Adad Thuan Thuan.do Luong.docthanh

7/30/2019 Adad Thuan Thuan.do Luong.docthanh

1/42

Bi tp ln o Lng TTCN Khoa in t

TRNG I HC K THUT CNG NGHIP

KHOA IN T

B mn: O LNG IU KHIN

BI TP LN

MN HC

O LNG V THNG TIN CNG NGHIP

Sinh vin :. .. . NGUYN TH THUN

Lp . . . .

K44KMT. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Gio vin hng dn . PHM VN THIM. . . . . . . . . . . . . . . . . . . . . . . .

. . .

GVHD: Phm Vn Thim - 1 - SVTH: Nguyn ThThun


2/42


Thi Nguyn 2011

TRNG HKTCN CNG HO X HI CH NGHA VIT NAM

KHOA IN T c lp - T do - Hnh phc

BI TP LN

MN HC: O LNG V THNG TIN CNG NGHIP

B MN : O LNG IU KHIN.............................

Sinh vin: ..... NGUYN TH THUN..................... Lp: K44KMT.................

Ngnh: K THUT MY TNH

..........................................................................................................................................

.................................

Gio vin hng dn: PHM VN THIM............

Ngy giao : 30/10/2011........................ Ngy hon thnh: 5/12/2011......................

Tn ti.........................................................................................................................

.........................................................................................................................................

Yu cu............................................................................................................................

.........................................................................................................................................

.........................................................................................................................................



3/42

Bi tp ln o Lng TTCN Khoa in t.........................................................................................................................................

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.........................................................................................................................................

.........................................................................................................................................

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GIO VIN HNG DN

(K v ghi r h tn)


4/42


5/42

Bi tp ln o Lng TTCN Khoa in t .I A(

.

U BC)

.I B(.

U CA)

.I C( .U AB)

Chng minh:

Cng t o nng lng tc dngXt cng sut tc thi trong mch ba pha l:

P3f= uAiA + uBiB + uCiC.

i vi mnh ba pha ba dy,v khng c dy trung tnh ln dng in

trung tnh bng khng ngha l:iA+ iB+ iC=0 => iC= -(iA+ iB).

Vy:

P3pha= uAiA + uBiB- uC(iA+ iB)

= uAiA + uBiB- uCiA uCiB

=iA (uA - uC) + iB (uB - uC)

=iAuAC + iBuBC

Vy cng sut tc dng ca ba pha l:

P3pha = UACIA cos(.

U AC,.

I A)+ UBCIBcos(.

U BC,.

I B)

Cng t o nng lng phn khngim o m thng l u ngun nn ta coi mch ba pha c ngun i

xng, ph ti mang tnh cht cm.

Ta c mmen quay tng ca cng t l:

Mq=K(IAUBCcos 1 + IBUCA cos 2+ ICUAB cos 3)

= K [IAUAcos(900-A) + IBUB cos(900-B) + ICUCcos(900-C)]



6/42

Bi tp ln o Lng TTCN Khoa in t= K (IAUAsinA + IBUB sin B + ICUC sin C)

= K Q3pha.

Ta thy mmen quay t l vi cng sut phn khng trong

mch ba pha cho nn s ch ca cng t s t l vi nng lngphn khng tiu th trong mch ba pha.

Hnh 1.1

Cu 2:chn thang o cho cc thit b o trn s

1, Chn thang o bin dng in .

Dng in nh mc pha s cp l:



7/42

Bi tp ln o Lng TTCN Khoa in t ( )A

U

SI

dm

dm

dm 735.5710.3

1000

.3===

Vy ta chn bin dng in c h s bin dng : Iscm 735.57 (A)

Itcm= 5 (A)

Vy chn BI c : KI =5

75

2, Chn thang o cho bin in p.

in p nh mc pha s cp ca my bin p l: UdmBA=10 (KV)

Cho nn chn BU c h s bin in p: KU= 1,010

3, Chn thang o cho ampek

V dng in pha th cp ca BI nh mc l 5(A) nn chn ampeke c thang o

l: DA=5(A)Dng ampe k in t:

Ampe k in t c ch to da trn c cu ch th in t.Ampe k ny

o c c dng mt chiu v dng xoay chiu

B phn chnh l mt cun dy dn, c th quay quanh mt trc, nm trong t

trng ca mt nam chm vnh cu. Cun dy c gn vi mt kim ch gc quay

trn mt thc hnh cung. Mt l xo xon ko cun v kim v v tr s khng khi

khng c dng in

Nguyn l hot ng

Khi dng in mt chiu chy qua cun dy, dng in chu lc tc ng ca t

trng (do cc in tch chuyn ng bn trong dy dn chu lc Lorentz) v b



8/42

Bi tp ln o Lng TTCN Khoa in tko quay v mt pha, xon l xo, v quay kim. V tr ca u kim trn thc o

tng ng vi cng dng in qua cun dy

y l mt thit b thc t o dng in c bn ngoi th trng

=> DA=5(A)

2, Chn thang o cho volk.

V in p pha th cp ca BU nh mc l 100(V) nn chn volke c thang ol: DV=100(V)GVHD: Phm Vn Thim - 8 - SVTH: Nguyn ThThun


9/42

Bi tp ln o Lng TTCN Khoa in t. Mch t gm li t bng tn silic c c tnh t tt v cun dy cch in cao.C cu chnh Kim v 0 bng c kh. Cn du n nh kim bng nam chm vnhcu, bng nha baklit c cch in cao. V bng nha ABS. Ca s bngnha PC trong sut (hn ch chy). 2 ci v d cho vic lp t Vn mt vo

bng in.

=>DV=100(V)



10/42

Bi tp ln o Lng TTCN Khoa in t3, Chn thang o cho cosk .

Do gc lch pha gia in p v dng in khng ph thuc vo h s bin i

ca BU v BI nn ta chn thang o cho Cos vi Cos [-11]

Nguyn l hot ng

Gc quay ca c cu l:



11/42


4,o tn s

Do tn s gia in p v dng in khng ph thuc vo h s bin i ca BUv BI nn ta chn thang o cho thit b o tn s c thang o 0 f 60 (Hz)

Nh Cung Cp Munhean c nhp t Singapre.

CngDng

ng h o tn s



12/42

Bi tp ln o Lng TTCN Khoa in tPhotoHnh nh

Cat.NoM s

TypePhn loi

RangeThang o

NG H O TN S - HIU MUNHEAN SINGAPORE Mt 96 x96 mm

CP-96FQP

Kim 220V ; 45 55Hz

ng h o tn s met cng hng

Nguyn l hot ng:

Tn s mt cng hng gm mt nam chm in,to ra bi cun

dy qun trn li st t hnh ch U,mt ming thp nm trong t trng ca nam



13/42

Bi tp ln o Lng TTCN Khoa in tchm in,gn cht vo thanh l co l thp rung c tn s dao ng ring khc

nhau. Tn s dao ng ring ca hai l thp k nhau hn km nhau l 0,25 hoc

0,5Hz.in p ca tn hiu cn do tn s s c a vo cun dy ca nam

chm in s to ra s dao ng ca tt c cc l thp.Tuy nhin l thp no c

tn s dao ng ring bng tn s f th s dao ng cc i do cng hng

ring,cn cc thanh khc khng cng hng th khng dao ng cc i.Nh vy

ta s c kt qu ti tr s tng ng vi thanh rung cc i.

5,Cng t o nng lng

A, Cng t o nng lng tc dng

V cho thi gian lm vic trong mt thng t=720h

Ta c Wtd = .Utcdm.Itcdm.Cos.t

Vi dng in p th cp cu BU l:

Utcdm =0,1 (KV) = 100 (V)

Dng in th cp ca BI l :5 (A)

Nu Wtdmax Cosmax = 1

=>Wtdmax = .0,1.5.1.720=623,54 (KWh)

Vy chn cng t o nng lng tc dng c thang o l 650 (KWh)

B, Cng t o nng lng phn khng



14/42

Bi tp ln o Lng TTCN Khoa in tTng t ta c t=720h

Wpk= .Utcdm.Itcdm.Sin.t

Vi dng in p th cp cu BU l:

Utcdm =0,1 (KV) = 100 (V)

Dng in th cp ca BI l :5 (A)

Nu Wpkmax Sinmax = 1

=>Wpkmax = .0,1.5.1.720=623,54 (KVArh)

Vy chn cng t o nng lng phn khng c thang o l 650 (KVArh)



15/42




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17/42




18/42


19/42




20/42


8 Thit b o tn s

III-Tnh s ch ca mi cng t trong thi gian mt thng, bit rng c 80% thigian my bin p lm vic ch nh mc , 20% thi gian my bin p lm

vic ch non ti. Xc nh cos tb ca ph ti.

1, Tnh s ch ca cng t o nng lng tc dng.mc 3pha 2 phn t

a, S ch ca cng t tc dng trong ch nh mc.

Thi gian nh mc:

Tdm = 0,8.24.30 = 576(h)

Idm = IAdm = IBdm = ICdm = 735.57 (A)

Udm= 10(KV)

cos dm = 0,9 dm = 25.84o

Ta c:

dmBdmBC

I

Bdm

U

BCAdmAC

I

Adm

U

AC TIU

K

I

K

UIU

K

I

K

U.,cos..cos..W

...

,

.

cttddm

+

=

= dmBdmBCAdmACI

dm

U

dm TIUIUK

I

K

U.,cos,cos..

....

+



21/42


= ( ) ( )[ ] 576.84,2530cos84,2530cos.75

5.735.57.

10

1,0.10 0000 ++

= ( )KWh605,345

I

UA

A

UAC

UBC

UBUCIB



22/42


Hnh 1.1

b, S ch ca cng t tc dng trong ch non ti.

Thi gian non ti:

Tnt = 0,2.24.30 = 144(h)

Int = IAnt = IBnt = ICnt = 0,7.Idm =0,7.57,735 = 40,41(A)

Unt = Udm = 10(KV)

cos nt = 0,75 nt = 41,4o

Ta c:

ntBntBC

I

Bnt

U

BCAntAC

I

Ant

U

AC TIUK

I

K

UIU

K

I

K

U.,cos..,cos..W

....

cttdnt

+

=

= ntBntBCAntACI

nt

U

dm TIUIUK

I

K

U.,cos,cos..

....

+

= ( ) ( )[ ] 144.4.4130cos4.4130cos.75

5.41,40.

10

1,0.10 0000 ++

= ( )KWh4,50



23/42


I

UA

A

UAC

UBC

UBUCIB

Hnh 1.2

c, S ch ca cng t tc dng trong mt thng.

( )Wh005,3964,50605,345WWW cttdntcttddmcttd K=+=+=



24/42

Bi tp ln o Lng TTCN Khoa in t2, Tnh s ch ca cng t o nng lng phn khng.

a, S ch ca cng t phn khng trong ch nh mc.

Thi gian nh mc:

Tdm = 0,8.24.30 = 576(h)

Idm = IAdm = IBdm = ICdm = 57,735(A)

Udm= 10(KV)

cos dm = 0,9 sindm= sin(arccos dm) = 0,436

Ta c:

( KVArhhTKI

KU

dmdm

I

dm

U

dm 424,167576.436,0.75

5.735,57.10

1,0.10.3.sin...3Wctpkdm ===

b,S ch ca cng t phn khng trong ch non ti.

Thi gian non ti:

Tnt = 0,2.24.30 = 144h

Int = 0,7Idm = 0,7.57,735 = 40,41(A)

Unt = Udm = 10(KV)

Cosnt = 0,75 sindm= sin(arccos dm) = 0,661

Ta c:

( )KVarhhT

K

I

K

Untnt

I

nt

U

nt 41,44144.661,0.75

5.41,40.

10

1,0.10.3.sin...3Wctpknt ===

c,S ch ca cng t pkn khng trong mt thng.



25/42


( )KVarhWW ctpkntctpkdm 834,21141,44424,167Wctpk =+=+=

3, Xc nh costb ca ph ti.

Ta c :

( )9,0

)424,167(605,345

605,345cos

2222=

+=

+=

ctpkdmcttddm

cttddmtbdm

WW

W

( )

75,0)41,44(4,50

4,50cos

2222=

+=

+=

ctpkntcttdnt

cttdnttbnt

WW

W

IV-Tnh sai s tuyt i v sai s tng i ca cc php o trong hai trnghp ph ti.

Cha sa

1, Sai s ca cc php o ch nh mc.

**C s l thuyt tnh sai s tng i , tuyt i

chnh xc l l tiu chun quan trong nht ca thit b o

xx dii =

Trong :xi l kt qu o ca ln o th i

xd l gi tr ng ca i lng o

i l sai lch ca ln o th i



26/42

Bi tp ln o Lng TTCN Khoa in tSai s php o l sai lch gia gi tr thc v gi tr o c , sai s

php o ch c th xc nh 1 cch tng i v ta khng th bit c gi tr

thc ca i lng cn o

Sai s ca php c th biu din lm 2 dng

+Sai s tuyt i

+Sai s tng i

Sai s tuyt i ca mt thit b o c nh ngha l gi tr ln nht ca sai

lch gy nn bi thit b trong khi o.

]max[ i=

Tuy nhin sai s tuyt i cha nh gi c chnh xc v yu cu ca cng

ngh thit b o.Thng thng chnh xc ca mt php o hoc mt thit b

o c nh gi bng sai s tng i

Sai s tng i c tnh vi mt php o

= vi x l gi tr i lng o

Sai s tng i c tnh vi mt thit b o

D=

vi D l khong gi tr nh nht ti gi tr ln nht

Gi tr % gi l sai s tng i quy i dng xp xp cc thit b o thnh

cp chnh xc

Theo quy nh hin hnh ca nh nc, cc dng c o c in c cp chnh

xc :0,05;0,1;0,2;0,5;1;1,5;2;2,5 v 4

Thit b o s c cp chnh xc: 0,005;0,01;0,02;0,05;0,1;0,2;0,5;1



27/42

Bi tp ln o Lng TTCN Khoa in tVy khi bit cp chnh xc ca mt thit b o ta c th xc nh c sai s

tng i quy i v suy ra sai s tng i ca thit b trong php o c th

X

D =

*Tnh ton sai s ca php o gin tip:

Gi s c 1 php o gin tip i lng y thng cc php o trc tip 1x , 2x ,..

nx . y=f( 1x , 2x ,.. nx ).

Ta c:

dy= 1 2. . ... . ny y ydx dx dxx x x + + +

- Sai s tuyt i ca php o gin tip c nh gi:

y= 2 2 2 21 211 2

( . ) ( . ) ... ( . ) ( . )n

n k

kn k

y y y yx x x x

x x x x=

+ + + =

1x , 2x ,.. nx : Sai s tuyt i ca php o cc i lng trc tip 1x , 2x ,.. nx .

- Sai s tng i ca php o gin tip c nh gi:

y =y

y

= 2 2 2 2 2 2 2 2 21 2 1 2

1 2

( ) .( ) ( ) .( ) ... ( ) .( ) ...n nn

xx xy y yx x x

y x y x y x

+ + + = + + +

1x , 2x ,.. nx : Sai s tng i ca php o trc tip cc i lng 1x , 2x

,.. nx .

-Bng tnh sai s ca hm y= 1x . 2x



28/42


Hm y Sai s tuyt i y Sai s tng i y = y/y

1x . 2x

2 2 2 2

1 2 2 1( ) ( )x x x x +

2 21 2

1 2( ) ( )

x x

x x

+

a, Sai s ca php o dng in.

Dng in qua cc pha A,B,C ( ba pha nh nhau ) c xc nh.

Idm = KI.IAdmVi IAm l dng m ampek o c ch nh mc.

( )AK

II

I

dmAdm 849,35.

75

735,57 ===

Ta c:

Sai s tuyt i ca ampek l:

( )ADI AAAdm 075,05.100

5,1. ===

Sai s tuyt i ca bin dng in l:

( )AKK IKII 225,05

75.

100

5,1. ===

Vy sai s tuyt i ca php o dng in ch nh mc l :

( ) ( ) 22 .. AdmIAdmIdm IKIKI +=



29/42


( ) ( )

( )A37,1

075,0.15849,3.225,022

=+=

Sai s tng i ca php o dng in ch nh mc l:

%37,2%100.735,57

37,1%100.% =

=

=

dm

dmI

I

Idm

b, Sai s ca php o in p.

Sai s tuyt i ca bin in p:

( )VKK UKUU 11,0

10.

100

1. ===

Sai s tuyt i ca volk :

( )VDU VVV 2100.100

2. ===

Kt qu o in p dy :



30/42

Bi tp ln o Lng TTCN Khoa in t ( )KVUKU VUd 10. ==

Vi UVddm l dng m volke o c ch nh mc.

VU =ddV m

U

U

K = ( )V1001,010

10000=

Vy sai s tuyt i ca php o in p dy l:

( ) ( )2 2dd. .d U V m U V U K U K U = +

( ) ( )

( )V84,142

2.100100.122

=+=

Sai s tng i ca php o in p dy:

142,84

.100% .100% 1, 43%% 10000

ddm

ddm d

U

U U

= = =

Sai s tuyt i ca php o in p pha:

( ) ( )2 2

. .fdm U fdm U V

U K U K U = +

( )2

2. .

3

Vddmfdm U U V

UU K K U

= +



31/42


( )

( )

2

21001. 100.2

3

208,17 V

= +

=

Sai s tng i ca php o in p pha :

%

208,17.100% .100% 2, 0817%

10000

f

Ufdmf

U

U

= = =

c, Sai s ca php o nng lng tc dng.

Nng lng tc dng ca ph ti ch nh mc l:

( )Wh5,548407605,365.

5

75.

1,0

10W..W cttddmtddm KKK IU ===

Sai s tuyt i ca cng t tc dng ch nh mc trong vng mt thng

cttddmcttd D.Wcttddm =

chn cng t o nng lng tc dng c thang o l 650 (KWh)

Ta c : =cttddmD ( )Wh650 KDcttdnt =

( )KWh75,9650.100

5,1Wcttddm ==



32/42

Bi tp ln o Lng TTCN Khoa in tVy sai s tuyt i ca php o nng lng tc dng ch nh mc l :

( ) ( ) ( ) 222tddm ......W cttddmIUcttddmIUcttddmIU WKKWKKWKK ++=

( ) ( ) ( ) 222 75,9.15.100605,365.225,0.100605,365.15.1 ++=

( )KWh18,17653=

Sai s tng i ca php o nng lng tc dng l :

%22.3%100.5,548407

18,17653%100.

tddmW=

=

=

tddm

tddm

W

W

d, Sai s ca php o nng lng phn khng .

Nng lng phn khng ca ph ti ch nh mc l :

( )KVarhKK IU 251136424,167.15.100W..W ctpkdmpkdm ===

Sai s tuyt i ca cng t phn khng :

ctpkdmctpk D.Wctpkdm =

chn cng t o nng lng phn khng c thang o l 650 (KVArh)

Ta c:

( )KVarhDD ctpkmtctpkdm 650==

( )KVarhWctpkdm 75,9650.100

5,1==

Vy sai s tuyt i ca php o nng lng phn khng l :



33/42

Bi tp ln o Lng TTCN Khoa in t ( ) ( ) ( )222pkdm ......W ctpkdmIUctpkdmIUctpkdmIU WKKWKKWKK ++=

( ) ( ) ( ) 222 75,9.15.100424,167.225,0.100424,167.15.1 ++=

( )KVarh74,15309=

Sai s tng i ca php o nng lng phn khng l:

%1,6%100.

251136

74,15309%100.

pkdmW =

=

= pkdm

pkdm

W

W

2, Sai s ca cc php o ch non ti.

a, Sai s ca php o dng in

Dng in ch non ti c xc nh :

Int = KI.IAnt

C : IAnt = 0,7.IAdm = 0,7.3,849 = 2,694(A)

Int = 15.2,694 = 40,41(A)

Sai s tuyt i ca ampek :

( )AII AdmAnt 1,0==

Vy sai s tuyt i ca php o dng in ch non ti l :



34/42

Bi tp ln o Lng TTCN Khoa in t ( ) ( ) 22 .. AntIAntInt IKIKI +=

( ) ( )

( )A28,1

075,0.15649,2.225,022

=

+=

Sai s tng i ca php o dng in ch non ti l :

%17,3%100.41,40

28,1%100.% =

=

=nt

ntI

I

Int

b, Sai s ca php o nng lng tc dng .Nng lng tc dng ca ph ti ch non ti l:

( )Wh756004,50.5

75.

1,0

10W..W cttdnttdnt KKK IU ===

Sai s tuyt i ca cng t tc dng ch non ti :

cttdntcttd D.Wcttdnt =

Ta c :

( )KWhDcttdnt 650=

( )KWh75.9650.100

5,1Wcttdnt ==

Vy sai s tuyt i ca php o nng lng tc dng ch non ti l



35/42

Bi tp ln o Lng TTCN Khoa in t( ) ( ) ( ) 222tdnt ......W cttdntIUcttdntIUcttdntIU WKKWKKWKK ++=

( ) ( ) ( ) 222 75,9.15.1004,50.225,0.1004,50.15.1 ++=

( )KWh37,14688=

Sai s tng i ca php o nng lng tc dng ch non ti:

%42,19%100.75600

37,14688%100.

tdnt

W =

=

=tdnt

tdnt

W

W

c, Sai s ca php o nng lng phn khng.

Nng lng phn khng ca ph ti ch non ti l :

( )KVarhKK IU 615,6641,44.15.100W..W ctpkntpknt ===

Sai s tuyt i ca cng t phn khng ch non ti :

ctpkdmctpk D.Wctpkdm =

Ta c :

( )KVarhDD cttdntctpknt 650==

( )KVarhWctpknt 75,9650.100

5,1 ==

Vy sai s tuyt i ca php o nng lng phn khng ch non ti l :



36/42

Bi tp ln o Lng TTCN Khoa in t ( ) ( ) ( )222

pknt ......W ctpkntIUctpkntIUctpkntIU WKKWKKWKK ++=

( ) ( ) ( ) 222 75,9.15.10041,44.225,0.10041,44.15.1 ++=

( )KVarh22,14674=

Sai s tng i ca php o nng lng phn khng ch non ti l:

%03,22%100.66615

22,14674%100.

pkntW=

=

=

pknt

pknt

W

W

Phn II:

Thu thp tn hiu truyn dn cho trm bin p.Vi s lng knh thu thp12.

S dng m hnh ni tip song song thu thp tn hiu.

*S lng knh 12 tng ng:S1,S2,..S12.

*Chn cc Mux: Chn Mux CD4051B.

Thit k b Mux chn tn hiu dng in trn cc pha

+ S1=D0 : Dng in qua pha A

S2=D1: Dng in qua pha B

S3=D2: Dng in qua pha C

S dng b Mux 1 c 2 chn a ch m ha 3 chn d liu vo



37/42


Bng trng thi

A1 C1 Q1

O O 1 D0

O 1 1 D1

1 O 1 D2

u ra:

1 1 0 1 0 0 1 1 0 1 2Q C [A .A .D A .A .D A .A .D= + +

Thit k b Mux chn tn hiu in p

in p pha A S4=D4

in p pha B S5=D5

in p pha C S6=D6

in p pha AB S7=D7

in p pha BC S8=D8

in p pha CA S9=D9



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Bi tp ln o Lng TTCN Khoa in tChn b mux c 3 chn a ch m ha 6 chn d liu

Bng trng thi

A3 A4 C2 Q1

0 0 0 1 D3

0 0 1 1 D4

0 1 0 1 D5

0 1 1 1 D6

1 0 0 1 D7

1 0 1 1 D8

u ra:

2 2 2 3 4 3 2 3 4 4 2 3 4 5 2 3 4 6Q C [A .A .A .D A .A .A .D A .A .A .D A .A .A .D= + + + +

. . . . . ]2 3 4 7 2 3 4 8A A A D A A A D+

Thit k b mux chn tn hiu cos, tn s, cng t do nng lng

+1 tn hiu cos do nng lng mt pha: S10=D9

+ 1 tn hiu tn s 3 pha: S11=D10



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Bi tp ln o Lng TTCN Khoa in t+ 1 cng t do nng lng: S12=D11

S dng b mux c 2 chn a ch m ha 3 chn d liu

Q1=D12

Q2=D13

Q3=D14

Bng trng thi

A6 C3 Q1

0 0 1 D9

0 1 1 D10

1 0 1 D11

u ra:

3 3 5 6 9 5 6 10 5 6 11Q C [A .A .D A .A .D A .A .D= + +

Thit k b MUX chn 3 tn hiu t 3 b mux1,mux2,mux3



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A8 C Q

0 0 1 D12

0 1 1 D131 0 1 D14

u ra:

7 8 12 7 8 13 7 8 14[A .A .D A .A .D A .A .DQ C = + +



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*Lp bng trng thi tng th

Tn hiu vo Chn clock Tn hiu ra

A0 A1 A2 A3 A4 A5 A6 A7 A8 C1 C2 C3 CQ1 Q2 Q3 Q

S1 0 0 X X X X X 0 0 1 0 0 1 D0 X X D12

S2 0 1 X X X X X 0 0 1 0 0 1 D1 X X D12

S3 1 0 X X X X X 0 0 1 0 0 1 D2 X X D12

S4 X X 0 0 0 X X 0 1 0 1 0 1 X D3 X D13

S5 X X 0 0 1 X X 0 1 0 1 0 1 X D4 X D13

S6 X X 0 1 0 X X 0 1 0 1 0 1 X D5 X D13

S7 X X 0 1 1 X X 0 1 0 1 0 1 X D6 X D13

S8 X X 1 0 0 X X 0 1 0 1 0 1 X D7 X D13

S9 X X 1 0 1 X X 0 1 0 1 0 1 X D8 X D13

S10 X X X X X 0 0 1 0 0 0 1 1 X X D9 D14

S11 X X X X X 0 0 1 0 0 0 1 1 X X D10 D14

S12 X X X X X 0 1 1 0 0 0 1 1 X X D11 D14



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