Do Luong Dien Tu Cua DU QUANG BIONH

8/14/2019 Do Luong Dien Tu Cua DU QUANG BIONH

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KHOA IN T - VIN THNGB MN IN T

O LNG IN T

Bin soan: D Quang Bnh

A NNG 2000


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BIN SON DQB, B/M TVT-HKT CHNG I: PHP O V K THUT O IN T

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CHNG 1: PHP O V KTHUT O IN T

o lng in t l phng php xc nh tr s ca mt thng s no mt cukin in t hay h thng in t. Thit b dng xc nh gi trc gi l "thit bo", chng hn, ng ho nhiu chc nng [multimeter] dng o tr s ca in tr,in p, v dng in trong mch in.Kt quo tu thuc vo hn ch ca thit bo. Cc hn ch s lm cho gi tro

c (hay gi tr biu kin) hi khc nh vi gi trng (tc l gi tr tnh ton theothit k). Do vy, quy nh hiu sut ca cc thit bo, cn phi c cc nh ngha v chnh xc [accuracy], r [precision], phn gii [resolution], nhy [sensitivity]v sai s [error] .1.1 CHNH XC [accuracy]. chnh xc schmc gn ng m gi tro c st so vi gi trng cai lng cn o. V d, khi mt tr s no c c trn ng h o in p[voltmeter] trong khong t 96V n 104V ca gi trng l 100V, th ta c th nirng gi tro c gn bng vi gi trng trong khong 4%. Vy chnh xc cathit bo s l 4%. Trong thc t, gi tr 4% ca v d trn l ' khng chnh xc

php o' ng hn l chnh xc, nhng dng biu din trn ca chnh xc tr

thnh chun thng dng, v cng c cc nh sn xut thit bo dng quy nh khnng chnh xc ca thit bo lng. Trong cc thit bo in t s, chnh xc bng 1 sm cng thm chnh xc ca khi pht xung nhp hay ca b gc thi gian.a) chnh xc ca lch y thang.Thng thng, thit bo in t tng t thng c chnh xc cho di dng phntrm ca lch ton thang o [fsd - full scale deflection]. Nu o in p bng ng ho in p [voltmeter], t thang o 100V (fsd), vi chnh xc l 4%, ch th soin p l 25V, so s c chnh xc trong khong 25V 4% ca fsd, hay (25 - 4)Vn (25 + 4)V, tc l trong khong 21V n 29V. y l chnh xc 16% ca 25V.iu ny c gi lsai sgii hn.V d trn cho thy rng, iu quan trng trong khi o l nn thc hin cc php o gnvi gi tr ton thang o nu c thc, bng cch thay i chuyn mch thang o. Nukt quo cn phi tnh ton theo nhiu thnh phn, th sai s gii hn ca mi thnh

phn sc cng vi nhau xc nh sai s thc t trong kt quo. V d, vi intrR c sai s 10% v dng inIc sai s 5%, th cng sutI2R s c sai s bng 5+ 5 + 10 = 20%. Trong cc ng h s, chnh xc c quy nh l sai sgi troc 1 ch s. V d, nu mt ng h c kh nng o theo 3 ch s hoc 3 ch s,th sai s s l 1/103 = 0,001 = (0,1% + 1 ch s).b) chnh xc ng v thi gian p ng.Mt s thit bo, nht l trong cng nghip dng o cc i lng bin thin theothi gian. Hot ng ca thit bo cc iu kin nh vy c gi l iu kin lm

vic ng. Do vy, chnh xcngl gn ng m gi tro c s bng gi trng m n s dao ng theo thi gian, khi khng tnh sai s tnh.Khi thit bo dng o i lng thay i, mt thut ng khc gi l p ng thi

gianc dng ch khong thi gian m thit bo p ng cc thay i ca ilng o. tr hon p ng ca thit bo c gi l tr[lag].1.2 R [precision]. r ca thit bo l php o mc ging nhau trong phm vi mt nhm cc sliuo. V d, nu 5 php o thc hin bng mt voltmeter l 97V, 95V, 96V, 94V, 93V, thgi tr trung bnh tnh c l 95V. Thit bo c r trong khong 2V, m chnhxc l 100V - 93V = 7V hay 7%. r c tnh bng gi tr cn trung bnh bnh phng ca cc lch. v d trn,


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2

cc lch l:+ 2, 0, + 1, - 1, - 2. Nn gi tr lch hiu dng l:

25

41104=

++++

Do mc trung bnh sai lch l 2. Nh vy, r s phn nh tnh khng i (hay khnng lp li - repeatability) ca mt s kt quo, trong khi chnh xc cho bit lch ca gi tro c so vi gi trng. r ph thuc vo chnh xc. chnh

xc cao hn s c r tt hn. Nhng ngc li s khng ng. chnh xc khngph thuc vo r. r c th rt cao nhng chnh xc c th khng nht thit lcao. Khi chnh xc gn lin vi lch thc t ca ng ho (hoc s hin th thctng h s), th r gn lin vi sai ssc ca gi tro. Sai s nh vy cth tng ln do th sai cc ng ho tng t hoc khng n nh cc b ch th s.1.3 PHN GII [resolution]. phn gii l s thay i nh nhtcc gi tro c (khng phi l gi tr 0) mmt thit bo c thp ngcho mt so xc nh. phn gii thng l gi trvch chia nh nht trn thang o lch. Nu mt ammeter c 100 vch chia, th i vithang o t 0 n 1mA, phn gii s l 1mA/100 = 10A. cc ng ho s,

phn gii l 1 ch s. phn gii cn phi c cng thm vi sai s do so nm

trong khong gia hai vch chia ln cn khng thc mt cch chnh xc. phn giicng c phn nh theo sai s ca r ngoi cc yu t khc nh th sai.1.4 NHY [sensitivity]. nhy l t sca thay i nh nhtp ng ra ca thit bo theo thay inh nhti lngu vo. V d, nu lch y thang ca mt ammeter A cho

bng 50A, v bng 100A ammeter B, th ammeter A nhy hn so vi ammeter B. nhy c th hin cho voltmeter di dng ohm / volt. Mt ng ho c lchy thang (fsd) l 50A s c in tr l 20 000 mc ni tip cho fsd mc 1V,trong khi mt ng h c fsd l 100A s c in trl 10 000 cho fsd mc 1V.Vy voltmeter 20 000/V c nhy cao hn so vi voltmeter 10 000/V.a) Ngng nhy.

Ngng nhy l mc tn hiu nh nht c thc pht hin di dng c nhiu vtp m. Cc tn hiu rt nh c th ln trong tp m, do vy khng th tng nhy camt h thng o v cng. Thng thng s dng php o i vi ngng nhy l

bin ca tn hiu vo m t s tn hiu trn nhiu bng n v hoc 0dB.b) Yu cu rng bng tn. rng bng tn chn lc c dng ci thin mc ngng. Khi tn s nhiu caohn ph tn ca tn hiu cn o, th phi s dng mch lc thng thp tn hiu truynqua vi mc nhiu khng ng k. Nu nhiu c tn s thp hn ph tn ca tn hiu o,th s dng b lc thng cao. T hp b lc thng thp v b lc thng cao s suy ra rng bng tn chn nhiu. Nu nhiu chim rng trong phm vi ph tn ca tn

hiu cn o, th b lc chn c th nn nhiu cng vi mt phn nh tn hiu o.1.5 CC LOI SAI S [errors].Mi thit bo c th cho chnh xc cao, nhng c th c cc sai s do cc hn chca thit bo, do cc nh hng ca mi trng, v cc sai s do ngi o khi thu nhncc s liu o. Cc loi sai s c ba dng: Sai s th, sai s h thng, sai s ngu nhin.a) Sai s th.Cc sai s th c th quy cho gii hn ca cc thit bo hoc l cc sai s do ngi o.Gii hn ca thit bo. V d nhnh hng qu ti gy ra bi mt voltmeter c nhy km. Voltmeter nh vy s r dng ng k t mch cn o v v vy s t lmgim mc in p chnh xc. nh hng do qu ti sc gii thch chi tit mc 1.7.

Sai sdo c.L cc sai lch do quan st khi c gi tro. Cc nhm ln nh vy c


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th do th sai, hay do nh gi sai khi kim nm gia hai vch chia. Cc thit bo skhng c cc sai s do c.b) Sai s h thng.Sai lch c cng dng, khng thay i c gi l sai s h thng. Cc sai s h thng chai loi: Sai s do thit bo v sai s do mi trng o.

Sai sca thit bo.Cc sai s do thit bo l do ma st cc b phn chuyn ng ca h thng o hay do

ng sut ca l xo gn trong c cu o l khng ng u. V d, kim ch th c thkhng dng mc 0 khi khng c dng chy qua ng h. Cc sai s khc l do chunsai, hoc do dao ng ca ngun cung cp, do ni t khng ng, v ngoi ra cn do sgi ho ca linh kin.

Sai sdo mi trngo l sai s do cc iu kin bn ngoi nh hng n thit botrong khi thc hin php o. S bin thin v nhit , m, p sut, t trng, c thgy ra cc thay i v dn in, r, cch in, in cm v in dung. Binthin v t tnh c th do thay i m men quay (tc lch). Cc thit bo tt s chocc php o chnh xc khi vic che chn cc dng cn mc ti a, s dng cc mnchn t trng, v. v. . . Cc nh hng ca mi trng o cng c th gy ra dchchuyn nhkt qu, do thay i nh v dng in.

c) Sai s ngu nhin.Cc sai s ngu nhin do cc nguyn nhn cha bit, xut hin mi khi tt c cc sai sth v sai s h thng c tnh n. Khi mt voltmeter, c hiu chun chnhxc v thc hin php o in p cc iu kin mi trng l tng, m ngi o thyrng cc so c thay i nh trong khong thi gian o. bin thin ny khng thhiu chnh c bng cch nh chun, hay hiu chnh thit bo, m ch bng phng

php suy lun cc sai s ngu nhin bng cch tng s lng cc php o, v sau xcnh gi tr gn ng nht ca i lng cn o.1.6 GII HN CA THIT BOMt thit bo c th c cc gii hn v thang o, cng sut (hay kh nng ti dng), tns, trkhng v nhy (nh hng qu ti). Cc vn c gii thch nh sau.

- Gii hn v thango. Mi thit bo c khong o ln nht v mt thng s cn o.Khong o sc chia thnh cc thang o nh thch hp. V d, mt voltmeter c tho cao nht l 300V chia thnh 5 thang o ph: 3V, 10V, 30V, 100V v 300V.Chuyn mch thang o s thit lp ti cc v tr chnh xc tu thuc vo gi tro yucu. Gi s php o in p l 9V th chng ta s s dng thang o 10V. Cc thang ocn phi c cho tt c cc thng s cn o. Cn phi chn thang o ng cho mi thngso thch hp. Nu o in p trn thang o dng in, th ng ho s h hng.- m rng thango. L thut ngc s dng ch s chnh lch gia gi tr lnnht v gi tr nh nht ca mt thang o. i vi gi tro ca ng hmc nh nhtl 10mA v 100mA mc cao nht, th m rng ca thang o l 100mA - 10mA =90mA. Mt ng ho in p c mc 0V gia, vi + 10V mt bn v - 10V pha

khc, s c mrng thang o l 20V.- Gii hn vcng sut. Mi thit bo u c kh nng x l cng sut ln nht, nncng sut ca tn hiu vo khng c vt qu gii hn cng sut o. Cng sut vtqu c th lm hng ng ho hay mch khuych i bn trong ng ho.- Gii hn v tn s. Phn ln ccu ng ng ho tng t c vai tr nh mtin cm mc ni tip v do vy s suy gim di tn s cao. Trong cc thit bo sdng cc mch chnh lu v cc mch khuych i, cc in dung ca tip gip c chol mt hn chi vi tn hiu o di tn s cao.Ccu o in ng c th chc s dng o tn hiu c tn s ln n 1000Hz (doin cm ni tip), cc ccu o tin (c b chnh lu) c th s dng o tn hiu


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c tn s ln n 10 000Hz, millivoltmeter xoay chiu c tho cc tn hiu c tn s lnn mt vi MHz. Cc hn ch tn s khc c th gy ra do cc in dung song song.My hin sng c th s dng o cc tn hiu c tn sdi megahertz, nhng githnh s tng khi cn rng bng tn cao hn. My hin sng khng s dng cun dyv h thng ch th kim, do vy nh hng bt li phn ln cc ccu o sc hnch v loi b.- Gii hn vtrkhng. Cc thit bo c dng o cc tn hiu ac, c trkhng ra

ph thuc vo mch ra ca transistorc s dng. Mt my pht tn hiu tn s cao cth c trkhng l 75 hay 50 ph hp vi trkhng vo ca h thng cn o. Ccthit bo in p nh voltmeter v my hin sng c trkhng vo cao. Mt voltmetertt va phi c th c trkhng vo khong 20000 / V, trong khi mt my hin sng vng ho s hay ng ho in t c th c trkhng vi megohm. Thit bo inp c trkhng cao hn s cho chnh xc ca php o cao hn, hay c nh hng quti t hn. Trkhng ca cc ccu o cun dy ng tu thuc vo nhy ca ngh, cn trkhng ca my hin sng kiu ng tia ph thuc vo trkhng vo ca bkhuych i dc s dng trong my hin sng.1.7 NH HNG DO QU TInh hng do qu ti c ngha l ssuy gim vtr sca thng smch cn o khi

mc thit bo vo mch. Thit bo s tiu th cng sut t mch cn o v s lm tica mch cn o. in trca ng ho dng s lm gim dng in trong mch cno. Tng t, mt voltmeter khi mc song song vi mch c in trcao, thc hin vaitr nh mt in trsong song [shunt], nn s lm gim in trca mch. iu ny tora mc in p thp trn ti c c trn ng ho. Do , ng h s ch th mcin p thp hn so vi in p thc, ngha l cn phi ly mc in p cao hn c lch ng. Nh vy, nh hng do qu ti s hn ch nhy v do cng c gi l

gii hn nhy. nh hng qu ti sc biu hin ng ho in p [voltmeter]nh sau.

Cho in trti l RL v ni trca ng h lRM. Cng vi mt in trmc ni tipvi tiRL lRS (hnh 1.1). in p thc t trnRL l VL khi khng mc ng ho vomch, v VM l in p trn ti khi c ng ho c tnh theo phng trnh (1.1) v(1.2) tng ng.

LS

LL

RR

REV

+

= (1.1)

)//(

)//(

MLS

MLM

RRR

RREV

+

= (1.2)

nh hng qu ti tnh theo phn trm c th tnh bng (VL - VM) x 100 / VL, nhv d1.1 v 1.2.V d 1.1:Vi hai ng ho in p, mtng h c nhy l 20 000/V, v ngh cn li c nhy l 1000/V, o in p trn RL trong mch hnh 1.2, trn thango 10V ca ng h. Tnh sai sdo qu ti cho c hai ng h.


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Trng hp thnht: =

= k3

200300

200100// ML RR

in p thc t khi cha c ng h = 9,1V11

100110

10010==

in p o c = 8,7V23

200

3

20010

3

20010

==

+

, Vy, sai s theo phn trm l 4,4%

Trng hp th2:in p thc t l 9,1V (nh tnh trn)=

= k

11

100

110

10100// ML RR

in p o c = 4,8V21

100

11

10010

11

10010

==

+

, Vy, sai s theo phn trm l 47,3%

V d 1.1, l i vi ngun in p hng. V d 1.2, cho thy nh hng khi ngun cungcp cho ti l c cung cp t mt ngun dng hng.V d 1.2:Mt ngun dngin khngi scung cp dngin l 1,5mA cho ti intrl 100k. Tnh in p ng v in p gn ng trn ti khi sdngng ho c

in tr l 1000 / Vo in p trn thango 100V. Tnh sai sdo qu ti theophn trm.in p ng = 1,5mA x 100k = 150Vin trca ng ho = 100V x 1000/V = 100kin trtng ng = 100k // 100k = 50kin p trn in tr50k = 1,5mA x 50k = 75VVy in p o c = 75VSai s % do qu ti = (150V - 75V) x 100 / 150V = 50%1.8 CAN NHIU PHP O.So vi tp nhiu bn trong c to ra bi cc gn sng ca ngun cung cp, hay bng

s di chuyn ln mt cch ngu nhin v c s lng v vn tc ca cc in t trongcc cu kin chng v thng (gi l nhiu Johnson hay nhiu trng, nhiu vch),hoc do cc qu trnh qu gy ra bi s gim t ngt thng lng qua mt in cm,cc thit bo c th b can nhiu t bn ngoi c gii thch nh sau.1. Can nhiu tn sthp. Khi cc dy dn in ngun cung cp chnh ac chy song songgn vi cc u dy tn hiu o, th nhiu mnh ac (tn s 50Hz) s can nhiu vo u tnhiu o do hiu ng in dung gia cc dy dn.2. Can nhiu tn scao. Cc tn hiu tn s cao c to ra bt c khi no c s pht ratia la in vng xung quanh thit bo. Tia la in c th to ra khi chuyn mchngun cung cp, do cc h thng nh la, do cc ng cin mt chiu, do cc myhn, do s phng in ho quang (tc s ion ho khng kh gn cc mch in p cao),


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v do h quang in trong cc n hunh quang. Tia chp l cc ngun tn s cao trongt nhin. Pht thanh qung b t cc i thu pht v tuyn v cc i pht thanh di ngcng sut cao, c lp t gn cc thit bo cng to ra cc tn hiu tn s cao. Cc tnhiu cao tn u c th can nhiu vo thit bo, cc tn hiu cao tn c thcchnh lu bng cc cu kin bn dn c trong cc thit bo, v nh vy s tc ng ncc kt quo do in p khng mong mun th hin di cc dng khc nhau trong

php o, lm cho kt quo sai hon ton. Mt s php o dc tin hnh cc im o

trong mch c cin p dc v in p ca cc tn hiu tn s cao. Cc php o in pdc s khng chnh xc nu khng lc bin p cao tn trc khi tn hiu o c chnhlu trong thit bo.Cc cch phng nga v khc phc cc php o loi b can nhiu cao tn.1. Trc tin l bao bc c hiu qu thit bo khng b can nhiu ngoi trc tip vothit bo.2. Thit bo phi c ni t.3. Cn phi lc cc tn hiu khng mongmun ti mch vo, dy o v dy nguncung cp cc tn hiu cao tn sc lc

b trc khi chnh lu, phi c mch chn

bng tn tn hiu o loi b nhiu v cannhiu tn s cao. Mch ni vi b my cnphi m bo. Mi hn b nt hay thiu ktni, s to ra mt in tr gia u vo vt i vi cc tn hiu tn s cao, nn inp cao tn s xm nhp ti u vo nh minhhohnh 1.3. Tin trong hnh 1.3, dng lc b cc tn hiu cao tn, c vai tr nh mt ngn mch i vi tn s cao. Nu thmch, hay im G khng kt ni vi t (do p lc no hay mi hn b nt), th tnhiu tn s cao s c ti im A sc a n u vo ca mch khuych i bngtransistor, nn sc khuych i v chnh lu (phn phi tuyn ca c tuyn) v s c

ti u ra di dng in p dc. Cc i pht thanh qung b a phng thnh thongnghe c trong ng nghe in thoi do can nhiu .4. Khi thc hin php o dc ti im c cin p dc cng nhin p cao tn, in pcao tn c th gy ra mc dng in ln chy qua u que o bi v u que o gn nhc ngn mch vi b my i vi tn hiu cao tn thng qua nh hng in dung, cth lm nng u que o (thc t ny xy ra khi o cc in p dc trong my pht). Mcni tip cun cm RF vi u que o loi b tnh trng trn.5. S dng mch khuych i thut ton ch vi sai s lm gim cc tn hiu nhiung knh rt cbn, c th loi b nhiu ng knh ln n mc 100dB. (Nu mc d c cc d phng nhiu cao tn trn, h hng h thng c th t tng ny n tngkhc, th nguyn nhn c th l v bo v, ni t, mch lc v cun cm cao tn, cn

phi kim tra k cc vn ).1.9 V BO V.V bo v l lp chn bng vt liu dn in c lp phn c tn hiu nhiu. Hiu quca lp bo v tu thuc vo: (i) kiu lp bo v, (ii) cc c tnh ca vt liu lm lp

bo v v (iii) hca lp bo v.Trng nhiu c th l in trng hoc t trng. Cc lp bo v bng t tnh s dngvt liu st t nh st. Cc lp bo v tnh in s dng vt liu dn in khng nhim tnh nhm. Cc vt liu dn in c c tnh in mi km nn s hp th cc nhiu doin trng tnh. Ngoi vic hp th, nhiu cng s gim do s phn x ca in trngkhi lp bo v. hp th nhiu t l vi dy ca vt liu. S phn x s xy ra khi


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c gin on trkhng c trng gia lp bo v v mi trng xung quanh lp bo v.1.10 NI TC ng dn trli mc t trn bng mch in, thng l ng mch rng v c intr rt thp. Dy tn hiu cn phi c t gn vi ng ni t gim nh hngin cm. ng mc t trn mch bng mch sc ni vi ng t hiu dng.Mc t, nh mch hnh 1.4, l khng ng, bi v in p c bc l trn chiu di

Zp do phn t II sc ni trli phn t I. nh hng s xu nu phn t I c nhy

cao, hoc nu phn t II l thit b cng sut ln.

Cc cch ni t nh mch hnh 1.5a, v 1.5b, l thch hp, nht l i vi tn hiu ctn s trn 10MHz, nu ch chn trnh vic hnh thnh cc vng t.

1.11 SO SNH THIT BO TNG TV THIT BO S.Cc thit bo tng t s dng lch ca kim ch th do tng tc gia dng in vt trng, hoc gia hai t trng. a s cc b phn ccu ng u c ma st, nn c

nhiu hn ch (nh gii hn tn s cao, nhy, sai s do qu ti) v cc sai s. Trongcc ng ho s, khng lin quan n s lm lch, s ch thc c b hin th(hin th bng tinh th lng hay bng LED), nn cc ng ho s khng c cc sai snh ca cc ng ho tng t.Cc u im ca thit bo s so vi cc loi ng ho tng t nh sau.a) u im ca ng ho s so vi ng ho tng t.1. chnh xc cao (thng dng l 0,0005% hay 5ppm)2. r cao (khi s lng o c th hin bng ch s, nn s khng thay i gi trca n) (in hnh l 1ppm).3. phn gii tt hn (tnh trng khng r rng ch b gii hn nhiu nht l mt chs).

4. Khng c sai s do th sai.5. Khng c sai s do c. Khng c sai s trong vic chuyn i s liu o.6. Trkhng vo rt cao (in hnh l 10M v in dung vo thp l 40pF) v v vysai s do qu ti khng ng k.7. Trkhng vo hu nh khng thay i trn tt c cc thang o.8. Snh chun t cc ngun mu bn trong ng h l hon ton n nh.9. Khng c sai s do dng sng tn hiu.10. Hin th cc tnh tng, c kh nng tng chnh 0 v tng chuyn thang o.Cc thang o thay i theo cc nc thp phn thay v thang o 10 , nn c s lngthang o t hn, kh nng mrng thang o ln hn.


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11. C kh nng x l so bng my tnh. Cc s liu o c thc lu tr v truysut bt k lc no.12. C kh nng x l cc tn hiu o di tn s rng hn.13. Thao tc o n gin, ch cn n nt n thit lp li tng chnh xc thit bocho cc s liu o mi.14. C kh nng kt hp nhiu thit bo vo mt thit b bng k thut s. C th lptrnh php o d dng.

15. Thit bo gn v kt cu chc chn hn.b) Cc nhc im ca ng ho s.1. Cn phi c ngun cung cp do s dng cc vi mch (IC).2. Cc i lng thay i chm, nh khi np t khng th quan st c. Cc ng htng t c th quan st cc bin thin nh khi o th tin phn.3. Khi o th diode khng th thc hin nh cch thng thng, nn c b sung mchchuyn dng dnh ring cho mc ch o th diode mt sng ho s (tc chcnng o mc st p trn tip gippn).4. Gi thnh cao, nhng gi thnh s gim xung theo s pht trin ca cng ngh chto cc IC mi.Vn cn nhiu tranh lun gia cc li th ca thit bo tng t so vi cc hin th s.

Tuy nhin, cc u im ca thit bo s c phn c ch trng hn cc loi thit botng t, nn thit bo s ngy cng trnn thng dng hn, nht l khi gi thnh cathit bo s gim xung. Trong cc h thng o rt phc tp, ccu o tng t ch thkim c th th hin bng hnh v trn my tnh ngoi hin th s.1.12 CHN KHONG O TNG V O TNGKhong o tng snh v du chm thp phn mt cch tng nhn c

phn gii ti u. Nu s ch th di 200, thit bo s 3 - ch s s tng cchuyn mch n thang o c nhy cao hn, cn nu gi tr hin th cao hn 1999, ththang o c nhy t hn tip theo sc chn. Bm v b gii m s thay i vtr du chm thp phn khi yu cu khong o tng.Mt ng ho tng hon ton ch cn tn hiu cn o c ti hai u vo ca ng h

o v iu chnh o thng s no, cn sau ton b cc tin trnh o (chnh 0, ch thcc tnh, thang o, hin th) sc tin hnh tng.i vi cc thit bo tinh vi, khuynh hng l kt hp nhiu thit bo vo mt thit

b. V d, b gim st thng tin c cc thit bo nh sau:

1. My to tn hiu RF 2. My to tn hiu AF3. ng ho cng sut RF 4. Voltmeter s5. ng ho cng sut AF 6. ng ho nhy7. ng ho h s mo dng 8. Bm tn s9. My phn tch ph 10. My hin sng nhs

Bt k thit bo no trong s cc thit bo trn c th hnh thnh hot ng theo lptrnh. Ch lm vic c chn, thit bo sc chn, loi php o yu cu c lp trnh theo lnh, nn tn hiu ra sc hin th hay c in, ton bc iukhin bng bn phm. Php o theo chng trnh trn my tnh cng gi l o tng.1.13 O TRONG MCH (ICT)Vic o th trong mch c tho th IC mc nh hay trung bnh m khng cn thoIC ra khi mch. im mu cht ca ICT l giao din BON. Cc u kp l cc u queo b giao tip s c bt gn c ti, ni chc chn n im cn o th.Chng trnh o th tng s cung cp d liu vo o th linh kin. V d, oth mt IC, bo th trong mch s truy xut bng trng thi cho IC t RAM ca thit


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9

bo th tng (ATE), v s so snh vi d liu ra ca IC cn o th vi bng trngthi chnh xc.1.14 K THUT O IN TPhp o cn phi c thc hin mt cch cn thn v s th hin cc s liu o phi

ph hp sau khi c tnh ton n cc gii hn v nhy, chnh xc v kh nngca thit bo. i khi so c thng nhng nu th hin kt qu sai, ngi ta c thhiu mch ang tt l c sai hng v ngc li. Hn na, vic s dng thit bo sai c

th to ra cc nguy him cho s an ton ca ngi o v thit bo. Cc k thut o sauy cn phi tun theo khi o th hay thc hin cc php o trong vic chn on hhng, sa cha v bo dng cc thit bin t.1. Ni thit bn ngun in li, tt hn ht l thng qua u ni ba chn, v thc hin

bt ngun cho h thng theo trnh t sau:Cc im quan trng c chuyn mch ON u tin, tip theo l ng [ON] ngun cungcp, sau ng [ON] thit bo, v cui cng ng ngun cung cp cho mch cn oth. Khi tt (chuyn mch sang OFF), th trnh t l ngc li, th trnh t phi c thchin ngc li: trc tin tt ngun cung cp cho mch cn o, tip theo l tt thit bo,sau tt ngun cung cp v cui cng l ngt in li. iu ny s bo v thit bov thit b cn o khi cc xung qu . Khng hn hay tho mi hn linh kin khi ngun

cung cp ang bt.2. Bt k lc no cng phi tt thit bo cn nu thit bo c chuyn mch sangng [on] ngay sau th cn phi c khong thi gian ng k cho php cc t trongthit x.3. Cc thit bo th cn phi c ni t mt cch hiu qu gim thiu cc binthin ca nhiu.4. Chn thang o ph hp theo tham s cn o, tu theo gi tro yu cu. Nu khng

bit gi tro yu cu, th hy chn thang o cao nht v sau o gim dn thang o choph hp, trnh cho thit bo b qu ti v b h hng. Thang o c chn cui cngs cho kt quo gn vi lch ln nht c th c i vi php o in p v dngin, v gn mc trung bnh i vi php o in tr, c chnh xc ti u i vi

h thng o.5. Khi gi tro bng 0, th ng ho cn phi ch th bng 0, nu khng th cn phic chnh 0 ph hp.6. Khng s dng cc u que o nhn c kch thc ln v chng c th gy ngn mch.Cc u que o cn phi nhn nht nu c thc.7. iu quan trng ca vic ni cc im o th: cc hng ch to thit b thng quynh cc im o th ti cc v tr thun tin trn bng mch in. in tr, mc in p dc,mc in p tn hiu v cc dng sng ca tn hiu sc quy nh cho mi im oth. (im o th thng l cc lp ng trn bng mch in). Cc im o th scm tt nht trnh nguy him qu ti cho mch cn o. Cc im o thc thit k

bi cc nh chuyn mn c kinh nghim, khi cn kho st thit b, khng c b qua

cc im o th nh vy trong qu trnh sa cha.8. Thng thng cc u que o mang du dng v m i vi cc php o in p vdng in trong mch. Ngun pin bn trong ng ho s c cc tnh ngc li, tc lu que o m ca ngun pin trong ng ho sc ni u que c nh dudng (que o mu en) v ngc li, nh th hin hnh 1.6. Thc t ny cn phi nhkhi o th cc diode, cc tin phn, cc transistor v cc vi mch.9. Nu cc im o th l khng cho trc, hoc nu cc php o l c thc hin ticc im khc nhau, th cn phi ch cc im nh sau:a) Khi o cc in p dc, php o cn phi c thc hin ngay ti cc linh kin thc t,v i vi vi mch o trc tip trn cc chn.


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b) S dng u kp o th IC thc hin cc php o trn cc chn ca IC.c) Khi cn o tn hiu trn mch in trong bng mch, nn kp u o trn chn ca cukin in tc ni vi ng mch in.d) Khi thc hin cc php o trn bng mch, cn phi m bo rng cc IC khng bin tch tnh do thit bo.e) Khi kim tra hmch, hy tho mt u ca cu kin in t ri thc hin php o.

Nu cu kin khng c tho mt u, th cc cu kin khc mc song song vi cukin nghi ngs ch th khng ng tin cy. C th kim tra cu kin nghi ngbng cuo. Khi tho mi hn ra khi bng mch in l kh khn th c th ct ng mch in lin

quan, do d dng hn li vt ct hn so vi vic tho mi hn cu kin o ri hn li,nhng khi hn li vt ct, cn phng mi hn b nt khng xy ra.f) Vic tho v hn IC l mt qu trnh kh phc tp cn phi ht sc cn thn. Cn phitho mi hn cho IC o th ch khi xc minh chc chn cc php o trn bng mchcho thy IC thc s hng.10. Cn phi tun theo cc lu v an ton m bo an ton cho ngi o, thit bo.11. Cn phi tun theo cc ch dn t hng dn s dng thit bo th, cng nh trnhto th.12. Cn phi nghin cu k cch vn hnh thit bo thc hin php o v cn phitun theo tt c cc im lu c cp.TM TT NI DUNG CHNG 1.

Cc thit bo dng xc nh gi tr thng s ca mt thit b hay h thng in t.Cc thut ng chnh xc, r, phn gii v nhy dng quy nh mt thit bo.C th c cc kiu sai s khc nhau km theo trong cc kt quo l cc sai s th, ccsai s h thng v cc sai s ngu nhin.Thit bo c th c gii hn v thang o, nhy, tn s, trkhng, nh hng do quti v gi ho.Thit bo c th b can nhiu t bn ngoi do khng ni t thit bo, hay do khnglc tn hiu tn s cao. cc ng ho kiu tng t, do trng lng, s cn bng v ma st ca ccu okiu lch nn c hn ch v tn s, hn ch v nhy v cc sai s khc. i vi cc

thit bo kiu s, do khng s dng ccu o kiu lch, nn s c chnh xc cao, r cao, phn gii tt hn, khng c sai s do c, khng c sai s do dng sng vnh hng do qu ti khng ng k. Ngoi ra cn c cc u im khc v thang o vx l tnh ton kt quo tng ng ho s.o cc thng s mt cch chnh xc, cn phi tun theo cc lu nh trnh tng -mng, hiu chnh 0, ni t thit bo, chn thang o v cc tnh ng, v cc lu v an ton in thng thng.


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BIN SON DQB, B/M TVT-HKT CHNG II: OSCILLOSCOPES

11

CHNG 2: THIT BO V QUAN ST DNG TN HIU

Thit bo v quan st dng tn hiu hay my hin sng, gi tt l CRO [Cathode -Ray Oscilloscope], l thit bo in t rt a nng, dng o th trong cc h thngin t. My hin sng s hin th cc dng sng ca tn hiu trn mn hnh, nn c tho bin cng nh tn s ca tn hiu. V cbn, my hin sng dng o in p,nhng cng c tho dng in, nu dng in c bin i thnh in p khi cho

dng in chy qua mt in trcnh. Tng t, my hin sng c tho in trnudng in t mt ngun dng hng c chy qua in trcn o nhi vi DMM(chng 3). My hin sng hai vt c th dng so snh hai dng sng khc nhau, cnmy hin sng hai chm tia c th dng so snh cc thay i v pha lin quan haidng sng. My hin sng c thc s dng hiu qu quan st dng sng thc ttrn mn hnh v nh lng dng sng. My hin sng c th dng hin thilng bt k nu c th bin i c thnh in p. Do my hin sng l thit bo adng nn cng c s dng trong cc lnh vc o v quan st kh tng, sinh hc, y tv cng nghip.2.1 NGUYN L HOT NG CA MY HIN SNG TNG TMy hin sng bao gm ng tia cathode (CRT), v cc mch lm lch hin th dng

sng. Nguyn l hot ng ca ng tia cathode v cc mch lm lch c gii thch nhsau:a) ng tia cathode.B phn chnh ca my hin sng l ng tia cathode, l mt n pht xin t donhit cao bao gm mt sng in t, cc bn lm lch v mn hnh hunh quang. Ttcc bc trong v bng thu tinh, rt chn khng nhhnh 2.1.

Sng in t gm cathode c lm bng Vonfram sc t nng pht x cc int. S di chuyn ca cc in tc iu khin bi li iu khin c in p m hnso vi cathode. Cc in t di chuyn qua cc in trng to ra bi cc li hi t vli gia tc, to thnh tia hi t sc nt. Chm tia in t nng lng cao sp vomn hnh hunh quang, nn s lm cho cc phn t phosphor lo sng. phn loe cang thu tinh, c ph lp than ch c hai bn lp v thu tinh. Lp than ch bn trong sngn cc in t khi pht x th cp, cn lp than ph bn ngoi lm nhim v bo v, trnh s bc x nhiu tn hiu qut. Mt lp nhm mng cng c t gn st mnhnh chn cc ion dch chuyn khi s va chm mn hunh quang v cng dng

phn x nh sng trli pha mn hnh nhm ci thin pht sng ca tia sng. ng tiacn phi c in p vi kV (gi l i cao p hay in th EHT) t vo lp ph than ch


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12

bn trong. Cc li khc s ly cc mc in p dc thch hp tin p cao thng quamch phn p.S lm lch tia theo chiu ngang c c bng cch s dng tn hiu rng ca. S lmlch tia theo chiu dc nhtn hiu cn quan st. Cc mch iu khin lch tia myhin sng (ngoi ng tia), s dng cc transistor nn yu cu cc mc in p dc thp hot ng.b) Lm lch chm tia.

Nguyn l hotng ca b gc thi gian (lm lch ngang). Chm tia sc lm lchtheo chiu ngang bng cch p t mt in p rng ca (nhhnh 2.2a), ln cp bnlch (gi l cp bn lch ngang) theo kiu lm lch tnh in. Khi khng c in p tnhiu ln hai bn lch (im A' ca tn hiu rng ca hnh v), im sng do tia to ra tiim bt u A trn mn hnh. Khi mc in p ca bn lch bn phi tng dn so vi

bn lch bn tri, th im sng s di chuyn v bn phi nn ln lt qua n cc imB, C, D v E trn mn hnh, tng ng vi mc in p rng ca B', C', D' v E'. Sau in p rng ca s trv li mc 0 nn im sng s trli im A ban u.

Slm lch dc. Cp bn lch th hai gi l cp bn lm lch dc. Tn hiu vo cn osc t vo cp bn lch dc sau khi c khuych i. Do nh hng ca mcin p lch dc m chm tia in t s b lch theo chiu dc trong khong P v Q, nh

hnh 2.2b.Nh vy, chm tia s chu hai s lm lch ngang v dc ng thi, nn nh hng hpthnh l ti to li tn hiu c bin thay i theo thi gian, nh th hin hnh 2.2c.Khi s lm lch theo chiu ngang iu khin im sng t A n B, th lm lch dc sko im sng n P, nn sau khong thi gian AB, im sng khng phi ti B m l tiP. Tng t, sau khong thi gian AC im sng l ti C; sau khong thi gian AD, imsng l ti Q, v sau khong thi gian AE, im pht sng l ti E, v. v. . . . Do vy, cc

phn t pht quang APCQE s ln lt pht sng v hin th dng sng vo. li khc trn mt my hin sng s cho php o khong thi gian trn trc ngang (X), v bin trn trc dc (Y).

Xo tia qut ngc hay tia quay v.


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13

Tn hiu rng ca gim rt nhanh t gi tr ln nht v 0, gi l tia quay v, hay tia qutngc. Tn hiu qut ngc s khng c nhn thy trn mn hnh, nu khng th dngsng c hin th s tr nn mo dng ln. Do trong sut khong thi gian qutngc, ng tia sc gitrng thi ngng pht sng, gi l xo tia, bng cch cungcp mc in p m cho li iu khin so vi cathode.c) ng b.ng bc s dng th hin qu trnh lm cho dng sng n nh. Dng sng sn

nh nu tn hiu qut bt u ti gi tr 0 ca tn hiu vo. Gi s thi gian qut th hin5 chu k ca tn hiu vo, tip theo sau khi qut ngc, vt sng s phi bt u vi imu l chu k th 6 ca tn hiu vo. iu ny c th thc hin hoc bng s kch khi

b to dao ng qut mt trng thi bn lin tc vi mt xung t tn hiu vo, hoc bngmt tn hiu ngoi bt k, hay nu tn hiu qut tun hon th bng cch iu chnh mchqut da trn vic tinh chnh nh thi. Tinh chnh bin thin thi gian, c th thchin bng cch cung cp mt phn nh tn hiu vo cho mch dao ng to tn hiu quttun hon.S lng chu k dng sng c hin th trn mn hnh s tu thuc vo khong thigian cn thit im sng di chuyn tim bt u (im tn cng bn tri ca mnhnh) n im tn cng bn phi, v chu k (hay tn s) ca tn hiu vo. Nu khong

thi gian ca tn hiu rng ca bng mt na chu k (T/2) ca dng sng vo, th mtna chu k dng sng vo sc hin th. Nu thi gian qut ca tn hiu rng ccbng mt chu k ca tn hiu vo th ton b chu k sc hin th. Nu thi gian qutca tn hiu rng ca bng 2 chu k tn hiu vo th hai chu k sc hin th, v. v. . .Do vy, khi bit khong thi gian to vt theo chiu ngang v s lng chu kc hinth trn mn hnh, th c th xc nh chu k hay tn s ca tn hiu vo. Khong thigian to vt ngang s c ch th trn chc nng iu khin thi gian / vch chia[Time/Div], tnh theo n v ms/div hay s/div.d) nhy ca slm lch.Bin ca dng sng vo sc xc nh bng cch m s vch chia theo chiu dctrn mn hnh tnh n nh ca dng sng. Suy ra mt na s vch chia s l bin

nh ca dng sng cn o. Gi tr ca mi vch chia theo chiu dc sc ch trnchuyn mch iu khin h s khuych i dc theo mV/div hay V/div. Chuyn mchiu khin dc c gi l nhy ca s lm lch. nhy lch tu thuc vo cc intrphn p v h s khuych i in p ca mch khuych i dc.V d 2.1:Vi tn hiu vo dc l 100mV (nh - nh) t vo u vo. Mch phn p slm gim tn hiu vo mc mt phn 10 ti u vo ca mch khuych i dc c h

skhuych i l 40dB. Tnh mc in p thc ttheo vch chia trn mn hnh v v tr nhy dc nu tn hiu o chim 5 vch chia trn mn hnh.Tn hiu ti u vo ca mch khuych i dc = 100 x 1/10 = 10mVH s khuych i 40dB c ngha l s khuych i in p ln 100 ln.Do vy, tn hiu ti u ra ca mch khuych i dc = 10mV x 100 = 1000mV.

Mc tn hiu ny chim 5 vch chia, nn mc in p thc t trn mt vch chia l200mV, nhng thang nhy sc t mc 20mV / div, c gi tr ch thng l100mV (nh - nh).e) My hin sng vt n.S khi ca my hin sng vt n (theo kiu qut kch khi) nhhnh 2.3. Tnhiu vo thng qua mch suy gim (nh trong voltmeterin t) theo cc thang o khcnhau i vi php o bin . Tn hiu sau sc khuych i bi mch khuychi dc (khuych i - Y ), v sc cung cp n cp bn lch dc lm lch theochiu dc. Mch dao ng qut s to ra tn hiu rng ca v c khuych i cungcp n cp bn lch ngang. Khong thi gian ca tn hiu qut c iu khin bi


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14

mch iu khin gc thi gian nn gi tr ca khong thi gian theo vch chia scch trn chuyn mch iu khin nh thi trn mt my hin sng. i vi mt sngdng o (chng hn nh cc mu hnh Lissajous hay cc php o iu ch), cn phit theo v tr qut ngoi v do vy chuyn mch S2 s cung cp tn hiu qut trong hayqut ngoi n mch khuych i tn hiu qut theo yu cu.

gin nh dng sng hin th, cn phi c cc thi im khi u qut ti cng mtv tr ca chu k tn hiu vo, tc l m bo sng b, tn hiu vt ngang c to ra

bi xung kch khi ly t mch khuych i dc (khuych i - Y) s kch khi mchdao ng qut bng ba hi n n. Khi cn kch khi ngoi, hay kch khi bng tnhiu in ac 50Hz (gn bn trong my hin sng) cng c th s dng bng chuynmch S1.Dy tr dng b tr gy ra do s khi ng mch qut sau khi kch khi. V vy,

dy tr s lm cho tn hiu o v tn hiu qut n cc cp bn lch trong ng tia mtcch ng thi.f) Qut lp li.My hin sng s dng mch a hi chy t do nn khng cn tn hiu kch khi. Mchqut s nhn c tn hiu lp li theo mi chu k ca mch dao ng a hi. dngsng n nh, cn phi c sng b gia tn s qut v tn s ca tn hiu cn o. iuny c th thc hin bng cch thay i tn s qut nhvic iu khin nh thi.

Skhc nhau gia qut lp li v qut kch khi. Tn s v pha ca tn hiu qut lpli cn phi c ng b vi tn hiu vo to ra dng sng hin thn nh. Nu tns sai lch, th s hin th dng sng s khng n nh.Qut kch khi s hin th vt theo chu k thi gian quy nh v v vy sn nh m

khng lin quan n tn s tn hiu vo.g) My hin sng qut tr.Do xung kch khi, s khi u qut s b tr, nn s khng th quan st vt sng trnmn hnh trong mt khong thi gian no . Tn hiu bn lch dc l lin tc, nn mt

phn ca tn hiu cn o s b mt. Do vy, cng cn phi lm tr tn hiu. V tn hiukhng c t trc tip vo cp bn lch dc m phi truyn qua mch dy tr, to rakhong thi gian cn thit cho mch qut khi u ti cp bn lch ngang trc khi tnhiu cn o n cp bn lch dc. Nu tr tn hiu l 200ns, v sng qut b trkhong 80ns, th tn hiu cn quan st sc hin th theo tn hiu qut ng khi btu qut, nh th hin hnh v 2.4.


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h) My hin sng vt kp. my hin sng vt kp hay hai vt, mt mch qut n sc hin th ti hai v trdc khc nhau trn mn hnh theo tng chu k rng ca lun phin. S khi my hinsng hai vt nhhnh 2.5.

Chuyn mch in t S3 s chn tn hiu vo dc (Y1), c a n mch khuych idc (Y) trong mt khong thi gian no , cn tn hiu Y2 sc cung cp n mchkhuych i dc trong khong thi gian tip theo. Chuyn mch S1 s cho php kch khihoc bng tn hiu Y1 hoc bng tn hiu Y2, hay bng tn hiu ngoi, hay kch khi bngtn hiu mng in 50Hz. Cu trc mch nhi vi my hin sng vt n. Chuynmch tng c tc nhanh c hai tn hiu c th quan st mt cch r rng trnmn hnh (do lu sng ca cht hunh quang v lu sng vng mc mt). my hin sng hai vt, ch c mt mch qut, nn i vi mt chu k tn hiu qut siu khin s lm lch dc ca mt dng sng vo, v i vi chu k qut tip theo cacng mt mch qut, tn hiu qut siu khin s lm lch dc ca tn hiu vo th hai.Vy hai dng sng ca hai tn hiu vo ring sc hin th, nhng khng c hin th

ng thi. Do khng th so snh lch pha gia hai dng sng. so snh quan hv pha, cn phi c hai dng sng c hin thng thi ti cng thi im, tc l cth thc hin bng my hin sng hai chm tia.i) My hin sng hai tia. my hin sng tia kp hay hai tia c hai sng in t, hai cp bn lch dc v hai cp

bn lch ngang. Hai cp bn lch ngang c t song song v c iu khin bi cngmt mt qut ngang (mch gc thi gian), tc l cng mt tn hiu qut s xut hinng thi ti hai cp bn lch ngang. S khi ca my hin sng hai tia nhhnh2.6, trong c mt mch dao ng qut c trong hnh v, nhng cc my hin sngt tin c hai mch to sng qut.


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k) Dy que o ca my hin sng.Dy que o l cc u ni my hin sng n thit b hay mch in t cn o th. Ngoichc nng dy que o n gin bng cp ng trc thng thng, cn phi c cc dyque o dng ring cho my hin sng m bo tn hiu o trung thc nht.

Dy que o DC 10:1. Mch khuych i dc (Y) c trkhng vo khong 1M mcsong song vi mt t khong 50pF. Cp ng trc c th c in dung k sinh khong50pF. iu ny s gy ra qu ti rt ln i vi mch in t cn o c trkhng cao.Gii php hn ch s qu ti l mc mt in tr9M ni tip nhhnh 2.7. Mchcn o s xem in trvo ca my hin sng l 10M thay cho 1M, nhng tn hiuti cc u vo ca my hin sng bng mt phn mi tn hiu t vo. suy gim cthc b bng mch khuych i dc.

Dy que o 10:1 tn scao. Vn v dung khng thp ca t 100pF i vi tn hiu odi tn s cao c gii quyt bng cch s dng mt tin nh c tr sC= 1/10 sovi tin song song nh mch hnh 2.8. Trimmer (t xoay) 10pF mc ni tip vi indung 100pF c in dung thc t vo khong 9pF l tng i thp nn s khng gyqu ti cho mch ca h thng c tn s cao, nhng s lm gim mc tn hiu ac vokhong mt phn mi. Cc nh hng ca que o sc kim tra bng cch cung cpxung vung vo que o quan st dng xung vung trn my hin sng, s cho tn hiu

tn s thp (phn bng phng) cng nh tn hiu tn s cao (cc cnh).


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17

Dy que o tch cc u o tch cc gm mt mch FET lp li cc ngun theo kiumch lp li emitterBJT, c in trvo cao v in trra thp nn s loi b squ ti cho mch cn o khi ni que o ca my hin sng vo mch. u o cng chomc in dung rt nh, do vy s ci thin p ng tn s cao (khi cn o cc xung tngnhanh). Ngoi ra, u o tch cc c suy gim thp hn nhiu so vi u o 10:1. Do u o tch cc s dng hiu quo cc tn hiu nh. Tuy nhin, u o tch ccc gi thnh cao nn t c s dng. u o 10:1 c s dng ph bin hn.

Dy que o kiu tch sngu o s dng mch tch sng bng diode tch tn hiuiu ch ra khi tn hiu cao tn (RF) c iu ch, v cng s chnh lu tn hiusng mang cao tn (RF) thnh mt chiu (dc). Bin nh ca sng mang sc hinth theo dng sng c chng chp trn tn hiu dc. Nh vy, u o s lm vic nhmch pht hin tn hiu cc my thu thanh v my thu thng tin, trong tn hiu cth trong di vi megahertz. Khi dng u o kiu tch sng bin i cc tn hiu tn scao thnh di m tn, nn c th s dng my hin sng c rng bng tn thp.

Dy que o cm ng dng. u o cm ng dng gm mt vng li t c th kp cdy dn o c dng in nhmch hnh 2.9.

Dy dn c dng in chy qua cn o ng vai tr nh mt cun dy scp ca tn hiuxoay chiu. Cun dy qun trn li t lm cun th cp. Khi c dng dc chy qua, dngin t trong cu kin hiu ng Hall gim xung, tc l lm tng mc chnh lch inth sc khuych i cung cp n my hin sng.2.2 MY HIN SNG S - NH.My hin sng s c chc nng nh- DSO [Digital Storage Oscilloscope], l thit boc gi thnh cao v phc tp, nht l my hin sng c chc nng x l tn hiu, cho khnng tnh ton cc gi tr trung bnh, hiu dng [r.m.s], bin i Fourier v phn tch ph.Kiu my hin sng s dng vi x l khng cn thit trong cc dch v sa cha, mthng dng hn l kiu my hin sng khng c vi x l, xc nh cc h hng v cc

xung chp chn.My hin sng nh- s s dng ng tia cathode thng thng (khng phi kiu ng tiac chc nng nh). Cc mu dng sng sc lu tr trong b nh, v c th hin thtrn mn hnh ca my hin sng thng thng. S khi ca my hin sng nh- skiu khng x l tn hiu cho hnh 2.10.Tn hiu cn o t vo my hin sng sc ly mu theo tng khong thi gian un. Mi mc mu sc chuyn n b bin i tng t sang s (ADC) to ra cctn hiu logic nh phn tng ng vi mc bin ca tn hiu c ly mu. Tnhiu nh phn sc lu tr trong b nhnn c th s dng khi cn thit. Khi cnquan st, tn hiu nh phn sc a n b bin i s - tng t, bin i tn hiunh phn thnh dng tn hiu tng t ban u cung cp cho ng tia cathode. Cc tn hiu


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18

iu khin v nh thi s kch hot b nhbt c lc no khi yu cu ghi v c d liu.Ngoi ra, mch iu khin v nh thi s cung cp tn hiu nh phn cho b gc thigian bin i thnh tn hiu gc thi gian tng ta n cp bn lm lch ngang(H), to ra vt sng trn mn hnh.

2.3 NG DNG O BNG MY HIN SNGa) Sdng my hin sng pht hin sai hngMy hin sng l thit bo c nhy rt cao, chnh xc v khng gy qu ti cho hthng cn o, do khng c c cu o kiu quay. My hin sng s hin th dng sngthc t ca tn hiu vo, nn c th bit mch c khuych i v mo dng hay khngmt cch d dng. My hin sng c th dng o mc in p dc, kho st cc tnhiu xung, cc tn hiu rng ca, tam gic, sng sin v cc tn hiu c dng phc tpkhc. My hin sng c tho tn s ca cc b dao ng v cc b to xung nhp. Myhin sng vt kp c th kim tra hai tn hiu vo (trong trng hp cc mch op - ampv cc cng), cng nh kim tra tn hiu u vo v u ra trong mch in t. Do vy,my hin sng c s dng ph bin trong vic o th, sa cha cc mch khuych i,cc mch dao ng, cc my pht, my thu v trong cc h thng mch s.b) Cc chc nng iu khin trn mt my hin sng.Cc chc nng iu khin trn mt my hin sng thng thng gm:

iu khin cng tia [Intensity control] dng iu chnh sng ca vt. iu khin hi t [Focus control] dng iu khin sc nt ca vt sng. Astigmatism iu khin nh thi.iu chnh khong thi gian / vch chia ca mch dao ng

qut (gc thi gian). iu khin h skhuych i dc (Y) dng iu chnh bin ca dng sng

hin th theo chiu dc, trong khong t 5mV/div n 20V/div. iu khin h skhuych i ngang(H) dng iu chnh di ca vt theo

chiu ngang. iu khin qutdng chn mch qut trong hay qut ngoi. iu khin kch khi [Trigger control] dng chn xung kch khi t b

khuych i dc (Y), hoc t tn hiu in li hay tn hiu ngoi (i vi cc loimy hin sng hin nay c thm chc nng iu khin ng b).

iu khin mc kch khi, dng iu chnh mc ca xung kch khi. iu khin v tr ngang, dng iu chnh v tr ca dng sng hin th theo

chiu ngang. iu khin v tr dc dng iu chnh v tr ca dng sng hin th theo chiu

dc.


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19

nh chuntrongs cung cp tn hiu tn s 1kHz, bin khng i (thng l2Vpp), kim tra vic nh chun que o.

c) Sdng my hin sng.1. Khi cha bt chuyn mch ngun cung cp, t cc nm chc nng iu khin

hi t [focus], cng chm tia [intensity] v iu khin h s khuych i[V/div] v tr thp nht (tn cng bn tri), v cc chc nng iu khin v trdc v ngang v tr gn im gia.

2.

Tip theo l bt chuyn mch ngun cung cp chnh ca my hin sng.3. Sau khong thi gian khi ng ca my hin sng cho cathode cn phi ct nng hon ton, to ra cng chm tia yu cu.

4. iu chnh chc nng iu khin cng chm tia c vt sng r rng xuthin trn mn hnh. iu chnh chc nng iu khin v tr dc v ngang nu cn.(i khi h s khuych i ngang c th biu hin thnh vt sng nu im sng

bt u ngoi khung mn hnh). Khi im sng c th nhn thy, di chuyn imsng vo trung tm v iu chnh hi t, nho lm cho im sng gn.Chc nng iu khin cng tia cn phi c iu chnh im sng khngqu chi, hoc khng qu m.

5. t ch qut theo v tr qut trong [Int.], v iu chnh h s khuych ingang mrng im sng thnh ng sng y ngang trn mn hnh.6. Kim tra s di chuyn theo chiu dc ca ng sng ngang. Mch khuych idc nh chun c sn trong thit bo.

7. t u que o vo h thng cn o. Chuyn mch ngun ca h thng cn o bt[ON].

8. iu chnh chc nng iu khin h s khuych i dc c cao ca dngsng yu cu trn mn hnh.

9. iu chnh dao ng qut (gc thi gian) c s chu k cn thit trn mn hnh.i vi my hin sng c kch khi, chu k cn phi n nh.

10.Kho st dng sng, o bin v kim tra c tnh ca tn hiu.11. c cc mu hnh Lissajous, a tn hiu ngoi c cung cp t my to sng

n u vo qut ngoi, dng cho php o tn s v pha.d) Cc php o vi my hin sng.o in p ca tn hiu vo Gi trnh - nh ca in p c o bng cch m svch chia theo chiu dc gia hai nh. Chng hn, nu bin nh - nh ca dngsng chim 4 vch chia trn thang nhy 500mV/div, th tr s nh - nh l500mV/div x 4div = 2V, vy bin nh l 1V.o khong thi gian ca chu k Chu k ca tn hiu o c bng cch tnh s chu ktrn b gc thi gian. Gi tr gc thi gian c trong mt chu k s l chu k ca tn hiu.V d, trn thang o 50s/div, c 2 chu k tn hiu chim 4 vch chia, th s vch chiachim bi mt chu k l 2 vch chia, nn chu k tn hiu l 100s. Tnh nghch o cachu k s cho tn s ca tn hiu, trong v d s tnh c l 1/100s = 10kHz.

o tn stheo mu hnh Lissajouso tn s tn hiu theo mu hnh Lissajous thc hinbng cch a tn hiu c tn s cn o vo u vo dc, v ni tn hiu c tn s bitvo u vo qut ngoi, s thu c cc mu hnh khc nhau trn mn hnh tu thuc vot s ca hai tn s v lch pha ca hai tn hiu. Cc mu hnh Lissajous nhhnh2.11.Khi hai tn s bng nhau, lch pha bng 0o s to ra mt ng thng nghing 45o sovi ng ngang; vi lch pha 180o, ng thng s to mt gc bng 135o so ving ngang. Khi lch pha l 90o, s to ra mt ng trn. i vi cc lch pha

bt k khc s to ra cc hnh ellipse.Khi hai tn s tn hiu khng bng nhau, th t s ca tn s cha bit (fv) i vi tn s


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20

bit (c tn s trn my to sng) (fh) sc xc nh bng t s ca s lng ccvng theo ng ngang i vi s lng cc vng theo ng dc.o chsiu chca tn hiu AMKhi tn hiu iu chc p t lm tn hiu qutngoi, v tn hiu c iu ch lm tn hiu dc (Y) nh th hin hnh 2.12a. Mu

hnh sc hin th nhhnh 2.12b. Ch siu ch sc tnh bng (p q)/(p + q).

o mo ca xungXung vo v xung ra c thc hin th trn my hin sng haivt. vng hoc vt qu ca phn nm ngang, v tng hay gim ca cccnh xung c th quan st trn mn hnh. Thi gian tng (ng vi mc thay i t 10%n 90% bin xung) v khong thi gian gim (ng vi mc thay i t 90% n10% bin xung) c tho c trn mu xung. rng ca xung sc o trongkhong t t mc 50% ca cnh tng n mc 50% ca cnh gim.e) Cc im lu khi sdng my hin sng.1. Ni v my hin sng vi t.2. Cng chm tia in t cn phi gimc thp c th quan st thun li. imsng khng c lu ti mt v tr trn mn hnh. Trong trng hp cn phi gi domt l do no , th hy mc cng thp.

3. Nn bt u php o vi mc nhy nh nht mch khuych i dc v tng dncho n khi t c mc thit lp thch hp.4. nh chun lch dc trc khi thc hin cc php o. C sn ngun in p trongmy hin sng cho vic nh chun.5. S dng que o ph hp khi thc hin php o trn cc tn hiu tn s cao, hay khi tnhiu vo qu ln.6. Khi tho my hin sng sa cha, hy cn thn c in p rt cao khong vikilovolt. Ngay c trong trng thi ngt chuyn mch ngun in li, cc t lc c inp cao c th gy nguy him cho ngi s dng, do vy t cn phi c x khi tinhnh cng vic trn my hin sng trng thi ct ngun.


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21

7. Phi cn thn khi s dng ng tia, h hng ngu nhin bt k s dn n hng mnhnh.8. Mn hnh pht quang c th pht x tia - x nh, khi cn thay th nn mua CRT tiuchun t nh sn xut c uy tn.


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O LNG IN T

BIN SON DQB, B/M TVT-HKT CHNG III: THIT BO IN TA NNG

22

CHNG 3: THIT BO IN TA NNG V CHUYN DNG

Thit bo in tc gii thiu trong chng ny l thit b cbn, rt cn thittrong vic ch to, sa cha, o th cc cu kin, mch in t v h thng in t. Srt bt li nu khng c cc thit bo o th mch, o gi tr ca cc thng s. Ccmc sau m t nguyn l cbn v ng dng ca mt s thit bo thng dng. Ccthit bo th BJT v thit b vc tuyn BJT sc gii thch mc 4.2, thit bo

th IC tuyn tnh v IC s sc m tchng 4.3 v 4.4 tng ng. Cc ch tiu kthut ca mt s thit bo gii thiu phn ph lc I.3.1NG HO KIU T- INa) Nguyn l hot ng ca ccu o kiu t- inng ho tng t thng dng trong o lng in in t trc y, s dng ccu cun dy di chuyn trong t trng ca nam chm vnh cu (PMMC), cn gi l ccu DArsonval, tc l ccu o kiu t - in. V cbn, ng ho kiu t - in lng ho dng mt chiu (dc), to nn bi cc thnh phn khc nhau nhhnh 3.1,vi ba b phn chnh l: (i) b phn to ra lc lm lch, (ii) b phn iu khin, v (iii)

b phn lm nht.

B phn to lc lm lch trong cc ng h t - in l tng tc gia t trng vdng in nh trong ng cin mt chiu. Khi cun dy mang dng c t trong ttrng, s to ra m men xon bngB xA xNxI(Newton-mt), trong B l mt tthng tnh theo Wb/m2,A l tit din ca cun dy tnh theo m2,Nl s vng dy trongcun dy, vIl dng in tnh theo ampere. M men s lm cho cun dy xoay. Dngin cao hn, s cho m men quay ln hn. Kim c gn trn cun dy, s di chuyntrn thang o. Cun dy qun trn mt khung nhm nh v c lp trn trc thng, khung dy c th xoay t do trong t trng u do m men quay.T trng u v mnh sc to ra bi nam chm hnh mng nga lm bng vt liut tnh.

B phn iu khin bao gm l xo c gn vo cun dy ng, cn li lc lm lch,nn s bng kx q, trong kl h s l xo (ty thuc vo cc kch thc v mm doca l xo), cn q l gc lm lch tnh theo . Khi lc iu khin bng vi m men xon,kim ch th s dng ti gi tr cn o. Khi dng in dng chy trong cun dy, lc xon

bng 0, l xo s bt u phc hi li v sa kim ch th v v tr mc dng bng 0.B phn lm nhtgm cc b to dng xoy khng kh, c vai tr n nh kim ch thti v tr ch th.b) ng ho dng in bng ccu t- inng ho kiu t - in v cbn l ng ho dng mt chiu (dc), c ch to cho lch ton thang ti cc gi tr dng thp, 1mA hoc thp hn (50A). Tuy nhin,ccu o c th dng o cc mc dng cao bng cch s dng cc in trc tr s


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23

thp mc song song vi cun dy ng gil cc in tr shunt. Gi s ta mun odng 100mA bng ng ho c lchton thang l 1mA, th in trshunt phic tr s sao cho mc dng 99mA chyqua shunt v ch 1mA chy qua cun dyng, nh th hin mch hnh 3.2.

Tr s in tr ca shunt c th tnh tphng trnh (3.1).

ShM

ShTM

RR

RII

+

= hay

MT

MMSh

II

RIR

= (3.1)

Trong ,IT l dng ton b,IM l dng c php chy qua ccu o,RM l in trca ccu o, vRSh l gi trin trca shunt. V d 3.1, cho cch tnh in trshunt.V d 3.1:in trca ccu o l 1000v dng c thchy qua ccu o ln nhtl 1mA. Gi tr caRShl bao nhiu cho php ng ho ch th 100mA ?Nu sdng cng ccu o o dng1A, th shunt ca ng h cn phi c l bao nhiu ?

11099

10001100

10001

MT

MMSh ,

II

RIR ==

=

=

Ccu o c thnh chun ch th mc dng 100mAthay cho 1mA khi mc shunt 10,1 vo mch o.Tng t, o mc dng 1A, cn phi c shunt vokhong 1 bng cch tnh nh sau:

1999

100011000

10001

MT

MMSh ==

=

=

II

RIR

ng h o c th c cc thang o dng khc bngchuyn mch n cc in trshunt khc nhau nhhnh3.3.V tr trng bn tri ca chuyn mch l thang o nh

nht (t 0 n 1mA) khi khng mc shunt vo php o.Cc v tr chuyn mch 2, 3, v 4 st in trR1,R2, vR3 mc song song vi ccu o cho cc thang cao hntng ng.Theo phng php trn, ccu o vn gi nguyn khngc shunt v tr thang o thp nht. Phng php o dngkhc l phng php shunt vn nng hay shunt Aryton.


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Shunt vn nng [shunt Aryton]Shunt vn nng gm hng lot in trc mc song song vi ccu o thng qua ccv tr ca chuyn mch thang o, nhhnh 3.4. v tr S-1 ca chuyn mch, shunt cang h lR1 +R2 +R3. v tr S-2, shuntR2 +R3 vR1 s trthnh mc ni tip vi ccu o. v tr S-3,R3 s song song cnR1 +R2 trnn mc ni tip vi ccu o. Vyshunt Aryton s hot ng theo hai cch. Th nht, dng r mch dng; th hai s lmgim nhy ca ccu o bng in trmc ni tip vi ccu o.

c) ng ho in p bng ccu o t- inng ho dng bng c cu o t - in cng c th s dng lm ng ho p[Voltmeter] bng cch mc ni tip mt in trln cng vi in trca ccu o. Gitr ca in trni tip c gi tr ln m bo ch mc dng chp nhn c chy quaccu o. Nu mc dng ca ccu o l IM v in p cn o l Vme Volt, gi tr cain tr ton bR (bng in trmc ni tip + in tr ca c cu o) sc tnh

bng phng trnh (3.2).

Mme I

R

V= (3.2)

V d 3.2: Ccu o t - in dng o 100V trn mt mch in, nu mc dng chyqua ccu o l 1mA, xc nh tr sin trmc ni tip. in trca ccu o l

1000.

1mA100V

Mme ===

RI

R

V, vyR = 100k, nn in trni tip = 100k - 1k = 99k.

Khi nhiu in trmc ni tip, c th chn bng mt chuyn mch c kt ni thitbo trthnh mt voltmeter nhiu thang o, nhhnh 3.5.d) ng ho in trbng ccu o t- in


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S dng ngun pin trong (pin kh), ccu o t - in c th dng lm ng ho intr[ohmmeter] o cc in trcha bit tr s nh mch hnh 3.6.Dng chy qua ccu o s chy qua in trcn o (RX). Gi tr ca dng in l lch ca kim ch th ca ccu o s ty thuc vo tr s ca in trcha bit. Thango ca ohmmeter c thnh chun v khc theo ohm (). Nu in tr qu ln,ngun pin c th khng cung cp do dng s qu nh, nn cn phi c ngun d

phng bng pin ln hn (E2 > E1) thc hin thng qua chuyn mch. Bin trR phi

c hiu chnh m bo rng khi in trcha bit bng 0 (tc l hai u que oc ngn mch vi nhau), ccu o phi ch th mc in trbng 0 ( lch ton b).Thang in trs th hin in trbng 0 ti lch y thang do in trbng 0 nghal mc dng ln nht chy qua ccu o. in trv cng ngha l khng c dng in,v l tn cng bn tri ca thang o (vch mc dng bng 0) phi c nh du bng trn thang o in tr. Cc thang o in trkhc nh thang 100, thang 10k, thang10M s c c bng cch s dng cc in trkhc nhau nh chuyn mch nhiuthang o nhhnh 3.7.o thang o in trthp nht, in trshunt phi l in trthp nht. i vi ccthang cao hn, phi tng tr s ca cc in trshunt. Theo hnh 3.7,R1 nh hn so vi

R2, vR2 nh hn so viR3, v. v. . .RZ l bin trchnh 0. Nu ccu o c lch y

thang l 1mA, RZ cn phi c iu chnh mch c dng 1mA khi ngn mch haiu que o vi nhau (tc l khiRX = 0).

e) Voltmet xoay chiu bng ccu o t- inCcu o t - in v cbn l ng ho dc. Nu a tn hiu xoay chiu (ac) nng h th kim ch th s dao ng xung quanh im 0 do qun tnh. Nn o in pac phi s dng mch chnh lu bng diode. Diode s chnh lu in p ac, bin i in

p ac thnh xung p mch dc. ng ho s ch th gi tr trung bnh nhin p dc.i vi b chnh lu bn k, mc in p dc trung bnh s bng vi Vm/ (trong Vm lmc in p nh ca xung p mch), cn i vi b chnh lu ton k, mc in p dctrung bnh l 2Vm/. Mc d kim ch th ca ng ho s lch ty theo tr s trung bnh,nhng thang o sc nh chun ch th gi tr hiu dng (rms) ca tn hiu ac.(vic nh chun theo cc mc tn hiu vo sng sin v do s ch th s khng nggi tr hiu dng i vi cc dng sng khc). Thng s dng mch chnh lu cu cho gi tr trung bnh cao hn, gn thp hn, v khng cn bin p im gia t tin,nh mch hnh 3.8.Cc in trR1,R2, v R3 c vai tr nh mch phn p. Cc diode ca mch chnh lu


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26

cu s chnh lu in p ac thnh dc. Mc in p dc trung bnh c to ra bng 90%ca tr s hiu dng (i vi b chnh lu bn k mc in p dc trung bnh bng 45%ca gi tr hiu dng). mch chnh lu cu s dng cc diode silicon, st p trn haidiode l 1,4V. Mc in p thc sc t ngang qua ccu o v in trnhn (RS).Chng hn, nu R1, R2 v R3mch hnh 3.8, l 9M; 0,9M; v 0,1M tng ng,in p a n mch chnh lu s l 10Vrms, nu in p t vo (nhc ghi ti ccv tr u cc ca chuyn mch) l 10V,; 100V; hay 1000V ngang qua mch phn p AB,

th tr s trung bnh dc ca in p chnh lu s l 2x10Vx1,4/ bng 9V. Sau khi tr stp 1,4V trn cc diode, in p dc thc ngang qua mch ccu o s l 7,6V nn cnphi c in tr7,6k k cin trca ccu o (i vi ccu o 1mA). Vy nhy ca voltmeterac khi dng mch cu l ch bng 76% ca nhy ca ccu o dc.(i vi mch chnh lu bn k, nhy s gim xung hn na n mc 38%)

f) ng ho dng xoay chiuChc nng o dng ac ch c mt sng ho. Dng in cn o chy qua mt intrcnh v o st p trn in trbng voltmeterac. in p ac s t l vi dng khiin trc tr s khng i. o dng ac, thng s dng mch bin i dng thnh p

bng IC op - amp. Trong mt sng ho gi thnh cao s dng cc b nhit ngu. Stp dc ngang qua tip gip ca nhit ngu s t l vi hiu ng nhit ty thuc vo cng hiu dng ca dng in. Do vy, so c gi trrms ca dng in bt k dngsng ca tn hiu.g) ng ho a nngKhi ccu o t - in hp thnh cc mch thng ammeter nhiu thang o, voltmeternhiu thang o, v ohmmeter nhiu thang o, ton b trong mt thit bo, th thit boc gi l ng ho a nng. ng ho a nng cng c gi l ng ho AVO(Ampere Volt Ohm). Khi s dng ng ho a nng thc hin cc php o cn phitun theo cc lu sau:

1. Chn chuyn mch thng so ng. Nu mun o in p, ng bao ging ho thang o dng in.2. Chn ng thang o ca mt thng so. Nu mun o gi trc cho l 80V,khng ng hthang o 0 10V, m ng ho thang o 0 100V.

3. Nu khng bit gi tr cn o, th hy ng ho thang o cao nht theothng so, v sau gim dn thang o theo cc nc gim dn cho n khi xcnh c thang o thch hp.

4. Thang o c chn cn phi c s ch th gn vi lch y thang (fsd) mcc thc i vi php o in p v dng in, v gn mt na thang o ivi php o in tr, bi v ng ho s cho sai s php o nh nht.

5. Nu kim ch th ca ng ho khng ti v tr 0 ngay khi khng c tn hiuvo, th phi hiu chnh bng b phn ckh ( cng ca l xo cn bng gn trn


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khung dy), c iu chnh 0 chnh xc.6. Khi o in tr, iu chnh bin trchnh 0 c lch y thang (fsd) khi ngn

mch hai u que o vi nhau.h) Sdng ng ho a nng d tm hhng.ng ho a nng thng c s dng o in tr, in p v dng in dc. Dnhin, i khi ng ho a nng cng c tho in p ac. Phn ln cc mch h hngc th xc nh c bng php o in p dc. Chng hn, trong mch hnh 3.9, nu in

trR hmch, th in p VC ti Cs bng 0. Nu cu kin (transistor) hmch, in pti Cs bng in p ngun cung cp. Nu cu kin b ngn mch, th in p ti Csbng 0. Khi o in p s th hin mt gi trin trhmch no , in trc thc kim tra bng chc nng o in trca ng ho a nng bng cch ngt kt nimt u in trra khi mch.

Chc nng o in trc th xc nh tin b r hay b ngn mch, hoc cun dy cb hmch hay khng. Cu kin bn dn c tho th bng cch o in trcc trngthi phn cc ngc hay phn cc thun ca tip gip bn dn. S thng mch khi thchin d mch c th kim tra bng ng h o in tr thang o th p nht caohmmeter.ng ho a nng l dng co thng thng, dng trong cc dch vo th, sacha do cch s dng n gin, cu trc chc chn, tng i chnh xc v khng yucu ngun cung cp ngoi, cng nh khng nh hng bi t trng k sinh.3.2 VOLTMETER S (DVM)Voltmeter s s dng nguyn l ca mch so in p tng t. Voltmeter s c ttc cc u im ca mch in t s khi so vi mch in t tng t.

a) Nguyn lSau khi mch suy gim cho vic chn thang o; tn hiu vo sc chuyn i thnhtn hiu s bi

b bin i tng t - s (ADC). Khi ADC c th s dng k thut tch phn n snhay hai sn dc. dng cbn nht, ADC s so snh tn hiu vo vi in p mu (cc

phng php nhn in p mu c th khc nhau). Ch cn in p vo ln hn so viin p mu, th tn hiu ra ca b so snh s cho mc logic 1, s gi cho cng AND mv cc xung nhp s truyn qua cng AND. Bm sm cc xung nhp . Ngay khiin p vo trnn bng vi in p mu, th tn hiu ra ca b so snh s bng 0. CngAND sng v dng vic m. Mc ra ca bm sc cht v cc LED hay tinhth lng s hin th gi tro. Hnh 3.10, l mch nguyn l cbn cng vi chuynmch thang o.Chuyn mch thang o hnh 3.10, s chn tn hiu ra t mch phn p. Cc tr s cacc in trphn p c th l 9M, 0,9M v 0,1M chn ra 1V ti u vo caADC cho cc u vo 1, 10 v 100V ca tn hiu cn o. Nu tn hiu cn o l 100V, thtn hiu vo a n b so snh s l (100/10) x (1/10) s l 1V do mch phn p. Nutn hiu cn o l 10V, th tn hiu vo a n b so snh s vn l 1V. Nh vy, b sosnh s ly Vin trong khong t 0 n 1V bt kin p thc t cn o. Mc in p vo(t 0 n 1V) sc bin i thnh tn hiu s m sc m v hin th.


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b) ng ho sa nng (DMM).V cbn, DMM l mt voltmeter s. Tt c cc thng s khc in p, nhin tr,dng in, in p acu c bin i thnh in p dc nhchuyn mch chn chcnng o nh hnh 3.11. Sau php o in p dc s cho gi tr ca thng s cn o.

o in tr, th in trphi c chuyn i thnh in p dc bng mc dng chyqua in tr cn o t mt ngun dng hng. Nu mc dng hng l 1mA, th suy ramc in p dcc to ra trn in trcha bit s t l trc tip theo mV. Nu intrchn l 1k, th mc in p c to ra s l 1V. i vi php o dng in, dngin sc bin i thnh in p dc bng cch cho dng in chy qua mt in trkhng i, chn l 1. Do vy mc in p dc st trn in trs bng mc dng in(in p = dng in x 1). i vi php o in p ac, in p ac trc ht phi cchnh lu v sau sc xem nhin p dc c tho c.Vic nh chun DMM sc kim tra bng php o in tr0 bng cch ngn mchhai u que o, khi t chc nng ca ng ho v tr chuyn mch o in tr, hocc tho in p khi bit mc in p dc trn thang o in p.c) ngha ca chs bn phn v ba phn t.B ch th 3 - ch sDVM cho thang o t 0 n 1V s ch cc gi tr t 0 n 999mV.Bc mc tng nh nht l 1mV. Vic b sung thm mt ch s (0 n 9 thnh nhm 4

bit) v thc cht s lm tng gi thnh, nn bin php tit kim nht l c th s dng chmt bit (0 hoc 1). Bit b sung s cho php DVM ch th cc tr s ln n 1999 thay cho999, tc mrng thang o ln gp i. Khi ch s th 4 c th ch c gi tr 0 hoc 1, ththang o c gi l ch s bn phn (1/2), nn gi l ng ho 3 ch s. ng ho c 4 ch s s ch th gi tro ln n 19999mV.


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Tng t, bng cch b sung hai bit 11, th ch s tn cng bn tri c th to ra l 3.ng ho 3 - ch s s cho php c 999 tip theo l 1999 hoc 2999 hoc 3999 (bngcch s dng 01, 10, v 11 tng ng), tc l tng thang o ln nht vo khong 4 ln.Vic b sung thang o nh vy c gi l ng ho 3 - ch s. ng ho 4 -ch s s cho s ch th ln n 39999. Vi vic b sung - ch s hoc - ch s slm cho chnh xc tng ln nhc minh hov d 6.3.V d 3.3: Cc schth12,375V v 32,375V sc hin th nhthno cc ng

ho (a) c 3 - chs, (b) c 3 - chs, (c) c 4 - chs, v (d) c 4 - chs?(a) ng ho c 3 - ch s s ch th trn thang o 99,9V, do kt quo l 12,375Vv 32,375V sc c l 12,37V v 32,3V tng ng.(b) ng ho 3 - ch s s ch th s 12,375V l 12,37V trn thang o 19,99V v so 32,375V l 32,3V trn thang o 099,9V.(c) ng ho 4 - ch s s ch th s 12,375V l 12,375V trn thang o 19,999V,nhng i vi sc 32,375V, th ng ho 4 - ch s s ch th l 32,37V trnthang o 99,99V.(d) ng ho 4 - ch s s ch th s 12,375V l 12,375V trn thang o 19,999V, vi vi so 32,375V s ch th l 32,375V trn thang o 39,999V.3.3 VOLTMETERIN T

Voltmeterin t s dng mch khuych i mt chiu, h s khuych i ca mchkhuych i s cho php s dng thit b o o cc in p di millivolt vmicrovolt. Trc y, cc voltmeter in t c gi l VTVM (vacuum tubevoltmeter). Hin ny thut ng VTVM vn c s dng mc d n chn khng c thay th bng cc transistor. Tn gi mi ca thit bo l voltmeterin t (EVM)hay voltmeter bng transistor (TVM).Voltmeterin t c th lp theo cc kiu mch khuych i khc nhau, chng hn nhmch khuych i bng transistorn hay mch khuych i cn bng, hoc bng op -amp. Mc ch cbn ca vic s dng mch khuych i dc l c h s khuych iv in trvo cao (tc l c tho c cc tn hiu yu), v cch ly ng ho vimch vo ca thit bo (tc l c th s dng ng h chc chn hn v nhy km

hn). Voltmeterin tin hnh nh mch hnh 3.12.EVM s dng mch khuych i bng FET lm tng u. Cu kin bng FET c trkhng vo rt cao, nn s khng gy ra qu ti cho mch cn o, cho chnh xc cao.


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Mch phn p c mc lm chuyn mch thang o, cho php iu khin cc thango in p khc nhau. Cc in trmch hnh 3.12, s bin i mc vo l 1V, 10V,100V v 1000V thnh 1V ti cng ca FET. Cc transistorQ1 v Q2 to thnh mch cu.Cu sc cn bng sao cho mt ch th ca ng h s ch th so bng 0 khi tn hiuvo bng 0.Khi s dng mch khuych i thut ton, ngun in p hng, khong 1V, s cung cptn hiu a n u vo khng o thng qua mt bin trc nh chun. Mc vo

dc sc gim xung n 0 i vi 1V nhcc mch suy gim thang o, s cung cptn hiu n u vo o. ng ho mc u ra quan st mc 0. in th k cain p mu sc hiu chnh c mc 0 trn ng h. Sau in p vo l cn

bng vi in p chun c chia thang, ch th trn mt s ca in th k. Mc d,phng php cn bng mc 0 c chnh xc cao hn, nhng cc thit bo gi thp,in p vo dc phi c chuyn i thnh millivolt, khong 10mV, h s khuych ica op - amp c iu chnh n mc 100 v sau mc ng ho vo u ra nhchun, cho php ch th trc tip mc in p vo.Mch khuych i dc hay b tri dng ra (do nhit ca tip gip), do mt s thit bo s dng mch ngt qung ngt in p dc thnh cc xung, c th s dng mchkhuych i ac nh mch hnh 3.13. Sau khi khuych i cc xung sc bin i

thnh mt chiu v c o bng cch s dng mch vi sai.

a) Cc u im ca EVM1. Trkhng vo ca EVM rt cao, nn nh hng do qu ti khng ng k trn

mch cn o th. Do vy, chnh xc ca php o cao.2. C tho in p ngay cc mc microvolt nhs khuych i tn hiu o.

kiu ccu o t - in, o mc in p di 100mV rt kh khn.3. in dung vo ca voltmeterin t l rt nh, vo khong vi picofarad. Do

c tho c cc mc in p tn hiu tn s cao sau khi chnh lu.4. ng h c nhy thp, 1000/V, v v vy r tin, th, c th s dng cho cc

php o.5. Ngun cung cp cho mch khuych i khng ly t mch cn o, m ly tVCC.6. kiu mch cu cn bng (nh th hin hnh 3.12), nh hng ca cc thay i

cc thng s ca FET v BJT l thp nht.7. dch mc 0 trong qu trnh hot ng khng xy ra.

b) Nhc im ca EVM1. Nhc im chnh l cn phi c khi ngun cung cp cho mch khuych i, nn

EVM c gi thnh cao v kch thc ln.2. Mch khuych i dc b tri mc tn hiu ra.

c) ng ho in ta nngng ho in ta nng (EMM) v cbn l voltmeter mt chiu, nhng s lm vicnh nhiu ng ho nh chuyn mch chc nng kt ni cc thit bo bin i ccthng so khc thnh in p dc, nh gii thch DMM.


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3.4 NG HO H S MO DNG TN HIUa) Nguyn lng ho h s mo dng dng o mo hi tng c trong tn hiu ra. Tn s c

bn sc trit nhmch lc thng cao, m tn s ct ca mch lc hi cao hn so vitn s cbn. Tn hiu tn s cbn s thot xung t v cc tn s cao hn so vi tns cbn (hi bc 2, hi bc 3, v. v. . .) sa n voltmeter to ra mc in p do ton

b cc hi c trong tn hiu. ng ho mo c th l kiu nhit ngu hoc voltmeter

in t. S khi ca ng ho mo in hnh cho hnh 3.14.

b) Thc hin php o moTn hiu vo sc cung cp t my to sng a n b khuych i cn o thmo. Tn hiu ra ca b khuych i l tn hiu vo ca thit bo. Khi chuyn mch St ti v tr B, s thu c ton b tn hiu ng ho u ra. Tn hiu ton b hoc sc iu chnh n mc lch y thang (fsd), hoc mc cnh no c ghi dutrn mt ch th ca ng h nhmch suy gim. Tip theo, chuyn mch t ti v tr A,nn b lc thng cao s loi b tn hiu tn s cbn cho n khi nhn c mc bin gim mt ch th ca ng h. Mc ch th l mc tng tri tn s cbn, tc llng cc sng hi. ng ho sc nh chun theo phn trm ca mo dng mc thc trc tip trn vch chia thp nht.c) Sdng ng ho mo trong vic d tm hhngC tho c mo xut hin trong thit b cn o th. Nguyn nhn gy mo dng

c thc xc nh bng ngn mch ln lt cc tng ca thit b, nhhnh 3.15.

Chng hn, nu tng III c lm ngn mch theo hnh v, v mo vn cn, tng no trong s cc tng III, hay IV, hoc V l tng gy mo dng. Nu mo dng c loi

b, th tng I hoc tng II c th b hng. Ton b cc tng c thc o th theo

phng php ny nhn din tng hng. Khi xc nh c tng hng theo phngphp trn, ta c tho th thm bit nguyn nhn gy mo dng v sa cha h hng.3.5 BM TN SBm tn l mt thit bo s c th dng o tn s, khong thi gian gia hai tnhiu, chu k xung, t s ca hai tn s v c thm s xung.a) Nguyn l hot ng. kiu o cbn nht l tn hiu cn o c t vo cng AND, cng ch mtrongkhong thi gian cnh bng mt xung mcng. Xung ra ca cng AND sc an bm 10 v cht s liu sau a n khi hin th nh th hin theo skhi hnh 3.16.


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Mch x l tn hiu gm mt khi khuych i v khi trigger Schmitt. Tn hiu dng

sng sin vo c khuych i v chuyn i thnh cc xung vung c a nmt u vo ca cng AND, u vo cn li ly xung nh thi t mch to xung nhp(khi gc thi gian).b) Cc php o bng bm tn.

Php o tn s: Cng AND s duy tr vic mcng theo chu k ca xung nh thi, nns cho cc xung tn hiu cn o tn s ti u ra ca cng AND trong khong thi gianmcng. Bm sm cc xung v sm sc lu tr vo b nhcng nhchin th. Xung nh thi kch khi bm ti thi im xut hin cnh trc v dng bm ti thi im xut hin cnh sau ca xung nhflip flop. Tng t, xung nh thicng siu khin b nh.

Nu khong thi gian ca xung nh thi l 1 giy, bm mcng trong khong thigian l 1 giy, bm cho s chu k tn hiu truyn qua cng trong mt giy, tc l o


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trc tip tn s ca tn hiu.i vi cc tn s cao, s dng mch chia 10 to ra cc xung nh thi t 1s n 1mstu theo cc v tr t ca chuyn mch nhiu v tr. Nu c 1000 xung ca tn hiu cno truyn qua cng AND trong khong thi gian 1ms, th tn s ca tn hiu l1000MHz. Cc bm khng thm cc tn s qu cao (di gigahertz), nn cc tns cao sc o bng k thut chia thang trc, tc l tn s tn hiu sc chia 2,4, 8, v. v. . . sao cho tn s sau khi chia thang trc ph hp vi thang o ca bm

tn. S khi bm tn c mch chia tn s xung nhp (mch chia gc thi gian) nhhnh 3.17.o chu k:o chu k, xung tn hiu vo c tc dng nh mt xung nh thi dng mcng truyn cc xung nhp qua cng AND. Tn hiu vo s mv ng cng AND,nn sm l s lng xung nhp c truyn qua cng s cho bit chu k thi gianca xung tn hiu cn o. S khi ca mch o chu k nhhnh 3.18a v b.o khong thi gian gia hai xung: C tho khong thi gian gia hai xung bngcch s dng mt tn hiu mcng v tn hiu cn li ng cng AND. Tn hiuxung nhp dng o khong thi gian khi mv ng cng. Php o c th hin hnh 3.19. Xung 1 lm cho u vo S= 1 vR = 0, nn Q = 1, khi xung 2 xut hin lmcho S= 0 vR = 1, nn Q = 0.

o t s tn sca hai tn hiu:Nu c hai tn hiu, th tn hiu tn s thp c sdng lm xung nh thi, cn tn hiu tn s cao sc m trong khong thi gian caxung nh thi. Php o cho bit t s ca hai tn s trn khi hin th nhs khi

hnh 3.20.

m xung: Mch n gin nhhnh 3.21, s cho bit l bm sm tng s xungtruyn qua cng trong sut khong thi gian chuyn mch gitrng thi kn mch.


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c) Bm tn sa nng.Bm tn sa nng kt hp tt c cc php o trong thit bo n nhcc v tr tca mch chuyn mch. Mch o cbn theo cc v tr chuyn mch nhhnh 3.22.

d) Cc u im ca bm tn s kiu s.Cc u im ca bm tn kiu s gm:

1. r cao (6 n 8 digit)2. chnh xc cao (do sn nh ca mch dao ng bng tinh th)3. Tc o cao.4. ng ho s khng p ng i vi nhiu v tp m mc thp khi t mc kch

khi cao.5. Dc s ch th.

e) Sdng my m tn s d tm trng thi hng ca thit b.Tn s ca cc b to xung nhp (trong my tnh), cc b dao ng ni (trong cc mythu) v cc b dao ng cao tn (RF) (trong cc my pht tn hiu v trong cc my thu -

pht), c thc o kim tra nu cc tng c h hng. My to xung v my tohm c thc kim tra tm sai hng bng cch o tn s, rng xung v khongthi gian ca cc xung c to ra. Cc im o th thng c sn trn bng mchca thit bin t c th ly tn hiu cn o bng ng ho tn s.3.6 MY PHT TN HIU CAO TN (RF)My pht tn hiu cao tn s to ra tn hiu tn s radio dng iu ch tn hiu cao tnvi tn hiu m tn theo kiu iu bin hay iu tn. Vic iu ch tn hiu m tn sc to ra bi thit bo. Cng c th cung cp tn hiu iu ch ngoi. S khi ca

my to tn hiu RF nhhnh 3.23.B suy gim sc nh chun cho mc suy gim mi nc l 20dB tng ng vi1V. Tn hiu m tn cng c th ly ra thng qua mch suy gim khc kim tra cctng m tn.a) Sdng my pht tn hiu RF


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My pht tn hiu RF phi c ni vi my thu cn o thng qua cp c bo v chngnhiu. Chn bng tn v tn s dao ng. Chn kiu iu ch v su iu ch. B suygim t ti v tr mc ra ca php o yu cu. u ra ca my to sng phi c ni

vi u vo ca thit b cn o th. Ni ngun cung cp v tip theo bt cng tc ngunca my pht tn hiu [ON]. My thu c iu chnh thu tn hiu. C th mcvoltmeterin t (EVM), hay ng ho mc cng sut m tn ti u ra ca my thu.

Nu khng c tn hiu ra my thu, th mc ra ca my pht tn hiu cn phi c tngthm, sao cho my thu c th nhn c tn hiu ra khng mo. Tt c cc php o thchin vi mc ra mc khng i v mc ra thay i ca my pht tn hiu.b) Sdng my to tn hiu trong vic chn on hng.My to tn hiu cao tn c s dng ph bin chn on tnh trng hng trong ccmy thu. Tn hiu phi c cung cp n my thu v s dng my hin sng quan stdng sng tn hiu c u vo v u ra ca cc tng khc nhau. Nu mt tng th hintn hiu ra bnh thng, th tt c cc tng trc tng l bnh thng v sai hng c

thtng sau k tip. Theo cch ny c th xc nh tng hng.My to tn hiu cng cn cho vic cn chnh my thu tn s dao ng ni v tn stn hiu RF l bng nhau (cng tn s) ti tt c mc thit lp trn nm tinh chnh trong

bng tn. Th tc v trnh t cn chnh o cc thng s ca my thu sc gii thchmc 5.4, chng 5.3.7 MY TO TN HIU M TN.My to tn hiu m tn bao gm b dao ng m tn (thng s dng b dao ng kiucu Wien), b khuych i m v b khuych i cng sut kt ni vi b suy gimnh chun nhs khi hnh 3.24.

S dng my to sng m tn chn on tnh trng hng trong cc mchkhuych i.My to tn hiu m tn dng cung cp tn hiu chun, khng nhiu o th hiu


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sut v o cc thng s ca mch khuych i m tn (nh h s khuych i, rngbng tn cng nh mo dng). Cch thit lp php o nhhnh 3.24. Trnh t ccbc thc hin php o nh sau:

1. Ni my to tn hiu m tn vi b khuych i. Mc ng ho mc cng sutm tn ti u ra ca b khuych i.2. iu chnh tn s ca my to tn hiu mc 1000Hz, v iu chnh suy gimca tn hiu ra ca my to sng mc m b khuych i c th cho tn hiu ra

khng mo. Ghi nhn mc ch th suy gim lx1, v mc ch th ca ng hocng sut pht ra l w1.3. Tip theo, mc my pht tn hiu trc tip vi ng ho cng sut song song vi

b khuych i cn o th. Mc ch th trn ng h c th gim. Tng dn mc cngsut pht ra ca my to tn hiu cho n khi s ch th ca ng ho bng trli trsw1. Ghi nhn s ch th mi ca my pht tn hiu l x2. Tnh t s ca hai s chth ca my pht tn hiux2/x1, biu din theo dB, s cho h s khuych i ca mchkhuych i.4. o rng bng tn, h s khuych i ca mch khuych i phi c o ticc tn s khc nhau, t 20Hz n 20kHz, v vc tuyn gia tn s (trn trc -xtheo thang logarithmic) theo h s khuych i theo dB (trn trc -y tuyn tnh), t

c tuyn ta c th xc nh rng bng tn mc 3dB.5. o mo, cn phi s dng ng ho mo dng, nh c gii thch mc 3.4. ng ho mo dng so mo hi tng to ra do mch khuych ii vi tn hiu cung cp t my to tn hiu. (phi m bo rng tn hiu pht ra camy to tn hiu khng b mo). Php o ny cng s cho bit mc tn hiu ra m tnln nht c th nhn c t mch khuych i trong gii hn mo cho php.

3.8 MY PHT XUNG.My pht xung phc tp hn so vi my to sng sin. Mt sng sin ch c hai thng s l

bin v tn s, trong khi sng xung c hng lot cc thng s nh bin xung, rng xung, tn s lp li ca xung, chu k, cng sut xung, chu k chuyn trng thi (qu) v. v. . . S khi ca my pht xung nhhnh 3.25.

B to xungB to xung gm mch dao ng cu Wien c ghp vi mch kch khiSchmitt. Tn s xung to ra ca mch kch khi Schmitt c thc iu khin hoc bntrong (iu khin trong - Int. control), hoc iu khin bn ngoi (iu khin ngoi - Ext.control). Xung n [Single] siu khin cc thng s ca xung bng tay. Cng ngoi[Ext. gate] s to ra cc cm xung.

Khinh thiKhi nh thi thc hin cc chc nng nh sau:1. Lm tr hay lm sm pha ca xung so vi xung kch khi.


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2. Mi xung sc to ra hai xung. Xung th nht s trng vi xung kch khi,xung th hai s thay i theo thi gian.

B pht tsMy pht xung c s dng ph bin trong cc php o th v chn onhng trong cc mch s. My pht t s thay th khi nh thi to ra d liu.

Biu khin dng xungBiu khin dng xung siu khin rng xung, chu kchuyn trng thi (thi gian tng v thi gian gim ca cc cnh xung), cc tnh caxung, bin xung v dch xung (t 0Vdc).

My pht xung c trkhng ra c trng l 50. My pht xung s ngn chn s hnhthnh sng dng trn ng truyn.My pht xung loi tt s to ra xung mn vi nh xung ngang v cc cnh ng. Tuynhin, nu kho st h hng, cc xung c th b suy bin th hin preshoot, qu mctrn [overshoot], dao ng tt dn [ringing], khng tuyn tnh [non linearity] v suy gim [droop] hay nghing [sag].Cc du hiu trn th hin hnh 3.26. Cc sai hng xung c th quan st bng myhin sng.

Cc cng dng ca my to xung.1. o th cc mch s bng cch cung cp cc xung th nghim cc cng logic.2. o nhy v t l bit li trong h thng thng tin s liu.3. My to xung dng pht hin li trn cc ng dy in thoi. Xung s truyn

qua ng dy in thoi tc nh sng (3 x 105km/s). Khi gp ng dy hmch, xung sc phn x v my pht. o khong thi gian trng nh trongradar, th c th tnh c chiu di ca cp khi bt.

4. Cc xung t my to xung c thc s dng o th h s khuych i vp ng tn s ca cc b khych i. Cc xung vung ngn s lm gim s tiutn cng sut cho mch.

5. My to xung cng c thc dng lm tn hiu iu chn cc b dao ng viba, radar.

6. Thng s thi gian hi phc ngc ca cc diode c th xc nh bng cch sdng cc xung t my to xung.

3.9 MY TO HM FUNCTION GENERATOR.Trong khi cc my to tn hiu ch to ra cc sng sin, v cc my to xung to ra ccxung vung hoc ch nht, th my to hm s to ra cc loi dng sng khc nhau. Ccdng sng m my to hm c th to ra l sng sin, cc xung vung hoc ch nht, ccsng tam gic v cc tn hiu rng ca.Cc dng sng khc nhau c to ra bng my to hm c thc ly ra ng thi.My to hm cng c thc kha pha vi tn hiu ngoi.Mch dao ng cbn ca thit b c th l mch dao ng a hi hay mch dao ng


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to sng sin nh kiu cu Wien. Cc dng dao ng, nu khng phi l sng sin c thc bin i t sng sin bng mch sa dng kiu in tr - diode. Cc dao ng cdng bt k c th bin i thnh cc xung bng mch kch khi Schmitt. Hnh 3.27, ls khi ca my to hm cbn.

Mch dao ng cu Wien c th to ra tn hiu sng sin c bng tn rng, t vi hertzn di megahertz. B khuych i m sm bo tn hiu dao ng khng b suygim. Mch khuych i cng sut v mch suy gim mc tn hiu (cc hp suy gimdB) s to ra sng sin ti u ra A. (mt s my to hm s dng cc mch a hi, tnhiu ra sc sa dng bang mch sa dng diode v in tr c sng sin).B to xung s dng mch kch khi Schmitt bin i sng sin thnh xung. Biuchnh dng xung to ra cc xung c rng , p.r.f, v cng sut xung theo yu cu tiu ra B.Tn hiu ra ca mch kch khi Schmitt sc cung cp n mch tch phn bng op amp v tip theo n mch iu ha tn hiu c sng tam gic ti u ra C.Chuyn mch bng UJT c th bin i sng tam gic thnh tn hiu rng ca, sau khiiu ha tn hiu s c ti u ra D.Cc cng dng ca my to hm.

1. Tn hiu sng sin c th dng o th h s khuych i ca mch khuych i.2. Sng vung c tho thp ng tn s thp v tn s cao ca mch khuych i

nhmy hin sng. nghing no ca phn nh ngang ca xung s cho bitp ng tn s thp ca mch khuych i km. S thay i thi gian tng vthi gian gim (tc sn xung) ca cc cnh xung s cho bit p ng tn s caoca mch khuych i km. Cc xung cng c th s dng o th cc cng s.

3. Cc sng tam gic c th dng o th tuyn tnh ca cc mch m sng tamgic truyn qua. Bt k s mo dng ca cc cnh tam gic, khi quan st trn mnhnh ca my hin sng, s cho bit khng tuyn tnh c to ra bi mchkhuych i.

4. Tn hiu rng ca c thc dng o th cc b to sng qut v cc mchkhuych i qut trong cc my thu hnh, cc my hin sng v cc monitor.

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