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Bai Giang Thiet Bi Do Va Labview

Jul 09, 2015

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i hc Khoa hc T nhinKhoa in t - Vin thng

Thit b o

GVGD: Trn Th Hunh Vn

Ni dungChng 1. Gii thiu v o lng Chng 2. o in p v dng in Chng 3. o in tr Chng 4. o cng sut v in nng Chng 5. Cc dng c o s Chng 6. Cc b to sng v oscilloscope Chng 7. Cc b phn tch sng v ha tn Chng 8. Cc bridge Chng 9. Dao ng k, tia m cc v my ghi XY. Chng 10. Cc b lc v Opamp. Chng 11. Lp trnh Labview

Chng 1

Gii thiu o lng

i lng o lng i lng in

i lng in tch cc: i lng mang nng lng in: in p, dng in, cng sut Cung cp nng lng cho cc mch o. Nng lng qu ln: mch phn p, phn dng. Khng mang nng lng: in tr, in cm, in dung, h cm. Cn cung cp in p, dng in

i lng in th ng:

i lng khng in:Nhit , p sut, tc , gia tc, pH o lng v iu khin t ng cn chuyn i sang i lng in bng cc b chuyn i hay cm bin

c tnh o lng Hiu bit c tnh ca thit b o gip ta la chn thit b o ph hp theo yu cu. c tnh tnh v ng c tnh tnh: thit b o c dng o cc i lng c iu kin khng thay i trong 1 qu trnh o. Tt c cc c tnh tnh c c nh qu trnh nh chun. Mt s c tnh tnh:

chnh xc (accuracy): mc chnh xc ca php o so vi gi tr mong mun. phn gii: khong chia nh nht thit b o p ng c chnh xc (precision): kh nng chnh xc ca thit b o khi lp li nhiu ln vi cng input. Sai s (error): sai lch gia gi tr o c v gi tr mong mun nhy: t s gia p ng ng ra tng ng vi s thay i u vo.

Sai s trong o lng

o: so snh i lng cha bit vi 1 i lng c chun ha. Kh xc nh c gi tr thc true value ca cc i lng o, nn tr s o c bng thit b o c gi l gi tr tin cy expected value. Kt qu o t khi phn nh ng tr s tin cy c v n b nh hng bi chnh thit b o v ngi s dng thit b. chnh xc ca php o c th hin bng hnh thc sai s.

Cc loi sai s trong o lng Sai s tuyt i: e = Yn - Xn Yn: tr s tin cy c (expected value) Xn: tr s o c Sai s tng i (%):

chnh xc tng i: chnh xc theo %: a =100% - er = (Ax100%)

Tnh chnh xc:n:

tr trung bnh ca n ln o

Sai s trong o lngV d: in p gia 2 u in tr c expected value bng 50 V. Dng vn k o c 49 V. Sai s tuyt i: e = 1V Sai s tng i: chnh xc:

Cc loi sai s trong o lng Sai s ch quan: do ngi s dng thit b khng ph hp; do vic c, tnh ton hoc ghi sai kt qu. Sai s h thng: ph thuc vo thit b o, iu kin mi trng

Sai s do thit b o: sai s do cng ngh ch to, s lo ha do s dng bo tr nh k Sai s do iu kin mi trng: nhit , p sut, m, in trng hay t trng tng. Sai s h thng chu nh hng khc nhau trng thi tnh v trng thi ng.

Trng thi tnh: ph thuc vo gii hn ca thit b o Trng thi ng: khng p ng theo tc thay i nhanh ca cc i lng o

Sai s ngu nhin

V d Mt vn k c nhy 1k/V, c ni vi 1 in tr Rx mc ni tip vi 1 miliampe k, c gi tr l 80V trn thang o l 150V. Khi miliampe k c gi tr l 10mA, tnh: a. gi tr in tr Rx b. sai s do hiu ng ti ca vn k

Gii:in tr tng cng ca mch: R = 80V/10mA = 8k in tr ca vn k bng: Rv = 1k/V x 150V = 150k in tr ca Rx = (RxRv)/(R-Rv) = 8.45k % error = (8.45 8) / 8.45 x100% = 5.3 %

Cc ngun sai s Khng nm vng cc thng s o v iu kin thit k

Thit k nhiu khuyt im Thit b o hot ng khng n nh

Bo tr thit b o km Do ngi vn hnh thit b o khng ng Do nhng gii hn ca thit k

c tnh ng C rt t thit b o p ng tc thi ngay vi s thay i ca i lng o: p ng chm hoc khng kp Nguyn nhn do cc c tnh ca thit b o: qun tnh, nhit dung, in dung Cc i lng o c 3 dng thay i sau:

Thay i theo dng hm bc theo thi gian Thay i tuyn tnh Thay i c dng hm hn sin Tc p ng Tnh chnh xc Tnh tr Sai s ng

Cc c tnh ng ca thit b:

p ng ng bc zero Lin h gia tn hiu cn o v tn hiu ra ca thit b o c biu din theo phng trnh sau:

an, bn: cc thng s ca h thng, gi s l khng i x0: tn hiu ra ca thit b o xi: tn hiu o Khi a0, b0 0, cc gi tr a, b khc = 0. Phng trnh vi phn cn li

p ng ng bc nht Trng hp trn cho thy i lng vo v i lng ra khng ph thuc vo thi gian, l iu kin l tng ca trng thi ng. V d: s thay i v tr tuyn tnh con chy bin tr tuyn tnh theo i lng o

p ng ng bc nht: Khi cc gi tr a1, b1, a0, b0 0 v cc gi tr khc = 0

Thit b no tha phng trnh ny c gi l thit b bc nht Chia 2 v phng trnh cho a0, ta c:

p ng ng bc nhthay

Vi Hm truyn ca thit b o bc nht: V d: nhit k thy ngn

p ng ng thit b bc hai

Phng trnh: Thit b o bc nht o cc i lng c nng lng

V d: lc k c nng lng l c nng, nhit k c nng lng l nhit nngV d: nng lng tnh in v t in trong mch LC.

Thit b o bc hai c s trao i gia hai nng lng

Phn tch thng k phn tch thng k c ngha cn phi tin hnh nhiu php o v yu cu sai s h thng phi nh hn sai s ngu nhin. ngha ca tr trung bnh s hc: Gi tr gn ng tt nht ca 1 i lng o chnh l tr trung bnh s hc ca nhiu ln o khc nhau. Tr trung bnh s hc ca n php o cho bi biu thc:

lch: sai bit gia mi gi tr o vi tr trung bnh

Phn tch thng k lch trung bnh: dng kim tra tnh chnh xc ca thit b o. lch trung bnh cng nh, thit b cng chnh xc. lch trung bnh:

lch chun: Vi s ln o (n < 30) th lch chun c dng:

Nu gim c lch chun th s ci tin hiu qu o lng

Phn tch thng k Sai s ngu nhin: tnh trn c s ng phn b Gauss ca lch chun.

Gii hn ca sai s ngu nhin: Tr s no c lch vt qu sai s ngu nhin s b loi b.

Gii hn ca sai s: cc nh sn xut thng xc nh sai s ca thit b bng sai s thang o.

V d: vn k c sai s thang o 2% thang o 300V. Tnh gii hn sai s o in p 120V. Sai s thang o: 2% x 300 = 6V Gii hn sai s 120V: 6/120 x 100% = 5%

V d 2: Vn k v ampe k c dng xc nh cng sut tiu th ca in tr. C hai c sai s thang o l 1%. Nu vn k c c thang o l 150 V c ch th 80V v ampe k c c thang o 100mA ch th 80 mA. Tnh gii hn sai s ca cng sut o c. Gii: Gii hn sai s ca thang o vn k: 150V x 1% = 1.5 V Gii hn sai s tr s 80 V: 1.5/80 x 100% = 1.86% Gii hn ca sai s thang o ampe k: 100 mA x 1% = 1 mA Gii hn sai s tr s 80 mA: 1/80 x 100% = 1.43% Gii hn sai s ca cng sut thu c: 1.86% + 1.43% = 3.29%

Nhng phn t trong thit b o in t

Cm bin: bin cc i lng khng in sang i lng in B ch bin tn hiu: bin i tn hiu in cho ph hp vi b ch th kt qu: mch iu hp, mch khuch i, mch cu o (in tr, in cm, in dung), mch lc, mch chnh lu , ADC, DAC B ch th kt qu: hin th di dng kim hoc s

Li ch ca in t trong o lng Cc tn hiu in c cc mch in t bin i cho ph hp vi mch o, mch thu thp d liu o lng u im ca mch in t:

nhy thch hp Tiu th nng lng t Tc p ng nhanh D tng thch truyn tn hiu i xa tin cy cao linh hot cao ph hp cc vn o lng

H thng o lng analog

H thng o lng s

iu khin t xa thit b o lng