co so do luong dien tu

8/9/2019 Co So Do Luong Dien Tu

1/93

1

CS O LNGIN T

Khoa Kthutin t1

Hc vin cng ngh bu chnh vin thng

2

Sch tham kho

1. Csk thut o lng in t, V Qu im, nh xutbn KHKT, 2001

2. o lng in-v tuyn in, VNh Giao v Bi VnSng, Hc vin k thut qun s, 1996

3. Electronic Test Instruments, Bob Witte, 2002

4. Radio Electronic Measurements, G.Mirsky, MirPublishers, Moscow, 1978


2/93

3

2.1 Cc phng php o:

1. Phng php o trc tip: dng my o hay cc mu o (cc chun)

nh gi s lng ca i lng cn o. Kt qu o chnh l tr s ca ilng cn o.

- VD: o in p bng vn-mt, o tn sbng tn s-mt, o cng sutbng ot-mt,...

- c im: n gin, nhanh chng, loi b c cc sai s do tnh ton

2. o gin tip: kt qu o khng phi l tr s ca i lng cn o, m l ccs liu cs tnh ra tr s ca i lng ny.

- VD: o cng sut bng vn-mt v ampe-mt, o h s sng chy bngdy o,...

- c im: nhiu php o v thng khng nhn bit ngay c kt qu o

Chng 1. Gii thiu chung v o lng in t

nh ngha: o lng l khoa hc v cc php o, cc phng php v cc cngc m bo cc phng php o t c chnh xc mong mun

aX

naaaFX ,...,, 21

4

3. Phng php o tng quan: dng o cc qu trnh phc tp, khi khngth thit lp mt quan h hm s no gia cc i lng ca mt qu trnhnghin cu- Php o tng quan c thc hin bng cch xc nh khong thi gianv kt qu ca mt s thut ton c kh nng nh c tr s ca i lngthch hp.

- VD: o tn hiu u vo v u ra ca mt h thng- c im: cn t nht hai php o m cc thng s t kt qu o ca chngkhng ph thuc ln nhau. chnh xc c xc nh bng di khongthi gian ca qu trnh xt.

4. Cc phng php o khc:- Phng php o thay th- Phng php hiu s (phng php vi sai, phng php ch th khng,

phng php b)- Phng php ch th s



3/93

5

2.2 Phng tin o v cc c tnh cbn

1. Phng tin o l phng tin kthut thc hin php o, chng c nhngc tnh o lng c qui nh.

- Phng tin o n gin: mu, thit b so snh, chuyn i o lng- Phng tin o phc tp: my o (dng c o), thit b o tng h pv h thngthng tin o lng.

+ Mu: phng tin o dng sao li i lng vt l cgi t r cho trc vi chnh xc cao. Chun l mu c cp chnh xc cao nht. Chun l phng tin om bo vic sao v gi n v.+ Thit bso snh:phng tin o dng so snh 2 i lng cng loi xemchng = , > , < .+ Chuyn i o lng:phng tin o dng bin i tn hiu thng tin olng v dng thun tin cho vic truyn tip, bin i tip, x l tip v gi linhng ngi quan st khng th nhn bit trc tip c (VD: b K o lng;bin dng, bin p o lng; quang in tr, nhit in tr,...)


6

+ Dng c o: phng tin o dng bin i tn hiu thng tin o lng vdng m ngi quan st c th nhn bit trc tip c (VD: vnmt, ampe mt,...)

+ Thit b o tng h p v h thng thng tin o lng: l cc phng tin o phctp dng kim tra, kim nh v o lng.


Dng c o

Mc tng ha

Dng c okhng t

ng

Dng c ot ng

Dng ca tnhiu

Dng c otng t

Dng co s

Phng phpbin i

Dng co bin

i thng

Dng co bini cnbng

Cc i lngu vo

Dng co dng

in

Dng co tn s

...


4/93

7

2. Cc c tnh cbn ca phng tin oCc c tnh tnh c xc nh thng qua qu trnh chun ho thit b.+Hm bin i: l tng quan hm s gia cc i lng u ra Y v cc ilng u vo X ca phng tin o, Y=f(X)

+ nhy: l t s gia bin thin ca tn hiu u ra Y ca phng tin ovi bin thin ca i lng o u vo X tng ng.

K hiu:

+Phm vi o: l phm vi thang oba ogm nhng gi tr m sai s cho php caphng tin o i vi cc gi tr o c qui nh+Phm v chth : l phm vi thang o c gii hn bi gi tr u v gi tr cui

ca thang o.+Cp chnh xc: c xc nh bi gi tr ln nht ca cc sai s trong thit b o.Thng c tnh ton bng i s tng i quy i.+phn gii: Chnh l chia ca thang o hay gi tr nh nht c thphn bitc trn thang o (m c thphn bit c sbin i trn thang o).


dYS

dX

8

3. Phn loi cc my o:

a) My o cc thng s v c tnh ca tn hiu:VD: Vn mt in t, tn s mt, MHS, my phn tch ph, ...S khi chung:

- Tn hiu cn o a ti u vo my

- Mch vo: truyn dn tn hiu t uv o ti Thit b bin i. Mch vo thngl b K ph ti catt (Zvo cao), thc hin phi hp trkhng.


Mch vo Thit bbin i Thit bchth

Nguncung cp

uvo y(t)


5/93

9

- Thit b bin i: thc hin so snh v phn tch.C th to ra tn hiu cn thit so snh tn hiu cn o vi tn hiu mu.C thphn tch tn hiu o vbin , tn s, hay chn lc theo thi gian.Thng l cc mch K, tch sng,bin i dng in p tn hiu, chuyn i

dng nng lng,...- Thit b chth:biu th kt qu o di dng thch hp vi gic quan giao tipca sinh l con ngi hay vi tin tc a v obphn iu chnh, tnh ton,...

VD: ng h o ch th kim, ng tia in t, h thng n ch th s, thit b nh,...Ngun cung cp: cung cp nng lng cho my, v lm ngun to tn hiu chun.

b) My o c tnh v thng s ca mch in:

Mch in cn o thng s: mng 4 cc, mng 2 cc, cc phn t ca mch in.S khi chung: cu to gm c ngun tn hiu v thit b ch th, (hv)VD: my o c tnh tn s, my o c tnh qu , my o h sphm cht, oRLC, my th n in t, bn dn v IC,...


10


Mch o itngo

Thit bchth

Nguntn hiu(b)

Nguncung cp

Nguntn hiu

Thit bchth

itngo(a)


6/93

11

c) my to tn hiu o lng:

S khi chung:

B to sng ch: xc nh cc c tnh ch yu ca tn hiu nh dngv tn sdao ng, thng l b to sng hnh sin hay xung cc loi

B bin i: nng cao mc nng lng ca tn hiu hay tng thm xc lp ca

dng tn hiu, thng l b K in p, K cng sut, b iu ch, thit b todng xung,...Cc my pht tn hiu siu cao tn thng khng cB bin i t gia B tosng ch v u ra, m dng B iu chtrc tip khng ch dao ng ch


Mchra

Thit bo

Nguncung cp

B iuch

B tosng ch

B bini

x(t)

12

Mch ra: iu chnh cc mc tn hiu ra, bin i Zra ca my. N thng lmchphn p, bin pphi hp trkhng, hay bph ti Catt.Thit b o: kim tra thng s ca tn hiu u ra. N thng l vn mt in t,thit b o cng sut, o h s iu ch, o tn s,...

Ngun: cung cp ngun cho cc bphn, thng lm nhim vbin i in pxoay chiu ca mng li in thnh in p 1 chiu c n nh cao.

d) Cc linh kin o lng:gm cc linh kin l, ph thm vi my o to nn cc mch o cn thit.Chng l cc in tr, in cm, in dung mu; hay cc linh kin ghp giacc bphn ca mch o (VD: b suy gim, b dch pha, bphn mch nhhng,...)



7/93

13

2.1 Khi nim & nguyn nhn sai s:* Khi nim sai s: l chnh lch gia kt qu o v gi tr thc ca i lngo. N ph thuc vo nhiu yu t nh: thit b o,phng thc o, ngi o* Nguyn nhn gy sai s:

- Nguyn nhn khch quan: do dng c o khng hon ho, i lng o bcan nhiu nn khng hon ton c n nh,...- Nguyn nhn ch quan: do thiu thnh tho trong thao tc, phng phptin hnh o khng hp l,...

2.2 Phn loi sai s* Theo cch biu din sai s:

-Sai s tuyt i: l hiu gia kt qu o c vi gi tr thc ca i lng o

-Sai s tng i chn thc: l gi tr tuyt i ca t s gia s a i s tuyt i vgi tr thc ca i lng o

Chng 2. nh gi sai s o lng

.100%ctthuc

X

X

thucdo XXX

14

-Sai s tng i danh nh:

-Sai s tng i qui i: l gi tr tuyt i ca t s gia s a i s tuyt i v gitr nh mc ca thang o.

cp chnh xc ca i lng o

Xdm= Xmax -Xmin : gi tr nh mc ca thang oNu gi tr thang o: 0Xmax Xdm=Xmax

* Theo sph thuc ca sa i svo i lngo:

-Sa i s im 0 (sai s cng) l sai s khng ph thuc vo gi tr i lng o.-Sa i s nhy (sai s nhn) l sai sph thuc vo gi tr i lng o


.100%dddo

X

X

%100.dm

qd

X

X


8/93

15

* Theo v trsinh ra sai sta c sai sphng php v sai sphng tin o:

-Sa i sphng php l sai s do phng php o khng hon ho- Sai sphng tin o l sai s do phng tin o khng hon ho. Gm: sai

s h thng, sai s ngu nhin, sai s im 0, sai s nhy, sai s cbn, sai sph, sai s ng, sai s tnh.Sai scbn ca phng tin o l sai s ca phng tin o khi s dng trongiu kin tiu chunSai sph ca phng tin o l sai s sinh ra khi s dngphng tin o iu kin khng tiu chunSai stnh l sai s ca phng tin o khi i lng o khng bin i theo thigianSai s ngl sai s ca phng tin o khi i lng obin i theo thi gian


16

Chng 2. nh gi sai s o lng* Theo qui lutxut hinsai s:

- Sai s h thng- Sai s ngu nhin

2.2.1 Sai shthng

- Do cc yu t thng xuyn hay cc yu t c qui lut tc ng.- Kt qu o c sai s ca ln o no cng u ln hn hayb hn gi tr thc cai lng cn oVD:

+ Do dng c, my mc o ch to khng hon ho+ Do chnphng php o khng hp l, hoc li trong qu trnh x l kt qu

o,...+ Do kh hu (nhit , m,...) khi o khng ging vi iu kinkh hu

tiu chun theo qui nh


9/93

17

2.2.2 Sai sngu nhin

- Do cc yu tbt thng, khng c qui lut tc ng.VD: + Do in p cung cp ca mch o khng n nh

+ Do bin thin kh hu ca mi trng xung quanh trong qu trnh o

Trs o sai: l kt qu cc ln o c cc gi tr sai khc qu ng, thng do sthiu chu o ca ngi o hay do cc tc ng t ngt ca bn ngoi.

Xlsai ssau khio:- i vi sai s h thng: x l bng cch cng i s gi tr ca sai s hthng vo kt qu o, hoc hiu chnh li my mc, thit b o vi my mu

- i vi sai s ngu nhin: khng x l c, ch c th nh lngcgi tr sai s ngu nhin bng l thuyt xc sut & thng k.


18

2.3 ng dng phng php phn b chun nh gi sai sYu cu: - tt c cc ln o u phi thc hin vi chnh xc nh nhau

- phi o nhiu ln2.3.1 Hm mtphn bsai s

- Tin hnh o n ln mt i lng no , ta thu c c c k t qu o c ccsai s tng ng l x1, x2, ...,xn

- Sp xp cc sai s theo gi tr ln ca n thnh tng nhm ring bit,vd: n1 sai s c tr s t 00,01; n2 sai s c tr s t 0,010,02; ...

- , ,... l tn sut ( hay tn s xut hin) cc ln o c cc

sai s ngu nhin nm trong khong c gi tr gii hn - Lp biu phn b tn sut:


n

n11 n

n22

limn

(x)=p(x)


10/93

19

p(x) l hm sphn b tiu chun cc sai s (hm s chnh tc). Thc t th phnln cc tr ng hp sai s trong o lng in t u thch hp vi qui lut ny

(hm Gauss) (1)


22

)( xheh

xp

h : thng s o chnh xch ln ng cong hp v nhn (xcsut cc sai s c tr sb th ln hn) thit b o c chnh xc cao

Qui tc phn bsai s:

a. Xc sut xut hin ca cc sai s c tr sb th nhiu hn xc sut xuthin ca cc sai s c tr s ln.b. Xc sut xut hin sai s khngph thuc du, ngha l cc sai s c tr sbng nhau v gi tr tuyt i nhng khc du nhau th c xc sut xut hinnh nhau.

20

2.3.2 Sdng cc c sphn b nh gi kt qu o v sai s o

1. Sai strung bnh bnh phng:+ o n ln mt i lng X, cc kt qu nhn c l n tr s sai s c gi tr

nm trong khong gii hn x1 xn

+ h khc nhau xc sut ca chng khc nhau+ h = const vi mt loi tr s o xc sut sai s xut hin ti x1 v ln cnca x1 l:

tng t ta c:


11

21

2

dxeh

dp xh

22

22

2

dxeh

dp xh

nxh

n dxeh

dp n22

.......................x


11/93

21

Xc sut ca n ln o coi nh xc sut ca mt s kin phc hp, do :Pph= dp1. dp2... dpn

Tm cc tr ca h:


02

22222

1

ii xhinn

xh

n

nph exh

he

hn

dh

dP

02 22 ixhn

Sai s TBBP ():

n

xxxh

n

dxdxdxeh

n ...21... 222

21

2

(2)

n

x

hi2

2

1(3)

n

xn

ii

12

(4)

22

Hm phn b tiu chun: (5)

Xc sut xut hin cc sai s c tr s < :

(6)

* Ly nh gi sai s ca KQ o tin cy cha m bo.

ly M=3 (sai s cc i).

(8)

Chng 2. nh gi sai s o lng2

2

2

2

1)(

x

exp

it t

dtexP

0

2

2

2

2

122 hxht ii

dtedtexPth t

1

0

2

2

0

2

22

2

2

2

2

997,02

2 3

0

2

2

dteMxPt

3/2638,0 xP (7)


12/93

23

2. Trstrung bnh cng:- o X, thu c n cc kt qu o: a1, a2, ..., an- Cc sai s ca c c ln o ring bit: x1= a1-X, x2= a2-X, ..., xn= an-X-C c xi cha bit X cn o cha bit- Thc t ch xc nh c tr s gn ng nht vi X (tr s c xc sut ln nht):

(9)

3. Sai sd:- Sai s mi ln o: xi =ai x cha bit v x cha bit.- Sai s d l sai s tuyt i ca gi tr cc ln o ai vi :

- Thc t:


n

a

n

aaaa

n

ii

n

121...

aaii

Xa

a

0.1111

n

ii

n

ii

n

ii

n

ii aaana (10)

24

(11)

- Sai s TBBP ca : (12)

4. Sai sTB: (13)

5. tin cy v khongtin cy:Xc sut ca cc sai s c tr s khng vt qu 1 gi tr cho trc no , bng:


11

2

1

2

nn

xn

ii

n

ii

n

a

a

)1(1

nnd

n

ii

a

dteXaPtt

i

/

0

2

2

2

2

ait

,


13/93

25

Nu bit P, da v obng hm s trong s tay tra cu v ton hay

(16)

l khong tin cy, khong ny c xc sut cha ng tr s thc ca i lng

cn o X l . P l tin cy ca php nh gi.Kt qu o:

(17)

m bo tin cy P =0,997 th ly t=3 ta c:

Quan h gia tin cy P, t, vi n >10 (bng 1)


ta

t

at

atXa

aa taXta

tP

10nataX

aaX 3 (18)

26


102 n

astaX

6. Sai scc i v sai sth:Sai s cc i (n >10)

Sai s th: sai s |i| ca ln quan st no ln hn sai s cc i th l sai

s th.7. Phn bstudent:Khong tin cy:

Gi tr ca ts c cho trong bng 2

2.4 Cch xc nh kt qu o:

Thc hin o n ln thu c c c k t qu o: a1, a2, ..., an

1. Tnh tr s trung bnh cng:

stM 3M

102 n

102 ns a s aa t X a t

n

aa

n

ii

1


14/93

27

2. Tnh sai s d:

Kim tra:

hay khng?

3. Tnh sai s TBBP:

4. Kim tra xem c sai s th?nu c sai s th th loi b kt qu o tng ng v thc hin li bc 1-4

5. Tnh sai s TBBP ca tr s TB cng:


01

n

ii

aaii

11

2

n

n

ii

na

28

6. Xc nh kt qu o: vi

nu :

* Cch vit hng chsca KQ o:

- Ly ch cn ly vi 2 s sau du phy.- Ly phi ch ly ch s sao cho bc ca s cui ca n bc ca haicon s ca .

VD: kt qu o l X = 275,24 1,08 th phi vit li l: X = 275,2 1,1


ataX 10n

102 n astaX

ata

at


15/93

29

Bng 2

Bng 1 Gi tr t theo gi tr xc sut cho trc

30

2.5 Sai s ca php o gin tip:

Gi s X l i lng cn o bngphp o gin tip; Y,V,Z l cc i lng oc bng php o trc tip

X = F(Y,V,Z)

Y, V, Z l cc sai s h thng tng ng khi o Y, V, Z ; X l sai s hthng khi xc nh X

X + X = F(Y+ Y,V+ V,Z+ Z )

Cc sai s c gi tr nh nn:


F F F

+ X=F Y,V,Z + + +Y

X Y V Z V Z

F F FX= + +Y

Y V ZV Z


16/93

31

TH1: X = aY + bV + cZ

X = a Y + bV + cZ

TH2:

Thc t dng sai s tng i:

Xc nh sai s TBBP ca php o gin tip thng qua sai s TBBP ca cc phpo trc tip thnh phn


X

X Y V Z

= + +Y

= + +

X Y V Z

X V Z

=KY V ZX

1 1 1

=K Y V Z +K Y V Z +K Y V ZX Y V Z

2 2 2

X = + +Y V ZF F F

Y V Z

32

3.1 Nguyn tc hot ng chung ca ccu o

Bao gm 2 thnh phn cbn : Tnh v ng.- Hot ng theo nguyn tcbin i lin tc in nng thnh cnng lm quayphn ng ca n. Trong qu trnh quay lc csinh cng chc mt phnthng lc ma st, mt phn lm bin i th nngphn ng.

- Qu trnh bin i nng lng trong CC c th hin theo chiubini: dng in Ix (hoc Ux ) nng lng in t Wt,Wt s tng tc vi phn ng v phn tnh to ra F (lc) to mmen quay(Mq) gc quay ; t l vi f(Ix) hoc = f(Ux)Gi s ccu o c n phn tnh in (mang in tch) v n cun dy.Thng thng in p c a vo cun dy. Nng lng in t sinh ra cxc nh nh sau:

Chng 3. Cc b ch th trong my o

1 12 2

1 1 11 1

1 1 1

2 2 2

i n i nj n j nn

dt ij ij i i ij i ji i ij i j i

W C U LI M I I


17/93

33

i : cun dy j : phn t mang in tch

: in dung v in pg ia 2 phn t tch in i v j.: dng in trong cc cun dy i v j.

: in cm ca cund y ih cm gia hai cun dy i v jNng lng in t sinh ra ph thuc vo in p, in dung, dng in,cun cm v h cm.Tng tc gia phn tnh v phn ng to ra 1 momen quaybng sbin thinca nng lng t trn sbin thin gc quay.

: sbin thin ca nng lng t: sbin thin ca gc quay


:ijM

,i jI I,ij ijC U

iL

ddtdW

qM

dtdWd

34

to r a s ph thuc gia gc quay v gi tr o; trong khi o ngi t a sdng thm l xophn khng to ra momen phn khng chng li s chuynng ca phn ng.

D: l h sphn khng ca lxoKim ch th s dng li v tr cn bng khi


DpkM

qMpkM

ddtdW

DddtdWD

1

dtW : ph thuc vo in p, dng in t vo cun dy.


18/93

35

3.2 Ccu ch th kim:Mt s dng c o lch:

- Dng c o t in kiu nam chm vnh cu (TNCVC).

- Dng c o in ng.- Dng c o kiu in t.

3.2.1 B chthkiu t in: hot ng theo nguyn tc bin i in nngthnh cnng nhs tng tc gia t trng ca mt nam chm vnhcu v t trng ca dng in qua 1 khung dy ng


36


1. Cu to:- Phn tnh: gm 1 nam chm vnh cu (1), hai

m cc t (2), 1 li st t (3). Gia (2) v (3)to thnh 1 khe h p hnh vnh khuyn chophp 1 khung dy quay xung quanh v c ttrng u hng tm (B)

- Phn ng: gm 1 khung dy nh (4) c thquay xung quanh trc ca 1 li st t, 1 kimch th (5) c gn vo trc ca khung dy, 1l xo phn khng (6) vi 1 u c gn votrc ca khung dy, u cn li c gn viv my.

nh v kim ng im `0` khi cha o th mt u ca l xo phnkhng trc c lin h vi mt vt chnh `0` chnh gia mt trcca ccu o.

(5)

(1)

(2)

(3) (6)

(2)

(4)

I N S


19/93

37

2. Hotng:

- Dng in trong cun dy ca ccu TNCVC phi chy theo mt chiunht nh cho kim dch chuyn (theo chiu dng) t v tr `0` qua sut

thang o.- o chiu dng in cun dy quay theo chiu ngc li v kim blch vpha tri im `0`. Do cc u ni ca dng c TNCVC cnh du `+` v `-` cho bit chnh xc cc cn ni. Ccu TNCVCc coi l c phn cc.

- Phng trnh m men quay v thang o:Khi c dng in I chy qua khung dy s to ra 1 t trng tng tc vit trng B ca NCVC to ra 1 mmen quay:

: bin thin ca t thng qua khung dyB: t trng NCVC


dd

Id

dWM

eq

dSNBd ...

38

N: s vng dyS: din tch khung dyd: bin thin gc quay ca khung dy

Mq= I.B.N.SM men quay Mq lm quay khung dy, khi mmenphn khng do l xo phnkhng tc ng vo khung dy tng

Mpk= D. (3.5)D - h sphn khng ca l xo - gc quay ca kim

Khi mmen quay Mq cn bng vi mmen phn khng Mp ca l xo th kim sdng li trn mt s ng vi mt gc no .

Mq = -Mpk (3.6)


ISID

SNB

DSNBI

...

....

0

D

SNBS

..0 l nhy ca ccu o


20/93

39

3. c im ca ccu o t in:+ u im: Thang o tuyn tnh c th khc thang o ca dng in I theo gc

quay ca kim ch th

nhy ccu o ln Dng ton thang (Itt) rt nh (cA) chnh xc cao, c th to ra cc thang o c cp chnh xc ti 0,5% t chu nh hng ca in t trngbn ngoi.

+ Nhc im: Cu tophc tp, db h hng khi c va p mnh Chuqu ti km do dy qun khung c ng knh nh

Ch lm vic vi dng 1 chiu, mun lm vic vi dng xoay chiu phi cthm it nn in

* ng dng: dng rt nhiu lm c cu ch th cho cc dng c o in nhVnmt, Ampemt,, cc php o cuc nbng


40

3.2.2 Ccu in t: hot ng theo nguyn l: nng lng in t c bin ilin tc thnh cnng nhs tng tc gia t trng ca cund y tnh khic dng in i qua vi phn ng ca ccu l c c l st t

1. Cu to: c 2 loi- Loi cun dy hnh trn.- Loi cun dy hnh dt


+ Loi cun dy hnh trn:

-Phn tnh: l 1 cun dy hnh tr trn, phatrong thnh ngc gn l st t mm un quanh

- Phn ng: gm 1 l st t cng c uncong v gn vo trc quay nm i din. Trntrc quay gn kim ch th v l xophn khng

Cun dytrn

L stng

L st tmm tnh

kim ch th

I1


21/93

41

+ Loi cund ydt:- Phn tnh: gm 1 cund ydt, gia c 1 khe hp.- Phn ng: gm 1 a st t c gn lch tm, ch

mt phn nm trong khe hp v c th quay quanhtrc. Trn trc ca a st t c gn kim ch th v l

xo phn khng2. Nguyn l hotng chung:

Khi c dng I in chy qua cun dy tnh s to ramt nng lng t trng


Cun dydt

kim ch th

I

ttq

dWM

d

21

2ttW LI

vi L l in cm cun dy, c gi tr tu thuc vo v tr tng i ca lst t ngv tnhSbin thin nng lng t trng theo gc quay to ra mmen quay trc quay kim ch th quay

42

Khi kim ch th quay mmenphn khng tng: Mpk=D.Ti v tr cn bng: Mpk= - Mq

Gc quay ca kim ch th t l vi bnh phng ca I qua cun dy

3. c im ca CC in t:+ u im:

CC t inc th lm vic vi dng xoay chiu. C cu to vng chc, kh nng chu ti tt.


21

2ttdW dLD I

d d

21 12tt

dWdLID d D d

20IS

d

dL

DS

2

10

,


22/93

43


+ Nhc im: nhy km do t trng phn tnh yu Thang ophi tuyn chnh xc thp do d nh hng ca t trng bn ngoi do tn hao st

t lnTuy nhin vn c dng nhiu trong cc ng h o in p ln

3.3 Ccu ch th s:

Vt cno

Tr s oc

Khong tcha cc xung

c tn s f

m xungtrong t

Hin thdi dng

ch s

kt qubin i

bin io

44

Chng 3. Cc b ch th trong my o1. Nguyn l hotng chung: cc ccu o hin th s thng dng phngphp bin i tr s ca i lng o ra khong thi gian c lu t phthuc tr s o cha y cc xung lin tip vi tn s nht nh.Thit b ch th m s xung trong khong thi gian t v th hin kt quphp m di dng ch s hin th.2. Cc c im:

(a) Cc u im: chnh xc o lng cao. Ch th kt qu o di dng ch s nn d c. C kh nng t chn thang o v phn cc Trkhng vo ln. C th lu li cc kt qu o a vo my tnh. Dng thun tin cho o t xa.

(b) Cc nhc im: S phc tp Gi thnh cao bn vng nh


23/93

45


3.3.1 B chthsdngit pht quang (LED_ Light Emitting Diode)-Do s ti hp ca cc phn t mang in (in t v l trng) ca lp tipxc p-n khi nh thin thun (cc e- vt t pha n v ti hp vi cc l trngti pha p), cc phn t mang in spht ra nng lng di dng nhit vnh sng.

-Nu vt liu bn dn trong sut th nh sng c pht ra v lp tip xc p-nl ngun sng. N l ngun itpht quang (LED).-Khi nh thin thun, phn t trng thi ng v pht sng.-Khi nh thin ngc, phn t trng thi ngt.

46

- S ti hp ca ccphn t mang in xy ra trong vt liu loi p, nn min pl b mt ca phn t it. c spht sng ti a, mng ant kim loic cho kt ta quanh mp ca vt liu loi p. u ni ca catt ca phnt loi ny l mng kim loi y ca min loi n.

* LED 7 on:

- Cc dng c o hin th s thng dng b ch th 7 on sng ghp li vinhau theo hnh s 8. Cc on sng l cc itpht quang. Khi cho dngin chy qua nhng on thch hp c th hin hnh bt k s no t 0-9.

- C 2 loi: LED 7 on sng Ant chungLED 7 on sng Catt chung

LED 7on sng Catt chung: catt ca tt c cc it u c ni chung viim c in th bng 0 (hay cc m ca ngun). Tc ng vo u vo

(ant) ca it mc logic 1 it sng.LED 7on sng Ant chung: cc ant c ni chung vi cc dng ca ngun(mc logic 1). Tc ng vo u vo (Catt) ca it mc logic 0 itsng.



24/93

47

st p khi phn cc thun it l 1,2V v dng thun khi c chi hp l l20mA.Nhc im: cn dng tng i ln.u im: ngun in p mt chiu thp, kh nng chuyn mch nhanh, bn, kchtc b.

3.3.2 B chthsdng tinh thlng (LCD):

-Tinh th lng l tn trng thi ca mt vi hp cht hu c c bit. Cc cht nynng chy 2 trng thi: lc u trng thi nng chy lin tc, sau nu nhit tip tc tng th chuyn sang cht lng ng hngbnh thng.-Pha trung gian gia hai trng thi ny l trng thi tinh th lng (va c tnhcht lng va c tnh cht tinh th).


48


-B ch th dng tinh th lng (LCD) thng cb tr cng theo dng s 7 on nh b ch thLED.-Trn 2 tm thu tinh c ph mt lp kim loidn in lm nn 2 in cc trong sut, gia 2 lpkim loi l lp cht lng tinh th.

-Khi ch th ch s, ngoi in p t vo 2 incc ca phn t cn cn ngun sng t pha trchay pha sau ca b ch th v phng.

(a) ngun sngt trc: khi c tnhiu th tinh th lng c nh sng

phn x t gng.

Cu to mi thanh


25/93

49


+ u chung ca cc phn t ch thLCD ni vi +E qua R.+ Cc in cc ring ni vi cc u raiu khin.

+ Khi transistor T6 tt, U6a = 0 phnt 6 khng ch th.+ Khi T6 thng, U6a = +E kchthch phn t 6 tr nn trong sut,cho nh sng i qua.

-(b) ngun sngt sau: khi c tn hiu th tinhth lng c nh sng i qua to nn hnh s trnmn hnh. Mn hnh l tm phng en.-Ngun in cung cp l ngun 1 chiu hoc lngun in p xung. VD:

50


u im ca chth tinh thlng:Ngun cung cp n gin, tiu th cng sut nh, cmW Kch tc b, ph hp vi cc thit b o dng mch t hp, kthut vi in t. Hnh ch s kh r rng, ch to n gin.

Nhc im: di nhit lm vic hp (100C-550C) tui th cha tht cao

Tuy vy cc u im l cbn nn loi ny ngy cng c dng nhiu, bit ltrong thit b o y t v mu sc c th thay i theo nhit bnh nhn.


26/93

51

Chng 4. My hin sng (xil)

4.1 Nguyn l quan st tn hiu trn MHS:

1. Phng php vdao ng ca tn hiu- Mt tn hiu thng c biu din di 2 dng:

+ Hm theo thi gian: u = f(t)+ Hm s theo tn s: u =(f)

- quan st dng sng, o cc c tnh v cc tham s ca tn hiu dng mt my o a nng l MHS (xil).

- MHS l mt loi my v di ng theo 2 chiu X v Y hin th dng tnhiu a vo cn quan st theo tn hiu khc hay theo thi gian. `Kim btv ` ca m y l mt chm sng, di chuyn trn mn hnh ca ng tia int theo qui lut ca in p a vo cn quan st.

Ff

f-F f+F f f

UAM

0

52


* Cc loi xil:-xil tn thp, xil tn cao, xil siu cao tn-xil xung (/T b)-xil 2 tia; xil nhiu knh-xil c nh(loi tng t v loi s)-xil s; xil c ci t VXL

2. Cng dng, tnh nng ca xil:xil l mt my o vn nng, n c cc tnh nng:-Quan st ton cnh tn hiu-o cc thng s cng ca tn hiu:

+ o in p, o dng in, o cng sut+ o tn s, chu k, khong thi gian ca tn hiu+ o di pha ca tn hiu

+ v t ng v o c c tnh ph ca tn hiu+ v c tuyn Vn-Ampe ca linh kin+v t ng, o c tuyn bin -tn s ca mng 4 cc


27/93

53


3. Cc chtiu kthut ch yu ca xil:

-Phm vi tn s cng tc: c xc nh bng phm vi tn s qut.- nhy (h s li tia theo chiu dc): mV/cmL mc in p cn thit a n u vo knh lch dc bng bao nhiu mV tia in t dch chuyn c di 1 cm theo chiu dc ca mn sng. nhy cng c th c tnh bng mm/V.-ng knh mn sng: xil cng ln, cht lng cng cao th ng knhmn sng cng ln (thng thng khong 70mm-150 mm).-Ngoi ra cn c h s li tia theo chiu ngang, trkhng vo,...

4. Ch qut tuyn tnh lin tc

a) Nguyn l qut ng thng trong MHS-a in p ca tn hiu cn nghin cu ln cp phin lch Y, v in pqut rng ca ln cp phin lch X.

54

Chng 4. My hin sng (xil)-Do tc dng ng thi ca c hai in trng ln 2 cp phin m tia in tdch chuyn c theo phng trc X v Y.-Qu o ca tia in t dch chuyn trn mn s vch nn hnh dng ca inp nghin cu bin thin theo thi gian.Ch : in p qut l hm lin tc theo thi gian qut lin tc

in p qut l hm gin on theo thi gian quti

b) Nguyn l qut tuyn tnh lin tc-in p qut tuyn tnh lin tc c tc dngli tia in t dch chuyn lp i lp li 1 cchlin tc theo phng ngang t lbc nht vithi gian.- qut tuyn tnh lin tc cn phi dng in

p bin i tuyn tnh lin tc (tng tuyn tnhhay gim tuyn tnh)

Qut t2 lin tc vi Tq = Tth


28/93

55


Chu k qut: Tq = tth + tng

Thng thng: tng 15% tth tc l tng rt nh hn tth nn c th coi Tq tth, ltng: tng = 0 (Tq = tth)

-Nu tn s qut cao, mn hunh quang c d huy mc cn thit th khimi ch c Uq t v ocp phin X c mt ng sng theo phng ngang. Khic c Uth t v o cp phin Y v nu Tq = nTth th trn mn xut hin dao

ng ca mt hay vi chu k ca in p nghin cu (Uth).-Nu Tq nTth th dao ng khng ng yn m lun di ng ri lon khquan st. Hin tng ny gi l khng ng b (khng ng pha gia Uq v Uth).

tngtth

Uq

t

Nn

56


-Thc t, tng 0. V tng


29/93

57


-H s khng ng thng ():

c nh quan st vi cht lng cao cn:- tng


30/93

59


6. Nguyn l ng b:-Khi quan st dng tn hiu trn MHS, i khi nh b tri, nhy,... l do mtng b.

*

Minh ha :

nh I, II, III l cc dao ng tng ng ti cc chu k qut tng ng. Nphn b ln lt t tri qua phi, do tnh cht lu nh ca mn hnh cc nh smdn theo th t tng ng cm gic dao ng chuyn ng t tri quaphi.

* tng t, dao ng c cm gic chuyn ng tphi qua tri

Nnthqth nTTTn

141

thqth TTT 143

thqth TnTnT

4

12

b

a

T

T

q

th 2

,

60


* (minh ha ):

Dao ng ng yn nhng khng phn nh ng dng tn hiu cn quan stm ch gm nhng on tn hiu khc nhau cn quan st m thi.

* Tq = nTth (minh ha Tq = Tth ),Dao ng n nh v phn nh ng dng tn hiu cn quan st. iu kin ng b: Tq = nTthQu trnh thit lp v duy tr iu kin ny l qu trnh ng b ca MHS

-Cc ch ng b:

+ ng b trong: tn hiu ng b ly t knh Y ca MHS

+ ng b ngoi (EXT)+ ng b li (LINE)

b

a

T

T

q

th 3 3

4

3

q

th

T

T

Nn


31/93

61


0

0

Uq1

Utht

t

Tq1

Tth 2Tth 3Tth

Tth

0

Uq4t

Tq4

0

Uq2t

Tq2

0

Uq3t

Tq3

62


4.2 S cu to mt MHS in hnh

1. Cu to MHS:- ng tia in t- Knh lch ng Y- Knh lch ngang X v ng b

- Knh Z (khng ch sng)

* ng tia in t:+ l bphn trung tm ca MHS, sdng loi ng 1 tia khng chbngin trng+ C nhim v hin th dng sng trnmn hnh v l i tng iu khin

chnh (Uy, Ux, UG).

Mchvo vphnp Y

Tinkhuch

i

Dytr

Khuchi Y

ixng

Toxungchun

K/ing bv todng

Toxung

ng b

Toin p

qut

i

lintc

Mchvo vK X

K/i Xi

xng

Chncc tnh K/iZ

Knh lch ng Y

Knh lch ngang X v ng b

Knh ZTi G ca CRT

UZ

UxUqut

S3

S2

S1AC

DC

GND

Uth

Vpp

CH

EXTLINEAC

50Hz

Ub Ux

Uxb

1

2

3

CRTX1

X2

Y1

Y2

Hnh: S khi MHS 1 knh dngng tia in t


32/93

63

* Knh lch ng Y: c nhim v nhn tn hiu v o cn quan st, bin i v tora in pphhp cung cp cho cp li ng Y1, Y2. Gm cc khi chc nng:

+ Chuyn mch kt ni u v o S 1: cho php chn ch hin th tn hiu.S

1

ti AC: ch hin th thnh phn xoay chiu ca Uth

.S1 ti DC: hin th c thnh phn mt chiu v xoay chiu ca Uth.S1 ti GND: ch quan st tn hiu ni t (0V).

+ Mch vo phn p Y: c nhim vphi hp trkhng v phn p tn hiu vo tng kh nng o in p cao. Thng dng cc khu phn p R-C mc lintip nhau, h s phn p khng ph thuc vo tn s. Chuyn mch phn pc a ra ngoi mtmyvkhiu l Volts/Div.+ Tin khuch i: c nhim v khuch i tn hiu, lm tng nhy chung ca

knh Y. Thng dng cc mch K c trkhng vo ln v c h s K ln.+ Dy tr: c nhim v gi chm tn hiu trc khi a ti K Y i xng,thng dng trong cc ch qut i trnh mt mt phn sn trc ca tnhiu khi quan st. Thng dng cc khu L-C mc lin tip.


64


+ K Yi xng: c nhim v K tn hiu, lm tng nhy chung ca knh Y,ng thi to ra in p i xng cung cp cho cp li ng Y1Y2.

+ To in p chun: to ra in p chun c dng bin , tn s bit trc,dng kim chun li c c h s lch tia ca MHS

* Knh lch ngang X v ng b: c nhim v to ra in p qut ph hp vdng v ng b vpha so vi UY1, Y2 cung cp cho cp li ngang X1X2

+ Chuyn mch ng b S2: cho php chn cc tn hiu ng b khc nhau.S2 ti CH: t ng b (Ub = Uth)S2 ti EXT: ng b ngoi (Ub=UEXT), tn hiu ng b c a qua u voEXT.S2 ti LINE: ng b vi li in AC 50Hz (Ub=UAC50Hz) ly t ngun nui.

+K ng b v to dng: k/i tn hiu Ubph h p v to r a dng xung nhnn cc tnh c chu k: Tx=Tb


33/93

65

+ To xung ng b: chia tn Ux v to ra xung ng b c chu k:Txb=nTx=nTb. Xung ny s iu khin b to in p qut to ra Uq rngca tuyn tnh theo ch qut i hoc qut lin tc v c c huk Tq=Txb.

+K Xi xng: K in p qut v to ra in p i xng a ti cp li

ngang X1X2.+ Mch vo v KX: nhn tn hiu UX v k/i, phn p ph hp.

+ Chuyn mch S3: chuyn mch la chn ch qut (qut lin tc, qut i)

+ B to in p qut: to in p qut lin tc (hoc qut i) a n cpphin X

* Knh iu khin ch sng Z: c nhim v nhn tn hiu iu ch sngUZ vo, thc hin chn cc tnh v k/i phhp ri a ti li iu ch G caCRT.


66


2. Cu to ca ng tia in t:

ng tia in t CTR (Cathode Ray Tube) l 1 ng thu tinh hnh tr c chnkhng cao, u ng c cha cc in cc, pha cui loe ra hnh nn ct, mt yc ph 1 lp hunh quang to thnh mn hnh. Cu to gm 3 phn:

Sng in t H thngli tia Mn hnh

Y2

Si tF

KG A1 A2

Y1

X1

X2R1 R2

RfocusRbright

-E A hu

Lpthan ch

Mn hunh quang

Hnh : S cu to ca ng tia in t


34/93

67


a) Mn hnh: - Lp hunh quang thng l hp cht ca Phtpho. Khi c in tbn ti mn hnh, ti v tr va p, in t s truyn ng nng cho cc in t lpngoi cng ca nguyn t Phtpho, cc in t ny s nhy t mc nng lngthp ln mc nng lng cao v tn ti trong 1 thi gian r t ngn ri t nhy vmc nng lng thp ban u v pht ra photon nh sng.- Mu sc nh sng pht ra, thi gian tn ti ca im sng ( d huy ca mnhnh) sph thuc v ohp cht ca Phtpho (t vi s n vi s).

b) Sngin t: gm si t F, catt K, li iu ch G (M), cc ant A1,A2.Nhim v: tog ia tc v hi t chm tia in t- Cc in cc c dng hnh tr, lm bng Niken, ring Katt c ph mt lp xitkim loi y tng kh nng bc x in t.- Cc in cc pha sau thng c vnh rng hn in cc pha trc v c nhiu

vch ngncc chm in t khng i qu xa tr c ng vic hi t s d dnghn. Vi cu to c bit ca cc in cc nh vy s to ra 1 t trng khngu c bit c th hi t v gia tc chm tia.

68


Ngun cp: UK = -2kVUKG = 0-50VUA2 = 0VUA1 = 300V

+ Li iu ch G c cung cp in p m hn so vi K v c ghp st K d dng cho vic iu chnh cng ca chm in tbn ti mn hnh.+ Chit p trn G (iu chnh in p) thng c a ra ngoi mt my v khiu l Bright hoc Intensity dng iu chnh sng ti ca dao ng trn mn hnh.+ Ant A2 (Ant gia tc) thng c ni t trnh mo dao ng khiin p cung cp cho cc in cc khng phi l in p i xng.+ Ant A1 (Ant hi t) cng c chit p iu chnh a ra ngoi mt my, khiu l Focus, dng iu chnh hi t ca chm tia in t trn mn hnh.


35/93

69


c) H thng li tia: c nhim v lm lch chm tia in tbn ti mn hnh theochiu ng hoc chiu ngang ca mn hnh.Cu to gm 2 cp phin l mlch c t trc, sau v bao quanh trc ca ng.+ Cp li ng Y1Y2.+ Cp li ngang X1X2 .

y a

70

*Xt qu o ca chm tia in tkhi i qua in trng ca 2 ant A1, A2:

+ UA2 > UA1 ng sc in trng c chiu i t A2 n A1+ e- chuyn ng theo chiu t A1 ti A2 nn n ng thi chu tc ng ca 2thnh phn lc, 1 thnh phn theo phng vung gc vi chm tia v 1 thnhphn dc theo chm tia.

+ Ti im A: chm e- c khuynh hng chuyn ng dc theo phng trc ng,ng thi hi t vi nhau theo phng bn knh ca chm tia

+ Ti B: thnh phn lc theo phng bn knh i chiu ngc li chm e- ckhuynh hng phn k khi tm theo phng bn knh.


Tuy nhin do cu to ca cc in cc, sphn b ca ng sc B t b cong hnphn v tr im A phn lng vntc theo phng bn knh B < im A khuynh hng hi t ca chm e- >khuynh hng phn k.


36/93

71


*Xt lch ca tia in ttheo chiu ng:

Khi Uy= 0, tia in tbn ti chnh gia mn hnh ti im O.Khi Uy 0, in trng gia cc phin l mlch s lm lch qu o ca tiain t theo chiu ngv bn ti mn hnh ti v tr M, lch 1 khong l y.

Ly: khong cch t cp li ng n mn hnh.ly: chiud ica c c cp phin l mlch.dy: khong cch gia 2 phin l mlch.UA: in pg ia tc ca ng tia (ph thuc v oUA2 v K).

dy

A2

Uy Y1

Y2

ly

Ly

y

M

O

chm e-+

-

Mn hnh

2y y y

oy yy A

U l Ly S U

d U

2y y

oyy y A

l LyS

U d U nhy ca ng

tia in t

72

Tng t, lch ca tia in t theo chiu ngang:

* Nguyn l to nh trn mn my hin sng:

iu khin ng thi tia in t theo 2 trc: trc thng ng v tr c nmngang, ngha l ng thi a vo n ng tia in t 2 in p iu khin UYv UX . Gi s a vo knh Y v a ti cp li ng Y1Y2;

a ti cp li ngang X1X2 in p trn cc cp li tia nh sau:

2x x x

ox xx A

U l Lx S U

d U


sin .th mU U ttaqU .

ySthUyyUyU 21 oySyKyS


37/93

73

* Mch to in p quti (mch Boostrap)

Nhn xt sb:

- D thng, tng T1 hot ng nh 1 mch TF n gin, T1- kho in t.

- in p trn t C T2 C*. in p b a v im X b mo do sgim dng in trn R gy ra.

Hot ng:

1. Trng thi ban u (t1 t 0):

- D thng: UX = EC UD EC

- T1 thng bo ho: UC = UCbh 0V

- T2 k/i C chung nn: URE = Uq 0V

- UC* = UX - Uq EC


74


2. Trng thi to Uq (t3 t t1):

t = t1: Ukc rng = t3 - t2T1 tt t C np in theo 2 giai on:

Giai on 1: t1 t > C nn UX tng tng ng.UR= UX UC = const IR =const dng n p cho t C

khng i, vi thi gian np tq = t3 t2.


38/93

75


3. Trng thi hi phc:t = t3: kt thc xung u vo. T1bo ho, t C phng in qua T1 UC gim

nn Uq gim UX gim.t = t4: UX EC D thng trli.

D thng, t C* c np b sung, khi C* c np y th thi gianhi phc kt thc.

Nhn xt:

-Trong giai on to qut tq, UC tng tuyn tnh nhngun np chnh l C* ctr s cc ln tch in (lm nhim v ging nh ngun 1 chiu np cho C).- gim mo phi tuyn tng K T2phi c iu chnh sao cho kU 1(R ln).

-T C* phi ln nhng khng qu ln v nu qu ln th thi gian hi phc camch tng.-T C phi nh nhng khng c chn qu nh v nu qu nh th n c gi trtng ng nh t k sinh mch lm vic khng n nh.

76


39/93

77

78


3. Mt sch lm vic:

a. Qut lin tc ng b trong (ngoi)-Dng quan st nh ca tn hiu lin tc theo thi gian v o cc tham sca chng.-S2 v tr CH (hoc EXT nu l ng b ngoi), S3 v tr 2

-Tn hiu t li vo knh Y, qua Mch vo v bphn p Yc khuch iti mt mc nht nh, sau c gi chm li ri a qua B K Y i

xng to 2 tn hiu c bin ln, o pha nhau a ti 2 phin ng


40/93

79

b. Quti ng b trong

-Dng quan st v o tham s ca dy xung khng tun hon hoc dyxung tun hon c hng ln.

-S2 v tr CH, S3 v tr 1

-Qu trnh hot ng: ging ch 1

c. Ch khuch i

-Dng o tn s, gc lch pha, su iu ch, v c tuyn Vn-Ampeca it hoc dng lm thit b so snh. Hnh nhn c trn mn MHS gil hnh Lixazu

- S3 v tr 3-B to qut trong c ngt ra khi qu trnh hot ng. MHS lm victheo 2 knh c lp X,Y v u vo X cng l u vo tn hiu


80

Chng 4. My hin sng (xil)4.3 MHS nhiu tia

Dng quan st ng thi nhiu qu trnh (tn hiu)

1. MHS 2 tia c li vo cp phin lch ngtch bit(knh A, knh B):

-Mi knh c mch K lm lch ring-Mt b to gc thi gian chung cho c 2 knh

2. MHS 2 knh dngng tia in t1 tiav CMin t-Hai b K tn hiu vo ring cho knh A,knh B

-Mt b K lch ng cho c 2 knh.Tn hiuv ob K ny c chuynmch lun phin gia 2 knh.-B to gc thi gian (b to sng qutngang) iu khin tn s chuyn mch


41/93

81


a) Phng php dng chuyn mch in t kiulun phin (ALT mode):

+ Tn hiuv ob K lch ng c chuynmch lun phin gia cc knh A v B.

+ B to gc thi gian iu khin tn s chuynmch+ 0t1: tn hiu t knh A c ni ti b K lchng, to thnh vt trn mn hin sng+ t1t2: tn hiu t knh B c ni ti b K lchng, to thnh vt trn mn hin sng+ Hai tn hiu hai knh c cng chu k T vc ng b vi nhau.

+ Dch chnh DC: dch chuyn tn hiu knh A(knh B) trn mn theo phng thng ng bngin p mt chiu+ cc chu k ti p theo: qu trnh lp li nhtrn. Tn s l p cao n mc m cc dng sngnh c hin ng thi.

c)

Dngsng

hin

82

Chng 4. My hin sng (xil)b)Phng php dng chuyn mch in tkiu ngt qung (Chop mode switching):S dng tn s chuyn mch cao hn nhiu so vi ch lun phin.

+ T1, T3, T5, T7,... tn hiu vo knh A c tora trn mn+ T2, T4, T6,... tn hiu vo knh B c to ratrn mn

+ Cc dng sng knh A v B c hin hnhnh nhng ng t nt. Khi tn s chuynmch l cao tnkhng th nhn ra nhng cht nt+ fth nh: nh hin trn mn MHS gn nh lintc+ fth ln; nfcm mfth : cc on ngt b lp do d huy ca ng v lu nh ca mt.

Ch : i vi tn hiu cao tn th kiu lunphin l tt nht, cn i vi tn hiu tn s thpth nn dng chuyn mch ngt qung


42/93

83


Chuyn mch in tphn ng theo thi gian:

(a)

(b)

(c)

(d)UZ

84

Chng 4. My hin sng (xil)-Mi knh tn hiu c cng thm mt lng in p 1 chiu E khc nhau cc ng biu din trn mn hnh MHS c tch ring tng ng, hnh (a).-Sau tn hiu c a n mch ca, v ch qua c ca khi c tn hiu mca tbPht sng chuyn mch.-Tn hiu mca l cc xung vung c thi gian xut hin xen k v ln ltcho tng ca mt, hnh (b).

-Ti mi thi im ch c duy nht 1 ca c mv cho tn hiu ca mt knhi qua.-B tngcng cc tn hiu u ra cc ca, UY c dng xung vi bin t lvi gi tr ca cc tn hiu cn quan st ti thi im c xung mca tng ngvi cc knh, hnh (c).-Sau khi khuch i Y, MHS c c hnh biu din tn hiu ca cc knh didng ng nt t, hnh (c).

-MHS lm vic ch ng b vi chu k ca tn hiu cn quan st v khngng b vi tn hiu chuyn mch.-Dng nhng xung c rng rt nh (UZ) c to ra t mch vi phn t ccxung mca a vo knh Z iu ch sng ca nh, hnh (d).


43/93

85


Chuyn mch in tphn ng theo mc:

En

86


-Phn hin th l ng truyn hnh lm lch bng ttrng-Nguyn l hot ng: chuyn cc gi tr tc thi ca tn hiu cc knh thnh ccchui xung rt hp xut hin ti cc thi im m tu thuc vo in p tn hiunghin cu. Cc xung ny c a vo khng ch sng ca ng hinhnh.

+ Mi knh tn hiu c cng thm mt lng in p 1 chiu E khc nhau,ri a n so snh vi tn hiu l xung rng ca a ti t b K lch ngca MHS (tn hiu qut dng).+ Mi khi URC = Uth, th u ra ca bso snh s xut hin mt xung hp. Ccxung hp ny c cng vi nhau ri a vo khng ch sng ca ng hinhnh.+ Ti thi im c xung, trn mn hnh xut hin mt chm sng trong khi bnhthng th ti. Vt ca chm sng trn mn hnh biu din hnh in p ca cctn hiu cn quan st.


44/93

87


4.4 xil in ts:

1.u im:- Duy tr nh ca tn hiu trn mn hnh vi khong thi gian khng hn ch.

- Tc c c th thay i trong gii hn rng-C th xem li cc on hnh nh lu gi vi tc thp hn nhiu- Hnh nh tt hn, tng phn hn so vi loi xil tng t- Vn hnh n gin- S liu cn quan st di dng s c th c x l trong xil hoc truyntrc tip vo my tnh khi ghp xil vi my tnh.

88

Chng 4. My hin sng (xil)2. Cu to v hotng:

Chuyn mch Sv tr 1: xil a nng thng thngChuyn mch Sv tr 2: xil c nhs.-in p cn quan st c a ti u vo Y, ti b ADC. Lc b iukhin gi 1 lnh ti u vo iu khin ca bADCv khi ng qu trnh bin

i. Kt qu l in p tn hiu c s ho. Khi kt thc qu trnh bin i, bADCgi tn hiu kt thc ti b iu khin.-Mi s nh phn c chuyn ti b nhv c nh v tr nhring bit.


45/93

89


-Khi cn thit, mt lnh t b iu khin lm cho cc s nhphn ny sp xptheo chui li theo th t xc nh v c a n bDAC-DACbin i cc gi tr nhphn thnh in p tng t a qua b khuchi Y ti cp phin l mlch Y ca ng tia in t.

-Do b nh c lin tip qut nhiu ln trong 1 giy nn mn hnh c snglin tc v h in dng sng l hnh v cc im sng.Nhc im: di tn b hn ch (khong 1-10MHz) do tc bin i ca bADC thp.Hin nay, cc xil c nhs c di tn rng c pht trin nhci t VXL,cc bbin i ADC c tc bin i nhanh hn.

90

Chng 5. o tn s, khong thi gian v o di pha5.1 Khi nim chung-Tn s l s chu k ca 1 dao ng trong mt n v thi gian.-Tn s gc: biu th tc bin i pha ca dao ng

l tn s gc tc thi v tn s tc thi

-Quan h gia tn s v bc sng:

hay

-Quan h gia chu k v tn s:

c im ca php o tn s:+ l php o c chnh xc cao nht trong kthut o lng nhspht trin

vt bc ca vic ch to cc mu tn s c chnh xc v n nh cao.+Lng trnh o rng (n 3.1011 Hz). Lng trnh o c phn thnh cc ditn s khc nhau.

t

dt

dt

tft 2 tft ,

cf

f

c

Tf

1


46/93

91

- Di tn thp: < 16Hz- Di tn s m thanh: 16 Hz < f < 20 KHz- Di tn s siu m: 20 KHz < f < 200 KHz- Di tn s cao: 200 KHz < f < 30 MHz

- Di tn s siu cao: 30 MHz < f < 3000 MHz- Di tn s quang hc: > 3GHzCc di tn s khc nhau c cc phng php o tn s khc nhau.

Bao gm:+ Nhm phng php o tn sbng cc mch in c tham sph thuc tn s+ Nhm phng php so snh+ Nhm phng php s

Php o tn s thngc s dng kim tra, hiu chun cc my to tnhiu o lng, cc my thu pht; xc nh tn scng hng ca cc mch daong; xc nh di thng ca b lc; kim tra lch tn s ca cc thit bang khai thc

Chng 5. o tn s, khong thi gian v o di pha

92

5.2 o tn s bng cc mch in c tham s ph thuc tn s:5.2.1 Phng php cu

Dng cc cu o m iu kin cn bng ca cu ph thuc vo tn s cangun in cung cp cho cu.

*Mch cu tng qut:iu kin cn bng cu:

VD1:


0.. 4231 ABUZZZZ

3

Z1

Z3Z4

Z2

iu kin cn bng cu:

1 3 2 4

3 3 33

. .

1

R Z R R

Z R j L C

Hnh 5-1

Hnh 5-2

BA

L3


47/93

93

iu chun nhnh cng hng ni tip cho cng hng ti tn s cn o fx (iuchnh C3).

Khi

B ch th cn bng l vn mt chnh lu, vn mt in t.

Nhc im:-Kh o c tn s thp do kh ch to cun cm c L ln tn s thp.-Kh thc hin ch th 0 do c tc ng ca in t trng ln cun cm


4231 .. RRRR 33 RZ

33 3 3

1 1

2x x

x

L fC L C

94


VD2:iu kin cn bng cu:

Chn

''1 32 4

3 3 4

. 1

1 x x

R RR R

j R C j C

4

3

3

4'2

'

1CCRRRR

4334

1RRCC xx v

4343

12

CCRRfxx

CCC 43RRR 43 v ta c:

Hnh 5-3

V

RCfx

2

12

'2

'1

R

R;

VR1,VR2 l phn in trca bin trVR trn nhnh 1,2 tng ng

11'1 RVRR 22

'2 RVRR


48/93

95

VD3: Cu T kpiu kin cn bng cu:

Khi :

Thang ca bin trR1 c khc trc tip theo n v tn s.

Phng php cu dng o tn s t vi chc Hz n vi trm Khz.Sai s: (0,5-1)%


12

2

212

12

2122

2

RRC

CCR

x

x

12 2RR 12 2CC v

112

1

CRx

114

1

CRfx

Hnh 5-4

96

Chng 5. o tn s, khong thi gian v o di pha5.2.2 Phng php cng hng

- Dng o tn s cao v siu cao- Nguyn tc chung: da vo nguyn l chn lc tn s ca mch cng hng.- Khi cbn ca s ny l mch cng hng. Mch ny c kch thchbng dao ng ly t ngun c tn s cn o thng quaKhi ghp tn hiu.

- Vic iu chnh thit lp trng thi cng hng nhdngKhi iu chun.- Hin tng cng hng c pht hin bng Khi ch th cng hng. Khiny thng l Vnmt tch sng.-Tu theo di tn s m cu to ca mch cng hng khc nhau. C 3 loi mchcng hng:

+ Mch cng hng c L, C tp trung+ Mch cng hng c L, C phn b+ Mch cng hng c L phn b, C tp trung.

U(fx) Khighp tn

hiu

Ch thCH

MchCH

iuchun

Hnh 5-5


49/93

97


1. Tn smt cng hng c tham stp trung.

+ y C v L u l cc linh kin c thng stp trung. Bphn iu chnh cng hng chnhl tbin i C c thang khc theo n v tn s.+ Ufx c ghp vo mch cng hng thng quacun ghp Lg.+ Mch ch th cng hng l mch ghp h cmgia cund yL2 v L v c tch sng bng itv ch thbng ccu o t in.+ Khi o ta a Ufx vo v iu chnh t C mchcng hng. Khi ccu o s ch th cc i.

Lg L

L2

Ufx C

D

T iuchnh

Ch th cnghng

+ Tn s mt loi ny thng dng trong di sng: 10 kHz 500 MHz.+ Sai s: (0,25-3)%

Hnh 5-6

LCfx 21

98


+ Cc ch ghp u gn v tr ni tt c nh sao cho cc v tr ny gn vi vtr bng sng khi c chiu di tng ng ltd=/2 th thit b ch th s chcc i.+ Khi dch chuyn pt tng vi di bng bi s nguyn ln /2 s t cnghng c th xc nh bc sng bng cch ly 2 im cng hng ln cn

l1=n/2; l2=(n-1) /2 l1-l2=/2+ Kt qubc sng o c ca tn hiu siu cao tn xbi cng thc:=2(l1-l2)

Hnh 5-7

2. Tn smt cng hng c tham sphn bdngcp ng trc.

+mch cng hng l 1 on cp ng trcc ni tt 1 u, u kia c ni bng 1 pt tng Pc th dch chuyn dc trc bi h thng rng caxon c c khc .+ vng ghp Vg a t/h vo, cn vng ghp Vghp t/h ra mch ch th cng hng.

P

Vg

V lt


50/93

99


+Bc sng (hoc tn s) c khc trc tip trn h thng iu chnh pttng.+Tn s mt loi ny thng dng trong di sng t 3cm - 20cm+ Do c h sphm cht cao (khong 5000) nn sai s ca n khong 0,5%.

3. Tn smt cng hng c tham sphn bdngng dn sng

+ ng dn sng c th l loi ng dn sng chnht hay ng dn sng trn.+ Piston P c th iu chnh dc theo ng bi hthng rng ca xon c c khc tn s. Nnglng kch thch hc cng hng c ghp qua lhng G trn thnh c ni tt ca ng.+ Khi iu chnh piston P c l

td=n/2 th thit b

ch th s ch cc i.+ Tn s mt vi hc cng hng thch hp vi disng nh hn 3cm.+ Do c h sphm cht cao (khong 30000) nnsai s ca nnh khong (0,010,05)%.

Hnh 5-8

PVlt

G

D

100


- Khi :vi nY, nX nguyn dngnY : s giao im ca ng ct dc vi nhnX : s giao im ca ng ct ngang vi nh

- Tng qut:

fX : tn s a vo knh lch ngang XfY : tn s a vo knh lch ng Y

X

Y

x

m

n

n

f

f

X

Y

Y

X

n

n

f

f

5.3 Phng php so snh

Phng php qut sin:- MHS t ch khuch i.- in p c tn s cn o Ufx c a vo knh Y,in p c tn s mu Ufm a vo knh X.

- Hnh nh nhn c trn mn l hnh Lixazu. Thayi fm sao cho trn mn nhn c hnh Lixazu nnh nht. nY=4, nX=2

Hnh 5-9


51/93

101


5.4o tn sbng phng php s-L p2 hin i v thng dng nht o tn su im:+ chnh xc cao+ nhy ln+ Tc o ln, t ng ho hon ton trong qu trnh o+ Kt qu o hin th di dng s

1. Phng php xc nh nhiu chu ka. S khib. Chc nng cc khi:- Mch vo: thc hin tin x l nhphn p, lc nhiu... hoc bin i t/h

tun hon dng bt k u vothnh hnh sin cng chu k vi t/h vo.- Mch to dng xung: bin i t/hhnh sin c chu k Tx thnh t/h xungnhn n cc tnh c chu k Tx.

Hnh 5-10

Toxungchun

B mxung

Gii mv ch

th

Mchvo

Todngxung

Chiatn

KhoUfx

fch Uct Nx

xungcht

Toxungiukhin

Ux U

xung xoUk

102


- To xung chun: to ra cc xung chun c chnh xc cao vi tn s f0,xung chun ny c a qua b chia tn to ra xung c tn s lfCT=f0/n=10k(Hz)- To xungiu khin: nhn t/h f CT v to ra xung /kng m kho c rng t=TCT=10-k(s)- Mch gii m v chth: Gii m xung m c v a vo cc ccu ch

th s, c th l dng Led 7 on hoc LCD ch th kt qu cn o.-B m: m cc xung u ra.

c. Nguyn l lm vic:-Trong t/g c xung iu khin kho s cm, xung m qua kho kch thch cho bm xung.-Gi s trong 1 chu k m t, m c

Nx xung. S xung Nx ny s c a quamch gii m v ch th hin th kt qul tn s cn o t=NxTx=Nx/fx fx=Nx/t=10k.Nx vi k=0, 1, 2,... Gin thi gian

Hnh 5-11


52/93

103


d. nh gi sai s:- Sai s ca xung /k (t) do sai s ca b to xung chun v b to xung /k gyra.-Sa i s lng t: 1/Nx

fx tngNx tng 1/Nx gim.fx gimNx gim 1/Nx tng.

- Khi fx nh nh hng ca sai s lng t s ln trong TH ny ta s chuynsang p2 o x 1 chu k.

2. Phng php xc nh mt chu ka. S khib. Chc nng cc khi:

Nguyn l tng t nh trng hp nhiuchu k nhng khc ch in p c tn scn o s c bin i thnh xung /kng mkho, cn xung m ly tb toxung m chun.

Hnh 5-12

Toxungm

chun

B mxung

Giim vch th

s

Mchvo

Todngxung

Toxung

k

Kho

Ufx Ux

Uk

Uch U Nx

xungxo

xungcht

104

Chng 5. o tn s, khong thi gian v o di phac. Nguyn l lm vic:- T/h Ufx a qua Mch vo tiB to dng xung to ra xung nhn c chuk Tx. Xung ny s /kB to dng xung/k to ra xung /k c rngt=nTx (VD: n=1)- Trong t/gian c xung t, xung m chun Uch qua kho kch thch cho bm xung.

- Gi s m c Nx xung th s xung Nx ny s c a qua mch gii mv ch th t c kt qu l tn s hoc chu k cn o t=Tx=Nx.T0, viT0 l chu k ca xung m chun

fx=1/Tx = 1/NxT0 = f0/Nx

d. nh gi sai s:- Do sai s ca xung m- Do sai s lng t (1/Nx)

Kt hp 2 p2

o trn to ra 1 my mtn c di tn o rng v chnh xc cao.

Hnh 5-13 Gin thi gian


53/93

105


5.5 o di pha

Cc p2: phng php v dao ng ; pp bin i di pha thnh khong thigian; pp bin i di pha thnh in p; ...

111 sin tUu m

222 sin tUu m

21

106


2 /T T v

0 0

360T

T

2T

T

(rad) hay

5.5.1. o di pha bng pp o khong thi gian:

L phng php ph bin o pha

Nguyn l:

+ Bin i cc in p c dng hnh sin thnh

cc xung nhn tng ng vi cc thi im min p bin i qua gi tr 0 vi gi tr o hmcng du.

+ Khong thi gian gia 2 xung gn nhau ca 2in p o t l vi gc di pha ca chng.

Hnh 5-14


54/93

107


Pha mt dng mch a hi ng b

- Cc in p hnh sin cn o di pha c a vo 2 u vo I v II-in p hnh sin c bin i thnh cc xung vung nhMch K hn chv a hi ng b, ri c a n Mch vi phn phn b.(Cc chu k dao ng bn thn ca b a hi c chn sao cho n ln hn

chu k ca in p o c tn s thp nht)- u ra ca Mch vi phn phn bl cc xung nhn, c a ti khng chhai b a hi ng b I v II.- u ra ca 2 b a hi ny c a ti mt mch tng hp, mch ny cng h o thi gian lch gia cc xung, cng l gc di pha ca 2 in p.

Hnh 5-15

108


- Mch vi phn phn b:

+ u ra ca n a ti u vo B ahi ng b I ch cc xung nhn dng(hnh c) tng ng vi sn trc caxung vung ng th nht v cc xung

nhn m (hnh d) tng ng vi sn sauca xung vung ng th 2+ a ti B a hi ng b II ch ccxung nhn dng (hnh d) ca ng th2 v cc xung nhn m (hnh c) ngth nht+ xc nh rng ca cc xung a ra(hnh , e)

mI

I000 180T

TII m

20

Hnh 5-16


55/93

109


5.5.2 Pha mt chthsa.Chc nng cc khi:- Mch vo: thc hin tin x l tn hiu vo, lc nhiu.- To dng xung: bin i tn hiu vo, to ra cc xung o n cc tnh c chuk T=chu k tn hiu vo (Ux1, Ux2).- Trigger: to ra xung vung c rng T v chu k T chnh l nh Ux1,Ux2(Ux1 c a vo u thit lp S ca Trigger, Ux2 c a vo u xo Rca Trigger).- To xung m chun c chu k Tch .- To xung o: chia tn s xung m chun to ra xung o c rng To.

Mchvo 1

Mchvo 2

To dngxung

To dngxung

Trigger

To xungchun

To xungo

Gii mv

ch th

B mxungKho

1Kho

2

U1(t)

U2(t)

Ux1

Ux2

UT

Uch

Uo

xung cht

xungxo

Nx

UU

nx

Hnh 5-17: S khi ca Phamt s

110


b/ Nguyn l lm vic:

- Xung UT t Trigger s iu khin ngmkho 1. Mi khi c xung, xung m Uchtb to xung m chun s c a quakho 1 v u ra ca kho 1 l xung Unx l

1 chui gm nhiu nhm xung m vc a vo kho 2.- Xung o Uo iu khin ng mkho 2trong thi gian c xung o To .- Gi s c h nhm xung c a quakho 2 vo kch thch cho b m xung,tng s xung m c l Nx, s xung Nxny c a qua mch gii m v ch th

hin th kt qu l gc lch pha cn o.- Ta c gc lch pha gia 2 tn hiu U1(t)v U2(t) l

t

t

t

t

t

t

Nx xung

Ux1

Ux2

UT

Uch

Unx

Uo

t

tU

Uth

U1U

2

To

n xung

TT

Hnh 5-18 Gin thi gian


56/93

111


,chnTT

0360 .T

T

xdo

ch NT

T.360

h

N

nx

(n l s xung ca 1 nhm xung, Tch l chu k xung m chun).

To

=h.T,

c. nh gi sai s:- Do sai s ca Tch.-Do sai s lng t :

-Sai s do khng ng nht ca knh 1, knh 2 l-Khc phc:+ a tn hiu U1(t) hoc U2(t) vo c 2 knh, gi s Phamt ch th gi tr l

, ta c:+ Qu trnh hiu chnh ny c th c thc hin nhb m xung thunnghch.

h

1

n

1 ,

'' do'dodo

'' do

112


5.5.3 o di pha bng phng php vdao ng1. Phng php dng qut tuyn tnh:

2. Phng php Lixazu:Gi thit o di pha ca t/hiu qua 1 M4C. Phng php ny c th s dngOxilo 1 knh hoc 2 knh. Gi s ta s dng xil 2 knh, s nh hnh 5-20.

+iu chnh xil lm vic ch qut Lixazu:Chn chuyn mch X-Y

Vert.Mode CH2 = UCH2Knh YSource CH1 = UCH1Knh X

111 sin tUu m

222 sin tUu m0

1 2 360 .

T

T

U1

U2

T

T

Hnh 5-19

M4CUV Ur

U1(t) U2(t)CH1 CH2

Hnh 5-20


57/93

113


+iu chnh h s lch pha nhn c dao ng Lixazu nm chnh giav trong gii hn mn hnh.

Volts/div (CH1 v CH2)POS-Y (CH1)

POS-XDao ng s c dng ng thng hoc ng Elip.

A

B

x

y

YMAX

XMAX

Hnh 5-21

+Xc nh gc trung tm ca dao ng : a ccchuyn mch kt ni u vo ca c 2 knh v v trGND, trn mn hnh s l 1 im sng, dch chuyn imsng v chnh gia mn hnh.

+a cc chuyn mch kt ni u vo v v tr AC, khi s nhn c daong c dng ng thng hoc Elip.

+Xc nh gc lch pha:

maxmaxmaxmax

arcsinarcsinsinX

B

Y

A

X

B

Y

A

114

Chng 6. o dng in v in p

6.1 o dng in6.1.1 o dngin 1 chiu bng Ampe mt t in

-Dng c o: Ampe mt t in, c mc ni tip vi mch c dng in cn osao cho ti cc dng dng i vo v ti cc m dng i ra khi ampe mt.-Yu cu: ni trRA nh m bo ampe mt nh hng rt t n n tr s

dng in cn o-Ampe mt t in: lch ca kim t l thun vi dng in chy qua cun dy.- o I ln mc in trsn vo mch o:

Io max = IA max + IS max

Ta c: IS max.RS = IA max.RA

S

SA

A

AS

S

A

A

S

R

RR

I

II

R

R

I

I

max

maxmax

max

max

S

A

A

do

R

R

I

I1

max

max

max

max

A

do

I

In ; :hsmrng thango

Hnh 6-1


58/93

115


Thay i RSbng cc gi tr khc nhau cc thang o khc nhau

V d

Ampe mt nhiu thango

-Thay i v tr CM ( B, C, D) o c cc dng c tr s khc nhauCh : s dng cng tc ng ri ct dng c khng b mt sn trnh dng qua qu ln gy hng

1

n

RR AS

Hnh 6-2

116

-Sn Ayrton: bo v cun dy ca khi b dng qu ln khi CM gia cc sn-Phn tch:

CM B: RA // (R1 nt R2 nt R3)CM C: (RA nt R3) // (R1 nt R2)CM D: (RA nt R2 nt R3) // R1


Hnh 6-3


59/93

117


Sai sdo nhit:-Cun dy trong dng c o TNCVC c qun bng dy ng mnh, v intrca n c th thay i ng k theo nhit - I chy qua cun dynung nng nRcun dy thay i sai sphp o dng-Khc phc: mc R

b

bng Mangan hoc Constantan vi cun dy (Mangan hocConstantan c h s in trph thuc t0bng 0)

nu Rb = 9 Rcun dy RA = Rb + Rcun dy = 10Rcun dy th khi Rcun dy thay i1% s khin cho RA thay i 0,1%RS cng c lm bng Mangan hoc Constantan trnh s thay i in trtheo t0

Hnh 6-4

118

6.1.2 o dngin xoay chiu hnh sin

Ccu o in t c dng ph bin mrng gii hn o dng bin p dng in (bbin dng)Bbin dng bin i I cn o c tr s ln sang dng in c tr s nh m ccu o in t c th lm vic c.


Hnh 6-5

Cun dy W1 mc nt vi mch c dngin cn oCun dy W2 mc vi ampe mt in tS vng W2 > s vng W1

Io = n.IA

nW

W

I

I

A

do 1

2

max

max

1

2

WWn vi l hsbin dng


60/93

119


Ch : dng qua ccu o khngc vt qu IA max- c cc thang o khc nhau cu to bbin dng vi cun th cp c nhiuu ra.

Hnh 6-6

120

6.2 o in p

6.2.1 c im & yu cu

- Php o d tin hnh, thc hin nhanh chng, chnh xc cao.

- Khong gi tr in p cn o rng (vi V-vi trm KV), trong di tn s rng(vi % Hz hng nghn MHz), v di nhiu dng tn hiu in p khc nhau

- Thit b o in p phi c Zvo ln

* Cc trs in p cn o

- tr s nh (Um), tr s hiu dng(Uhd, U),, tr s trung bnh(Utb, U0)

in p c chu k dng khng sin:


2

0

1T

U u t dt T

2 2 2 2 20 1 2

0

...n

kk

U U U U U


61/93

121

Quan h gia Um, U, U0 :

VD:h(6-7a) l in p hnh sin:


0 01 T

U u t dt T

mb

Uk

U

0

dU

k

U

kb : hsbin ca tn hiu in p; kd: hsdng ca tn hiu in p

2. ;mU U 0 0,9U U

1, 41;bk 1,11dk

Hnh 6-7a

122

h(6-7b) l in p xung rng ca:

2

22

0

1

3

Tm m mU U Uu t t U t dt

T T T

0 2

mU

U 1,73;bk 1,15dk


Hnh 6-7b


62/93

123

h(6-7c) l in p xung vung gc:

U = Um v U0 = Um kb = kd = 1


: 0

2

:2

m

m

TU t

u tTU t T

Hnh 6-7c

124

Chng 6. o dng in v in p6.2.2 o in p 1 chiu

(a) Dng vn mt t in:

- Dng c o: Vn mt t in, c mc // vi mch c in p cn o sao chocc dng ca Vn mt ni vi im c in th cao v cc m ca Vn mt

ni vi im c in th thp hn.- Yu cu: in trvo ca vn mt RV ln m bo vn mt nh hng rt tn n tr s in p cn o

- o in p ln mc in trph vo mch o:

Uo max = IV(Rp + RV)

V

VP

V

do

R

RR

U

U

max

max

nR

R

U

U

V

P

V

do 1max

max ; n : hsmrng thangoHnh 6-8


63/93

125


VD* Vn mt nhiu thango

- c cu to t mt dng c o lch, mt s in trph v mt cngtc xoay- 2 mch vn k nhiu khong o thng dng:(H6-9a) 1 thi im ch c 1 trong 3 in trph c mc ni tip vi myo. Khong o ca vn k: Uo = IV (RV + RP)RP c th l RP1, RP2, RP3

VP RnR 1

Hnh 6-9a

126


(H6-9b) cc in trph c mc ni tip v mi ch ni c ni vi mttrong cc u ra ca cng tcKhong o ca vn k: Uo = IV (RV + RP)RP c th l RP1, RP1+RP2, RP1+RP2+RP3VD* nhy ca vn mt-l t s gia in trton phn v ch s in p ton thang ca vn mt

n v: /V, nhy cng ln th vn mt cng chnh xcVD: mt vn mt c: Rtp = RV + RP= 1M, dng c o 100V trn ton thang

nhy ca vn mt: ?1M/100V = 10k/V

Hnh 6-9b

AB

C D


64/93

127

(b) Dng vn mt s

- S khi n gin:

+ T.B.V gm: * b lc tn thp cho Uo khng cn thnh phn sng hi* bphn p: thay i thang o

* b chuyn i phn cc in p: thay i cc tnh ca Uo+ Bbin i in p - khong thi gian: bin i tr s Uo ra khong thi gian

t iu khin cng ng m+ Cng: bin i khong thi gian t thnh cng


Hnh 6-10

128

+ B to xung m: to ra cc xung m c tn s nht nh a ti Cng.Ch cc xung m xut hin trong khong thi gian t ng vi cng mmithng qua c cng ti b m xung

+ B m xung: m cc xung trong khong thi gian t+ Thit b hin th s: chuyn i t s xung m c thnh ch s hin th- S khi chi tit:


Ngunin p

mu

To xungm chun

Sosnh

Trigger KhoB m

xung

Gii mv ch th

B iu khin

Ux

K2

K2

+ -CCM

in t

E0

R Uss

RS

UTUch

U0Nx

xungxo

xungcht

sn

p+

-

Hnh 6-11 S khi Vnmt sthi gian xung

K1

Mchvo


65/93

129

b/ Nguyn l lm vic:

- Khi cha o, kho S h(khng v tr np hoc phng).- Qu trnh bin i c thc hin theo 2 bc tch phn sau:

+ Bc 1: Ti t1, b iu khin a ra xung iu khin K1 a kho S v v trn, in p Ux qua mch vo qua R np cho C UC tng.

+ Bc 2: n thi im t2, b iu khin a ra xung iu khin K2 a S vv tr p v kt thc qu trnh np, C sphng in qua ngun in p mu (ngunin p khng i, 1 chiu E0), UC gim n thi im t3 UC= 0, b so snha ra xung so snh USS.Xung K2 v xung USS s c a vo u vo thit lp (S) v xo (R) caTrigger u r a ca Trigger l xung vung c rng Tx, xung ny s iu

khin ng mkho cho php xung m chun qua kho kch thch cho bm xung.Gi s trong thi gian Tx c Nx xung qua kho, s xung Nx c a qua mchgiim v c h th biu th kt qu UDC cn o.


130

* Xc nh Ux=f(Nx):-Qu trnh C np:

Kv: h s truyn t ca mch vo.Gi s trong thi gian bin i, Ux=const:

vi T1 = t2-t1

- Qu trnh C phng:

vi Tx=t3-t2


2

1

..1

)( 1

t

t

xvcn dtUKRCtUU

RCTUKttUK

RCU xvxvn

112 ..)(..10

xxv

c

n

t

t

cc

TERCRC

TUKtU

ttERC

U

dtERC

tUtU

..1..)(

)(.1

.1

)()(

01

3

230

023

3

2


66/93

131


chxxv

xc TNE

TUKTtU .

..0)(

0

13

Tch : chu k ca xung m chun.

constTK

ETS

NSNTK

ET

U

v

ch

xxv

ch

x

1

00

01

0

.

....

.

xk

x NU .10

(thng chn S0=10kvi k=0, 1,)

c/ Gin thi gian: hnh 6-12

t

t

t

t

t

t

Nx xung

Tx

C phngC np

t1 t2 t3

T1

K1 K2

t1 t2Un

Uk

Uc

Uss

UT

Uch

U

Hnh 6-12

132


d/nh gi sai s:-Sa i s Tch, Kv, E0, T1.-Sa i s lng t (do xp x Tx vi Nx).-Sa i s do tr ca cc Trigger.- Sai s do nhiu tc ng t u vo. Tuy nhin, vi phng php tch phn 2ln, c th loi tr hon ton nhiu chu k nu chn T1= n.Tnh vi Tnh l chu k

nhiu.


67/93

133

6.2.3 o in p xoay chiu

S khi ca vn mt o in p xoay chiu c tr s ln

S khi ca vn mt o in p xoay chiu c tr s nh

Thit b vo: gm cc phn t bin i in p o u vo nh bphn p,

mch tng trkhng vo... vi mc ch l ghp Uo mt cch thch hp vimch o l vn mt.

B tch sng: bin i in p xoay chiu thnh dng in hay in p 1 chiu.


Thit bvo Tchsng K dng1 chiu Thit b chth kim

Thit bvo

K in pxoay chiu

Tchsng

Thit b chth kim

134

Cc loi mch tch sng:a) Tch sng nh (bin )- L tch sng m Ura trc tip tng ng vi tr s bin ca Uvo. Phn t gim gi li tr s bin ca Uo l t in. T in c np ti gi tr nh caUo thng qua phn t tch sng.- Mch c th dng diode hoc Transistor. y ta dng mch tch sng nh

dng diode.+mch tch sng nh c uv om:


sinx mU t U t

InIp

Rt

CUX

D

-Um

Um

CnpCphng

UC=Um

t

UX(t)

Hnh 6-13


68/93

135


Nguyn l lm vic:

- Trong na chu k (+) u tin, D thng, C c np in nhanh qua trR thng vi hng s np n=R thng.C v UC tng n khi UC Ux(t). Lc ny

D tt v t C sphng in qua Rt vi hng sphng p=Rt.C-Khi UC gim nkh iUC < Ux(t) th t li c np.Nu chn nUX(t). Lc ny D tt v t C sphng in quaRt vi hng sphng p=Rt.C ; v UC gim nkhi UC < UX(t) t li c np.Nu chn n


69/93

137


b) Mch tch sng trung bnh:C nhim vbin i in p xoay chiu thnh in p 1 chiu c gi tr trungbnh t l vi tr s in p trung bnh ca in p vo.Thng dng cc mch chnh lu c chu k hoc na c huk .

+ Mch chnh lu na c huk :D1 l chnh lu na c huk . D2 l ngn khng inp ngc qu ln nh thng D1 t ln Vn k vlm cho in trtrong mch tch sng ng u trongc chu k.Chn Rt>> RD1thunRt+ RD1th=R2+RD2th

RtD1

D2

R2

Kiu mc 2 diode //

RtD

Kiu mc 1 diode

Utb=Um/

Um

0

URt

Hnh 6-15

138


+Chnh lu c chu k:

Utb=2Um/

Um

0

URV

t

D1 D2

D3D4

Ux AC

V

Hnh 6-16

c) Mch tch sng hiu dngNhim v: Bin i in p xoay chiu thnh 1 chiu c gi tr t l vi gi tr hiu dng ca in p xoay chiu.

Tch snghiu dng

UX AC UR DC=k.UXhd


70/93

139

tch sng hiu dng cn phi nhng bc sau:

+Bnh phng in p: dng mch bnh phng in p h oc dng mch c ctuyn Vn-Ampe bc 2 (i=S0UX2)

+Ly tch phn, v dng cc mch khai cn hoc dng phng php khc thang o.

Ta xt mch tch sng hiu dng dng cc mch c c tuyn Vn-Ampe bc 2: tng kh nng o in ph iu dngxy dng cc mch c c tuyn Vn-Ampe bc 2 bng cch xp x c tuyn thnh nhng on tuyn tnh lin tip

nhau.Gi s xy dng mch c c tuyn xp x thnh 4 on nh hnh v (0U1),(U1U2), (U2U3), (U3).


Tt

t

dttxUThdU

0

0

21

140


4 on tng ng vi cc khu diode mc lin tip vi nhau nh s :

-Diode Di c phn cc bi cp in trRi v Ri im lm vic ca chngl Ui (i=1, 2, 3)-Tnh ton mch nh sau:

Gi s cn xp x c tuyn: i=S0U2

mA

D1

R1R

2 R3

U3

R3

0

R2R1

UX iD0 iD2R0

iD1 iD3

En

iA

D2 D3

U1 U2

Hnh 6-17


71/93

141

x 1 1 2 3

xA D0

00

2 xx 1 A 1 0 1

0

1 x 2 1 2 3

A D0 D1 x0 1 1

2x 2 A 2 0 2

2 x 3 1 2 3

A

Voi 0 U U : D ,D ,D tat

Ui i

RR ?

UU U i i S U

R

Voi U U U : D thong,D ,D tat

1 1i i i U

R R R ?

U U i i S U

Tuong tu voi U U U : D , D thong, D tat

i

D0 D1 D2 2

3 x 1 2 3

A D0 D1 D2 D3

x 30 1 2 3

n. ii

i i

i i

i i i R ?

voi U U : D , D ,D thong

i i i i i

1 1 1 1U R ?R R R R

E RU

R R '

Biet R Tinh duoc R '


iA

S0U32

S0U22

S0U12 iD0

iD0+

iD1

iD0+iD1+iD2

iD0+iD1+iD2+iD3

UX

UXiA

Hnh 6-18

142

Chng 7. o cng sut

7.1 Khi nim

Cng sut: nng lng in t trng tiu th trn ti trong mt n v thi gian.Mch in mt chiu: P = U.IMch in xoay chiu: p = u.iMch in c dng iu ho:

: CS thc hin

CS phn khng: Q = U.I sin

Mch in hot ng ch xung CS xung (Pxung): l tr s CS trung bnh

trong khong t/g c xung ()

oscIUpdtTPT

0

..1

Z

Rcos 22 XRZ ;

T

xung uidtP0

1T

tb uidtTP

0

1

TPP xungtb

;


72/93

143

Chng 7. o cng sut

- Lng trnh o CS: 10-6 W 107 W- n v o CS: ot (W)- n v o CS tng i: dBW, dBmW: dng so snh cc mc CS cc vtr khc nhau.

7.2 Cc phng php o cng sut

c im o CStn scao:- Bin i CS v i lng trung gian ri o i lng - Sai s ca php o ph thuc vo sphi hp trkhng gia ngun pht vph ti, ph thuc vo tn s v cc tc ng ca mi trng.

Cc phng php o CStn scao:-o CS dng chuyn i Hall (dng cho c t/s thp v t/s cao)-o CS bng cch o in p trn ti thun tr-o CS bng in trnhit

1

lg10PP : CS tng i; P: tr s CS W(mW) ti mt v tr no ;

P1: tr s CS ban u (1W hoc 1mW)

144

Chng 7. o cng sut

o CStn sthp: dng phng php nhnPhng tin o cng sut: ot mt, gm ot mt o CS hp th v ot mt oCS truyn thng.- Ot mt o CS hp th: l phng tin o CS tiu tn trn ti phi hp cachnh phng tin o (hnh 7-1). N hp th ton b CS ca ngun pht khingun pht khng mc ti ngoi

- Ot mt o CS truyn thng: l phng tin o CS truyn theo ng truynti ti (hnh 7-2). N ch h p th mt phn nng lng ca ngun pht cnphn ln nng lng truyn ti ti ring ca n.

Hnh 7-1

Ti hpth

Bin i

nng lng

Thit b

chth

Ot mtP

Hnh 7-2

Tithc

Bin inng lng

Thit bchth

Ot mt

P


73/93

145

Chng 7. o cng sut

7.2.1. Phng php nhn:

- Khi dng in l iu ho th CS tc dng cn o trn ti:-o CS trn ti c th thc hin trc tip bng cch dng 1 thit b nhn

nhn in p v dng in trn ti.-S khi:

cosUIP

22122121 41

xxxxxx

tIxtUx sin,sin 21

Hnh 7-3

Btng

B bnhphng

Btng

ng ht in

B occ

B occ

B bnhphng

Btng

x1

x2

x1

-x2

x1-x2 (x1-x2)2

x1+x2(x1+x2)2

-(x1-x2)2

4x1x2

x2

146

Chng 7. o cng sut

- in p c o bng mt ng h t in mc song song vi 1 t in. Chs ca ng h l thnh phn 1 chiu: 2UIcos, l CS cn o trn ti.- Phn t c c tuyn bc 2: ly phn u ca c tuyn V-A ca it hoc

transistor. (yu cu n phi c c tuyn ng nht).- Sai s: (5-10)%

7.2.2. o CS dng chuyn i Hall:

ttUIxx sinsin44 21 tUIUI 2cos2cos2

- Chuyn i Hall c cu to bngbn mng cht bn dn n tinh th(Si hoc Ge) vi 2 cp cc t vung

gc vi nhau v nm trn cc thnhhp ca bn tinh th (Hnh 7-4).

Hnh 7-4


74/93

147

Chng 7. o cng sut

-cp cc dng D c cp I mt chiu hoc xoay chiu, cp cc p A cho rain p t l vi tch ca I v t cm tc ng vung gc ln b mt ca tinh th.-Sc in ng Hall:

eH = KH.B.i (*)

KH: h s chuyn i, ph thuc vo vt liu, kch thc, hnh dng ca tm bndn v nhit mi trng.-Nu B ~ Ut; i ~ It eH = KH.KI.Ut.It KI : h s t l

mch in 1 chiu, sc in ng Hall :eH = KH.KI.Pt

mch in xoay chiu hnh sin, sc in ng Hall:eH = KH.KI.Um.Im.sin(t).sin(t - )

= KH.KI.U.I.cos - KH.KI.U.I.cos(2t - )

Nu mc vo 2 cc p 1 dng c t in th ch s ca dng c t l vi Ptbtrong mch dng xoay chiu thang o ca dng c c th c khc trctip theo n v CS.

148

Chng 7. o cng sut

T (*) mun eH tng th hoc i tng, hoc B tng; thng tng B v i tngtng nhit ca bn dn (t dng) tng B cn nh hng v gn chuyn i Hall v tr thch hp trong ngsng v cp ng trc, hoc s dng mi trng c t thm cao.

u im:

- Khng c sai s do mt phi hp trkhng.- Qun tnh nh- Di tn rng- Cu trc n gin

Nhc im:- eHph thuc mnh vo nhit


75/93

149

Chng 7. o cng sut

7.2.3 o CS bng pp o in p trn ti thun tr:

- Phng php o cng sut bng cch o in p trn ti thun trl cs ch to ot-mt o cng sut hp th ca ngun pht vi 1 ti mu thun tr.

Ti mu l 1 in trb mt hoc dng khi c cu trc c bit.

Cu to: phn in tr c dng hnh tr libng gm, trn ph lp than ch c bit; mnchn phi hp nm dc theo theo chiu di caphn in tr, c ng knh bin thin theohm m.Vi kt cu nh vy sng in t lan truyn t

ngun pht ti khng b mo, ti l thun trphi hp trkhng tt vi ngun pht. phi h p tr khng, in tr b mt ton

phn Rt theo dng 1 chiu phi bng trkhngsng ca cp ng trc vi mc ch

Ot mt Vn mtt

Rt

=RtRt

Ticp

ngtrc

vi trkhng

sng

mn chn phi hp

Hnh 7-5

150

Chng 7. o cng sut gim vic mt phi h p tr khng khi mch vo ca Vn mt mc songsong vi ti. mrng phm vi o cng sut, Vn mt ch o mt phn in p trn ti. mrng di tn ta thng dng Vn-mt in t loi tch sng bin cu vo m.Khi c phi hp trkhng cng sut tiu th trn ti c xc nh thng qua

gi tr bin Um v gi tr hiu dng U ca in p ri trn ti Rt

Cng sut trn ti Rt thng qua gi tr in p Um m vn mt o c bng:

- Vn mt c khc thang o theo n v cng sut.Sai s: sai s do lch phi hp trkhng; sai s ca Rt ; sai s ca Vn mt.

Sai s tng 20% cng sut o.Ot mt loi ny o c CS n hng chc nghn W di tn n vi GHz

t

m

t R

U

R

UP

2

22

2'2'2 mt

t UR

RP


76/93

151

Chng 7. o cng sut

7.2.4o cng sut dngin trnhit

Ot mt dng in trnhit c xy dng trn csmch cu in tr, 1trong nhng nhnh ca chng mc in trnhit. Thng dng o CS nht hng chc mW (t/s hng chc GHz).

a/ Cu to ca in trnhit:* Cu to ca Blmt: l 1 si dy in trrt mnh lm bng bch kim hayvnfram, c t trong bnh thu tinh (hnh 7-6).

Hnh 7-6

+ Trong bnh c cha kh trhay c chn khngcao gim s truyn nhit ra mi trng v tngtc t nng dy in tr.

+ Chiu di ca si dy in trphi tho mn k:

c ng u; min : di cc tiu ca bc sngin t ca ngun cng sut cn o.

8minl , s phn b dn in trn si dy

152

Chng 7. o cng sut

+ Quan h gia in trca Blmt v cng sut cn o (hnh 7-7):Rb = R0 + aPb

R0 :in trca Blmt khi P = 0;a,b : h s t l, ph thuc kch thc, vt liu ca blmt

+ Di in tr ca blmt: hng chc n vi trm m vi nhy(312)/mW

Hnh 7-7


77/93

153

Chng 7. o cng sut

* Cu to ca Tesmitor: l in trcn bng bn dn c h s nhit m .

Hnh 7-8

+ Hai dy bch kim hoc iridian c ng knh (20 30) m ni vi nhau ti ht cu lm bng bndn, tt c c t trong bnh thu tinh.

+ in trca Tesmitor khong (100 3000) .+ Quan h gia in trca Tesmitor v cng sutcn o (hnh 7-9)

* So snh gia blmt v tesmitor:+ Blmt c u im l d ch to, c tnh t phthuc nhit mi trng; nhc im: db quti, kch thc ln nn hn ch s dng on

sng cm, Zvo nh nn kh thc hin phi hp tr khng vi ng truyn.+ Tesmitor c u im l nhy cao, t b qu ti,tr s R ln, tr s L,C bn thn nh, kch thcnh, bn cao; nhc im: kh ch to, c tnhph thuc t0 mi trng.

Hnh 7-9

154

Chng 7. o cng sut

b/Otmt dngin trnhit

* Otmt xy dng trn mch cu n khng cn bng:

+ Otmt c nui bng ngun in p 1 chiu vi chitp Rc dng iu chnh dng qua cc nhnh cu, vi

ch dng mt cn bng trong nhnh ch th.+ 1 nhnh cu ta mc in tr nhit, trc khi o cnthay i in trTesmitor bng nhit nng ca dng inqua chuyn i (/chnh chit p Rc) cu cn bng.Lc ny MicroAmpemet ch "0".+ Khi c ngun cng sut cao tn tc ng ln RT lm chon gim tr mt cn bng cu xut hin dng inqua vi thang o khc trc tip theo cng sut.

+ Sai s: khong 10%, ph thuc ch yu vo s thay inhit mi trng, s khng phi h p tr khng caOtmt vi ng truyn v sai s ca thit b ch th.

A

A

Ngunin p 1

chiu

R1

R3 R2

RT

A

Hnh 7-10

Px

Rc


78/93

155

Chng 7. o cng sut

* Otmt xy dng trn mch cu n cn bng:

+ ch th cn bng cu, cho bit tr s ca cng sut.RT mc vo 1 nhnh cu, chn R1=R2= R3=RTPx= 0 = R.

+ Khi cha c ngun CS t/ng ln RT, tng t nh THtrn ta iu chnh dng in trong mch thay i RT vthit lp cn bng cu. thi im cu cn bng, ch"0", cn ch dng in I0 .+ Khi c ngun CS t/ng ln RT lm cho RT, cu mt cnbng. cu cn bng ta phi tng /trbng cch dngin trong mch. thi im cn bng ch .+ Qua hai bc /chnh cn bng cu, RT ca Tesmitorkhng i nn CS tiu th trn Tesmitor trong 2 bc nh

nhau do :

A mA

A

mA

mA '0I

2'0202'

020

444II

RPP

RIRIP Txx

TTt

Ngunin p 1

chiu

R1

R3 R2

RT

A

Hnh 7-11

Px

mA Rc

156

Chng 7. o cng sut

+ u im: m bo c s phi h p tr khng v RT ca Tesmitor khngthay i di tc ng ca cng sut Px cc thi im cn bng cu. Tuynhin thang o ca khng khc trc tip theo cng sut v dng I0 lunthay i theo nhit mi trng khi Px = 0.

mA


79/93

157

Chng 8. o cc tham s iu ch & c tnh ph ca t/hiu

8.1 Phn tch ph ca tn hiu

-C th dng MHS quan st v nghin cu ph ca tn hiu. Dao ng cc l theo quan h ph thuc gia bin cc thnh phn sng hi ca tnhiu theo tn s.

-Khi trc X ca MHS l trc thang tn s, cn trc Y l tr c thang bin.- v thphbin -tn s ca dao ng tn hiu, o tn s v t sbin cc phn lng ring bit ca ph dng my phn tch ph-Da vo cc thph ta c thphn tch c tnh v o lng c cc thngs ca tn hiu

VD:

+ o c h s iu ch bin thngqua thph ca dao ng iu bin

ff0 -F f0 +Ff0

U0

20mU

20mU

Hnh 8-1: Phca dao ngiu bin

158

Chng 8. o cc tham s iu ch & c tnh ph ca t/hiu* Nguyn l ca thit bphn tch ph: da trn csdng hin tng cnghng chn lc tn s.+ i vi cc mch cng hng c di thng tn hp (h sphm cht Q cao)

th bin ca dao ng cng bc s l cc i nu tn s tc ng trnghp vi tn sbn thn (tn s cng hng) ca mch cng hng v bin l r t nh khi c lch cng hng.

+ Do , mch cng hng c tc dng nh mt b lc, b lc ny c kh nngtch ring c cc phn lng sng hi khc ca tn hiu vi phn lngsng hi c tn s trng vi tn sbn thn ca mch (tn s cng hng).

* My phn tch phc 2 loi:+ Loi phn tch song song+ Loi phn tch ni tip

8.1.1 My phn tch phtheo phng php p/tch song songGi s c mt h thng b lc di hp c sp xp lin tip k st nhau theo

thang tn s trong di tn t fminfmax. Mi ng cong cng hng ca b lcc biu th n gin bng mt hnh CN, di thng tn ca b lc l f (hnh8-2.a). Trong di tn ca thit bphn tch c n b lc.


80/93

159


-Nu tn hiu c phn tch c ph nm trong di tn s cng tc ca b lctrn (hnh 8-2.b) th khi c tn hiu vo, mi b lc s c tc ng i vi

ring tng thnh phn ph m tn s ca thnh phn ph ny tng ng vi tns ca bn thn b lc.-in p u ra ca mi b lc s t l vi bin ca thnh phn ph tngng. Cc in p ny c o bi cc Vn mt (hnh 8-2.c)-Tr s ch th ca cc vn mt v tn s cng hng ca mi b lc cu toc thph ca tn hiu in p nghin cu.

f

ffn

minmax

Hnh 8-2

160

Chng 8. o cc tham s iu ch & c tnh ph ca t/hiu8.1.2. My phn tch phtheo phng php p/tch ni tip-Ch c mt b cng hng.-B cng hng ny c th iu chnh c tng ng vi tng tn s mttrong di tn sphn tch t fminfmax.1. S khi: gm 1 b lc di hp iu chnh c v mt MHS2. Nguyn l hotng:

+in p tb To in p qut rng ca c a ti cp phin lm lch Xca ng tia in t, ng thi c a ti b To sngiu tn iu ch tnsb ch sng ca n.

Hnh 8-3 My phn tch phni tip


81/93

161

+Ti bBin tn c hai tn hiu c a ti l tn hiu cn nghin cu ph vin p ca b To sngiu tn. y tn s ca b To sngiu tn ngoisai cng vi mt trong cc thnh phn sng hi ca tn hiu s to ra mt tn smi bng hiu ca 2 tn s trn.

+Khi tn s hiu ny bng tn s cng hng ca bKhuch i trung tn thphn lng in p c tn s c khuch i, sau c tch sng ri lic khuch i bng bKhuch i tn thp trc khi a ti cp phin lmlch Y ca ng tia in t.

+Tia in tb lch i so vi ng nm ngang (v tr ban u) mt tr s t lvi tr s trung bnh ca in p tn hiu nghin cu trong di thng tn f.+Mi khi tr s tc thi ca tn sb To sngiu tnbin i to nn mt tns hiu bng trung tn vi ln lt 2 thnh phn sng hi k tip nhau ca tn

hiu th ng thi tia in t c dch chuyn theo trc ngang v trn mn lixut hin mt vch sng khc theo trc dc.+Bin ca cc vch ny tng ng vi in p (hay cng sut) ca cc phnlng thnh phn ca ph.+Sau mt chu k qut, ton b cc vch ph ca tn hiu nghin cu c vtrn mn MHS.


162


VD: tn hiu phn tch ph l mt xung vung bini c chu k v c h s /T ln (hnh 8-4.a)-Mi thnh phn ph c biu thbng 1 vch sngtrn mn hnh. Khong cch gia 2 vch trn thangtn sbng tn s lp li ca xung tn hiu F = 1/T.-Yu cu: b To sngiu tnphi c tn s trung

tm n nh. Nu khng n nh s lm dch chuyntt c cc ph theo trc tn s (khi tn sbin i tt) hoc l lm dch chuyn tng thnh phn ringbit ca ph (khi tn sbin i nhanh) kh quanst & lm gim chnh xc khi o lng cc thngsph

Hnh 8-4Hnh 8-5


82/93

163

-c tuyn iu ch ca b to dao ngiu tn phi thng (hnh 8-6)

-Khi c tuyn thng th ph c hnh dng

nh hnh 8-7(a), nu khng thng th thang tn s s khc nhau theo ng qutngang & ph s b mo dng theo chiungang ,hnh 8-7(b).


Hnh 8-6

Hnh 8-7

164

Chng 8. o cc tham s iu ch & c tnh ph ca t/hiu-B to in p qut: to ng qut ngang trn ng tia in t v iu chtn s.-Bpht sngiu tn: c tc bin i tn s sao cho in p tn hiu tngti c mc in p cc i trong khong thi gian ng vi di thng tn cab khuch i trung tn (KTT).-Cc thng s ca khi KTT: di thng tn, tn s cng hng, h s K.

Di thng tn: tu thuc vo mc ch, cng dng ca my phn tch ph.My phn tch ph tn s thp chn di thng tn sao cho c th phn bitc r rng 2 thnh phn ph cnh nhau.Nu my phn tch ph c bng tn rng, v gm nhiu thnh phn ch cnv ng bao ca ph.-Cc xung u ra ca b K c bin t l vi nng lng ca tng bphnca ph.-Chn tn s trung tn sao cho loi b c s cho qua tn hiu tn s nh (giiphp: tng tn s trung tn). Nu khng th trn mn MHS s xut hin ng

thi 2 dng ph: mt ph thc v mt ph nh.Mu thun gia tng tn s trung tn v gim nh di thng tn gii php:dng 2 bbin tn v K trung tn.


83/93

165


-Chn h s khuch i da trn yu cu v bin cc tiu ca tn hiu nghincu v bin a vo b tch sng.-o b rng phbng cch so snh ph cn o vi ph chun.-Ph chun thng dng l ph ca tn hiu iu tn m tn s iu ch c dngiu ho.

S khi ca bphn to tn hiu c phchun:

+B to sngiu chpht ra in phnh sin c tn s 1-10 Mhz a tiiu chb to sng chun.

+Tn hiu iu tn t b to sngchun c a vo b trn tn cngvi tn hiu nghin cu.

+Trn mn ca MHS xut hin phca tn hiu nghin cu vph ca tnhiu iu tn chun. Khong cch giacc thnh phn ca ph chun l bit.

Hnh 8-8 B to tn hiucphchun

166

+Bit tn s iu ch v cc s lng cc thnh phn ca ph chun th c thxc nh c ng cc phn on ca ph cn o.

Tm li, bin i bin in p iu ch bin i s lng cc thnh phnca ph chun. Bin i tn s iu ch bin i c khong cch giacc thnh phn ca ph chun. Do c th o c b rng ca bt k phno.

VD:

(a) Cc thnh phn ca ph cn o(b) Cc thnh phn ca ph chun

Hnh 8-9 Cc vch phkhi so snh



84/93

167

Chng 9. o cc tham s ca mch in

-Mt s thng s: R, L, C, Q, gc tn hao tg.Cc phng php o tham smch: phng php Vn-Ampe, phng php sosnh bng mch cu, phng php cng hng, phng php o dng cc thitb ch th s.

9.1 o tham s mch bng pp vn-ampe:

Theo s hnh 9-1a , gi tr in tr o c l:

RA: in trtrong ca ampe mt

: sai s phng php

AxA

AR

A

Vx RRI

UU

I

UR x

'

x

App R

RHnh 9-1a

168

Theo s hnh 9-1b:

RV: i

co so do luong dien tu

Documents