h b.Th d i s N = 5 F5, 6816 (h 24) sang h nh phn (2) ta dng 4 bit vit cho mi s hng ca

s ny:N = 0101 1111 0101 , 0110 10002

4.1.4 i mt s t h bk sang h bp

Qua trung gian ca h b, ta c th i t h bk sang h bp. Mun i s N t h bk sang h bp, trc nhti s N sang h b ri t h b tip tc i sang h bp.

Th d:- i s 1234,678 sang h 161234,678 = 001 010 011 100,110 1112 = 0010 1001 1100,1101 11002 = 29C,DCH- i s ABCD,EFH sang h 8ABCD,EFH = 1010 1011 1100 1101,1110 11112 = 1 010 101 111 001 101,111 011 1102 = 125715,7368Di y l bng k cc s u tin trong cc h khc nhau:

10 CHNG 4. BIN I QUA LI GIA CC H THNG S

Figure 4.1

Bng 1.1

Chng 5

Cc php tnh trong h nh phn1

5.1 Cc php tnh trong h nh phn

Cc php tnh trong h nh phn c thc hin tng t nh trong h thp phn, tuy nhin cng c mts im cn lu

5.1.1 Php cng

L php tnh lm c s cho cc php tnh khc.Khi thc hin php cng cn lu :0 + 0 = 0 ;0 + 1 = 1 ;1 + 1 = 0 nh 1 (em qua bt cao hn).Ngoi ra nu cng nhiu s nh phn cng mt lc ta nn nh :- Nu s bit 1 chn, kt qu l 0;- Nu s bit 1 l kt qu l 1- V c 1 cp s 1 cho 1 s nh (b qua s 1 d, th d vi 5 s 1 ta k l 2 cp)Th d: Tnh 011 + 101 + 011 + 0111 1 [U+F0AC] s nh1 1 1 [U+F0AC] s nh0 1 1+ 1 0 10 1 10 1 11 1 1 0

5.1.2 Php tr

Cn lu :0 - 0 = 0 ;1 - 1 = 0 ;1 - 0 = 1 ;0 - 1 = 1 nh 1 cho bit cao hnTh d: Tnh 1011 - 01011 [U+F0AC] s nh


11

12 CHNG 5. CC PHP TNH TRONG H NH PHN

1 0 1 1- 0 1 0 10 1 1 0

5.1.3 Php nhn

Cn lu :0 x 0 = 0 ;0 x 1 = 0 ;1 x 1 = 1Th d: Tnh 1101 x 1011 1 0 1x 1 0 11 1 0 10 0 0 01 1 0 11 0 0 0 0 0 1

5.1.4 Php chia

Th d: Chia 1001100100 cho 11000Ln chia u tin, 5 bit ca s b chia nh hn s chia nn ta c kt qu l 0, sau ta ly 6 bit ca

s b chia chia tip (tng ng vi vic dch phi s chia 1 bit trc khi thc hin php tr)

Figure 5.1

Kt qu : (11001.1) 2 = (25.5)10

Chng 6

M ha1

6.1 M ha

6.1.1 Tng qut

M ha l gn mt k hiu cho mt i tng thun tin cho vic thc hin mt yu cu c th no .Mt cch ton hc, m ha l mt php p mt i mt t mt tp hp ngun vo mt tp hp khc

gi l tp hp ch.

Figure 6.1

(H 1.1)Tp hp ngun c th l tp hp cc s, cc k t, du, cc lnh dng trong truyn d liu . . . v tp

hp ch thng l tp hp cha cc t hp th t ca cc s nh phn.Mt t hp cc s nh phn tng ng vi mt s c gi l t m. Tp hp cc t m c to ra

theo mt qui lut cho ta mt b m. Vic chn mt b m ty vo mc ch s dng.Th d biu din cc ch v s, ngi ta c m ASCII (American Standard Code for Information

Interchange), m Baudot, EBCDIC . . .. Trong truyn d liu ta c m d li, d v sa li, mt m. . ..

Vn ngc li m ha gi l gii m.Cch biu din cc s trong cc h khc nhau cng c th c xem l mt hnh thc m ha, l cc

m thp phn, nh phn, thp lc phn . . . v vic chuyn t m ny sang m khc cng thuc loi biton m ha.

Trong k thut s ta thng dng cc m sau y:1This content is available online at .

13

14 CHNG 6. M HA

6.1.2 M BCD (Binary Coded Decimal)

M BCD dng s nh phn 4 bit c gi tr tng ng thay th cho tng s hng trong s thp phn.Th d:S 62510 c m BCD l 0110 0010 0101.M BCD dng rt thun li : mch in t c cc s BCD v hin th ra bng n by on (led hoc

LCD) hon ton ging nh con ngi c v vit ra s thp phn.

6.1.3 M Gray

M Gray hay cn gi l m cch khong n v.Nu quan st thng tin ra t mt my m ang m cc s kin tng dn tng n v, ta s c cc

s nh phn dn dn thay i. Ti thi im ang quan st c th c nhng li rt quan trng. Th d gias 7(0111) v 8 (1000), cc phn t nh phn u phi thay i trong qu trnh m, nhng s giao honny khng bt buc xy ra ng thi, ta c th c cc trng thi lin tip sau:

0111 0110 0100 0000 1000Trong mt quan st ngn cc kt qu thy c khc nhau. trnh hin tng ny, ngi ta cn m

ha mi s hng sao cho hai s lin tip ch khc nhau mt phn t nh phn (1 bit) gi l m cch khongn v hay m Gray.

Tnh k nhau ca cc t hp m Gray (tc cc m lin tip ch khc nhau mt bit) c dng rt chiu qu rt gn hm logic ti mc ti gin.

Ngoi ra, m Gray cn c gi l m phn chiu (do tnh i xng ca cc s hng trong tp hp m,ging nh phn chiu qua gng)

Ngi ta c th thit lp m Gray bng cch da vo tnh i xng ny:- Gi s ta c tp hp 2n t m ca s n bit th c th suy ra tp hp 2n+1 t m ca s (n+1) bit

bng cch:- Vit ra 2n t m theo th t t nh n ln- Thm s 0 vo trc tt c cc t m c c mt phn ca tp hp t m mi- Phn th hai ca tp hp gm cc t m ging nh phn th nht nhng trnh by theo th t ngc

li (ging nh phn chiu qua gng) v pha trc thm vo s 1 thay v s 0 (H 1.2).

Figure 6.2

15

(H 1.2) thit lp m Gray ca s nhiu bit ta c th thc hin cc bc lin tip t tp hp u tin ca s

mt bit (gm hai bit 0, 1).Di y l cc bc to m Gray ca s 4 bit. Ct bn phi ca bng m 4 bit cho gi tr tng ng

trong h thp phn ca m Gray tng ng (H 1.3).

Figure 6.3

(H 1.3)Nhn xt cc bng m ca cc s Gray (1 bit, 2 bit, 3 bit v 4 bit) ta thy cc s gn nhau lun lun

khc nhau mt bit, ngoi ra, trong tng b m, cc s i xng nhau qua gng cng khc nhau mt bit.

16 CHNG 6. M HA

Chng 7

Bi tp chng 1-k thut s1

7.1 Bi tp

1. i cc s thp phn di y sang h nh phn v h thp lc phn :a/ 12 b/ 24 c/ 192 d/ 2079 e/ 15492f/ 0,25 g/ 0,375 h/ 0,376 i/ 17,150 j/ 192,18752. i sang h thp phn v m BCD cc s nh phn sau y:a/ 1011 b/ 10110 c/ 101,1 d/ 0,1101e/ 0,001 f/ 110,01 g/ 1011011 h/ 101011010113. i cc s thp lc phn di y sang h 10 v h 8:a/ FF b/ 1A c/ 789 d/ 0,13 e/ ABCD,EF4. i cc s nh phn di y sang h 8 v h 16:a/ 111001001,001110001 b/ 10101110001,00011010101c/ 1010101011001100,1010110010101 d/ 1111011100001,010101110015. M ha s thp phn di y dng m BCD :a/ 12 b/ 192 c/ 2079 d/15436 e/ 0,375 f/ 17,250


17

18 CHNG 7. BI TP CHNG 1-K THUT S

Chng 8

HM LOGIC1

8.1 HM LOGIC

Nm 1854 Georges Boole, mt trit gia ng thi l nh ton hc ngi Anh cho xut bn mt tc phm vl lun logic, ni dung ca tc phm t ra nhng mnh m tr li ngi ta ch phi dng mt tronghai t ng (c, yes) hoc sai (khng, no).

Tp hp cc thut ton dng cho cc mnh ny hnh thnh mn i s Boole. y l mn ton hcdng h thng s nh phn m ng dng ca n trong k thut chnh l cc mch logic, nn tng ca kthut s.

Chng ny khng c tham vng trnh by l thuyt i s Boole m ch gii hn trong vic gii thiucc hm logic c bn v cc tnh cht cn thit gip sinh vin hiu vn hnh ca mt h thng logic.

8.1.1 HM LOGIC C BN

8.1.1.1 Mt s nh ngha

- Trng thi logic: trng thi ca mt thc th. Xt v mt logic th mt thc th ch tn ti mt tronghai trng thi. Th d, i vi mt bng n ta ch quan tm n ang trng thi no: tt hay chy. Vytt / chy l 2 trng thi logic ca n.

- Bin logic dng c trng cho cc trng thi logic ca cc thc th. Ngi ta biu din bin logic bimt k hiu (ch hay du) v n ch nhn 1 trong 2 gi tr : 0 hoc 1.

Th d trng thi logic ca mt cng tc l ng hoc m, m ta c th c trng bi tr 1 hoc 0.- Hm logic din t bi mt nhm bin logic lin h nhau bi cc php ton logic. Cng nh bin logic,

hm logic ch nhn 1 trong 2 gi tr: 0 hoc 1 ty theo cc iu kin lin quan n cc bin.Th d, mt mch gm mt ngun hiu th cp cho mt bng n qua hai cng tc mc ni tip, bng

n ch chy khi c 2 cng tc u ng. Trng thi ca bng n l mt hm theo 2 bin l trng thi ca2 cng tc.

Gi A v B l tn bin ch cng tc, cng tc ng ng vi tr 1 v h ng vi tr 0. Y l hm ch trngthi bng n, 1 ch n chy v 0 khi n tt. Quan h gia hm Y v cc bin A, B c din t nh bngsau:


19

20 CHNG 8. HM LOGIC

Figure 8.1

8.1.1.2 Biu din bin v hm logic

8.1.1.2.1 Gin Venn

Cn gi l gin Euler, c bit dng trong lnh vc tp hp. Mi bin logic chia khng gian ra 2 vngkhng gian con, mt vng trong gi tr bin l ng (hay=1), v vng cn li l vng ph trong gitr bin l sai (hay=0).

Th d: Phn giao nhau ca hai tp hp con A v B (gch cho) biu din tp hp trong A v B lng (A AND B) (H 2.1)

Figure 8.2

(H 2.1)

8.1.1.2.2 Bng s tht

Nu hm c n bin, bng s tht c n+1 ct v 2n + 1 hng. Hng u tin ch tn bin v hm, cc hngcn li trnh by cc t hp ca n bin trong 2n t hp c th c. Cc ct u ghi gi tr ca bin, ct cuicng ghi gi tr ca hm tng ng vi t hp bin trn cng hng (gi l tr ring ca hm).

Th d: Hm OR ca 2 bin A, B: f(A,B) = (A OR B) c bng s tht tng ng.

21

Figure 8.3

8.1.1.2.3 Bng Karnaugh

y l cch biu din khc ca bng s tht trong mi hng ca bng s tht c thay th bi mt m ta (gm hng v ct) xc nh bi t hp cho ca bin.

Bng Karnaugh ca n bin gm 2n . Gi tr ca hm c ghi ti mi ca bng. Bng Karnaugh rtthun tin n gin hm logic bng cch nhm cc li vi nhau.

Th d: Hm OR trn c din t bi bng Karnaugh sau y

Figure 8.4

8.1.1.2.4 Gin thi gian

Dng din t quan h gia cc hm v bin theo thi gian, ng thi vi quan h logic.Th d: Gin thi gian ca hm OR ca 2 bin A v B, ti nhng thi im c mt (hoc 2) bin c

gi tr 1 th hm c tr 1 v hm ch c tr 0 ti nhng thi im m c 2 bin u bng 0.

22 CHNG 8. HM LOGIC

Figure 8.5

(H 2.2)

8.1.1.3 Qui c

Khi nghin cu mt h thng logic, cn xc nh qui c logic. Qui c ny khng c thay i trong sutqu trnh nghin cu.

Ngi ta dng 2 mc in th thp v cao gn cho 2 trng thi logic 1 v 0.Qui c logic dng gn in th thp cho logic 0 v in th cao cho logic 1Qui c logic m th ngc li.

8.1.1.4 Hm logic c bn (Cc php ton logic)

Hm NOT (o, b) : Y = ABng s tht

Figure 8.6

8.1.1.4.1 Hm AND [tch logic, ton t (.)] : Y = A.B

Bng s tht

23

Figure 8.7

Nhn xt: Tnh cht ca hm AND c th c pht biu nh sau:- Hm AND ca 2 (hay nhiu) bin ch c gi tr 1 khi tt c cc bin u bng 1hoc- Hm AND ca 2 (hay nhiu) bin c gi tr 0 khi c mt bin bng 0.

8.1.1.4.2 Hm OR [tng logic, ton t (+)] : Y = A + B

Bng s tht

Figure 8.8

Nhn xt: Tnh cht ca hm OR c th c pht biu nh sau:- Hm OR ca 2 (hay nhiu) bin ch c gi tr 0 khi tt c cc bin u bng 0hoc- Hm OR ca 2 (hay nhiu) bin c gi tr 1 khi c mt bin bng 1.Hm EX-OR (OR loi tr) Y = A BBng s tht

24 CHNG 8. HM LOGIC

Figure 8.9

Nhn xt: Mt s tnh cht ca hm EX - OR:- Hm EX - OR ca 2 bin ch c gi tr 1 khi hai bin khc nhau v ngc li. Tnh cht ny c dng

so snh 2 bin.- Hm EX - OR ca 2 bin cho php thc hin cng hai s nh phn 1 bit m khng quan tm ti s nh.- T kt qu ca hm EX-OR 2 bin ta suy ra bng s tht cho hm 3 bin

Figure 8.10

- Trong trng hp 3 bin (v suy rng ra cho nhiu bin), hm EX - OR c gi tr 1 khi s bin bng 1l s l. Tnh cht ny c dng nhn dng mt chui d liu c s bit 1 l chn hay l trong thit kmch pht chn l.

8.1.1.5 Tnh cht ca cc hm logic c bn:

8.1.1.5.1 Tnh cht c bn:

C mt phn t trung tnh duy nht cho mi ton t (+) v (.):

25

A + 0 = A ; 0 l phn t trung tnh ca hm ORA . 1 = A ; 1 l phn t trung tnh ca hm ANDTnh giao hon:A + B = B + AA . B = B . ATnh phi hp:(A + B) + C = A + (B + C) = A + B + C(A . B) . C = A . (B . C) = A . B . CTnh phn b:- Phn b i vi php nhn: A . (B + C) = A . B + A . C- Phn b i vi php cng: A + (B . C) = (A + B) . (A + C)Phn b i vi php cng l mt tnh cht c bit ca php ton logicKhng c php tnh ly tha v tha s:A + A + . . . . . + A = AA . A . . . . . . . . A = ATnh b:

Figure 8.11

8.1.1.5.2 Tnh song i (duality):

Tt c biu thc logic vn ng khi [thay php ton (+) bi php (.) v 0 bi 1] hay ngc li. iu ny cth chng minh d dng cho tt c biu thc trn.

Th d:

Figure 8.12

8.1.1.5.3 nh l De Morgan

nh l De Morgan c pht biu bi hai biu thc:

26 CHNG 8. HM LOGIC

Figure 8.13

nh l De Morgan cho php bin i qua li gia hai php cng v nhn nh vo php o.nh l De Morgan c chng minh bng cch lp bng s tht cho tt c trng hp c th c ca

cc bin A, B, C vi cc hm AND, OR v NOT ca chng.

8.1.1.5.4 S ph thuc ln nhau ca cc hm logic c bn

nh l De Morgan cho thy cc hm logic khng c lp vi nhau, chng c th bin i qua li, s bini ny cn c s tham gia ca hm NOT. Kt qu l ta c th dng hm (AND v NOT) hoc (OR vNOT) din t tt c cc hm.

Th d:Ch dng hm AND v NOT din t hm sau: Y = A.B+B.C+A.CCh cn o hm Y hai ln, ta c kt qu:

Figure 8.14

Nu dng hm OR v NOT din t hm trn lm nh sau:

Figure 8.15

Chng 9

cc dng chun ca hm logic1

9.1 CC DNG CHUN CA HM LOGIC

Mt hm logic c biu din bi mt t hp ca nhng tng v tch logic.Nu biu thc l tng ca nhng tch, ta c dng tngTh d :

Figure 9.1

Nu biu thc l tch ca nhng tng, ta c dng tchTh d :

Figure 9.2

Mt hm logic c gi l hm chun nu mi s hng cha y cc bin, dng nguyn hay dngo ca chng.

Th d :

Figure 9.3


27

28 CHNG 9. CC DNG CHUN CA HM LOGIC

l mt tng chun.Mi s hng ca tng chun c gi l minterm.

Figure 9.4

l mt tch chun.Mi s hng ca tch chun c gi l maxterm.Phn sau y cho php chng ta vit ra mt hm di dng tng chun hay tch chun khi c bng s

tht din t hm .

9.1.1 Dng tng chun

c c hm logic di dng chun, ta p dng cc nh l trin khai ca Shanon.Dng tng chun c c t trin khai theo nh l Shanon th nht:Tt c cc hm logic c th trin khai theo mt trong nhng bin di dng tng ca hai

tch nh sau:

Figure 9.5

(1)H thc (1) c th c chng minh rt d dng bng cch ln lt cho A bng 2 gi tr 0 v 1, ta c

kt qu l 2 v ca (1) lun lun bng nhau. Tht vy

Figure 9.6

Vi 2 bin, hm f(A,B) c th trin khai theo bin A :

Figure 9.7

29

Mi hm trong hai hm va tm c li c th trin khai theo bin B

Figure 9.8

f(i,j) l gi tr ring ca f(A,B) khi A=i v B=j trong bng s tht ca hm.Vi 3 bin, tr ring ca f(A, B, C) l f(i, j, k) khi A=i, B=j v C=k ta c:Nhc li tnh cht ca cc hm AND v OR: b1.b2.... bn = 1 khi b1, b2..., bn ng thi bng 1 v a1

+ a2 + ... + ap = 1 ch cn t nht mt bin a1, a2, ..., ap bng 1Tr li th d trn, biu thc logic tng ng vi hng 1 (A=0, B=0, C=1) c vit

Figure 9.9

ng thi.Biu thc logic tng ng vi hng 2 l ng thi

Figure 9.10

Tng t, vi cc hng 3, 5 v 7 ta c cc kt qu:

Figure 9.11

Nh vy, trong th d trnZ = hng 1 + hng 2 + hng 3 + hng 5 + hng 7


Figure 9.12

Tm li, t mt hm cho di dng bng s tht, ta c th vit ngay biu thc ca hm di dng tngchun nh sau:

- S s hng ca biu thc bng s gi tr 1 ca hm th hin trn bng s tht- Mi s hng trong tng chun l tch ca tt c cc bin tng ng vi t hp m hm

c tr ring bng 1, bin c gi nguyn khi c gi tr 1 v c o nu gi tr ca n = 0.

9.1.2 Dng tch chun

y l dng ca hm logic c c t trin khai theo nh l Shanon th hai:Tt c cc hm logic c th trin khai theo mt trong nhng bin di dng tch ca hai

tng nh sau:

Figure 9.13

(2)Cch chng minh nh l Shanon th hai cng ging nh chng minh nh l Shanon th nht.Vi hai bin, hm f(A,B) c th trin khai theo bin A

Figure 9.14

Mi hm trong hai hm va tm c li c th trin khai theo bin B

Figure 9.15

31

Figure 9.16

Vy:

Figure 9.17

Cng nh dng chun th nht, f(i,j) l gi tr ring ca f(A,B) khi A=i v B=j trong bng s tht cahm.

Vi hm 3 bin:

Figure 9.18

S s hng trong trin khai n bin l 2n. Mi s hng l tng (OR) ca cc bin v tr ring ca hm.- Nu tr ring bng 0 s hng c rt gn li ch cn cc bin (0 l tr trung tnh ca php cng logic)A + B + C + f(0,0,0) = A + B + C nu f(0,0,0) = 0- Nu tr ring bng 1, s hng trin khai = 1A + B +

Figure 9.19

+ f(0,0,1) = 1 nu f(0,0,1) = 1v bin mt trong biu thc ca tch chun.Ly li th d trn:


Figure 9.20

Cc tr ring ca hm nu trn.- Hm Z c gi tr ring f(0,0,0) = 0 tng ng vi cc gi tr ca bin hng 0 l A=B=C=0 ng thi,

vy A+B+C l mt s hng trong tch chun.- Tng t vi cc hng (4) v (6) ta c cc t hp

Figure 9.21

- Vi cc hng cn li (hng 1, 2, 3, 5, 7), tr ring ca f(A,B,C) = 1 nn khng xut hin trong trinkhai.

Tm li, ta c

Figure 9.22

- ngha ca nh l th hai:Nhc li tnh cht ca cc hm AND v OR: b1.b2.... bn =0 ch cn t nht mt bin trong b1, b2,...,

bn =0 v a1 + a2 + ... + ap =0 khi cc bin a1, a2, ..., ap ng thi bng 0.Nh vy trong th d trn:Z = (hng 0).(hng 4).(hng 6)

33

Figure 9.23

Tht vy, hng 0 tt c bin = 0: A=0, B=0, C=0 ng thi nn c th vit (A+B+C) = 0. Tng tcho hng (4) v hng (6).

Tm li,Biu thc tch chun gm cc tha s, mi tha s l tng cc bin tng ng vi t hp

c gi tr ring =0, mt bin gi nguyn nu n c gi tr 0 v c o nu c gi tr 1. Stha s ca biu thc bng s s 0 ca hm th hin trn bng s tht.

9.1.3 i t dng chun ny sang dng chun khc:

Nh nh l De Morgan, hai nh l trn c th chuyn i qua li.Tr li th d trn, thm ctZ ngangvo bng s tht:

Figure 9.24

Din t Z ngag theo dng tng chun:

Figure 9.25

Ly o hai v:


Figure 9.26

Dng nh l De Morgan mt ln na cho tng tha s trong biu thc, ta c:

Figure 9.27

Din t Z ngang theo dng tch chun:

Figure 9.28

Ly o hai v:

Figure 9.29

Figure 9.30

Figure 9.31

35

9.1.4 Dng s

n gin cch vit ngi ta c th din t mt hm Tng chun hay Tch chun bi tp hp cc sdi du tng (tn) hay tch (). Mi t hp bin c thay bi mt s thp phn tng ng vi tr nhphn ca chng. Khi s dng cch vit ny trng lng cc bin phi c ch r.

Th d :Cho hm Z xc nh nh trn, tng ng vi dng chun th nht, hm ny ly gi tr ca cchng 1, 2, 3, 5, 7, ta vit

Figure 9.32

Tng t, nu dng dng chun th hai ta c th vit Z =f(A,B,C)= (0,4,6).Ch : Khi vit cc hm theo dng s ta phi ch r trng s ca cc bit, th d ta c th ghi km theo

hm Z trn 1 trong 3 cch nh sau: A=MSB hoc C=LSB hoc A=4, B=2, C=1

Figure 9.33

Chng 10

RT GN HM LOGIC 11

10.1 RT GN HM LOGIC

thc hin mt hm logic bng mch in t, ngi ta lun lun ngh n vic s dng lng linh kint nht. Mun vy, hm logic phi dng ti gin, nn vn rt gn hm logic l bc u tin phithc hin trong qu trnh thit k. C 3 phng php rt gn hm logic:

- Phng php i s- Phng php dng bng Karnaugh- Phng php Quine Mc. Cluskey

10.1.1 Phng php i s

Phng php ny bao gm vic p dng cc tnh cht ca hm logic c bn. Mt s ng thc thng cs dng c nhm li nh sau:

Figure 10.1

Chng minh cc ng thc 1, 2, 3:1This content is available online at .

37

38 CHNG 10. RT GN HM LOGIC 1

Figure 10.2

Cc ng thc (1), (2), (3) l song i ca (1), (2), (3).Cc qui tc rt gn:-Qui tc 1:Nh cc ng thc trn nhm cc s hng li.Th d:Rt gn biu thc

Figure 10.3

Theo (1)

Figure 10.4

Vy

Figure 10.5

Theo (3)

Figure 10.6

39

V kt qu cui cng:

Figure 10.7

-Qui tc 2:Ta c th thm mt s hng c trong biu thc logic vo biu thc m khng lm thayi biu thc.

Th d:Rt gn biu thc:

Figure 10.8

Thm ABC vo c:

Figure 10.9

Theo (1) cc nhm trong du ngoc rt gn thnh: BC + AC + ABVy:

Figure 10.10

- Qui tc 3: C th b s hng cha cc bin c trong s hng khcTh d 1: Rt gn biu thc AB +


Figure 10.11

C + ACBiu thc khng i nu ta nhn mt s hng trong biu thc vi 1, v d (B+

Figure 10.12

):

Figure 10.13

Trin khai s hng cui cng ca v phi, ta c:

Figure 10.14

Tha s chung:

Figure 10.15

Tm li:

41

Figure 10.16

Trong bi tan ny ta n gin c s hng AC.Th d 2: Rt gn biu thc (A+B).(

Figure 10.17

+C).(A+C)Biu thc khng i nu ta thm vo mt tha s c tr =0, v d B.[U+F042]

Figure 10.18

Theo (2)

Figure 10.19

Vy:

Figure 10.20

Trong bi tan ny ta b s hng A+C- Qui tc 4:C th n gin bng cch dng hm chun tng ng c s hng t nht.Th d:Hm f(A,B,C) = (2,3,4,5,6,7) vi trng lng A=4, B=2, C=1Hm o ca f:


Figure 10.21

Vy f(A,B,C) = A+B

10.1.2 Dng bng Karnaugh

Dng bng Karnaugh cho php rt gn d dng cc hm logic cha t 3 ti 6 bin.

10.1.2.1 Nguyn tc

Xt hai t hp bin AB v A

Figure 10.22

hai t hp ny ch khc nhau mt bit, ta gi chng l hai t hp k nhau.Ta c: AB + A

Figure 10.23

= A , bin B c n gin .Phng php ca bng Karnaugh da vo vic nhm cc t hp k nhau trn bng n gin bin c

gi tr khc nhau trong cc t hp ny.Cng vic rt gn hm c thc hin theo bn bc: V bng Karnaugh theo s bin ca hm Chuyn hm cn n gin vo bng Karnaugh Gom cc cha cc t hp k nhau li thnh cc nhm sao cho c th rt gn hm ti mc ti gin Vit kt qu hm rt gn t cc nhm gom c.

10.1.2.2 V bng Karnaugh

- Bng Karnaugh thc cht l mt dng khc ca bng s tht, trong mi ca bng tng ng vimt hng trong bng s tht.

v bng Karnaugh cho n bin, ngi ta chia s bin ra lm i, phn na dng to 2n/2 ct, phnna cn li to 2n/2 hng (nu n l s l, ngi ta c th cho s lng bin trn ct ln hn s lng bincho hng hay ngc li cng c). Nh vy, vi mt hm c n bin, bng Karnaugh gm 2n , mi tngng vi t hp bin ny. Cc trong bng c sp t sao cho hai k nhau ch khc nhau mt n v nh

43

phn (khc nhau mt bit), iu ny cho thy rt thun tin nu chng ta dng m Gray. Chnh s sp tny cho php ta n gin bng cch nhm cc k nhau li.

Vi 2 bin AB, s sp t s theo th t: AB = 00, 01, 11, 10 (y l th t m Gray, nhng cho dta dng s nh phn tng ng c th t ny: 0, 1, 3, 2)

Th d: Bng Karnaugh cho hm 3 bin (A = MSB, v C = LSB) (H 2.3)

Figure 10.24

(H 2.3)Vi 3 bin ABC, ta c: ABC = 000, 001, 011, 010, 110, 111, 101, 100 (s nh phn tng ng: 0, 1, 3,

2, 6, 7, 5, 4)Lu l ta c th thit lp bng Karnaugh theo chiu nm ngang hay theo chiu ng.Do cc t hp cc ba tri v phi k nhau nn ta c th coi bng c dng hnh tr thng ng v cc

t hp ba trn v di cng k nhau nn ta c th coi bng c dng hnh tr trc nm ngang. V 4 thp bin 4 gc cng l cc t hp k nhau.

Hnh (H 2.4) l bng Karnaugh cho 4 bin.

Figure 10.25

(H 2.4)

10.1.2.3 Chuyn hm logic vo bng Karnaugh.

Trong mi ca bng ta a vo gi tr ca hm tng ng vi t hp bin, n gin chng ta c thch ghi cc tr 1 m b qua cc tr 0 ca hm. Ta c cc trng hp sau:

T hm vit di dng tng chun:Th d 1 :


Figure 10.26

Figure 10.27

(H 2.5)Nu hm khng phi l dng chun, ta phi a v dng chun bng cch thm vo cc s hng

sao cho hm vn khng i nhng cc s hng cha cc bin.Th d 2 :

Figure 10.28

Hm ny gm 4 bin, nn a v dng tng chun ta lm nh sau:

Figure 10.29

V Hm Y c a vo bng Karnaugh nh sau (H 2.6):

45

Figure 10.30

(H 2.6)T dng s th nht, vi cc trng lng tng ng A=4, B=2, C=1Th d 3: f(A,B,C) = (1,3,7). Hm s s ly gi tr 1 trong cc 1,3 v 7.T dng tch chun: Ta ly hm o c dng tng chun v ghi tr 0 vo cc tng ng vi t

hp bin trong tng chun ny. Cc cn li cha s 1.Th d 4 :

Figure 10.31

V bng Karnaugh tng ng (H 2.7).

Figure 10.32

(H 2.7)


T dng s th hai:Th d 5 : f(A,B,C) = (0,2,4,5,6)Hm s ly cc tr 0 cc 0, 2, 4, 5, 6. D nhin l ta phi ghi cc gi tr 1 trong cc cn li (H 2.7).T bng s tht:Th d 6 : Hm f(A,B,C) cho bi bng s tht

Figure 10.33

Ta ghi 1 vo cc tng ng vi cc t hp bin hng 1, 3 v 7, kt qu ging nh th d 1.Trng hp c mt s t hp cho gi tr hm khng xc nh: ngha l ng vi cc t hp ny

hm c th c gi tr 1 hoc 0, do , ta ghi du X vo cc tng ng vi cc t hp ny, lc gom nhmta s dng n nh s 1 hay s 0 mt cch ty sao cho c c kt qu rt gn nht.

Chng 11

RT GN HM LOGIC 21

Th d 7: f(A,B,C,D) = (3,4,5,6,7) vi cc t hp t 10 dn 15 cho hm c tr bt k (khng xc nh)(H 2.8).

Figure 11.1

(H 2.8)

11.1 Qui tc gom nhm

Cc t hp bin c trong hm logic hin din trong bng Karnaugh di dng cc s 1 trong cc , vy vicgom thnh nhm cc t hp k nhau c thc hin theo qui tc sau:

- Gom cc s 1 k nhau thnh tng nhm sao cho s nhm cng t cng tt. iu ny c ngha l s shng trong kt qu s cng t i.

- Tt c cc s 1 phi c gom thnh nhm v mt s 1 c th nhiu nhm.- S s 1 trong mi nhm cng nhiu cng tt nhng phi l bi ca 2k (mi nhm c th c 1, 2, 4, 8

... s 1). C mi nhm cha 2k s 1 th t hp bin tng ng vi nhm gim i k s hng.- Kim tra bo m s nhm gom c khng tha.


47


11.2 Qui tc rt gn

- Kt qu cui cng c ly nh sau:Hm rt gn l tng ca cc tch: Mi s hng ca tng tng ng vi mt nhm cc s 1 ni trn v

s hng ny l tch ca cc bin, bin A (hay__

A ) l tha s ca tch khi tt c cc s 1 ca nhm ch chatrong phn na bng trong bin A c gi tr 1 (hay 0). Ni cch khc nu cc s 1 ca nhm ng thinm trong cc ca bin A v

__

A th bin A s c n gin. Hnh di y minh ha vic ly cc thas trong tch

Th d i vi bng (H 2.9) ta c kt qu nh sau:- Hm Y l hm 4 bin A,B,C,D

Figure 11.2

(H 2.9)- Nhm 1 cha 2 s 1 (k=1), nh vy nhm 1 s cn 3 bin, theo hng, 2 s 1 ny 2 ng vi

__

A B vAB, bin A s c n gin v theo ct th 2 ny ng vi t hp

Figure 11.3

.Kt qu ng vi nhm 1 l:

Figure 11.4

49

- Nhm 2 cha 4 s 1 (4=22 , k=2), nh vy nhm 2 s cn 2 bin, theo hng, 4 s 1 ny 2 ng vit hp

__

A

Figure 11.5

v__

A B, bin B s c n gin v theo ct th 4 ny ng vi t hp CD v C

Figure 11.6

, cho php n gin bin D .Kt qu ng vi nhm 2 l:

__

AC.- Nhm 3 cha 4 s 1 (4=22 , k=2), nh vy nhm 2 s cn 2 bin, theo hng, 4 s 1 ny ng vi t

hp__

A B, theo ct 4 s 1 ny chim ht 4 ct nn 2 bin Cv D c n gin.Kt qu ng vi nhm 3 l:

__

AB.V hm Y rt gn l: Y =

Figure 11.7

Di y l mt s th dTh d 1 :Rt gn hmY =


Figure 11.8

(H 2.10)(H 2.10)cho

Figure 11.9

Th d 2 :Rt gn hmY =f(A,B,C,D) = (0,2,4,5,8,10,12,13) vi A=MSB

Figure 11.10

(H 2.11)(H 2.11) cho

51

Figure 11.11

Th d 3 :Rt gn hm S cho bi bng s tht:

Figure 11.12

Bng Karnaugh: (H 2.12)

Figure 11.13

(H 2.12)Kt qu


:

Figure 11.14

11.3 Rt gn cc hm nhiu bin bng cch dng bng Karnaugh4 bin:

rt gn cc hm nhiu bin (5 v 6 bin) ngi ta c th dng bng Karnaugh 4 bin. Di y l vith d:

Th d 4: Rt gn hm f(A,B,C,D,E) = [U+F0E5] (0,2,8,10,13,15,16,18,24,25,26,29,31) vi (7,9,14,30)khng xc nh

Trc nht v 2 bng Karnaugh cho 4 bin BCDE, mt ng vi__

A v mt vi A Bng ng vi

__

A dng cho cc s t 0 n 15 Bng ng vi A dng cho cc s t 16 n 31 Nhm cc s 1 c cng v tr hai bng, kt qu s n gin bin A

- Nhm cc s 1 ca tng bng cho n ht , kt qu c xc nh nh cch lm thng thng, nh A v__

A trong tng nhm (H 2.13).

Figure 11.15

(H 2.13)nhm (1) cho :

53

Figure 11.16

; (2) cho : BCE ; (3) cho :

Figure 11.17

Vy

Figure 11.18

Th d 5: Rt gn hmf(A,B,C,D,E,F)=[U+F0E5](2,3,6,7,8,9,12,13,14,17,24,25,28,29,30,40,41,44,45,46,56,57,59,60,61,63)Tng t nh trn nhng phi v 4 bng cho:__

A

Figure 11.19

cho cc s (0-15) ;__

A B cho cc s (16-31) ;AB cho cc s (48-63) v A

Figure 11.20

cho cc s (32-47).


Figure 11.21

(H 2.14)Kt qu: (1) cho

Figure 11.22

;

Figure 11.23

Vy:

55

Figure 11.24

11.4 Phng php Quine-Mc. Cluskey

Phng php Quine-Mc. Cluskey cng da trn tnh k ca cc t hp bin n gin s bin trong cc shng ca biu thc dng tng (minterm). Trong qu trnh n gin ny c th xut hin cc s hng gingnhau m ta c th b bt c.

Phng php c thc hin qua 2 giai an:Giai an 1: Da trn tnh k ca cc t hp bin n gin s bin trong cc s hng ca biu thc

dng tng (minterm).Giai an 2: Kim tra v thc hin vic ti gin .Th d di y minh ha cho vic thc hin phng php rt gn mt hm logic.Th d 1: Rt gn hm f(A,B,C,D) = (1,2,4,5,6,10,12,13,14)Giai an 1- Cc minterm c nhm li theo s s 1 c trong t hp v ghi li trong bng theo th t s 1 tng

dn:Trong th d ny c 3 nhm:Nhm cha mt s 1 gm cc t hp 1, 2, 4Nhm cha hai s 1 gm cc t hp 5, 6, 10, 12Nhm cha ba s 1 gm cc t hp 13, 14Bng 1:

Figure 11.25


Figure 11.26

Chng 12

RT GN HM LOGIC 31

- Mi t hp trong mt nhm s c so snh vi mi t hp trong nhm k cn. Nu 2 t hp ch khcnhau mt bin, ta c th dng biu thc AB +

___

A B = B n gin c 1 bin. Bin n gin cthay bi du -. nh du x vo cc t hp xt trnh sai st

Nh vy, t hp th nht ca nhm th nht 0001 so snh vi t hp th nht ca nhm th hai 0101v chng ch khc nhau bin B, vy chng c th n gin thnh 0-01. Hai s hng 1 v 5 c gomli thnh nhm (1,5) v c ghi vo bng 2.

Tip tc so snh t hp 0001 ny vi cc t hp cn li ca nhm 2 (0110, 1010, 1100), v chng khcnhau nhiu hn 1 bit nn ta khng c kt qu no khc. Nh vy, ta so snh xong t hp th nht,nh du x trc t hp ny ghi nh.

Cng vic tin hnh tng t cho nhm th hai v th ba.Lu : Nhn xt v vic so snh cc t hp vi nhau ta thy c th thc hin nhanh c bng cch

lm bi ton tr 2 s nh phn tng ng ca 2 t hp, nu kt qu l mt s c tr = 2k (1, 2, 4,8 ...) th 2t hp so snh c v bin c n gin chnh l bin c trng s =2k (th d 2 t hp 1 v 5 c hius l 4 nn n gin c bin B), nu hiu s 6= 2k th 2 t hp khng so snh c, tc khng c binc n gin.

Kt qu cho bng th hai- Bng th hai gm cc t hp c rt gn v ch cn li 2 nhm (gim mt nhm so vi bng 1).Bng 2


57


Figure 12.1

Thc hin cng vic tng t nh trn vi hai nhm trong bng th hai ny, cc s hng s c nhmli nu chng ch khc nhau mt bin v c v tr du - trng nhau. Ta c bng th 3.

Bng 3:

Figure 12.2

Quan st bng th 3 ta thy c cc t hp ging nhau, nh vy ta c th lai b bt cc t hp ny vch gi li mt.

Kt qu ca hm rt gn gm tng cc s hng tng ng vi cc t hp khng gom thnh nhm trongcc bng u tin, l t hp (1,5) trong bng 2, tr tng ng l

___

A__

C D vi cc t hp cn li trongbng cui cng, l cc t hp (2,6 ; 10,14) m tr tng ng l C

__

C__

D , (4,5 ; 12,13) cho B__

C v (4,6 ;12,14) cho B

__

D trong bng 3. Vy:

59

Figure 12.3

n y, nu quan st cc t hp cho cc kt qu trn, ta thy cc t hp cn cha cc s hng gingnhau (s 4 v s 12 chng hn), nh vy kt qu trn c th l cha ti gin.

Giai an 2: c th rt gn hn na ta lp mt bng nh sau:Ct bn tri ghi li cc t hp chn c trong giai on 1, cc ct cn li ghi cc tr thp phn c

trong hm ban u.Trn cng hng ca t hp ta nh du * di cc ct c s tng ng (v d hng cha t hp 1,5 c

cc du * ct 1 v 5). Tng t cho cc t hp khc.Bng 4

Figure 12.4

Xt cc ct ch cha mt du *, l cc ct 1,2,10 v 13, cc t hp cng hng vi cc du * ny sc chn, l cc t hp (1,5), (2,6 ; 10,14), (4,5 ; 12,13), tng ng vi

___

A__

C D + C__

D + B__

C . nhdu X di cc ct tng ng vi cc s c trong cc t hp chn. Nu tt c cc ct u c nh duth cc t hp chn din t hm ban u.

Trong trng hp ca bi ton ny, sau khi chn cc t hp ni trn th tt c ct c nh du do kt qu cui cng l (sau khi loai b t hp B

__

D ):

Figure 12.5

Th d 2: Rt gn hm f(A,B,C,D) = (3,4,6,7,8,11,12,15)Giai an 1Bng 1:


Figure 12.6

So snh cc t hp ca 2 nhm gn nhau ta c kt qu cho bng th hai- Bng th hai gm cc t hp c rt gn v ch cn li 3 nhm (gim mt nhm so vi bng 1).Bng 2

Figure 12.7

Bng 3:

61

Figure 12.8

Kt qu ca hm rt gn gm tng cc s hng tng ng vi cc t hp khng gom thnh nhm: (4,6),(4,12), (8,12), (6,7) v (3,7;11,15)

Figure 12.9

Giai an 2:Bng 4

Figure 12.10

Cc ct 3, v 8 ch cha mt du *, cc t hp cng hng vi cc du * ny s c chn, l cct hp (3,7;11,15) v , (8,12), tng ng vi CD v A

__

C__

D .nh du X di cc ct tng ng vi cc s c trong cc t hp chn.n y ta thy cn 2 ct 4 v 6 cha c du X, trong lc chng ta cn n 3 t hp chn. D nhin

trong trng hp ny ta ch cn chn t hp (4,6) (___

A B__

D ) thay v chn (4,12) v (6,7) th du X lp y cc ct.

Tm li:


Figure 12.11

Th d v bi ton y :Th d 1:Cho hm logic F(A, B, C) tha tnh cht: F(A,B,C) = 1 nu c mt v ch mt bin bng 1a- Lp bng s tht cho hm F.b- Rt gn hm F.c- Din t hm F ch dng hm AND v NOTGiia. Da vo iu kin ca bi ton ta c bng s tht ca hm F:

Figure 12.12

b. Rt gn hm FBng Karnaugh

Figure 12.13

63

Figure 12.14

c. Din t hm F ch dng hm AND v NOTDng nhl De Morgan, ly o 2 ln hm F:

Figure 12.15

Th d 2:Cho hm logic F(A, B, C, D) tha tnh cht: F(A,B,C,D) = 1 khi c t nht 3 bin bng 1a- Rt gn hm F.b- Din t hm F ch dng hm OR v NOTGiia- Rt gn hm FTa c th a hm v bng Karnaugh m khng cn v bng s tht.Ta a s 1 vo tt c cc cha 3 tr 1 tr ln

Figure 12.16

V kt qu ca hm rt gn l:F(A,B,C,D) = ABC + ABD + ACD + BCDb- Din t hm F ch dng hm OR v NOTDng nh l De Morgan cho tng s hng trong tng


Vit li hm F:

Figure 12.17

Figure 12.18

Chng 13

BI TP-chng 2-kts1

13.1 BI TP

1. Din t mi mnh di y bng mt biu thc logic:a/ Tt c cc bin A,B,C,D u bng 1b/ Tt c cc bin A,B,C,D u bng 0c/ t nht 1 trong cc bin X,Y,Z,T bng 1d/ t nht 1 trong cc bin X,Y,Z,T bng 0e/ Cc bin A,B,C,D ln lt c gi tr 0,1,1,02. Tnh o ca cc hm sau:

Figure 13.1

3. Chng minh bng i s cc biu thc sau:a/

Figure 13.2


65

66 CHNG 13. BI TP-CHNG 2-KTS

b/

Figure 13.3

c/

Figure 13.4

d/

Figure 13.5

e/

Figure 13.6

4. Vit di dng tng chun cc hm xc nh bi:a/ f(A,B,C) = 1 nu s nh phn (ABC)2 l s chnb/ f(A,B,C) = 1 nu c t nht 2 bin s = 1c/ f(A,B,C) = 1 nu s nh phn (ABC)2 >5d/ f(A,B,C) = 1 nu s bin s 1 l s chne/ f(A,B,C) = 1 nu c 1 v ch 1 bin s =15. Vit di dng tch chun cc hm bi tp 46. Vit di dng s cc bi tp 47. Vit di dng s cc bi tp 58. Rt gn cc hm di y bng phng php i s (A = MSB)

67

Figure 13.7

9. Dng bng Karnaugh rt gn cc hm sau: (A = MSB)a/ f(A,B,C) = (1,3,4)b/ f(A,B,C) = (1,3,7)c/ f(A,B,C) = (0,3,4,6,7)d/ f(A,B,C) = (1,3,4) . Cc t hp bin 6,7 cho hm khng xc nh

Figure 13.8

f/ f(A,B,C,D) = (5,7,13,15)g/ f(A,B,C,D) = (0,4,8,12)h/ f(A,B,C,D) = (0,2,8,10)i/ f(A,B,C,D) = (0,2,5,6,9,11,13,14)j/ f(A,B,C,D) = (0,1,5,9,10,15)k/ f(A,B,C,D) = (0,5,9,10) vi cc t hp bin (2,3,8,15) cho hm khng xc nhl/ f(A,B,C,D,E) = (2,7,9,11,12,13,15,18,22,24,25,27,28,29,31)m/ f(A,B,C,D.E) = (0,2,8,10,13,15,16,18,24,25,26,29,31) vi cc t hp bin (7,9,14,30) cho hm khng

xc nhn/ f(A,B,C,D,E,F) = (2,3,6,7,8,9,12,13,14,17,24,25,28,29,30,40,41,44,45,46,56,57,59,60,61,63)o/ f(A,B,C,D,E,F) = (9,11,13,15,16,18,20,22,25,27,29,31,32,34,36,38,41,43,45,47,48,50,52,54)10. Lm li cc bi tp t 9f bng phng php Quine-Mc Cluskey.

68 CHNG 13. BI TP-CHNG 2-KTS

Chng 14

CNG LOGIC1

14.1 CNG LOGIC

Cng logic l tn gi chung ca cc mch in t c chc nng thc hin cc hm logic. Cng logic c thc ch to bng cc cng ngh khc nhau (Lng cc, MOS), c th c t hp bng cc linh kin rinhng thng c ch to bi cng ngh tch hp IC (Integrated circuit).

Chng ny gii thiu cc loi cng c bn, cc h IC s, cc tnh nng k thut v s giao tip giachng.

14.1.1 CC KHI NIM LIN QUAN

14.1.1.1 Tn hiu tng t v tn hiu s

Tn hiu tng t l tn hiu c bin bin thin lin tc theo thi gian. N thng do cc hin tng tnhin sinh ra. Th d, tn hiu c trng cho ting ni l tng hp ca cc tn hiu hnh sin trong di tns thp vi cc ha tn khc nhau.

Tn hiu s l tn hiu c dng xung, gin on v thi gian v bin ch c 2 mc r rt: mc cao vmc thp. Tn hiu s ch c pht sinh bi nhng mch in thch hp. c tn hiu s ngi ta phis ha tn hiu tng t bng cc mch bin i tng t sang s (ADC)

14.1.1.2 Mch tng t v mch s

Mch in t x l cc tn hiu tng t c gi l mch tng t v mch x l tn hiu s c gi lmch s.

Mt cch tng qut, mch s c nhiu u im so vi mch tng t:- D thit k v phn tch. Vn hnh ca cc cng logic da trn tnh cht dn in (bo ha) hoc

ngng dn ca transistor. Vic phn tch v thit k da trn chc nng v c tnh k thut ca cc IC vcc khi mch ch khng da trn tng linh kin ri

- C th hot ng theo chng trnh lp sn nn rt thun tin trong iu khin t ng, tnh ton, lutr d liu v lin kt vi my tnh.

- t b nh hng ca nhiu tc c kh nng dung np tn hiu nhiu vi bin ln hn rt nhiu sovi mch tng t.

- D ch to thnh mch tch hp v c kh nng tch hp vi mt cao.Da vo s cng trong mt chip, ngi ta phn loi IC s nh sau:- S cng < 10: SSI (Small Scale Integrated), mc tch hp nh.- 10 < S cng < 100: MSI (Medium Scale Integrated), mc tch hp trung bnh.


69

70 CHNG 14. CNG LOGIC

- 100 < S cng < 1000: LSI (Large Scale Integrated), mc tch hp ln.- 1000 < S cng < 10000: VLSI (Very Large Scale Integrated), mc tch hp rt ln- S cng > 10000: ULSI (Ultra Large Scale Integrated), mc tch hp siu ln.

14.1.1.3 Biu din cc trng thi Logic 1 v 0

Trong h thng mch logic, cc trng thi logic c biu din bi cc mc in th. Vi qui c logicdng, in th cao biu din logic 1, in th thp biu din logic 0. Ngc li ta c qui c logic m.Trong thc t, mc 1 v 0 tng ng vi mt khong in th xc nh v c mt khong chuyn tip giamc cao v thp, ta gi l khong khng xc nh. Khi in p ca tn hiu ri vo khong ny, mch skhng nhn ra l mc 0 hay 1. Khong ny ty thuc vo h IC s dng v c cho trong bng thng sk thut ca linh kin. (H 3.1) l gin in th ca cc mc logic ca mt s cng logic thuc h TTL.

Figure 14.1

(H 3.1)

Chng 15

CNG LOGIC C BN1

15.1 CNG LOGIC C BN

15.1.1 Cng NOT

- Cn gi l cng o (Inverter), dng thc hin hm o

Figure 15.1

- K hiu (H 3.2), mi tn ch chiu di chuyn ca tn hiu v vng trn l k hiu o. Trong nhngtrng hp khng th nhm ln v chiu ny, ngi ta c th b mi tn.

Figure 15.2

15.1.2 Cng AND

- Dng thc hin hm AND 2 hay nhiu bin.- Cng AND c s ng vo ty thuc s bin v mt ng ra. Ng ra ca cng l hm AND ca cc bin

ng vo.- K hiu cng AND 2 ng vo cho 2 bin (H 3.3a)


71

72 CHNG 15. CNG LOGIC C BN

Figure 15.3

- Nhn xt:- Ng ra cng AND ch mc cao khi tt c ng vo ln cao.- Khi c mt ng vo = 0, ng ra = 0 bt chp cc ng vo cn li.- Khi c mt ng vo =1, ng ra = AND ca cc ng vo cn li.Vy vi cng AND 2 ng vo ta c th dng 1 ng vo lm ng kim sot (H 3.3b), khi ng kim sot =

1, cng m cho php tn hiu logic ng vo cn li qua cng v khi ng kim sot = 0, cng ng , ngra lun bng 0, bt chp ng vo cn li.

Vi cng AND c nhiu ng vo hn, khi c mt ng vo c a ln mc cao th ng ra bng ANDca cc bin cc ng vo cn li.

Hnh (H 3.4) l gin thi gian ca cng AND hai ng vo. Trn gin , ng ra Y ch ln mc 1 khic A v B u mc 1.

Figure 15.4

(H 3.4)

15.1.3 Cng OR

- Dng thc hin hm OR 2 hay nhiu bin.- Cng OR c s ng vo ty thuc s bin v mt ng ra.- K hiu cng OR 2 ng vo

73

Figure 15.5

(H 3.5)- Bng s tht

Figure 15.6

- Nhn xt: - Ng ra cng OR ch mc thp khi c 2 ng vo xung thp.- Khi c mt ng vo =1, ng ra = 1 bt chp ng vo cn li.- Khi c mt ng vo =0, ng ra = OR cc ng vo cn li.Vy vi cng OR 2 ng vo ta c th dng 1 ng vo lm ng kim sot, khi ng kim sot = 0, cng

m, cho php tn hiu logic ng vo cn li qua cng v khi ng kim sot = 1, cng ng, ng ra lunbng 1.

Vi cng OR nhiu ng vo hn, khi c mt ng vo c a xung mc thp th ng ra bng OR cacc bin cc ng vo cn li.

15.1.4 Cng BUFFER

Cn gi l cng m. Tn hiu s qua cng BUFFER khng i trng thi logic. Cng BUFFER c dngvi cc mc ch sau:

- Sa dng tn hiu.- a in th ca tn hiu v ng chun ca cc mc logic.- Nng kh nng cp dng cho mch.- K hiu ca cng BUFFER.


Figure 15.7

(H 3.6)Tuy cng m khng lm thay i trng thi logic ca tn hiu vo cng nhng n gi vai tr rt quan

trng trong cc mch s.

15.1.5 Cng NAND

- L kt hp ca cng AND v cng NOT, thc hin hm

Figure 15.8

( y ch xt cng NAND 2 ng vo, c gi t suy ra trng hp nhiu ng vo).- K hiu ca cng NAND (Gm AND v NOT, cng NOT thu gn li mt vng trn)- Tng t nh cng AND, cng NAND ta c th dng 1 ng vo lm ng kim sot. Khi ng kim

sot = 1, cng m cho php tn hiu logic ng vo cn li qua cng v b o, khi ng kim sot = 0, cngng, ng ra lun bng 1.

- Khi ni tt c ng vo ca cng NAND li vi nhau, n hot ng nh mt cng o

Figure 15.9

(H 3.7)

15.1.6 Cng NOR

- L kt hp ca cng OR v cng NOT, thc hin hm

75

Figure 15.10

K hiu ca cng NOR (Gm cng OR v NOT, nhng cng NOT thu gn li mt vng trn)

Figure 15.11

(H 3.8)Cc bng s tht v cc gin thi gian ca cc cng BUFFER, NAND, NOR, sinh vin c th t thc

hin ly

15.1.7 Cng EX-OR

- Dng thc hin hm EX-OR.

Figure 15.12

- Cng EX-OR ch c 2 ng vo v 1 ng ra- K hiu (H 3.9a)- Mt tnh cht rt quan trng ca cng EX-OR:+ Tng ng vi mt cng o khi c mt ng vo ni ln mc cao, (H 3.9b)+ Tng ng vi mt cng m khi c mt ng vo ni xung mc thp, (H 3.9c)


Figure 15.13

(H 3.9)

15.1.8 Cng EX-NOR

- L kt hp ca cng EX-OR v cng NOT- Cng EX-NOR c 2 ng vo v mt ng ra- Hm logic ng vi cng EX-NOR l

Figure 15.14

- K hiu (H 3.10)- Cc tnh cht ca cng EX-NOR ging cng EX-OR nhng c ng ra o li.

Figure 15.15

(H 3.10)

15.1.9 Cng phc AOI (AND-OR-INVERTER)

ng dng cc kt qu ca i s BOOLE, ngi ta c th kt ni nhiu cng khc nhau trn mt chip IC thc hin mt hm logic phc tp no . Cng AOI l mt kt hp ca 3 loi cng AND (A), OR (O)v INVERTER (I). Th d thc hin hm logic

77

Figure 15.16

ta c cng phc sau:

Figure 15.17

(H 3.11)

15.1.10 Bin i qua li gia cc cng logic

Trong chng Hm Logic chng ta thy tt c cc hm logic c th c thay th bi 2 hm duy nhtl hm AND (hoc OR) kt hp vi hm NOT. Cc cng logic c chc nng thc hin hm logic, nh vychng ta ch cn dng 2 cng AND (hoc OR) v NOT thc hin tt c cc hm logic. Tuy nhin, vcng NOT cng c th to ra t cng NAND (hoc NOR). Nh vy, tt c cc hm logic c th c thchin bi mt cng duy nht, l cng NAND (hoc NOR). Hm ny cho php chng ta bin i qua ligia cc cng vi nhau.

Quan st nh l De Morgan chng ta rt ra qui tc bin i qua li gia cc cng AND, NOT v OR ,NOT nh sau:

Ch cn thm cc cng o ng vo v ng ra khi bin i t AND sang OR hoc ngc li. D nhinnu cc ng c o ri th o ny s mt i.

Th d 1: Ba mch di y tng ng nhau:(H 3.12b) c c bng cch i AND - OR thm cc o cc ng vo v ra. T (H 3.12b) i sang (H

3.12c) ta b 2 cng o ni t ng ra cng NOR n ng vo cng AND


Figure 15.18

(H 3.12)Th d 2: V mch tng ng ca cng EX-OR dng ton cng NANDDng nh l De-Morgan, biu thc hm EX-OR vit li:

Figure 15.19

V mch tng ng cho (H 3.13)

Figure 15.20

(H 3.13)

Chng 16

THNG S K THUT CA IC

S1

16.1 THNG S K THUT CA IC S

s dng IC s c hiu qu, ngoi s chn v bng s tht ca chng, ta nn bit qua mt s thutng ch cc thng s cho bit cc c tnh ca IC.

16.1.1 Cc i lng in c trng

- VCC: in th ngun (power supply): khong in th cho php cp cho IC hot ng tt. Th d viIC s h TTL, VCC=50,5 V , h CMOS VDD=3-15V (Ngi ta thng dng k hiu VDD v VSS chngun v mass ca IC h MOS)

- VIH(min): in th ng vo mc cao (High level input voltage): y l in th ng vo nh nht cnc xem l mc 1

- VIL(max): in th ng vo mc thp (Low level input voltage): in th ng vo ln nht cn cxem l mc 0.

- VOH(min): in th ng ra mc cao (High level output voltage): in th nh nht ca ng ra khi mc cao.

- VOL(max): in th ng ra mc thp (Low level output voltage): in th ln nht ca ng ra khi mc thp.

- IIH: Dng in ng vo mc cao (High level input current): Dng in ln nht vo ng vo IC khing vo ny mc cao.

- IIL: Dng in ng vo mc thp (Low level input current) : Dng in ra khi ng vo IC khi ng vony mc thp

- IOH: Dng in ng ra mc cao (High level output current): Dng in ln nht ng ra c th cp choti khi n mc cao.

- IOL: Dng in ng ra mc thp (Low level output current): Dng in ln nht ng ra c th nhnkhi mc thp.

- ICCH,ICCL: Dng in chy qua IC khi ng ra ln lt mc cao v thp.Ngoi ra cn mt s thng s khc c nu ra di y


79

80 CHNG 16. THNG S K THUT CA IC S

16.1.2 Cng sut tiu tn (Power requirement)

Mi IC khi hot ng s tiu th mt cng sut t ngun cung cp VCC (hay VDD). Cng sut tiu tn nyxc nh bi in th ngun v dng in qua IC. Do khi hot ng dng qua IC thng xuyn thay igia hai trng thi cao v thp nn cng sut tiu tn s c tnh t dng trung bnh qua IC v cng suttnh c l cng sut tiu tn trung bnh

Figure 16.1

Trong

Figure 16.2

i vi cc cng logic h TTL, cng sut tiu tn hng mW v vi h MOS th ch hng nW.

16.1.3 Fan-Out:

Mt cch tng qut, ng ra ca mt mch logic i hi phi cp dng cho mt s ng vo cc mch logickhc. Fan Out l s ng vo ln nht c th ni vi ng ra ca mt IC cng loi m vn bo m mch hotng bnh thng. Ni cch khc Fan Out ch kh nng chu ti ca mt cng logic

Ta c hai loi Fan-Out ng vi 2 trng thi logic ca ng ra:

Figure 16.3

Thng hai gi tr Fan-Out ny khc nhau, khi s dng, an ton, ta nn dng tr nh nht trong haitr ny.

Fan-Out c tnh theo n v Unit Load UL (ti n v).

16.1.4 Thi tr truyn (Propagation delays)

Tn hiu logic khi truyn qua mt cng lun lun c mt thi gian tr.

81

C hai loi thi tr truyn: Thi tr truyn t thp ln cao tPLH v thi tr truyn t cao xung thptPHL. Hai gi tr ny thng khc nhau. S thay i trng thi c xc nh tn hiu ra. Th d tn hiuqua mt cng o, thi tr truyn c xc nh nh (H 3.14)

Ty theo h IC, thi tr truyn thay i t vi ns n vi trm ns. Thi tr truyn cng ln th tc lm vic ca IC cng nh.

Figure 16.4

(H 3.14)

16.1.5 Tch s cng sut-vn tc (speed- power product)

nh gi cht lng IC, ngi ta dng i lng tch s cng sut-vn tc l tch s cng sut tiutn v thi tr truyn. Th d h IC c thi tr truyn l 10 ns v cng sut tiu tn trung bnh l 50 mWth tch s cng sut-vn tc l:

10 ns x 5 mW =10.10-9x5.10-3 = 50x10-12 watt-sec = 50 picojoules (pj)Trong qu trnh pht trin ca cng ngh ch to IC ngi ta lun mun t c cc IC c cng sut

tiu tn v thi tr truyn cng nh cng tt. Nh vy mt IC c cht lng cng tt khi tch s cngsut-vn tc cng nh. Tuy nhin trn thc t hai gi tr ny thay i theo chiu ngc vi nhau, nn takh m t c cc gi tr theo mun, d sao trong qu trnh pht trin ca cng ngh ch to linh kinin t tr s ny lun c ci thin .

16.1.6 Tnh min nhiu (noise immunity)

Cc tn hiu nhiu nh tia la in, cm ng t c th lm thay i trng thi logic ca tn hiu do nhhng n kt qu hot ng ca mch.

Tnh min nhiu ca mt mch logic ty thuc kh nng dung np hiu th nhiu ca mch v cxc nh bi l nhiu. L nhiu c c do s chnh lch ca cc in th gii hn (cn c gi l ngnglogic) ca mc cao v thp gia ng ra v ng vo ca cc cng (H 3.15).


Figure 16.5

(H 3.15)Tn hiu khi vo mch logic c xem l mc 1 khi c tr >VIH(min) v l mc 0 khi VNH

u lm cho in th ng vo ri vo vng bt nh v mch khng nhn ra c tn hiu thuc mc logicno. Tng t cho trng hp ng ra mc thp tn hiu nhiu c tr dng bin >VNL s a mchvo trng thi bt nh.

16.1.7 Logic cp dng v logic nhn dng

Mt mch logic thng gm nhiu tng kt ni vi nhau. Tng cp tn hiu gi l tng thc v tng nhntn hiu gi l tng ti. S trao i dng in gia hai tng thc v ti th hin bi logic cp dng v logicnhn dng.

(H 3.16a) cho thy hot ng gi l cp dng: Khi ng ra mch logic 1 mc cao, n cp dng IIH chong vo ca mch logic 2, vai tr nh mt ti ni mass. Ng ra cng 1 nh l mt ngun dng cp cho ngvo cng 2

(H 3.16b) cho thy hot ng gi l nhn dng: Khi ng ra mch logic 1 mc thp, n nhn dng IILt ng vo ca mch logic 2 xem nh ni vi ngun VCC.

83

Figure 16.6

(H 3.16)Thng dng nhn ca tng thc khi mc thp c tr kh ln so vi dng cp ca n khi mc cao,

nn ngi ta hay dng trng thi ny khi cn gnh nhng ti tng i nh, v d khi ch cn thc cho mtled, ngi ta c th dng mch (H 3.17a) m khng th dng mch (H 3.17b).

Figure 16.7

(a) (H 3.17) (b)

16.1.8 Tnh Schmitt Trigger

Trong phn gii thiu l nhiu, ta thy cn mt khong in th nm gia cc ngng logic, y chnh lkhong in th ng vi transistor lm vic trong vng tc ng. Khong cch ny xc nh l nhiu v ctc dng lm gim rng sn xung (tc lm cho ng dc ln v dc xung ca tn hiu ra dc hn)khi qua mch. L nhiu cng ln khi vng chuyn tip ca ng vo cng nh, tn hiu ra thay i trng thitrong mt khong thi gian cng nh nn sn xung cng dc. Tuy nhin vn cn mt khong sn xungnm trong vng chuyn tip nn tn hiu ra khng vung hon ton. (H 3.18a) v (H 3.18b) minh ha iu


Figure 16.8

(H 3.18) ci thin hn na dng tn hiu ng ra, bo m tnh min nhiu cao, ngi ta ch to cc cng c

tnh tr in th (H 3.19a), c gi l cng Schmitt Trigger(H 3.19b) m t mi quan h gia Vout v Vin ca mt cng o Schmitt Trigger.

Figure 16.9

(H 3.19)(H 3.20a&b) l k hiu cc cng Schmitt Trigger.

Figure 16.10

85

(a) (b)(H 3.20)

Chng 17

H TTL1

17.1 H TTL

Trong qu trnh pht trin ca cng ngh ch to mch s ta c cc h: RTL (Resistor-transistor logic),DCTL (Direct couple-transistor logic), RCTL (Resistor-Capacitor-transistor logic), DTL (Diod-transistorlogic), ECL (Emitter- couple logic) v.v.... n by gi tn ti hai h c nhiu tnh nng k thut cao nhthi tr truyn nh, tiu hao cng sut t, l h TTL (transistor-transistor logic) dng cng ngh ch toBJT v h MOS (Cng ngh ch to MOS)

Di y, ln lt kho st cc cng logic ca hai h TTL v MOS

17.1.1 Cng c bn h TTL

Ly cng NAND 3 ng vo lm th d thy cu to v vn hnh ca mt cng c bn

Figure 17.1

(H 3.21)Khi mt trong cc ng vo A, B, C xung mc khng T1 dn a n T2 ngng, T3 ngng, ng ra Y

ln cao; khi c 3 ng vo ln cao, T1 ngng, T2 dn, T3 dn, ng ra Y xung thp. chnh l kt qu cacng NAND.


87

88 CHNG 17. H TTL

T CL trong mch chnh l t k sinh to bi s kt hp gia ng ra ca mch (tng thc) vi ng voca tng ti, khi mch hot ng t s np in qua R4 (lc T3 ngng) v phng qua T3 khi transistor nydn do thi tr truyn ca mch quyt nh bi R4 v CL, khi R4 nh mch hot ng nhanh nhng cngsut tiu th lc ln, mun gim cng sut phi tng R4 nhng nh vy thi tr truyn s ln hn (mchgiao hon chm hn). gii quyt khuyt im ny ng thi tha mn mt s yu cu khc , ngi ta ch to cc cng logic vi cc kiu ng ra khc nhau.

17.1.2 Cc kiu ng ra

Ng ra totempole

Figure 17.2

(H 3.22)R4 trong mch c bn c thay th bi cm T4, RC v Diod D, trong RC c tr rt nh, khng ng

k. T2 by gi gi vai tr mch o pha: khi T2 dn th T3 dn v T4 ngng, Y xung thp, khi T2 ngngth T3 ngng v T4 dn, ng ra Y ln cao. T CL np in qua T4 khi T4 dn v phng qua T3 (dn), thihng mch rt nh v kt qu l thi tr truyn nh. Ngoi ra do T3 & T4 lun phin ngng tng ng vi2 trng thi ca ng ra nn cng sut tiu th gim ng k. Diod D c tc dng nng in th cc B caT4 ln bo m khi T3 dn th T4 ngng.

Ng ra cc thu h c mt s li im sau:Mch ny c khuyt im l khng th ni chung nhiu ng ra ca cc cng khc nhau v c th gy h

hng khi cc trng thi logic ca cc cng ny khc nhau.Ng ra cc thu h

89

Figure 17.3

(H 3.23)- Cho php kt ni cc ng ra ca nhiu cng khc nhau, nhng khi s dng phi mc mt in tr t

ng ra ln ngun Vcc, gi l in tr ko ln, tr s ca in tr ny c th c chn ln hay nh tytheo yu cu c li v mt cng sut hay tc lm vic.

im ni chung ca cc ng ra c tc dng nh mt cng AND nn ta gi l im AND (H 3.24)- Ngi ta cng ch to cc IC ng ra c cc thu h cho php in tr ko ln mc vo ngun in

th cao, dng cho cc ti c bit hoc dng to s giao tip gia h TTL vi CMOS dng ngun cao.Th d IC 7406 l loi cng o c ng ra cc thu h c th mc ln ngun 24 V (H3.25)Mch (H 3.26) l mt cng o c ng ra 3 trng thi, trong T4 & T5 c mc Darlington cp

dng ra ln cho ti. Diod D ni vo ng vo C iu khin. Hot ng ca mch gii thch nh sau:- Khi C=1, Diod D ngng dn, mch hot ng nh mt cng o- Khi C=0, Diod D dn, cc thu T2 b ghim p mc thp nn T3, T4 & T5 u ngng, ng ra mch

trng thi tng tr cao.K hiu ca cng o ng ra 3 trng thi, c ng iu khin C tc ng mc cao v bng s tht cho

(H 3.27)Cng c cc cng o v cng m 3 trng thi vi ng iu khin C tc ng mc thp m SV c th

t v k hiu v bng s tht.(H 3.28) l mt ng dng ca cng m c ng ra 3 trng thi: Mch chn d liu

90 CHNG 17. H TTL

Figure 17.4

(H 3.28)

Chng 18

HO MOS1

18.1 HO MOS

Gm cc IC s dng cng ngh ch to ca transistor MOSFET loi tng, knh N v knh P . Vi transistorknh N ta c NMOS, transistor knh P ta c PMOS v nu dng c hai loi transistor knh P & N ta cCMOS. Tnh nng k thut ca loi NMOS v PMOS c th ni l ging nhau, tr ngun cp in c chiungc vi nhau do ta ch xt loi NMOS v CMOS.

Cc transistor MOS dng trong IC s cng ch hot ng mt trong 2 trng thi: dn hoc ngng.- Khi dn, ty theo nng pha ca cht bn dn m transistor c ni tr rt nh (t vi chc n

hng trm K ) tng ng vi mt kha ng.- Khi ngng, transistor c ni tr rt ln (hng 1010), tng ng vi mt kha h.

18.1.1 Cng c bn NMOS

Figure 18.1


91

92 CHNG 18. HO MOS

(a) (b) (c)(H 3.29)(H 3.29a), (H 3.29b) v (H3.29c) l cc cng NOT, NAND v NOR dng NMOSBng 3.2 cho thy quan h gia cc in th ca cc ng vo , ra cng NOT

Figure 18.2

Bng 3.2Ngoi ra vn hnh ca cng NAND v NOR c gii thch nh sau:Cng NAND:- Khi 2 ng vo ni ln mc cao, T2 v T3 dn, ng ra xung thp.- Khi c 1 ng vo ni xung mc thp, mt trong 2 transistor T2 hoc T3 ngng, ng ra ln cao. chnh l kt qu ca cng NAND 2 ng vo.Cng NOR:- Khi 2 ng vo ni xung mc thp, T2 v T3 ngng, ng ra ln cao.- Khi c 1 ng vo ni ln mc cao, mt trong 2 transistor T2 hoc T3 dn, ng ra xung thp. chnh l kt qu ca cng NOR 2 ng vo.

18.1.2 Cng c bn CMOS

H CMOS s dng hai loi transistor knh N v P vi mc ch ci thin tch s cng sut vn tc, mcd kh nng tch hp thp hn loi N v P. (H 3.30a), (H 3.30b) v (H 3.30c) l cc cng NOT, NAND vNOR h CMOS

93

Figure 18.3

(a) (b) (c)(H 3.30)Bng 3.3 cho thy quan h in th ca cc ng vo , ra cng NOT

Figure 18.4

Bng 3.3Ngoi ra vn hnh ca cng NAND v NOR c gii thch nh sau:Cng NAND:- Khi 2 ng vo ni ln mc cao, T1 v T2 ngng, T3 v T4 dn, ng ra xung thp.- Khi c 1 ng vo ni xung mc thp, mt trong 2 transistor T3 hoc T4 ngng, mt trong 2 transistor

T1 hoc T2 dn, ng ra ln cao. chnh l kt qu ca cng NAND 2 ng vo.Cng NOR:- Khi 2 ng vo ni xung mc thp, T1v T2 dn, T3 v T4 ngng, ng ra ln cao.- Khi c 1 ng vo ni ln mc cao, mt trong 2 transistor T3 hoc T4 dn, mt trong 2 transistor T1

hoc T2 ngng, ng ra xung thp. chnh l kt qu ca cng NOR 2 ng vo.

18.1.3 c tnh ca h MOS

Mt s tnh cht chung ca cc cng logic h MOS (NMOS, PMOS v CMOS) c th k ra nh sau:

94 CHNG 18. HO MOS

- Ngun cp in : VDD t 3V n 15V- Mc logic: VOL (max) = 0V VOH (min) = VDDVIL (max) = 30% VDD VIH (min) = 70%VDD- L nhiu : VNH = 30%VDD VNL = 30%VDDVi ngun 5V, l nhiu khang 1,5V, rt ln so vi h TTL.- Thi tr truyn tng i ln, khang vi chc ns, do in dung k sinh ng vo v tng tr ra ca

transistor kh ln.- Cng sut tiu tn tng i nh, hng nW, do dng qua transistor MOS rt nh.- S Fan Out: 50 ULDo tng tr vo ca transistor MOS rt ln nn dng ti cho cc cng h MOS rt nh, do s Fan

Out ca h MOS rt ln, tuy nhin khi mc nhiu tng ti vo mt tng thc th in dung k sinh tngln (gm nhiu t mc song song) nh hng n thi gian giao hon ca mch nn khi dng tn s caongi ta gii hn s Fan Out l 50, ngha l mt cng MOS c th cp dng cho 50 cng ti cng lot.

- Nh ni trn, CMOS c ci thin thi tr truyn so vi loi NMOS v PMOS, tuy nhin mt tch hp ca CMOS th nh hn hai loi ny. D sao so vi h TTL th mt tch hp ca h MOS nichung ln hn rt nhiu, do h MOS rt thch hp ch to di dng LSI v VLSI.

18.1.4 Cc lot CMOS

CMOS c hai k hiu: 4XXX do hng RCA ch to v 14XXX ca hng MOTOROLA, c hai lot 4XXXA(14XXXA) v 4XXXB (14XXXB), lot B ra i sau c ci thin dng ra.

Ngoi ra cn c cc lot :- 74C : CMOS c cng s chn v chc nng vi IC TTL nu c cng s. Th d IC 74C74 l IC gm

2 FF D tc ng bi cnh xung ng h ging nh IC 7474 ca TTL. Hu ht (nhng khng tt c) ccthng s ca lot 74C ging vi 74 TTL nn ta c th thay th 2 loi ny cho nhau c.

- 74HC (High speed CMOS), 74HCT: y l lot ci tin ca 74C, tc giao hon c th so snh vi74LS, ring 74HCT th hon ton tng thch vi TTL k c cc mc logic. y l lot IC CMOS cdng rng ri.

- 74AC v 74ACT (Advance CMOS) ci tin ca 74 HC v HCT v mt nhiu bng cch sp xp li tht cc chn, do n khng tng thch vi TTL v s chn

Chng 19

GIAO TIP GIA CC H IC S1

19.1 GIAO TIP GIA CC H IC S

Giao tip l thc hin vic kt ni ng ra ca mt mch hay h thng vi ng vo ca mch hay h thngkhc. Do tnh cht v in khc nhau gia hai h TTL v CMOS nn vic giao tip gia chng trong nhiutrng hp khng th ni trc tip c m phi nh mt mch trung gian ni gia tng thc v tng tisao cho in th tn hiu ra tng thc ph hp vi tn hiu vo ca tng ti v dng in tng thc phi cp cho tng ti.

Figure 19.1

Bng 3.4C th ni iu kin thc trc tip- Khi dng in ra ca tng thc ln hn hoc bng dng in vo ca tng ti c hai trng thi thp

v cao.- Khi hiu th ng ra ca tng thc hai trng thi thp v cao ph hp vi in th vo ca tng ti.Nh vy, trc khi xt cc trng hp c th ta xem qua bng k cc thng s ca hai h IC


95

96 CHNG 19. GIAO TIP GIA CC H IC S

19.1.1 TTL thc CMOS

- TTL thc CMOS dng in th thp (VDD = 5V):T bng 3.4 dng in vo ca CMOS c tr rt nh so vi dng ra ca cc lot TTL, vy v dng in

khng c vn Tuy nhin khi so snh hiu th ra ca TTL vi hiu th vo ca CMOS ta thy VOH(max) ca tt c

cc lot TTL u kh thp so vi VIH(min) ca TTL, nh vy phi c bin php nng hiu th ra ca TTLln. iu ny thc hin c bng mt in tr ko ln mc ng ra ca IC TTL (H 3.33)

- TTL thc 74 HCT:Nh ni trc y, ring lot 74HCT l lot CMOS c thit k tng thch vi TTL nn c th

thc hin kt ni m khng cn in tr ko ln.- TTL thc CMOS dng ngun cao (VDD = +10V)Ngay c khi dng in tr ko ln, in th ng ra mc cao ca TTL vn khng cp cho ng vo

CMOS, ngi ta phi dng mt cng m c ng ra h c th dng ngun cao (Th d IC 7407) thchin s giao tip (H 3.34)

Figure 19.2

(H 3.33) (H 3.34)

19.1.2 CMOS thc TTL

- CMOS thc TTL trng thi cao:Bng 3.4 cho thy in th ra v dng in ra mc cao ca CMOS cp cho TTL . Vy khng c

vn trng thi cao- CMOS thc TTL trng thi thp:Dng in vo trng thi thp ca TTL thay i trong khong t 100 A n 2 mA. Hai lot 74HC

v 74HCT c th nhn dng 4 mA . Vy hai lot ny c th giao tip vi mt IC TTL m khng c vn .Tuy nhin, vi lot 4000B, IOL rt nh khng giao tip vi ngay c mt IC TTL, ngi ta phi dngmt cng m nng dng ti ca lot 4000B trc khi thc vi IC 74LS (H 3.35)

- CMOS dng ngun cao thc TTL:C mt s IC lot 74LS c ch to c bit c th nhn in th ng vo cao khong 15V c th c

thc trc tip bi CMOS dng ngun cao, tuy nhin a s IC TTL khng c tnh cht ny, vy c thgiao tip vi CMOS dng ngun cao, ngi ta phi dng cng m h in th ra xung cho ph hp viIC TTL (H 3.36)

97

Figure 19.3

(H 3.35) (H 3.36)Vi th d dng cng thit k mch1. Dng cng NAND 2 ng vo thit k mch to hm Y = f(A,B,C) =1 khi tha cc iu kin sau:a. A=0, B=1 v C=1b. A=1, B=1 bt chp CGiiD vo iu kin ca bi ton ta c bng s tht ca hm Y

Figure 19.4

Rt gn hm:


Figure 19.5

Y =AB+BC (H 3.37) dng tan cng NAND to hm, ta dng nh l De Morgan, bin i hm Y:

Figure 19.6

V mch c dng (H 3.37)2. Cho mch

Figure 19.7

(H P3.38)a./ Vit biu thc hm Y theo cc bin A,B,C.b./ Rt gn hm logic nyc./ Thay th mch trn bng mt mch ch gm cng NAND 2 ng voGiia./ Ta c

99

Figure 19.8

b./ Rt gn

Figure 19.9

c./ V mch thay th dng cng NAND 2 ng voTrc nht ta v mch tng ng hm rt gn, sau dng bin i cng

Figure 19.10

(H P3.39)

Chng 20

BI TP CHNG 3-KTS1

20.1 BI TP

1. Thit k mch thc hin cc hm sau y dng ton cng NAND 2 ng vo:a./ f(A,B,C) = 1 nu (ABC)2 l s chn.b./ f(A,B,C) = 1 nu c t nht 2 bin = 1.c./ f(A,B,C) = 1 nu s nh phn (ABC)2 > 5.d./ f(A,B,C) = 1 nu s bin c gi tr 1 l s chn.e./ f(A,B,C) = 1 nu c mt v ch mt bin = 1.2. Thit k mch gm 2 ng vo D, E v 2 ng ra P, C tha cc iu kin sau y:- Nu E = 1 D = 0 P = 1, C = 0- Nu E = 1 D = 1 P = 0, C = 1- Nu E = 0 D bt k P = 1, C = 13. Hm logic F(A, B, C) tha tnh cht sau y :F(A,B,C) = 1 nu c mt v ch mt bin bng 1a- Lp bng s tht cho hm F.b- V mch logic to hm F.4. Thit K mch to hm

Figure 20.1

bng cc cng NAND 2 ng vo5. Hm F(A,B,C) xc inh bi bng s tht


101

102 CHNG 20. BI TP CHNG 3-KTS

Figure 20.2

6. Rt gn hm logic :f(A,B,C,D) = (0,1, 2, 4, 5, 8), A = MSB. Hm khng xc nh vi cc t hp bin (3, 7,10).Dng s cng NOR t nht thc hin mch to hm trn.7. Hm f(A,B,C) =1 khi s bin = 1 l s chn- Vit biu thc logic ca hm f(A,B,C) theo t hp bin A,B,C.- Dng cc cng EX-OR thc hin mch to hm trn.8. Mt mch t hp nhn vo mt s nh phn A=A3A2A1A0 (A0 l LSB) to ra ng ra Y mc cao

khi v ch khi 0010

Chng 21

MCH T HP1

21.1 MCH T HP

Cc mch s c chia ra lm hai loi: Mch t hp v Mch tun t.- Mch t hp: Trng thi ng ra ch ph thuc vo t hp cc ng vo khi t hp ny n nh.

Ng ra Q ca mch t hp l hm logic ca cc bin ng vo A, B, C . . ..Q = f(A,B,C . . .)- Mch tun t : Trng thi ng ra khng nhng ph thuc vo t hp cc ng vo m cn ph thuc

trng thi ng ra trc . Ta ni mch tun t c tnh nh. Ng ra Q+ ca mch tun t l hm logicca cc bin ng vo A, B, C . . . . v ng ra Q trc .

Q+ = f(Q,A,B,C . . .)Chng ny nghin cu mt s mch t hp thng dng thng qua vic thit k mt s mch n gin

v kho st mt s IC trn thc t.

21.1.1 MCH M HA

M ha l gn cc k hiu cho cc i tng trong mt tp hp thun tin cho vic thc hin mt yucu c th no . Th d m BCD gn s nh phn 4 bit cho tng s m ca s thp phn (t 0 n 9) thun tin cho my c mt s c nhiu s m; m Gray dng tin li trong vic ti gin cc hm logic . . ..Mch chuyn t m ny sang m khc gi l mch chuyn m, cng c xp vo loi mch m ha. Thd mch chuyn s nh phn 4 bit sang s Gray l mt mch chuyn m.

21.1.1.1 Mch m ha 2n ng sang n ng

Mt s nh phn n bit cho 2n t hp s khc nhau. Vy ta c th dng s n bit m cho 2n ng vo khcnhau, khi c mt ng vo c chn bng cch a n ln mc tc ng, ng ra s ch bo s nh phntng ng. l mch m ha 2n ng sang n ng.

(H 4.1) l m hnh mt mch m ha 2n ng sang n ng.- (H 4.1a) l mch c ng vo v ra tc ng cao : Khi cc ng vo u mc thp, mch cha hot

ng, cc ng ra u mc thp. Khi c mt ng vo c tc ng bng cch n kha K tng ng a ng vo ln mc cao, cc ng ra s cho s nh phn tng ng.

- (H 4.1b) l mch c ng vo v ra tc ng thp. Hot ng tng t nh mch trn nhng c mctc ng ngc li. (trong m hnh (H 4.1b) k hiu du o ng ra ch mc tc ng thp, cn ngvo khng c du o v l mch tht)


103

104 CHNG 21. MCH T HP

Trong trng hp ng ra c mc tc ng thp, mun c ng s nh phn ng ra, ta phi o ccbit c.

Figure 21.1

(a) (b)(H 4.1)D nhin, ngi ta cng c th thit k theo kiu ng vo tc ng thp v ng ra tc ng cao hay ngc

li. Trn thc t, ta c th c bt c loi ng vo hay ra tc ng theo bt c kiu no (mc cao hay thp).Ngoi ra, trnh trng hp mch cho ra mt m sai khi ngi s dng v tnh (hay c ) tc ng

ng thi vo hai hay nhiu ng vo, ngi ta thit k cc mch m ha u tin: l mch ch cho ra mtm duy nht c tnh u tin khi c nhiu ng vo cng c tc ng.

21.1.1.1.1 M ha u tin 4 ng sang 2 ng

Thit k mch m ha 4 ng sang 2 ng, u tin cho m c tr cao, ng vo v ra tc ng caoBng s tht v s mch (H 4.2)

Figure 21.2

105

Bng 4.1Nhn thy bin 0 trong bng s tht khng nh hng n kt qu nn ta ch v bng Karnaugh cho

3 bin 1, 2 v 3. Lu l do trong bng s tht c cc trng hp bt chp ca bin nn ng vi mt trring ca hm ta c th c n 2 hoc 4 s 1 trong bng Karnaugh. Th d vi tr 1 ca c 2 hm A1 v A0 dng cui cng a n 4 s 1 trong cc 001, 011, 101 v 111 ca 3 bin 123.

T bng Karnaugh, ta c kt qu v mch tng ng. Trong mch khng c ng vo 0, iu ny chiu l mch s ch bo s 0 khi khng tc ng vo ng vo no.

Figure 21.3

(H 4.2)

21.1.1.1.2 M ha 8 ng sang 3 ng

Chng ta s kho st mt IC m ha 8 ng sang 3 ng.Trn thc t khi ch to mt IC, ngoi cc ng vo/ra thc hin chc nng chnh ca n, ngi ta

thng d tr thm cc ng vo v ra cho mt s chc nng khc nh cho php, ni mch m rnghot ng ca IC.

IC 74148 l IC m ha u tin 8 ng sang 3 ng, vo/ ra tc ng thp, c cc ng ni mch m rng m ha vi s ng vo nhiu hn.

Di y l bng s tht ca IC 74148, trong Ei ng vo ni mch v cho php, Eo l ng ra ni mchv Gs dng m rng cho s nh phn ra.

Da vo bng s tht, ta thy IC lm vic theo 10 trng thi:- Cc trng thi t 0 n 7: IC m ha cho ra s 3 bit- Cc trng thi 8 v 9: dng cho vic m rng, s gii thch r hn khi ni 2 IC m rng m ha cho

s 4 bit


Figure 21.4

Bng 4.2(H 4.3) l cch ni 2 IC thc hin m ha 16 ng sang 4 ng

107

Figure 21.5

(H 4.3)- IC2 c Ei = 0 nn hot ng theo cc trng thi t 0 n 8, ngha l m ha t 0 n 7 cho cc ng ra

A2A1A0.- IC1 c Ei ni vi Eo ca IC2 nn IC1 ch hot ng khi tt c ng vo d liu ca IC2 ln mc 1 (IC2

hot ng trng thi 8)* m ha cc s t 0 n 7, cho cc ng vo 8 n 15 (tc cc ng vo d liu ca IC2) ln mc 1,

IC2 hot ng trng thi 8.Lc Ei1 = Eo2 = 0: kt qu l IC1 s hot ng trng thi t 0 n 7, cho php to m cc s t 0

n 7 (t 111 n 000) v IC2 hot ng trng thi 8 nn cc ng ra (A2A1A0)2= 111, y l iu kinm cc cng AND cho m s ra l B2B1B0 = A2A1A0 ca IC1, trong lc B3 = Gs2 = 1, ta c ktqu t 1111 n 1000, tc t 0 n 7 (tc ng thp).

Th d m s 4 , a ng vo 4 xung mc 0, cc ng vo t 5 n 15 ln mc 1, bt chp cc ngvo t 0 n 3, m s ra l B3B2B1B0=Gs2B2B1B0=1011, tc s 4

* m ha cc s t 8 n 15, cho IC2 hot ng trng thi t 0 n 7 (a ng vo ng vi s munm xung thp, cc ng vo cao hn ln mc 1 v cc ng vo thp hn xung mc 0), bt chp cc ngvo d liu ca IC1 (cho IC1 hot ng trng thi 9), nn cc ng ra (A2A1A0)1=111, y l iu kinm cc cng AND cho m s ra l B2B1B0= A2A1A0 ca IC2, , trong lc B3 = Gs2 = 0, ta c ktqu t 0111 n 0000, tc t 8 n 15.

Th d m s 14, a ng vo 14 xung mc 0, a ng vo 15 ln mc 1, bt chp cc ng vo t 0n 13, m s ra l B3B2B1B0 = Gs2B2B1B0 = 0001, tc s 14

Mun c ng ra ch s nh phn ng vi ng vo c tc ng m khng phi o cc bit ta c ththay cc cng AND bng cng NAND

21.1.1.2 Mch to m BCD cho s thp phn

Mch gm 10 ng vo tng trng cho 10 s thp phn v 4 ng ra l 4 bit ca s BCD. Khi mt ng vo(tng trng cho mt s thp phn) c tc ng bng cch a ln mc cao cc ng ra s cho s BCDtng ng

Bng s tht ca mch:


Figure 21.6

Bng 4.3Khng cn bng Karnaugh ta c th vit ngay cc hm xc nh cc ng ra:A 0 = 1 + 3 + 5 + 7 + 9 A 1 = 2 + 3 + 6 + 7A 2 = 4 + 5 + 6 + 7 A 3 = 8 + 9Mch cho (H 4.4)

Figure 21.7

(H 4.4) to m BCD u tin cho s ln, ta vit li bng s tht v dng phng php i s n gin cc

hm xc nh cc ng ra A3 , A2 , A1 , A0

109

Figure 21.8

Bng 4.4

Figure 21.9

Mch cho (H 4.5)


Figure 21.10

(H 4.5)

21.1.1.3 Mch chuyn m

Mch chuyn t mt m ny sang mt m khc cng thuc loi m ha.Mch chuyn m nh phn sang GrayTh thit k mch chuyn t m nh phn sang m Gray ca s 4 bit.Trc tin vit bng s tht ca s nh phn v s Gray tng ng. Cc s nh phn l cc bin v cc

s Gray s l hm ca cc bin .

111

Figure 21.11

Bng 4.5Dng bng Karnaugh xc nh X, Y, Z, T theo A, B, C, DQuan st bng s tht ta thy ngay: X = A,Vy ch cn lp 3 bng Karnaugh cho cc bin Y, Z, T (H 4.6 a,b,c) v kt qu cho (H 4.6 d)


Figure 21.12

(H 4.6 ) (d)

Chng 22

MCH GII M1

22.1 MCH GII M

22.1.1 Gii m n ng sang 2n ng

22.1.1.1 Gii m 2 ng sang 4 ng:

Thit k mch Gii m 2 ng sang 4 ng c ng vo cho php (cng c dng ni mch) n gin, ta xt mch gii m 2 ng sang 4 ng c cc ng vo v ra u tc ng cao .Bng s tht, cc hm ng ra v s mch:


113

114 CHNG 22. MCH GII M

Figure 22.1

(H 4.7)

22.1.1.2 Gii m 3 ng sang 8 ng

Dng 2 mch gii m 2 ng sang 4 ng thc hin mch gii m 3 ng sang 8 ng (H 4.8)

115

Figure 22.2

Quan st bng s tht ta thy: Trong cc t hp s 3 bit c 2 nhm trong cc bit thp A1A0 honton ging nhau, mt nhm c bit A2 = 0 v nhm kia c A2 = 1. Nh vy ta c th dng ng vo G chobit A2 v mc mch nh sau.

Figure 22.3

(H 4.8)Khi A2=G=0, IC1 gii m cho 1 trong 4 ng ra thp v khi A2=G=1, IC2 gii m cho 1 trong 4 ng ra

caoTrn th trng hin c cc loi IC gii m nh:- 74139 l IC cha 2 mch gii m 2 ng sang 4 ng, c ng vo tc ng cao, cc ng ra tc ng

thp, ng vo cho php tc ng thp.- 74138 l IC gii m 3 ng sang 8 ng c ng vo tc ng cao, cc ng ra tc ng thp, hai ng

vo cho php G2A v G2B tc ng thp, G1 tc ng cao.


- 74154 l IC gii m 4 ng sang 16 ng c ng vo tc ng cao, cc ng ra tc ng thp, 2 ngvo cho php E1 v E2 tc ng thp

Di y l bng s tht ca IC 74138 v cch ni 2 IC m rng mch gii m ln 4 ng sang 16ng (H 4.9)

Figure 22.4

Ghi ch G2 =G2A+G2B , H = 1, L =0, x: bt chp

Figure 22.5

(H 4.9)

117

Mt ng dng quan trng ca mch gii m l dng gii m a ch cho b nh bn dn.Ngoi ra, mch gii m kt hp vi mt cng OR c th to c hm logic.Th d, thit k mch to hm

Figure 22.6

Vi hm 3 bin, ta dng mch gii m 3 ng sang 8 ng. 8 ng ra mch gii m tng ng vi 8 thp bin ca 3 bin, cc ng ra tng ng vi cc t hp bin c trong hm s ln mc 1. Vi mt hm vit di dng tng chun, ta ch cn dng mt cng OR c s ng vo bng vi s t hp bin trong hmni vo cc ng ra tng ng ca mch gii m cng cc t hp bin c trong hm li ta s c hmcn to.

Nh vy, mch to hm trn c dng (H 4.10)

Figure 22.7

(H 4.10)D nhin, vi nhng hm cha phi dng tng chun, chng ta phi chun ha. V nu bi ton c yu

cu ta phi thc hin vic i cng, bng cch dng nh l De Morgan.

22.1.2 Gii m BCD sang 7 an

22.1.2.1 n 7 an

y l lai n dng hin th cc s t 0 n 9, n gm 7 an a, b, c, d, e, f, g, bn di mi an l mtled (n nh) hoc mt nhm led mc song song (n ln). Qui c cc an cho bi (H 4.11).


Figure 22.8

(H 4.11)Khi mt t hp cc an chy sng s to c mt con s thp phn t 0 - 9.(H 4.12) cho thy cc on no chy th hin cc s t 0 n 9

Figure 22.9

(H 4.12)n 7 on cng hin th c mt s ch ci v mt s k hiu c bit.C hai loi n 7 on:- Loi catod chung (H 4.13a), dng cho mch gii m c ng ra tc ng cao.- Loi anod chung (H 4.13b), dng cho mch gii m c ng ra tc ng thp.

Figure 22.10

119

(a) (H 4.13) (b)

22.1.2.2 Mch gii m BCD sang 7 on :

Mch c 4 ng vo cho s BCD v 7 ng

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