ky thuat do luong ch3
TRANSCRIPT
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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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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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- 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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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- 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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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
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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.