Ky Thuat Do Luong Ch3

7/30/2019 Ky Thuat Do Luong Ch3

1/21

Gio trnh: kthuto lng 1 Dnh cho sinh vin nhm ngnh k thut

Bin son: PGS.TS. Nguyn Hu Cng Khoa in t -i hc K thut CNVersion1 - Thng 8 nm 2009

47

Chng 3

MCH O LNG V GIA CNG THNG TIN O

3.1.Khi nim chung

3.1.1. nh ngha- Mch o lng l thit b k thut 1m nhim v bin i, gia cng thng tin ,

phi hp cc tin tc vi nhau trong mt h vt l thng nht.

- C th coi mch o nh l mt khu tnh ton, thc hin cc php tnh i s

trn s mch nhvo kthut in t theo yu cu k thut ca thit bo.

3.1.2. Phn loi mch o: Theo chc nng ca cc mch o m ta c th phn

thnh nhiu loi mch o nh sau:

- Mch t l: l mch thc hin mt php nhn (hoc chia) vi mt h s k.

Ngha l nu i lng vo l x th i lng ra l kx.

i din cho cc loi ny l: in trsun, b phn p, my bin p, ,my bin

dng, v.v...

-Mch khuch i: Cng ging nh mch t l mch khuch i lm nhim v

nhn thm mt h s K gi l h s khuch i. Tuy nhin mch khuch i th

thng thng, cng sut ra ln hn cng sut vo (iu ny ngc vi mch t l),

ngha l i lng vo iu khin i lng ra.

-Mch gia cng v tnh ton: Bao gm cc mch thc hin cc php tnh is nh cng, tr, nhn, chia, tch phn, vi phn,v.v...

-Mch so snh: l mch so snh gia hai in p. Mch ny thng c s

dng trong cc thit bo dng phng php so snh.

-Mch to hm: L mch to ra nhng hm s theo yu cu ca php o nhm

mc ch tuyn tnh ha cc c tnh ca tn hiu o u ra cc b phn cm bin.

-Mch bin i A/D, D/A: l loi mch bin i t tn hiu o tng t thnh

s v ngc li, s dng cho k thut o s v ch to cc mch ghp ni vi my

tnh.

-Mch o sdng kthut vi xll mch o c ci t vi x l to ra cc

cm bin thng minh, khc bng my tnh, nhv gia cng sb s liu o v,v...

3.2. Mch t l

3.2.1.Mch t l v dng

L loi mch thng dng nht. i vi mch mt chiu thng dng mch Sun,

cn i vi mch xoay chiu thng dng bin dng in.


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48

1.in trsun

- Sun l mt in trmc song song vi ccu ch th (h.3-1). Dng chy trong

mch chnh l I ; trong mch ch th l ICT.

Cc dng

Cc p

Ta c: ICT

IK

I= ; thng thng KI > 1 v c gi h s phn dng in.

in trSun Rsc tnh theo cng thc sau:

RS =RCT

K 1I (3-1)

- Sun c cu to nh l in tr4 cc: 2 cc dng v 2 cc p. Hai cc dng

c a dng in IS vo cn hai cc p s ly p ra mc vo ccu ch th hoc

a ti mch pha sau (h.3-2). Trn Sun thng ghi dng in IS c thi qua n v

in p u ra l: Us = IsRs = (I - ICT) Rs v cp chnh xc.

- t chnh xc cao mt Sun thng ch lm vic vi mt ch th nht nh

v phi c dy ni xc nh in tr. iu chnh in trSun ta c th x rnh

khc nhau.

- Mun dng Sun c h s chia dng khc nhau c th dng Sun vi nhiu cpkhc nhau nh hnh 3-3.

CCCT

ICT

RCT

IS RS

Hnh 3-1: Cch mc SunHnh 3-2: Cu to Sun

ICTRCT

R1 R2 R3 R4

I1 > I2 > I3 > I4

Hnh 3-3: Cu to Sun nhiu cp


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49

- tnh cc in trR1, R2, R3, R4 ta c th da vo biu thc (3-1) nh sau:

Lp h phng trnh ng vi cc dng khc nhau:

4

CTS 1 2 3 4

4

RR R R R R

n 1

= = + + +

; 44CT

In

I

=

3

CT 4S 1 2 3

3

R RR R R R

n 1

+= = + +

; 33

CT

In

I=

2

CT 4 3 2S 1 2 2

2 CT

R R R IR R R ;n

n 1 I

+ += = + =

1

CT 4 3 2 1S 1 1

1 CT

R R R R IR R ;n

n 1 I

+ + += = =

Ta c 4 phng trnh vi 4 n s. T gii ra ta tm c cc in trcn tm R1, R2,

R3, R4.

Trong cng nghip Sun c lm bng vt liu c in trrt t ph thuc nhit

nh Manganin. Thng ngi ta ch to Sun vi dng in t vi mA in p Sun

c60, 75, 100, 150 v 300 mV.

- ng dng: Sun c dng ch yu trong mch mt chiu. Trong mch xoay

chiu c th dng khi ti l thun tr. Cn khi ti l in khng th mc phi sai s v

gc pha.

in tr Sun ch yu s dng m rng thang o trong cc ampemet mtchiu.

2. Bin dng in

Bin dng in (BI) l mt my bin p c bit c

cun scp rt t vng cho dng ph ti trc tip chy

qua; cun th cp nhiu vng hn, dy nh v c ni

kn mch vi mt ampemet (hoc cun dng ca cng

t, wattmet ...). V in trca ampemet rt nh chonn c th coi my bin dng lun lm vic ch

ngn mch.

Ta c:

I1W1 I2W2

hay

I1/I2 = W2/W1 = KI,

KI l h s my bin dng hay cn gi l h s mrng thang o.

A

BII1

I2

W1

W2

Hnh 3-4. S sdng BIo dng in


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50

Thng thng, d dng cho vic ch to v s dng, W1 ch c mt vng, ng vi

dng in I1chnh mc theo mt dy su tin no ; W2 nhiu vng hn

ng vi dng I2chnh mc l: I2m = 1A hoc I2m = 5A.

V d: my bin dng: 100/5 ; 200/5; 300/ 5Trong trng hp ampemet ni hp b vi bin dng in th s ch ca ampemet

c khc vch theo gi tr dng in I1 pha scp.

3.2.2. Mch t l v p

1. Mch phn p

L mch phn in p, thng U1 ln hn U2 tc l cng sut vo ln hn cng sut

ra. Ta phn bit mt s mch nh sau.

a, Mch phn p in tr

Cc in tr R1, R2 c ni nh

hnh 3-5. H s phn p c tnh l:

m =2

1

U

U(3-2)

Ta phn bit hai trng hp:

- Khi khng c ti hay RT = ta c:

mo = 2

1

2

21

2

1

R

R

1RI

)RR(I

U

U

+=

+

= (3-3)

- Khi c ti RT ta c:

mT =

T 21

2 T1 1 2 T 2 T

2 2 TT 2

2 T

R RIR I

R RU R (R R ) R R

U R RR RI

R R

+

+ + + = =

+

=

= 1 + 1 1 10

2 T T

R R Rm

R R R+ = + (3-4)

Lc ti l nhng ccu o c in trkhng i, ngi ta dng R2 l in tr

ca ngay bn thn ccu, trong trng hp ny phn p ch c in trR1 v c gi

l in trph (h.3-6).

in trphc tnh nh sau:

RP = RCT(m - 1) (3-5)

y m = xCT

U

Ut s gia in p cn o v in p trn ccu ch th.

U1

U2R2

R1

RT

Hnh 3-5: Mch phn p in tr


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51

Nu mt Volmet c nhiu thang o th cch

tnh cc in trph nh sau:

1PR = R1 = RCT(m1 - 1)

m1 =1

CT

U

U

2PR = R1 + R2 = RCT(m2 - 1)

m2 =2

CT

U

U

3PR = R1 + R2 + R3 = RCT(m3 - 1)

m3 =3

CT

U

U

Ty ta c th tm c cc in trR1, R2,

R3 tng ng.

Ngi ta cn ch to phn p c h s phn p thay i tu .

Thng thng l mt bin tr trt c gn thm mt thang chia , trny c khc h s phn p tng ng vi v tr ca n. Nhng bin tr trt phn in

p vi chnh xc khng cao (thng t 1 - 5%).

Cc phn p c cp chnh xc cao (0,05 - 0,1) c ch to theo kiu nhy cp

hoc b tr thnh tng cp thp phn (Hnh 3-7). Vt liu thng lm bng dy in

trmanganin c h s nhit in trthp. in p vo U1 cnh, cn in p ra bin

thin t 0,0001 U1n 0,9999U1.

U3

U2

U1

UcT

R3

R2

R1

RcT

Hnh 3-6: Mrng thang o

ca Volmet

. . .

0,1 0,09 0,05 0,01

U1

1,0 0,9 0,8 0,7 0,2 0,1

U2

Hnh 3-7: B phn p c cp chnh xc cao


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52

b,Mch phn p in dung

Phn p in dung c th dng trong mch xoay chiu, cc tin C1, C2c

ghp ni tip vi nhau v c c trng bng in dung C1, C2 cng vi cc in tr

r R1, R2 theo m hnh thay th song song(Hnh 3-8).H s phn p l:

m =

+

+

+=

11

1

22

2

2

1

RCj

11C

RCj

11C

1U

U(3-6)

Nh vy m ph thuc tn s.

Nu tn s ln th:

22RC

1

v

11RC

1


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53

u im ca phn p ny l h s phn p K t thay i lc ti u ra thay i.

Nhng nhc im l tn s thay i s gy ra sai s tn s.

2. Mch bin in p o lng(Bu)

Mch bin in p cng l hnh thc ca mch phn p in cm, ch khc l U1c th nh hn hay ln hn U2 (tc l Ku c th ln hn hay nh hn 1) khi U2 > U1 ta

c bin p tng p, U2 < U1 ta c bin p h p. Cc in p U1, U2 c th lin h vi

nhau c vin v t (bin p t ngu) hoc c th ch ni nhau bng t v cch in

vi nhau.

H s bin p:

Ku =

2

1

2

1

W

W

U

U= (3-9)

Trong hng dn s dng ca bin in p thng ch r cng sut nh mc,

in p vo U1 v in p ra U2 v h s Ku, cc u ni ca cun scp c l hiu

l A, X cun th cp l a, x.

Ngc vi bin dng o lng bin p o lng s dng ch hmch

cun th cp. V th cun th cp thng c ni vi Volmt c in tr vo ln

(Hnh 3-10) o in p U2 sau nhn vi Ku ta c U1.

in p nh mc ca cun th cp

thng l 100V. Cn in p nh mc cacun scp chnh l in p cn i hay kim

tra.

Cp chnh xc ca bin p o lng l

0,05; 0,1; 0,2; 0,5.

3.3. Mch khuch i o lng

ng v phng din gia cng tin tc,

mch khuch i (K) cng xem nh mt

mch t l, ngha l: Xr = KXv

Tuy nhin mch K c iu c bit

ngc vi mch t l l mch K thng c cng sut ra ln hn cng sut vo. C

th coi i lng vo iu khin i lng ra. y l u vit ca mch in t. Nhc

khuch i, ngi ta tng nhy ca cc thit bo ln rt nhiu cho php o

c nhng i lng o rt nh m trc kia khng o c.

Hnh 3-10: Mc volmet vobin p o lng

V

W2

W1

a x

A X

U~


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54

3.4.1. Mch K lp li

Trong cc thit bo tn hiu o c ly ra t cc b cm bin c cng sut ra

rt nh. Mun khuch i c nhng tn hiu nh vy i hi in tr vo ca b

khuch i phi rt ln. to c iu thng ta s dng cc mch lp li uvo (mch colectchung) (h.3-11).

Mch lp li s dng tranzito bnh thng (h.3-11a). c im ca mch ny l

in trvo ln in trra nh. Nhc phn hi m su h s K theo in p Ku = 1

(thng b l ln hn 1 mt t).

Ngc li h s K v dng kh ln

KI =b

C

I

I

= 1 + (3-10)

- l h s khuch dng ca tranzito.

- tng in tr vo ngi ta s dng K thut ton vi phn hi m su

(Hnh 3-11b).

Hnh 3-11: Mch lp li

a) dng tranzitor

b) dng khuch i thut ton

b)

Uvo

-EC

Ura

+EC

+EC

UvoR Ura

a)


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55

3.3.2. Mch Ko lng

Trong cc mch o lng thng s dng b Ko lng. l kt hp gia

cc b lp li v cc b Kin p. Hnh 3-12 ch r mt b Ko in p ra camch cu.

tng u l hai b lp li dng mch K thut ton IC1, IC2. H s K ca

tng u l:

K1 = 1 +2

31

R

RR +

C thiu chnh bng cch thay i in trR2. in p vo c t vo hai

u ca hai mch lp li, nh th to kh nng o hiu in th gia hai u vo ny

so vi t.

tng th hai s dng IC3 c h s K l:

K2 =4

5

R

R

Do h s K ca c mch l:

K = K1.K2 =4

5

R

R.(1 +

2

31

R

RR +) (3-11)

IC2

IC1

IC3Uvo Ura

R5

R4

R4

R5

R1

R2

R3

Hnh 3-12: B Ko lng


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56

3.4. Mch gia cng tnh ton

3.4.1. Mch cng

L loi mch thc hin php cng. Thng

thng l cng in p.Hnh 3-13 l s mch cng in

p dng K thut ton (KTT) mc theo

mch o du. Tn hiu ra Ura t l vi tng

i s ca cc tn hiu vo.

R

R)U...UU(U 2

n21ra+++= (3-12)

Nu R2 = R th =

=n

1iira

UU (3-13)

Trng hp mc theo s khng o du, ta c mch hnh 3-14.

)U...UU(R

R1

n

1U

n21

2

ra+++

+= (3-14)

Nu n = 1 +1

2

R

R

th =

=n

1iira

UU (3-15)

3.4.2. Mch trMch K hiu hai in p u vo

thng s dng KTT, trong mch t hp gia

mch o du v khng o du (h.3-15).

1

1

4

2

132

341

raU

R

RU

RRR

RRRU

+

+= (3-16)

Khi R1 = R2 v R3 = R4

Ura

U1

U2

Un

R

R

R

R2

R1

Hnh 3-13: Mch cng in p viK thut ton o du

Ura

U1

U2

Un

RR

R

R1

R2

Hnh 3-14: Mch cng in pvi KTT khng o du

Ura

U1R1

R4

R3

Hnh 3-15: Mch trdng KTT

U2R2


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57

th )UU(R

RU

12

1

4

ra= (3-17)

Trong trng hp ny th sai s ca mch l t nht. Vi mch ny th in tr

vo ca hai u vo khng nht thit phi ging nhau. in trvo mch o du

ging nhmch Ko du, cn in trvo mch khng o du bng tng cc

in trR2 v R3.

3.4.3. Mch nhn

C rt nhiu trng hp phi s dng mch nhn nh khi o cng sut P = UI

cos hoc phi nhn hai in p, v.v... v th mch nhn trong o lng l rt quantrng.

Trong o lng phn t nhn c dng rng ri l phn tin ng, st in

ng, v cm ng c xt chng 2. Cc mch nhn ny c dng ch to

cc Wtmt o cng sut tc dng v phn khng; ch to cng tin. Ngoi ra

mch nhn cn c th s dng chuyn i Hn (Hall) v cc b nhn in t s dng

cc b KTT (do gii hn ca chng trnh nn ta khng xt)

3.4.4. Mch chia

Mch chia c s dng rng ri trong cc php o gin tip. Kt qu php

chia c khi l mt i lng nhng cng c khi l con s khng th nguyn thng

c trng cho phm cht. Thng dng nht l cc phng php: lgmt, mch cu,

mch chia in t, ...

* Mch chia bng ccu ch th logomet c gc quay t l vi t s hai dng

in(chng 2)

* Mch chia da trn mch cu cn

bng l mch ly t s gia hai in trca

hai nhnh ca cu (h.3-16).

Khi cu cn bng ta c phng trnh:

RxR4 = RyR3 (3-18)

R3 thng l mt bin tr c in tr

ton b l tm ca gc quay = f1(R3).

EC

Rx R3

Ry R4

Hnh 3-16: Mch cu cn bng sdng lm mch chia


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58

y

x

43R

RRR = (3-19)

x1 3 2y

Rf (R ) f

R

= =

(3-20)

3.4.5. Mch tch phn

Mt mch tch phn c biu din nh hnh 3-17

Quan h gia in p vo v ra ca b tch phn nh sau:

=

T

ov

1ra dtUCR

1

U (3-21)

Trong mch ny tc thay i

ca in p ra t l nghch vi hng s

thi gian = R1C.

* Trng hp khi tn hiu vo thay

i theo kiu bc thang th tc thay i

ca tn hiu ra s l:

tU ra

= -

CRU

1

v

Nh thu ra s c tn hiu tuyn tnh tng dn theo thi gian.

3.4.6 Mch vi phn

Mch vi phn n gin c th thc hin bng in cm hay in dung (h.3-18 a, b).

R2

Uv

R1

C

Ura

Hnh 3-17: Mch tch phn

Ura

RC

Uvo

C

UraL Ura Uvo R

Uv

RP Rf

a) b) c)

Hnh 3-18- Cc mch vi phn

a)Vi phn bng phn tRL; b) Vi phn bng CR;

c) Vi phn sdng KTT.


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59

in p ra mch 3-18 a:

LL

diU L

dt= (3-22)

Dng in qua in dung C mch 3-18b l:C

cdu

i Cdt

= (3-23)

Tuy nhin trong thc t nng cao chnh xc ta thng kt hp mch CR

vi KTT (h.3-18c). y l mch K c phn hi bng in tr. in p ra ca mch

vi phn nh vy l:

dt

dUCRU v

1ra= (3-24)

3.5. Mch so snh

Trong k thut o lng ngi ta s dng rt rng ri mch so snh

(Comparator). Dng mch so snh l pht hin thi im bng nhau ca hai i

lng vt l no (in p chng hn). Trong phng php o kiu so snh thng

s dng mch so snh pht hin lch khi im khng ca in k.

Mch so snh ngy nay thng c s dng vi b KTT (khuch i thut

ton) mc theo kiu mt u vo hay hai u vo, hoc c thm phn hi dng nh

to ra c tnh tr ca b so snh. Mch so snh cng c th to ra bng cc in tr

mu chng hn mch cu, mch in th k vi thit b ch th lch khng bng in

k.Do gii hn ca chng trnh, ta ch xt mt vi mch so snh cbn


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60

3.5.1. B so snh cc tn hiu khc du bng KTT (khuch i thut ton) mc

theo mch mt u vo

B so snh tn hiu mt u vo c

biu din hnh 3-19a so snh hai inp vo khc du nhau. thi im cn bng

in p ra Ura ca mch so snh s chuyn

sang mt trng thi khc.Qu trnh chuyn

trng thi c biu din hnh 3-19b.

Trc thi im in p Uc nh hn in

p t Esp (Set point) v th m Esp s quyt

nh trng thi u ra. Trong trng hp

ny Esp > 0 v th Ura E

K.

Khi in p Uct n gi tr2

1sp

R

R.E

th in p ra sc quyt nh bi Uc v

trnn bng E +K

.

Ti thi im cn bng Uc = Esp (R1/R2) th

K ri vo ch khng n nh tuyn tnh.Vic chuyn trng thi s xy ra vi mt

thi im tr no y l thi gian phng

ca mt tin k sinh no y ca KTT

(h.3-19b).

5.5.2. Mch cu o

Mch cu c xem nh l mt mch so snh in tr. Tuy nhin thc cht l

ta bin thnh s so snh hai mc in th. Cu to ca mch cu gm c: 4 in trR1,R2, R3, R4 mc theo mch nh hnh 3-20.

thi im cu cn bng

th in th 2 im c, d bng

nhau tc l Uc = Ud.Lc ta c

h thc:

R1R4 = R2R3 (3-25)c)

Hnh 3-20:Mch cu o

U

d

c

a b

R4

R1 R2

R3

Uv

R1

R2

EK

EK

Ura

Uc(t) ESP

R1 R2

a)

0

0 t

ESP

Uc = -ESP2

1

RR

Uc(t)

t +

EK

EKEK

b)

0

0 t

ESP

Uc = -ESP2

1

RR

Uc(t)

t +

b)

U

Hnh 3-19. B so snh mt u vo vi cc tnhiu khc du


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61

t c trng thi cn bng ngi ta thng iu chnh mt trong 4 nhnh

(v d R2 chng hn). Qu trnh tm cn bng l qu trnh iu chnh so snh hai

in th Uc v Ud . Khi bng nhau th in k ch 0 ta c h thc(3-25) khng ph

thuc vo in p ngun. Nu ch to cc in trchnh xc (bng vt liu manganin)th c th s dng mch cu o in trvi chnh xc cao.

Tht vt gi s thay R1 bng in trcn o Rx. trng thi cn bng ta c:

2

4

3

xR.

R

RR = (3-26)

Nu chn R3 = R4 th Rx = R2. y l php o in trvi chnh xc

cao. Nguyn l ca php o thc hin theo phng php so snh cn bng.

5.5.3. Mch in th k

L mch o da trn phng php so snh cn bng gia hai in p: in p

cn o Ux v in p mu EN. Hnh 3-21a l s khi ca mch o in th k. in

p Uxc so snh vi in p EN. thi im bng nhau. c EN suy ra Ux. Hnh 3-

21b l s nguyn l c th.

u tin bt cng tc K sang v tr 1 iu chnh R1 sao cho km in k ch 0.

Lc ta c:

EN = INRN hay IN = EN/RN l i lng chnh xc v EN v RN l pin mu vin trmu. By gibt cng tc K sang v tr 2 iu chnh Rx sao cho in k ch 0

ta c Ux = INRx = xN

N RR

E.


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62

Hnh 3-22. To hm bngbin tr

Nu ta ch to sao cho

n

N

N

10R

E

= th ta c Ux = 10

n

Rx

y cng l mt php o

in p chnh xc, bi v kt qu

o ph thuc vo chnh xc

ca pin mu EN v ca in tr

mu RN v Rx. chnh xc ca

php o cn ph thuc vo

ngng nhy ca in k ch0 na. Thng ta phi chn in

k t in hoc in t nhy

(trong khong t 10-6 - 10-9

A/vch).

3.6. Mch to hm

Mch to hm rt thng dng trong k thut o lng to ra mt quan hhm gia tn hiu vo v tn hiu ra no . Ta s xt mt s mch to hm cbn.

3.6.1. Mch to hm bng bin tr

Bin tr ca mch to hm c thit din

c ch to theo hm s mong mun(h.3-22).

Trong cc bin tr to hm ny di chuyn ca

con chy t l vi i lng vo l = k1X. Li ca

bin trc hnh theo hm s cn to thnh. Gisin tr ca ton b bin tr l Rbt, in p

ton bt ln n l Ubt. Ta c in p ra:

x

bt

bt

raR

R

UU =

Nu Rx = f(l) th Ura =f(k1X) = k1f(1)

Hnh 3-21: Mch in thk

Ux

EN

SS

K

a)

1 2 Ux

RxRN

IN

R1B

b)


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63

3.6.2. Mch to hm bng dit bn dn

Dit l tng c xem nh ch dn in mt chiu, c th coi in trngc l

v cng ln, in trthun bng 0.

S mch to hm n gin nht cho trn hnh 3-23a. Nh b phn p ABtrn dy t in p nn Uo. cc katt ca cc dit c in p Uo1, Uo2, ...

Khi thay i gi tr Ux ta c th phn tch nh sau: Lc 0 < Ux < Ux1 tt c cc

dit u kho, khng c dng in i qua mch phn p, in p Uxc t trn in

trR v RN ni tip nhau.

N

N

xNRR

RUU

+=

Lc Ux1 < Ux < Ux2, dit D1 mcn cc dit khc vn ng ta c:

++=

N1

N1

xRR

RRRIU

xN x x

E

RUU U IR U

R R= =

+

viN1

N1

ERR

RRR

+=

Ux

RR1D1

U01

R2D2U02

R3D3U03

R4D4U04

A

B +U0

RN

Ura

a)

Ura

Ux

b)

U01U02

U03

U04

Ux1 Ux2 Ux3 Ux4

Hnh 3-23: Mch to hm bng it bn dn


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64

Cng nh vy, lc Ux2 < Ux < Ux3, cc dit D1, D2u m, dng trong mch

chnh tng ln, in p ri trn ti nh hnh 3-23b gm nhng on thng c gc

khc nhau ni li vi nhau. Kt qu l ta nhn c ng cong theo hm s mong

mun. hiu chnh cong ta c th thay i cc in trR1 R2 R3 R4 cho ph hp.

3.7. Cc b chuyn i tng ts v s tng t

3.7.1. Mu

Do s pht trin nhanh chng ca k thut s, c bit l sng dng ph bin

ca my tnh in t s, nn ngi ta thng dng mch s x l tn hiu tng t.

Mun dng h thng s x l tn hiu tng t th phi bin i tn hiu tng t

thnh tn hiu s tng ng ri a vo h thng s x l. Mt khc thng c yu

cu cn bin i tn hiu s (kt qu x l) thnh tn hiu tng t tng ng a ra

s dng. Ta gi s chuyn i t tn hiu tng t sang tn hiu s l chuyn i AD

v mch in thc hin cng vic l ADC (Analog to Digital Converter). Gi s

chuyn i t tn hiu s sang tn hiu tng t l chuyn i DA v mch in tng

ng l DAC (Digital to Analog Converter).

3.7.2. Mch chuyn i s - tng t(DAC)

Mch chuyn i s tng tc dng chuyn i cc tn hiu s thnh

tn hiu tng t.

Bn cht ca qu trnh chuyn i DAC l qu trnh nhn mt nhm xung di

dng m nh phn sau bin i thnh mt mc in th hay cng dng in

tng t no . Mc (hay ln) ca tn hiu p (dng) ny t l vi gi tr su

vo nhn c.

Ngi ta thng s dng 3 phng php chnh trong mch DAC l:

Bchuyn

i

DAC

D0D1

Dn

Tn hiu ratng t

Dliusuvo

.

.

.

.

.

.

.

Hnh 3-24. S khi ca mt b DAC


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65

- Phng php to ra in th.

- Phng php to ra dng in.

- Phng php nhn.

Gii hn chng trnh ta ch xt mt loi minh ha.

Phng php to ra in th vi in trc trng s khc nhau:

Mch gm mt ngun in p chun Uch , cc chuyn mch K0 , K1 , Kn-1 , cc

in trc gi tr ln lt l R/20 , R/21 ,R/2n-1 v mt KTT. Vi mch in nh

hnh v trn khi mt kho in no c ni vo ngun in p chun Uch th s cp

cho mch KTT mt dng in c cng :

Cng dng in ny c lp vi cc kho cn li. Trong trng hp c

nhiu kho K cng ni vo Uch ta s c nhiu dng in cng chung chy qua Rf to

thnh in p ra. Ta thy tr sin p ra ph thuc vo ch kho in no c ni

vi Uch tc l ph thuc vo gi tr ca bt tng ng trong tn hiu sc a vo

mch chuyn i.

Mt cch tng qut mt DAC n bit (t B0n Bn-1) ch to theo phng php in tr

c trng s khc nhau, ta c th tnh in p tng t ra theo cng thc sau:

Vi: B0n Bn-1 c gi tr 1 hoc 0.

Bi= 0 kho Ki ni mass; B

i= 1 kho Ki ni vi Uch

R/20

R/21

R/2n-1

Rf

Ura

Uch+

_K0

K1

Kn-1

BNN

BLN

U0+

_

Hnh 3-25. Mch DAC vi in trc trng skhc nhau

Ii =Uch

R/2i

Ura = -Uch [20.B0 +21.B1 ++2n-1.Bn-1]

Rf

R


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66

3.7.3. Mch chuyn i tng t- s (ADC)

Chc nng ca ADC l bin i tn hiu tng t thnh tn hiu s.

Bn cht ca qu trnh bin i ADC l nhn vo mt gi trin th (tng t)sau mt khong thi gian xc nh n sinh ra trn u ra mt m nh phn (s) t l vi

gi tr tng tu vo. Qu trnh chuyn i ny phc tp v mt thi gian hn so

vi s chuyn i trong DAC.

Qu trnh chuyn i A/D nhn chung c thc hin qua 4 bc cbn, l: ly

mu; nhmu; lng t ha v m ha. Cc bc lun lun kt hp vi nhau trong

mt qu trnh thng nht. V d: ly mu v nhmu l cng vic lin tc v cng mt

mch in; lng t ha v m ha l cng vic ng thi thc hin trong qu trnh

chuyn i vi khong thi gian cn thit l mt phn ca thi gian nhmu

1.nh l ly mu

a, Khi nim vly mu tn hiu

Ly mu (Sampling) ca mt tn hiu

x(t) l thay th n bng mt tn hiu xung

u(t) c rng rt nh, tun hon vi tn s

Fcgi l tn s ly mu nh Hnh 3-26.

Trong iu kin l tng, rng xung

0, gi tr ly mu l gi tr tc thi ca

ng cong x(t) cn gia cc on th

u(t) s bng 0. Nh vy qu trnh ly mu tn

hiu thc cht l qu trnh ri rc ho u tn

hiu.

i lng nghch o vi Fc l Tc =

CF

1 c gi l chu k ly mu hay l bc

ri rc ho.

Tuy nhin trong thc t, tn ti mc

d rt nh so vi Tc.

Hnh 3-26. Ly mu tn hiu


21/21


Bin son: PGS TS Nguyn Hu Cng Khoa in t - i hc K thut CN 67

b,nh l ly mu

Hin nay tn ti cc nh l ly mu ni ting l nh l Kotelnikov (Nga),

nh l ly mu ca Shanon (M) v nh l Nyquist (M). Ta gii thiu mt

trong ba nh l:

Shanon, nh bc hc ngi M pht biu v chng minh nh l ly mu

nh sau:

Mt tn hu lin tc x(t) c phgii hn trong khong 0fmax c thc biu

din hon ton bng cc mu cch u nhau, tn sFc sao cho Fc2fmax.

Vi Fc gi l tn sly mu

2. Lng tha v m ha

Tn hiu s khng nhng ri rc trong thi gian m cn khng lin tc trong bin i

gi tr. Mt gi tr bt k ca tn hiu su phi biu th bng bi s nguyn ln gi

trn v no , gi tr ny l nh nht c chn. Ngha l nu dng tn hiu s biu

thin p ly mu th phi bt in p ly mu ha thnh bi s nguyn ln gi tr

n v. Qu trnh ny gi l lng tha. n vc chn theo qui nh ny gi l

n v lng t, k hiu . Vic dng m nh phn biu th gi tr tn hiu s l m

ha. M nh phn c c sau qu trnh trn chnh l tn hiu u ra ca chuyn iA/D.

Tn hiu tng t l lin tc nn khng nht thit l bi s nguyn ln ca , do

khng trnh khi sai s lng t ha. Do tn ti nhng cch khc nhau khi phn chia

cc mc lng t dn n sai s lng t ha khc nhau.

3.Ly mu v nhmu

Khi ni trc tip in th tng t vi u vo ca ADC, tin trnh bin i c th btc ng ngc nu in th tng t thay i trong tin trnh bin i. Ta phi ci

thin tnh n nh ca tin trnh chuyn i bng cch s dng mch ly mu v nh

mu ghi nhin th tng t khng i trong khi chu k chuyn i din ra.

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