GT Cam Bien Dien Cong Nghiep

7/30/2019 GT Cam Bien Dien Cong Nghiep

1/179

****************************************************************i hc nng

Trng i hc bch khoa-----------------------------------

Th.S. Hong Minh Cng

Gio trnh

Cm bin cng nghip

- Nng 2004 -****************************************************************


2/179

Li mu

Cm bin c nh ngha nh mt thit b dng cm nhn v bin i cc

i lng vt l v cc i lng khng mang tnh cht in thnh cc i lng in

c tho c. N l thnh phn quan trng trong mt thit bo hay trong mt hiu khin tng.

t lu cc b cm bin c s dng nh nhng b phn cm nhn v

pht hin, nhng ch t vi ba chc nm tr li y chng mi th hin vai tr quan

trng trong k thut v cng nghip c bit l trong lnh vc o lng, kim tra v

iu khin tng. Nhcc tin b ca khoa hc v cng ngh trong lnh vc vt liu,

thit bin t v tin hc, cc cm bin c gim thiu kch thc, ci thin tnh

nng v ngy cng mrng phm vi ng dng. Giy khng c mt lnh vc no m khng s dng cm bin. Chng c mt trong cc h thng tng phc tp,

ngi my, kim tra cht lng sn phm, tit kim nng lng, chng nhim mi

trng. Cm bin cng c ng dng rng ri trong lnh vc giao thng vn ti, sn

xut hng tiu dng, bo qun thc phm, sn xut t ... Bi vy trang b nhng kin

thc cbn v cm bin trthnh mt yu cu quan trng i vi cc cn b k thut.

i vi sinh vin ngnh cin t cng nh cc ngnh tng ho trong cc

trng i hc k thut, mn hc cm bin cng nghip l mt mn hc bt buc trong

chng trnh o to, nhm trang b nhng kin thc cbn v cm bin hc tt

cc mn hc chuyn ngnh. Gio trnh cm bin cng nghip c vit cho chuyn

ngnh cin t gm 10 chng, gii thiu nhng kin thc cbn v cm bin, cu

to, nguyn l hot ng, cc c trng cbn v s mch o ca nhng cm bin

c s dng ph bin trong cng nghip cng nh trong th nghim, nghin cu v

c sp xp theo cng dng ca cc b cm bin.

Do ni dung gio trnh bao qut rng, ti liu tham kho hn ch v trnh c

hn ca ngi bin son nn chc chn gio trnh khng trnh khi sai st. Tc gi

mong mun nhn c s gp ca bn c v ng nghip gio trnh c hon

thin hn. Cc nhn xt, gp xin gi v Khoa Ckh Trng i hc Bch Khoa,

i hc Nng.

Tc gi


3/179

Chng I

Cc Khi nim v c trng cbn

1.1. Khi nim v phn loi cm bin

1.1.1. Khi nim

Cm bin l thit b dng cm nhn bin i cc i lng vt l v cc i

lng khng c tnh cht in cn o thnh cc i lng in c tho v x l

c.

Cc i lng cn o (m) thng khng c tnh cht in (nh nhit , p sut

...) tc ng ln cm bin cho ta mt c trng (s) mang tnh cht in (nhin tch,

in p, dng in hoc trkhng) cha ng thng tin cho php xc nh gi tr ca

i lng o. c trng (s) l hm ca i lng cn o (m):

( )mFs =

(1.1)

Ngi ta gi (s) l i lng u ra hoc l phn ng ca cm bin, (m) l i

lng u vo hay kch thch (c ngun gc l i lng cn o). Thng qua o c (s)

cho php nhn bit gi tr ca (m).

1.1.2. Phn loi cm bin

Cc b cm bin c phn loi theo cc c trng cbn sau y:

- Theo nguyn l chuyn i gia p ng v kch thch (bng 1.1).Bng 1.1

Hin tng Chuyn i p ng v kch thch

Hin tng vt l

- Nhit in

- Quang in

- Quang t

- in t

- Quang n hi

- Tin- Nhit t...

Ho hc

- Bin i ho hc

- Bin i in ho

- Phn tch ph ...

- Bin i sinh ho


4/179

Sinh hc - Bin i vt l

- Hiu ng trn cth sng ...

- Phn loi theo dng kch thch (bng 1.2)

Bng 1.2

m thanh

- Bin pha, phn cc

- Ph

- Tc truyn sng ...

in

- in tch, dng in

- in th, in p

- in trng (bin, pha, phn cc, ph)

- in dn, hng sin mi ...

T

- T trng (bin, pha, phn cc, ph)

- T thng, cng t trng

- t thm ...

Quang

- Bin, pha, phn cc, ph

- Tc truyn

- H s pht x, khc x

- H s hp th, h s bc x ...

C

- V tr

- Lc, p sut

- Gia tc, vn tc- ng sut, cng

- M men

- Khi lng, t trng

- Vn tc cht lu, nht ...

Nhit

- Nhit

- Thng lng

- Nhit dung, t nhit ...

Bc x- Kiu- Nng lng

- Cng ...


5/179

- Theo tnh nng ca b cm bin (bng 1.3)

Bng 1.3

- nhy

- chnh xc

- phn gii

- chn lc

- tuyn tnh

- Cng sut tiu th

- Di tn

- tr

- Kh nng qu ti

- Tc p ng

- n nh

- Tui th

- iu kin mi trng

- Kch thc, trng lng

- Phn loi theo phm vi s dng ( bng 1.4).

Bng 1.4

- Cng nghip

- Nghin cu khoa hc

- Mi trng, kh tng

- Thng tin, vin thng

- Nng nghip

- Dn dng

- Giao thng

- V tr

- Qun s

- Phn loi theo thng s ca m hnh mch thay th :

+ Cm bin tch cc c u ra l ngun p hoc ngun dng.

+ Cm bin thng c c trng bng cc thng s R, L, C, M .... tuyn tnh hoc

phi tuyn.

1.2. ng cong chun ca cm bin1.2.1. Khi nim

ng cong chun cm bin l ng cong biu din s ph thuc ca i lng in

(s) u ra ca cm bin vo gi tr ca i lng o (m) u vo. ng cong chun c

th biu din bng biu thc i s di dng ( )mFs = , hoc bng th nh hnh 1.1a.

s

si

s


6/179

Da vo ng cong chun ca cm bin, ta c th xc nh gi tr mi cha bit ca m

thng qua gi tro c si ca s.

d s dng, ngi ta thng ch to cm bin c s ph thuc tuyn tnh gia

i lng u ra v i lng u vo, phng trnh s= F(m) c dng s = am +b vi a,

b l cc h s, khi ng cong chun l ng thng (hnh 1.1b).

1.2.2. Phng php chun cm bin

Chun cm bin l php o nhm mc ch xc lp mi quan h gia gi tr s o

c ca i lng in u ra v gi tr m ca i lng o c tnh n cc yu t

nh hng, trn cs xy dng ng cong chun di dng tng minh ( th

hoc biu thc i s). Khi chun cm bin, vi mt lot gi tr bit chnh xc mi

ca m, o gi tr tng ng si ca s v dng ng cong chun.

a) Chun n gin

Trong trng hp i lng o ch c mt i lng vt l duy nht tc ng ln

mt i lng o xc nh v cm bin s dng khng nhy vi tc ng ca cc i

lng nh hng, ngi ta dng phng php chun n gin. Thc cht ca chun

n gin l o cc gi tr ca i lng u ra ng vi cc gi xc nh khng i ca

i lng o u vo. Vic chun c tin hnh theo hai cch:

m1 m2

s1

s2

s

m

Hnh 1.2 Ph ng php chu n c m bi n

Hnh 1.1 ng cong chu n c m bi na) D ng ng cong chu n b) ng cong chu n c a c m bi n tuy n tnh


7/179

- Chun trc tip: cc gi tr khc nhau ca i lng o ly t cc mu chunhoc cc phn t so snh c gi tr bit trc vi chnh xc cao.

- Chun gin tip: kt hp cm bin cn chun vi mt cm bin so snh csn ng cong chun, c hai c t trong cng iu kin lm vic. Khi tc ng ln

hai cm bin vi cng mt gi tr ca i lng o ta nhn c gi tr tng ng ca

cm bin so snh v cm bin cn chun. Lp li tng t vi cc gi tr khc ca i

lng o cho php ta xy dng c ng cong chun ca cm bin cn chun.

b) Chun nhiu ln

Khi cm bin c phn t b tr (tr choc tr t), gi tro c u ra ph

thuc khng nhng vo gi tr tc thi ca i lng cn o u vo m cn ph

thuc vo gi tr trc ca ca i lng ny. Trong trng hp nh vy, ngi ta

p dng phng php chun nhiu ln v tin hnh nh sau:

- t li im 0 ca cm bin: i lng cn o v i lng u ra c gi trtng ng vi im gc, m=0 v s=0.

- o gi tru ra theo mt lot gi tr tng dn n gi tr cc i ca ilng o u vo.

- Lp li qu trnh o vi cc gi tr gim dn t gi tr cc i.Khi chun nhiu ln cho php xc nh ng cong chun theo c hai hng o

tng dn v o gim dn.

1.3. Cc c trng cbn

1.3.1. nhy ca cm bin

a) Khi nim

i vi cm bin tuyn tnh, gia bin thin u ra s v bin thin u vo m

c s lin h tuyn tnh:

m.Ss =

(1.2)

i lng S xc nh bi biu thcmsS

= c gi l nhy ca cm bin.

Trng hp tng qut, biu thc xc nh nhy S ca cm bin xung quanh gi tr mi

ca i lng o xc nh bi t s gia bin thin s ca i lng u ra v bin thin m

tng ng ca i lng o u vo quanh gi tr:


8/179

imm

m

sS

=

=

(1.3)

php o t chnh xc cao, khi thit k v s dng cm bin cn lm sao

cho nhy S ca n khng i, ngha l t ph thuc nht vo cc yu t sau:

- Gi tr ca i lng cn o m v tn s thay i ca n.- Thi gian s dng.- nh hng ca cc i lng vt l khc (khng phi l i lng o) ca mi

trng xung quanh.

Thng thng nh sn xut cung cp gi tr ca nhy S tng ng vi nhng iu

kin lm vic nht nh ca cm bin.

b) nhy trong ch tnh v t s chuyn i tnh

ng chun cm bin, xy dng trn cso cc gi tr siu ra tng ng vi ccgi tr khng i mi ca i lng o khi i lng ny t n ch lm vic danh nh

c gi l c trng tnh ca cm bin. Mt im Qi(mi,si) trn c trng tnh xc nh mt

im lm vic ca cm bin ch tnh.

Trong ch tnh, nhy S xc nh theo cng thc (1.3) chnh l c ca c

trng tnh im lm vic ang xt. Nh vy, nu c trng tnh khng phi l tuyn tnh th

nhy trong ch tnh ph thuc im lm vic.

i lng ri xc nh bi t s gia gi tr siu ra v gi tr miu vo c gi l

t s chuyn i tnh:

iQ

im

sr

=

(1.4)

T (1.4), ta nhn thy t s chuyn i tnh ri khng ph thuc vo im lm vic Qi v

ch bng S khi c trng tnh l ng thng i qua gc to.

c) nhy trong chng

nhy trong chng c xc nh khi i lng o bin thin tun hon theo

thi gian.

Gi s bin thin ca i lng o m theo thi gian c dng:

tcosmm)t(m 10 += (1.5)

Trong m0 l gi tr khng i, m1 l bin v tn s gc ca bin thin i lng

o.

u ra ca cm bin, hi p s c dng:


9/179

)tcos(ss)t(s 10 ++=

Trong :

- s0 l gi tr khng i tng ng vi m0 xc nh im lm vic Q0 trn ng congchun ch tnh.

- s1 l bin bin thin u ra do thnh phn bin thin ca i lng o gy nn.- l lch pha gia i lng u vo v i lng u ra.Trong chng, nhy S ca cm bin c xc nh bi t s gia bin ca

bin thin u ra s1 v bin ca bin thin u vo m1ng vi im lm vic c xt Q0,

theo cng thc:

0Q1

1

m

sS

=

nhy trong chng ph thuc vo tn si lng o, )f(SS = . S bin thin

ca nhy theo tn s c ngun gc l do qun tnh c, nhit hoc in ca u o, tc l

ca cm bin v cc thit b ph tr, chng khng th cung cp tc thi tn hiu in theo kp

bin thin ca i lng o. Bi vy khi xt s hi p c ph thuc vo tn s cn phi xem

xt s mch o ca cm bin mt cch tng th.

1.3.2. tuyn tnh

a) Khi nim

Mt cm bin c gi l tuyn tnh trong mt di o xc nh nu trong di ch

, nhy khng ph thuc vo i lng o.

Trong ch tnh, tuyn tnh chnh l s khng ph thuc ca nhy ca

cm bin vo gi tr ca i lng o, th hin bi cc on thng trn c trng tnh

ca cm bin v hot ng ca cm bin l tuyn tnh chng no i lng o cn nm

trong vng ny.

Trong chng, tuyn tnh bao gm s khng ph thuc ca nhy ch

tnh S(0) vo i lng o, ng thi cc thng s quyt nh s hi p (nh tn

s ring f0 ca dao ng khng tt, h s tt dn cng khng ph thuc vo i lng

o.

Nu cm bin khng tuyn tnh, ngi ta a vo mch o cc thit b hiu chnh

sao cho tn hiu in nhn c u ra t l vi s thay i ca i lng o u

vo. S hiu chnh c gi l s tuyn tnh ho.

b) ng thng tt nht


10/179

Khi chun cm bin, t kt qu thc nghim ta nhn c mt lot im tng

ng (si,mi) ca i lng u ra v i lng u vo. V mt l thuyt, i vi cc

cm bin tuyn tnh, ng cong chun l mt ng thng. Tuy nhin, do sai s khi

o, cc im chun (mi, si) nhn c bng thc nghim thng khng nm trn cng

mt ng thng.

ng thng c xy dng trn cscc s liu thc nghim sao cho sai s l

b nht, biu din s tuyn tnh ca cm bin c gi l ng thng tt nht.

Phng trnh biu din ng thng tt nht c lp bng phng php bnh phng

b nht. Gi s khi chun cm bin ta tin hnh vi N im o, phng trnh c dng:

bams +=

Trong :

( )2

i

2

i

iiii

mm.N

m.sm.s.N

a

=

( )

=

2

i2i

iii2ii

mm.N

m.s.mm.sb

c) lch tuyn tnh

i vi cc cm bin khng hon ton tuyn tnh, ngi ta a ra khi nim lch tuyn

tnh, xc nh bi lch cc i gia ng cong chun v ng thng tt nht, tnh bng

% trong di o.

1.3.3. Sai s v chnh xc

Cc b cm bin cng nh cc dng co lng khc, ngoi i lng cn o (cm

nhn) cn chu tc ng ca nhiu i lng vt l khc gy nn sai s gia gi tro c v

gi tr thc ca i lng cn o. Gi x l lch tuyt i gia gi tro v gi tr thc x

(sai s tuyt i), sai s tng i ca b cm bin c tnh bng:

100.x

x= [%]

Sai s ca b cm bin mang tnh cht c tnh bi v khng th bit chnh xc gi tr

thc ca i lng cn o. Khi nh gi sai s ca cm bin, ngi ta thng phn chng

thnh hai loi: sai s h thng v sai s ngu nhin.

- Sai s h thng: l sai s khng ph thuc vo s ln o, c gi tr khng i hocthay i chm theo thi gian o v thm vo mt lch khng i gia gi tr thc v gi tr

o c. Sai s h thng thng do s thiu hiu bit v ho, do iu kin s dng khng

tt gy ra. Cc nguyn nhn gy ra sai s h thng c th l:


11/179

Do nguyn l ca cm bin.

+ Do gi tr ca i lng chun khng ng.

+ Do c tnh ca b cm bin.

+ Do iu kin v ch s dng.

+Do x l kt quo.

- Sai s ngu nhin: l sai s xut hin c ln v chiu khng xc nh. Ta c th don c mt s nguyn nhn gy ra sai s ngu nhin nhng khng th don c

ln v du ca n. Nhng nguyn nhn gy ra sai s ngu nhin c th l:

+ Do s thay i c tnh ca thit b.

+ Do tn hiu nhiu ngu nhin.

+ Do cc i lng nh hng khng c tnh n khi chun cm bin.

Chng ta c th gim thiu sai s ngu nhin bng mt s bin php thc nghim thch

hp nh bo v cc mch o trnh nh hng ca nhiu, tng iu chnh in p ngun

nui, b cc nh hng nhit , tn s, vn hnh ng ch hoc thc hin php o lng

thng k.

1.3.4. nhanh v thi gian hi p

nhanh l c trng ca cm bin cho php nh gi kh nng theo kp v thi gian

ca i lng u ra khi i lng u vo bin thin. Thi gian hi p l i lng c s

dng xc nh gi tr s ca nhanh.

nhanh tr l khong thi gian t khi i lng o thay i t ngt n khi bin thin

ca i lng u ra ch cn khc gi tr cui cng mt lng gii hn tnh bng %. Thi

gian hi p tng ng vi % xc nh khong thi gian cn thit phi chi sau khi c s

bin thin ca i lng o ly gi tr ca u ra vi chnh xc nh trc. Thi gian hi

p c trng cho ch qu ca cm bin v l hm ca cc thng s thi gian xc nh

ch ny.

Trong trng hp s thay i ca i lng o c dng bc thang, cc thng s thi

gian gm thi gian tr khi tng (tdm) v thi gian tng (tm) ng vi s tng t ngt ca i

lng o hoc thi gian tr khi gim (tdc) v thi gian gim (tc) ng vi s gim t ngt ca

i lng o. Khong thi gian tr khi tng tdm l thi gian cn thit i lng u ra tngt gi tr ban u ca n n 10% ca bin thin tng cng ca i lng ny v khong thi

gian tng tm l thi gian cn thit i lng u ra tng t 10% n 90% bin thin bin

thin tng cng ca n. m

m0

t

0s

s


12/179

Tng t, khi i lng o gim, thi gian tr khi gim tdc l thi gian cn thit i

lng u ra gim t gi tr ban u ca n n 10% bin thin tng cng ca i lng

ny v khong thi gian gim tc l thi gian cn thit i lng u ra gim t 10% n

90% bin thin bin thin tng cng ca n.

Cc thng s v thi gian tr, tdm, tm, tdc, tc ca cm bin cho php ta nh gi v thi gian

hi p ca n.

1.3.5. Gii hn sdng ca cm bin

Trong qu trnh s dng, cc cm bin lun chu tc ng ca ng lc chc, tc ng

nhit... Khi cc tc ng ny vt qu ngng cho php, chng s lm thay i c trng lm

vic ca cm bin. Bi vy khi s dng cm bin, ngi s dng cn phi bit r cc gii hn

ny.

a) Vng lm vic danh nh

Vng lm vic danh nh tng ng vi nhng iu kin s dng bnh thng

ca cm bin. Gii hn ca vng l cc gi tr ngng m cc i lng o, cc i

lng vt l c lin quan n i lng o hoc cc i lng nh hng c th

thng xuyn t ti m khng lm thay i cc c trng lm vic danh nh ca cm

bin.

b) Vng khng gy nn hhng

Vng khng gy nn h hng l vng m khi m cc i lng o hoc cc i

lng vt l c lin quan v cc i lng nh hng vt qua ngng ca vng lm

vic danh nh nhng vn cn nm trong phm vi khng gy nn h hng, cc c

trng ca cm bin c th b thay i nhng nhng thay i ny mang tnh thun

nghch, tc l khi trv vng lm vic danh nh cc c trng ca cm bin ly li

gi tr ban u ca chng.


13/179

c) Vng khng ph hu

Vng khng ph hy l vng m khi m cc i lng o hoc cc i lng vt

l c lin quan v cc i lng nh hng vt qua ngng ca vng khng gy nn

h hng nhng vn cn nm trong phm vi khng b ph hy, cc c trng ca cm

bin b thay i v nhng thay i ny mang tnh khng thun nghch, tc l khi trv

vng lm vic danh nh cc c trng ca cm bin khng th ly li gi tr ban u

ca chng. Trong trng hp ny cm bin vn cn s dng c, nhng phi tin

hnh chun li cm bin.

1.4. Nguyn l chung ch to cm bin

Cc cm bin c ch to da trn c scc hin tng vt l v c phn

lm hai loi:

- Cm bin tch cc: l cc cm bin hot ng nh mt my pht, p ng (s)l in tch, in p hay dng.

- Cm bin thng: l cc cm bin hot ng nh mt trkhng trong p ng (s) l in tr, t cm hoc in dung.

1.4.1. Nguyn l ch to cc cm bin tch cc

Cc cm bin tch cc c ch to da trn csng dng cc hiu ng vt l

bin i mt dng nng lng no (nhit, choc bc x) thnh nng lng in.

Di y m t mt cch khi qut ng dng mt s hiu ng vt l khi ch to cm

bin.

a) Hiu ng nhit in

Hai dy dn (M1) v (M2) c bn cht ho hc khc nhau c hn li vi nhau

thnh mt mch in kn, nu nhit hai mi hn l T1 v T2 khc nhau, khi

trong mch xut hin mt sut in ng e(T1, T2) m ln ca n ph thuc chnh

lch nhit gia T1 v T2.

T1

T2

(M1)

(M2)

(M2)

e

Hnh 1.4. S hi u ng nhi t i n.

T1


14/179

Hiu ng nhit in c ng dng o nhit T1 khi bit trc nhit T2,

thng chn T2 = 0oC.

b) Hiu ng hoin

Mt s tinh th gi l tinh th hoin (v d tinh th sulfate triglycine) c tnh

phn cc in t pht vi phn cc ph thuc vo nhit , lm xut hin trn cc

mt i din ca chng nhng in tch tri du. ln ca in p gia hai mt ph

thuc vo phn cc ca tinh th hoin.

Hiu ng hoin c ng dng o thng lng ca bc x nh sng. Khi ta

chiu mt chm nh sng vo tinh th hoin, tinh th hp th nh sng v nhit

ca n tng ln, lm thay i s phn cc in ca tinh th. o in p V ta c th

xc nh c thng lng nh sng .

c) Hiu ng p inMt s vt liu gi chung l vt liu p in (nh thch anh chng hn) khi b bin dng

dc tc ng ca lc chc, trn cc mt i din ca tm vt liu xut

hin nhng lng in tch bng nhau nhng tri du, c gi l hiu ng p in. o V ta

c th xc nh c cng ca lc tc dng F.

d) Hiu ng cm ng in t

v

Hnh 1.5 ng d ng hi u ng ho i n

V F

Hnh 1.6 ng d ng hi u ng p i n

F


15/179

Khi mt dy dn chuyn ng trong t trng khng i, trong dy dn xut hin

mt sut in ng t l vi t thng ct ngang dy trong mt n v thi gian, ngha

l t l vi tc dch chuyn ca dy. Tng t nh vy, trong mt khung dy t

trong t trng c t thng bin thin cng xut hin mt sut in ng t l vi tc

bin thin ca t thng qua khung dy.

Hiu ng cm ng in tc ng dng xc nh tc dch chuyn ca vt

thng qua vic o sut in ng cm ng.

e) Hiu ng quang in

- Hiu ng quang dn: (hay cn gi l hiu ng quang in ni) l hin tng

gii phng ra cc ht dn t do trong vt liu (thng l bn dn) khi chiu vo chng

mt bc x nh sng (hoc bc xin t ni chung) c bc sng nh hn mt

ngng nht nh.

- Hiu ng quang pht xin t: (hay cn gi l hiu ng quang in ngoi)

l hin tng cc in tc gii phng v thot khi b mt vt liu to thnh dng

c th thu li nhtc dng ca in trng.

g) Hiu ng quang - in - t

Khi tc dng mt t trng B vung gc vi bc x nh sng, trong vt liu bn

dn c chiu sng s xut hin mt hiu in th theo hng vung gc vi t

trng B v hng bc x nh sng.

h) Hiu ng Hall

e

Hnh 1.7 ng d ng hi u ng c m ng i n t

B

B

V

Hnh 1.8 ng d ng hi u ng quang - i n - t


16/179

Khi t mt tm mng vt liu mng (thng l bn dn), trong c dng in chy

qua, vo trong mt t trng B c phng to vi dng in I trong tm mt gc , s xut

hin mt hiu in th VH theo hng vung gc vi B v I. Biu thc hiu in th c

dng:

= sin.B.I.KV HH

Trong KH l h s ph thuc vo vt liu v kch thc hnh hc ca tm vt liu.

Hiu ng Hall c ng dng xc nh v tr ca mt vt chuyn ng. Vt cn xc

nh v tr lin kt chc vi thanh nam chm, mi thi im, v tr thanh nam chm xc

nh gi tr ca t trng B v gc tng ng vi tm bn dn mng lm vt trung gian. V

vy, hiu in th VHo c gia hai cnh tm bn dn l hm ph thuc vo v tr ca vt

trong khng gian.

1.4.2. Nguyn ch to cm bin thngCm bin thng thng c ch to t mt trkhng c cc thng s ch yu nhy

vi i lng cn o. Gi tr ca trkhng ph thuc kch thc hnh hc, tnh cht in ca

vt liu ch to (nhin trsut , t thm , hng sin mi ). V vy tc ng ca

i lng o c thnh hng ring bit n kch thc hnh hc, tnh cht in hoc ng

thi c hai.

S thay i thng s hnh hc ca trkhng gy ra do chuyn ng ca phn t chuyn

ng hoc phn t bin dng ca cm bin. Trong cc cm bin c phn t chuyn ng, mi

v tr ca phn tng sng vi mt gi tr xc nh ca trkhng, cho nn o trkhng c

th xc nh c v tr ca i tng. Trong cm bin c phn t bin dng, s bin dng ca

phn t bin dng di tc ng ca i lng o (lc hoc cc i lng gy ra lc) gy ra

s thay i ca trkhng ca cm bin. S thay i trkhng do bin dng lin quan n lc

tc ng, do lin quan n i lng cn o. Xc nh trkhng ta c th xc nh c

i lng cn o.

B

SN

v

X

X

Hnh 1.9 ng d ng hi u ng Hall


17/179

S thay i tnh cht in ca cm bin ph thuc vo bn cht vt liu ch to tr

khng v yu t tc ng (nhit , chiu sng, p sut, m ...). ch to cm bin,

ngi ta chn sao cho tnh cht in ca n ch nhy vi mt trong cc i lng vt l trn,

nh hng ca cc i lng khc l khng ng k. Khi c th thit lp c s ph

thuc n tr gia gi tri lng cn o v gi tr trkhng ca cm bin.

Trn bng 1.1 gii thiu cc i lng cn o c kh nng lm thay i tnh cht

in ca vt liu s dng ch to cm bin.

Bng 1.1

i lng cn o c trng nhy cm Loi vt liu s dng

Nhit Kim loi (Pt, Ni, Cu)

Bn dn

Bc x nh sng Bn dn

Bin dng T thm ()Hp kim Ni, Si pha tpHp kim st t

V tr (nam chm) Vt liu tin tr:Bi, InSb

1.5. Mch o

1.5.1. S mch o

Mch o bao gm ton b thit bo (trong c cm bin) cho php xc nh chnh

xc gi tr ca i lng cn o trong nhng iu kin tt nht c th.

u vo ca mch, cm bin chu tc ng ca i lng cn o gy nn tn hiu inmang theo thng tin vi cn o.

u ra ca mch, tn hiu in qua x l c chuyn i sang dng c thc

c trc tip gi tr cn tm ca i lng o. Vic chun ho m bo cho mi gi tr ca

ch thu ra tng ng vi mt gi tr ca i lng o tc ng u vo ca mch.

Dng n gin ca mch o gm mt cm bin, b phn bin i tn hiu v thit b ch

th, v d mch o nhit gm mt cp nhit ghp ni trc tip vi mt milivn k.

V

Hnh 1.10 S m ch o nhi t b ng c p nhi t


18/179

Trn thc t, do cc yu cu khc nhau khi o, mch o thng gm nhiu thnh

phn trong c cc khi ti u ho vic thu thp v x l d liu, chng hn

mch tuyn tnh ho tn hiu nhn t cm bin, mch khin dung k sinh, cc b

chuyn i nhiu knh, b khuch i, b so pha lc nhiu, b chuyn i tng t -

s, b vi x l, cc thit b h tr... Trn hnh 1.11 biu din s khi mt mch in

o in th trn b mt mng nhy quang c lp rp t nhiu phn t

1.5.2. Mt s phn tcbn ca mch o

a) B khuch i thut ton (KTT)

B khuch i thut ton mch tch hp l b khuch i dng mt chiu c hai u vo

v mt u ra chung, thng gm hng trm tranzito v cc in tr, tin ghp ni vi

nhau. S b khuch i thut ton biu din trn hnh 1.12.

Cc c tnh cbn ca b khuch i thut ton:- B khuch i c hai u vo: mt u o (-), mt u khng o (+).- in trvo rt ln, chng trm Mn G.- in trra rt nh, cphn chc .- in p lch u vo rt nh, cvi nV.- H s khuch i hmch rt ln, c100.000.- Di tn lm vic rng.

FC(1)

D(2)

PA(3)

PC(4)

(5) ADC(6)

My in

CPU(7)

Mnhnh

Hnh 1.11 M ch o i n th b m t1) My pht ch c n ng 2) C m bi n i n tch 3) Ti n khu ch i

4) So pha l c nhi u 5) Khu ch i 6) Chuy n i t ng t s 7) My tnh

+

Hnh 1.12 S b khu ch i thu t ton

K

Ura = K.Uvo

Ura


19/179

- H s suy gim theo cch ni chung CMRR l t s h s khuch i ca b khuchi thut ton i vi cc tn hiu sai lch v h s khuch i theo cch ni chung ca cng

b khuch i thut ton. Thng thng CMRR vo khong 90 dB.

- Tc tng hn ch s bin thin cc i ca in p tnh bng V/s.b) B khuch i o lng IA

B khuch i o lng IA c hai u vo v mt u ra. Tn hiu u ra t l vi hiu

ca hai in p u vo:

UA)UU(AU ra == +

u vo vi sai ng vai tr rt quan trng trong vic kh nhiu ch chung v tng

in trvo ca KTT. in p trn Raphi bng in p vi sai u vo U v to nn dng

inaR

U

i

= . Cc in p ra t KTT U1 v U2 phi bng nhau v bin nhng ngc

pha. in p U3 ca tng th hai bin i u ra vi sai thnh u ra n cc. H s khuch i

tng ca IA bng:

1

3

a R

R

R

R21A

+=

c) Khin p lch

i vi mt b khuch KTT l tng khi hmch phi c in p ra bng

khng khi hai u vo ni mt. Thc t v cc in p bn trong nn to ra mt in pnh (in p phn cc) u vo KTT cvi mV, nhng khi s dng mch kn in

p ny c khuch i v to nn in p kh ln u ra. khin p lch c

th s dng s hnh 1.14, bng cch iu chnh bin trR3.

Hnh 1.13 S b khu ch i o l ng g m ba K TT ghp n i i n tr

+

+

+

R10k

R10k

Ra1k

R2 10k R3 10k

R2 10k

R3190k

U1

U2

U3

u ra

+

+ 9V

R310k

R2100k

- 9V

u ra

+

1

2

3

4 5

6714

7


20/179

d) Mch lp li in p

lp li in p chnh xc, ngi ta s dng b KTT lm vic ch khng

o vi h s khuch i bng 1 s nh hnh 1.15.

Trong b lp in p, cc dng ca KTT c ni trc tip vi tn hiu vo,

cn cc m c ni trc tip vi u ra, to nn in p phn hi 100% do h s

khuch i bng 1. Mch lp in p c chc nng tng in tr u vo, do vy

thng dng ni gia hai khu trong mch o.

e) Mch cu

Cu Wheatstone thng c s dng trong cc mch o nhit , lc, p sut,

t trng... Cu gm bn in trR1, R2, R3 cnh v R4 thay i (mc nh hnh

1.16) hot ng nh cu khng cn bng da trn vic pht hin in p qua ngcho ca cu.

Hnh 1.15 S m ch l p i n p

+ 9V

- 9V

u ra

+

2

3

4

6714

u vo

7

Hnh 1.15 S m ch c u

+

R1 R3

R2R4 = R(1+)

U Vra


21/179

Trong mch cu, in p ra l hm phi tuyn nhng i vi bin i nh

(> R2 hoc R2 >> R1 in p ra ca cu gim. t K = R1/R2 nhy

ca cu l:

( )2k1

K

R

U

+= .


22/179

Chng II

Cm bin quang

2.1.Tnh cht v n vo nh sng

2.1.1. Tnh cht ca nh sng

Nh chng ta bit, nh sng va c tnh cht sng va c tnh cht ht.

nh sng l mt dng ca sng in t, vng nh sng nhn thy c bc sng t 0,4 -

0,75 m. Trn hnh 2.1 biu din ph nh sng v s phn chia thnh cc di mu ca ph.

Vn tc truyn nh sng trong chn khng c = 299.792 km/s, trong mi trng vt cht

vn tc truyn sng gim, c xc nh theo cng thc:

ncv =

n - chit sut ca mi trng.

Mi quan h gia tn s v bc sng ca nh sng xc nh bi biu thc:

- Khi mi trng l chn khng :

=c

- Khi mi trng l vt cht :

=v

.

Trong l tn s nh sng.Tnh cht ht ca nh sng th hin qua s tng tc ca nh sng vi vt cht.

nh sng gm cc ht nh gi l photon, mi ht mang mt nng lng nht nh,

nng lng ny ch ph thuc tn s ca nh sng:

= hW

(2.1)

0,3

95

0,4

55

0,4

90

0,5

75

0,5

90

0,6

50

0,7

50

c c tm tm lam l cvng

da cam h ng ngo i

0,01 0,1 0,4 0,75 1,2 10 30 100

c c tm h ng ngo i h. ngo i xa

trngth

y

h.n.ng

n

( m)

Hnh 2.1 Ph nh sng


23/179

Trong h l hng s Planck (h = 6,6256.10-34J.s).

Bc sng ca bc x nh sng cng di th tnh cht sng th hin cng r,

ngc li khi bc sng cng ngn th tnh cht ht th hin cng r.

2.1.2. Cc n vo quang

a) n vo nng lng

- Nng lng bc x (Q): l nng lng lan truyn hoc hp th di dng bc xo

bng Jun (J).

- Thng lng nh sng (): l cng sut pht x, lan truyn hoc hp tho bng oat

(W):

dt

dQ=

(2.2)

- Cng nh sng (I): l lung nng lng pht ra theo mt hng cho trc ng

vi mt n v gc khi, tnh bng oat/steriadian.

=d

dI

(2.3)

- chi nng lng (L): l t s gia cng nh sng pht ra bi mt phn t b

mt c din tch dA theo mt hng xc nh v din tch hnh chiu dAn ca phn t

ny trn mt phng P vung gc vi hng .

ndA

dIL =

(2.4)

Trong dAn = dA.cos, vi l gc gia P v mt phng cha dA.

chi nng lng o bng oat/Steriadian.m2.

- ri nng lng (E): l t s gia lung nng lng thu c bi mt phn t bmt v din tch ca phn t.

dA

dE

=

(2.5)

ri nng lng o bng oat/m2.

b) n vo th gic


24/179

nhy ca mt ngi i vi nh sng c bc sng khc nhau l khc nhau.

Hnh 2.2 biu din nhy tng i ca mt V() vo bc sng. Cc i lng th

gic nhn c ti lng nng lng tng ng thng qua h s t l K.V().

Theo quy c, mt lung nh snh c nng lng 1W ng vi bc sng max

tng ng vi lung nh sng bng 680 lumen, do K=680.

Do vy lung nh sng n sc tnh theo n vo th gic:

( ) ( ) ( )= V680V lumen

i vi nh sng ph lin tc:

=

dd

)(d)(V680

2

1

V lumen

Tng t nh vy ta c th chuyn i tng ng cc n vo nng lng v

n vo th gic.

Bng 2.1 lit k cc n vo quang cbn.

Bng 2.1

i lng o n v th gic n v nng lng

Lung (thng lng) lumen(lm) oat(W)

Cng cadela(cd) oat/sr(W/sr)

chi cadela/m2 (cd/m2) oat/sr.m2 (W/sr.m2)

ri lumen/m2 hay lux (lx) oat/m2 (W/m2)

Nng lng lumen.s (lm.s) jun (j)

2.2.Cm bin quang dn

2.2.1. Hiu ng quang dn

Hnh 2.2 ng cong nh y t ng i c a m t

(m)

V()

0,3 0,4 0,5 0,6 0,7 0,8

0

0,5

1

max


25/179

Hiu ng quang dn (hay cn gi l hiu ng quang in ni) l hin tng gii

phng nhng ht ti in (ht dn) trong vt liu di tc dng ca nh sng lm tng

dn in ca vt liu.

Trong cht bn dn, cc in t lin kt vi ht nhn, gii phng in t khi

nguyn t cn cung cp cho n mt nng lng ti thiu bng nng lng lin kt Wlk.

Khi in tc gii phng khi nguyn t, s to thnh ht dn mi trong vt liu.

Ht dn c gii phng do chiu sng ph thuc vo bn cht ca vt liu b

chiu sng. i vi cc cht bn dn tinh khit cc ht dn l cp in t - l trng.

i vi trng hp bn dn pha tp, ht dn c gii phng l in t nu l pha tp

dono hoc l l trng nu l pha tp acxepto.

Gi s c mt tm bn dn phng th tch V pha tp loi N c nng cc donor

Nd, c mc nng lng nm di vng dn mt khong bng Wd ln nhit

phng v khi trong ti nng n0 ca cc donor b ion ho do nhit l nh.

+ l tr ng

- i n t

h

+ l tr ng

h-

- i n t

h+

Hnh 2.3. nh h ng c a b n ch t v t li u n h t d n c gi i phng

+ ++Wd

+ +++

h hVng d n

Vng ho

Hnh 2.4. T bo quang d n v s chuy n m c n ng l ng c a i n t

L

V

chi u sng

A


26/179

Khi trong ti, nng in t c gii phng trong mt n v thi gian t l

vi nng cc tp cht cha b ion ho v bng a(Nd -no), vi h s a xc nh theo

cng thc:

=kT

qWexpa d

(2.6)

Trong q l tr tuyt i ca in tch in t, T l nhit tuyt i ca khi

vt liu, k l hng s.

Sin t ti hp vi cc nguyn t b ion ho trong mt n v thi gian t l

vi cc nguyn t b ion ho n0 v nng in t cng chnh bng n0 v bng

r. 20n , trong r l h s ti hp.

Phng trnh ng hc biu din s thay i nng in t t do trong khi

vt liu c dng:

( ) 200d0 n.rnNa

dt

dn=

trng thi cn bng ta c : 0dt

dn0 =

Suy ra:

2/1

d

2

2

0r

N.a

r4

a

r.2

an

++=

(2.7)

dn trong ti c biu din bi h thc:

00 nq= (2.8)

Trong l linh ng ca in t.

Khi nhit tng, linh ng ca in t gim, nhng s tng mt in t

t do do s kch thch nhit ln hn nhiu nn nh hng ca n l nhn t quyt nh

i vi dn.

Khi chiu sng, cc photon s ion ho cc nguyn t donor, gii phng ra cc

in t. Tuy nhin khng phi tt c cc photon p ti b mt vt liu u gii phng

in t, mt s b phn x ngay b mt, mt s b hp th v chuyn nng lng cho

in t di dng nhit nng, ch phn cn li mi tham gia vo gii phng in t. Do

vy, sin t (g) c gii phng do b chiu sng trong mt giy ng vi mt n

v th tch vt liu, xc nh bi cng thc:


27/179

( )

==h

R1.

L.A

1

V

Gg

(2.9)

Trong :

G - sin tc gii phng trong th tch V trong thi gian mt giy.

V=A.L, vi A, L l din tch mt cnh v chiu rng tm bn dn (hnh 2.4).

- hiu sut lng t (sin t hoc l trng trung bnh c gii phng khi

mt photon b hp th).

R - l h s phn x ca b mt vt liu.

- bc sng nh sng.

- thng lng nh sng.

h - hng s Planck.

Phng trnh ng hc ca ti hp trong trng hp ny c dng:

( ) 2d n.rgnNadt

dn+=

Thng thng bc x chiu ti ln sin tc gii phng ln hn

rt nhiu so vi in tc gii phng do nhit:

( )nNag d >> v n>>n0

Trong iu kin trn, rt ra phng trnh ng hc cho mt in tiu kin cn

bng di tc dng chiu sng:2/1

r

gn

=

(2.10)

dn tng ng vi nng in tiu kin cn bng:

nq= .

(2.11)

T cng thc (2.9), (2.10) v (2.11) ta nhn thy dn l hm khng tuyn tnh

ca thng lng nh sng, n t l vi 1/2. Thc nghim cho thy s m ca hm

nm trong khong 0,5 - 1.

2.2.2. T bo quang dn

a) Vt liu ch to


28/179

T bo quang dn c ch to cc bn dn a tinh thng nht hoc n tinh

th, bn dn ring hoc bn dn pha tp.

- a tinh th: CdS, CdSe, CdTe.

PbS, PbSe, PbTe.

- n tinh th: Ge, Si tinh khit hoc pha tp Au, Cu, Sb, In.

SbIn, AsIn, PIn, cdHgTe.

Vng ph lm vic ca cc vt liu ny biu din trn hnh 2.5

b) Cc c trng

- in tr: Gi trin trti RC0 ca cc quang in trph thuc rt ln vo hnh

dng hnh hc, kch thc, nhit v bn cht ho l ca vt liu ch to. Cc cht

PbS, CdS, CdSe c in trti rt ln ( t 104 - 10925oC), trong khi SbIn,

SbAs, CdHgTe c in trti tng i nh ( t 10 - 10325oC). in tr Rc

ca cm bin gim rt nhanh khi ri tng ln. Trn hnh 2.6 l mt v d v s thayi ca in trcm bin theo ri sng.

Hnh 2.5. Vng ph lm vi c c a m t s v t li u quang d n

0,2 0,6 1 2 3 4 5 10 20 30

CdS

CdSeCdTe

PbSPbSe

PbTeGe

Si

GeCuSnIn

AsInCdHg

, m

i n tr ()

102

104

106

106


29/179

T bo quang dn c th coi nh mt mch tng ng gm hai in trRc0 v

Rcp mc song song:

cpco

cpco

cRR

RRR

+=

(2.12)

Trong :

Rco - in trtrong ti.

Rcp - in trkhi chiu sng:= aRcp .

a - h s ph thuc vo bn cht vt liu, nhit , ph bc x.

- h s c gi tr t 0,5 - 1.

Thng thng Rcp


30/179

Trong iu kin s dng thng thng I0


31/179


32/179

- Thu tn hiu quang: dng t bo quang dn thu v bin tn hiu quang thnhxung in. Cc xung nh sng ngt qung c th hin qua xung in, trn cs

c th lp cc mch m vt hoc o tc quay ca a.

2.2.3. Photoiot

a) Cu to v nguyn l hot ng

Xt hai tm bn dn, mt thuc loi N v mt thuc loi P, ghp tip xc nhau. Ti

mt tip xc hnh thnh mt vng ngho ht dn v ti vng ny tn ti mt in

trng v hnh thnh hng ro th Vb.

Khi khng c in th ngoi t ln chuyn tip (V=0), dmg in chy qua

chuyn tip i = 0, thc t dng I chnh l dng tng cng ca hai dng in bng nhau

v ngc chiu:

- Dng khuch tn cc ht cbn sinh ra khi ion ho cc tp cht (l trong trongbn dn loi P, in t trong bn dn loi N) do nng lng nhit ca cc ht dn c

bn ln vt qua hng ro th.

- Dng ht dn khng cbn sinh ra do kch thch nhit (in t trong bn dn P,l trng trong bn dn N) chuyn ng di tc dng ca in trng E trong vng

ngho.

Hnh 2.9 Dng t bo quang d n i u khi n r lea) i u khi n tr c ti p b) i u khi n thng qua tranzito khu ch i

+ +

Vng chuy n ti p

E

P N

Vng ngho

Vb

Hnh 2.10 S chuy n ti p P - N v hi u ng quang i n trong vng ngho

+

N

P

h

+

Ir

Vng ngho


33/179

Khi c in p t ln it, hng ro th thay i ko theo s thay i dng ht c

bn v b rng vng ngho. Dng in qua chuyn tip:

0d

0 IkT

qVexpII

=

Khi in p ngc ln (Vd


34/179

b) Ch hot ng

- Ch quang dn:

S nguyn l (hnh 2.12a) gm mt ngun Es phn cc ngc it v mt in

trRm o tn hiu.

Dng ngc qua it:

p0d

0r IIkT

qVexpII ++

=

(2.40)

Trong Ip l dng quang in:

( )( )Xexp

hc

R1qI 0p

=

(2.41)

Khi in p ngc Vd ln, thnh phn

kT

qVexp d 0, ta c:

P0R III +=

Thng thng I0


35/179

im lm vic ca it l im giao nhau gia ng thng ti v ng c tuyn

i-V vi thng lng tng ng. Ch lm vic ny l tuyn tnh, VR t l vi thng

lng.

- Ch quang th:

Trong ch ny khng c in p ngoi t vo it. it lm vic nh mt bchuyn i nng lng tng ng vi mt my pht v ngi ta o th hmch V0C

hoc o dng ngn mch ISC.

o th hmch: Khi chiu sng, dng IP tng lm cho hng ro th gim mt

lng Vb. S gim chiu cao hng ro th lm cho dng ht dn cbn tng ln, khi

t cn bng Ir= 0.

Ta c: 0IIkT

qVexpI p0

d0 =++

Rt ra:

+=

0

Pb

I

I1log

q

kTV

gim chiu cao Vb ca hng ro th c th xc nh c thng qua o in

p gia hai u it khi hmch.

+=

0

POC

I

I1log

q

kTV

Khi chiu sng yu IP I0 v ta c:

0

POC

I

Ilog

q

kTV =

Trong trng hp ny VOC c gi tr tng i ln (c0,1 - 0,6 V) nhng ph thuc

vo thng lng theo hm logarit.

0,2

0,4

VOC, V


36/179

o dng ngn mch:Khi ni ngn mch hai u it bng mt in trnh hn rd

no , dng on mch ISC chnh bng IP v t l vi thng lng (hnh 2.14):

PSC II =

c im quan trng ca ch ny l khng c dng ti, nhvy c th gimnhiu v cho php o c thng lng nh.

c) nhy

i vi bc x c ph xc nh, dng quang in IP t l tuyn tnh vi thng

lng trong mt khong tng i rng, c5 - 6 decad. nhy ph xc nh theo

cng thc:

( ) ( )

=

=hc

XexpR1qI)(S P

Vi s.

nhy ph ph thuc vo , hiu sut lng t, h s phn x R v h s

hp th.

Hnh 2.14 S ph thu c c a dng ng nm ch vo thng l ng nh sng

ISC,V

Thng l ng, mW

0

10

20

0,1 0,2

1,0

0,6

0,4

0,2

0,1

0,04

S()S(P)


37/179

Ngi s dng cn phi bit nhy ph da trn ng cong ph hi p

S()/S(P) v gi tr ca bc sng P ng vi nhy cc i. Thng thng S(P)

nm trong khong 0,1 - 1,0 A/W.

Khi nhit tng, cc i P ca ng cong ph dch chuyn v pha bc

sng di. H s nhit ca dng quang dndT

dI.

I

1 P

P

c gi tr khong0,1%/oC.

d) Sng dng photodiot

- S lm vic ch quang dn:

c trng ca ch quang dn:

+ tuyn tnh cao.

+ Thi gian hi p ngn.+ Di thng ln.

Hnh 2.16 trnh by so dng ngc trong ch quang dn.

S cs(hnh 2.17a):

r1

2m0 I

R

R1RV

+=

0,3 0,4 0,5 0,6 0,7 0,8 0,9 1,0 (m)

0

0,1

0,2

0,3

0,4

S (A/W)

Hnh 2.16 S ph thu c c a nh y vo nhi t

T1

T2>T1

T2


38/179

Khi tng in trRm s lm gim nhiu. Tng trvo ca mch khuch i phi

ln trnh lm gim in trti hiu dng ca it.

S tc ng nhanh (hnh 2.17b):

( ) r210 IRRV +=

in trca iot nh v bngK

RR 21 + trong K l h s khuch i tn s lm vic.

T C2 c tc dng b trnh hng ca t k sinh Cpl vi iu kin 22pl1 CRCR = . B

khuch i y phi c dng vo rt nh v s suy gim do nhit cng phi khng

ng k.

- S lm vic ch quang th:

c trng ca ch quang th:

+ C th lm vic ch tuyn tnh hoc logarit tu thuc vo ti.+ t nhiu.

+ Thi gian hi p ln.

+ Di thng nh.

+ Nhy cm vi nhit ch logarit.

S tuyn tnh (hnh 2.18a): o dng ngn mch Isc.

Trong ch ny:

scm0 I.RV =

S logarit(hnh 2.18b): o in p hmch Voc.

oc1

20 V

R

R1V

+=

ES

Rm

Ir

R1

R2

V0

CP1

Rm

ESR1+R2

R1

V0

R2

Hnh 2.17 S m ch o dng ng c trong ch quang d n

a) b)

++

C2+

VcoR1

R2

V0

+_IOC

R1=Rm

Rm

V0

_

+


39/179

2.2.4. Phototranzito

a) Cu to v nguyn l hot ng

Phototranzito l cc tranzito m vng baz c th c chiu sng, khng c

in p t ln baz, ch c in p trn C, ng thi chuyn tip B-C phn cc ngc.

in p t vo tp trung hu nh ton b trn chuyn tip B-C (phn cc

ngc) trong khi chnh lch in p gia E v B thay i khng ng k (VBE

0,6-0,7 V). Khi chuyn tip B-C c chiu sng, n hot ng ging nh photoiot

ch quang th vi dng ngc:

P0r III +=

Trong I0 l dng ngc trong ti, IP l dng quang in di tc dng ca thng

lng 0 chiu qua b dy X ca baz(bc sng < S):

( )0P hc

)Xexp(R1qI

=

Dng Irng vai tr dng baz, n gy nn dng colectIc:

( ) ( ) ( ) p0rc I1I1I1I +++=+=

- h s khuch i dng ca tranzito khi u chung emit.

C th coi phototranzito nh t hp ca mt photodiot v mt tranzito (hnh 2.19b).

Phodiot cung cp dng quang in ti baz, cn tranzito cho hiu ng khch i .

Hnh 2.19 Phototranzitoa) S m ch i n b) S t ng ng

c) Tch c p i n t l tr ng khi chi u sng baz

a) b) c)

|+

i n th

CBE


40/179

Cc in t v l trng pht sinh trong vng baz(di tc dng ca nh sng) s b

phn chia di tc dng ca in trng trn chuyn tip B - C.

Trong trng hp tranzito NPN, cc in t b ko v pha colecttrong khi l

trng b gi li trong vng baz(hnh 2.19c) to thnh dng in t t E qua B n C.

Hin tng xy ra tng t nh vy nu nh l trng phun vo bazt mt ngun bn

ngoi: in th baz tng ln lm gim hng ro th gia E v B, iu ny gy nn

dng in t IE chy t E n B v khuch tn tip t B v pha C.

b) nhy

Khi nhn c thng lng 0, iot baz-colectsinh ra dng quang in Ip,

dng ny gy nn trong phototranzito mt dng ( ) pcp I1I += , trong gi tr ca Icp

c rt ra t cng thc ca Ip:

( ) ( ) ( ) 0cp hc XexpR1q1I +=

i vi mt thng lng 0 cho trc, ng cong ph hi p xc nh bi

bn cht ca iot B-C: vt liu ch to (thng l Si) v loi pha tp (hnh 2.20). i

vi mt bc sng cho trc, dng colectIc khng phi l hm tuyn tnh ca thng

lng hoc chiu sng bi v h s khuch i ph thuc vo dng Ic (tc l cng

ph thuc thng lng), ngha l0

cI

ph thuc vo 0.

nhy ph S(p) bc sng tng ng vi im cc i c gi tr nm trong

khong 1 - 100A/W.

c) S dng phototranzito

Phototranzito c th dng lm b chuyn mch, hoc lm phn t tuyn tnh.

ch chuyn mch n c u im so vi photodiot l cho php s dng mt cch

S()S(p)(%)

100

80

60

40

20

0,4 0,6 0,8 1,0

(m)

Hnh 2.20 ng cong ph h i p c a photodiot


41/179

trc tip dng chy qua tng i ln. Ngc li, ch tuyn tnh, mc d cho

khuch i nhng ngi ta thch dng photoiot v n c tuyn tnh tt hn.

- Phototranzito chuyn mch:

Trong trng hp ny s dng thng tin dng nh phn: c hay khng c bc

x, hoc nh sng nh hn hay ln hn ngng. Tranzito chn hoc bo ho cho php

iu khin trc tip (hoc sau khi khuch i) nh mt rle, iu khin mt cng logic

hoc mt thyristo (hnh 2.21).

- Phototranzito trong ch tuyn tnh:

C hai cch s dng trong ch tuyn tnh.

- Trng hp th nht: o nh sng khng i (ging nh mt luxmet).

- Trng hp th hai: thu nhn tn hiu thay i dng:

( ) ( )tt 10 +=

( ) ( )t.SItI 10cc +=

2.2.5. Phototranzito hiu ng trng

Phototranzito hiu ng trng (photoFET) c s tng ng nh hnh

2.23.

++

++ +

Hnh 2.21 Photodiotzito trong ch chuy n m cha) R le b) R le sau kh ch i c) C ng logic d) Thyristo

+

Hnh 2.22 S nguyn l luxmet

Trong 1(t) l thnh ph n

thay i v i bin nh sao cho

khng d n t i phototranzito b ch n

ho c b o ho v c th coi hh y

khng i. Trong i u ki n , dng

colect c d ng:

G

D

SS

G

D

+


42/179

Trong phototranzito hiu ng trng, nh sng c s dng lm thay i in tr

knh. Vic iu khin dng mng IDc thc hin thng qua s thay i in p VGS gia

cng v ngun. Trong ch phn cc ngc chuyn tip P-N gia cng v knh, in p

ny s xc nh rng ca knh v do dng mng c dng:

2

P

GSDSSD

V

V1II

+=

Vi IDS - dng mng khi VGS = 0.

VP - in p tht knh.

Khi b chiu sng, chuyn tip P-N hot ng nh mt photodiot cho dng ngc:

P0r III +=

IP = Sg - dng quang in.

I0 - dng in trong ti.

Sg - nhy ca iot cng - knh.

- thng lng nh sng.

Dng Irchy qua in trRg ca mch cng xc nh in th VGS v v dng mng:

( ) gP0gGS EIIRV +=

Eg - th phn cc ca cng.

Phototranzito hiu ng trng c ng dng nhiu trong vic iu khin in

p bng nh sng.

2.3.Cm bin quang in pht x

2.3.1. Hiu ng quang in pht x

Hiu ng quang in pht x hay cn c gi l hiu ng quang in ngoi l

hin tng cc in tc gii phng khi b mt vt liu to thnh dng khi chiu


43/179

vo chng mt bc x nh sng c bc sng nh hn mt ngng nht nh v c th

thu li nhtc dng ca in trng.

Cch pht xin t khi chiu sng vt liu xy ra theo ba giai on:

- Hp th photon v gii phng in t bn trong vt liu.- in t va c gii phng di chuyn n b mt.- in t thot khi b mt vt liu.

Khi mt in t hp th photon v c gii phng, di chuyn ca n trong khi

vt liu mang tnh ngu nhin theo mi hng, do ch mt lng rt nh hng ti

b mt. Mt khc, trong qu trnh di chuyn, cc in t ny c th va chm vi cc

in t khc v mt i mt phn nng lng do ch mt lng nhin tc gii

phng ti c b mt. Mt khc, s pht x ca cc in t sau khi n c b

mt ch c th xy ra khi ng nng ca n thng c hng ro th phn cch vt

liu vi mi trng.

Vi tt c nhng iu kin trn, sin t pht x trung bnh khi mt photon b

hp th (hiu sut lng t ) thng nh hn 10% v t khi vt qu 30%.

Vt liu ch to: Ph thuc vo bc sng nh sng, vt liu ch to photocatot

c th chn trong cc loi sau:

- AgOCs nhy vng hng ngoi.

- Cs3Sb, (Cs)Na2KSb v K2CsSb: nhy vi nh sng nhn thy v vng t ngoi.

- Cs2Te, Rb2Te v CsTe ch nhy trong vng t ngoi.

Hiu sut lng t ca cc vt liu trn ~ 1 - 30%.

Ngoi ra cn dng cc hp cht ca cc cht thuc nhm III - V, l cc hp cht

GaAsxSb1-x , Ga1-xInxAs, InAsxP1-x, ngng nhy sng ca chng nm vng hng

ngoi ( ~1m), hiu sut lng tt ti 30%.

2.3.2. T bo quang in chn khng

T bo quang in chn khng gm mt ng hnh tr c mt ca s trong sut,

c ht chn khng (p sut ~ 10-6

- 10-8

mmHg). Trong ng t mt catot c khnng pht x khi c chiu sng v mt anot.

A

K

A

K

AK


44/179

S tng ng v s thay i ca dng anot Ia ph thuc vo in th anot - catot

Vakbiu din trn hnh 2.25.

c trng I - V c hai vng r rt:

+ Vng in tch khng gian c trng bi s tng mnh ca dng khi in p

tng.

+ Vng bo ho c trng bi s ph thuc khng ng k ca dng vo in

p.T bo quang in c s dng ch yu trong vng bo ho, khi n ging

nh mt ngun dng, gi tr ca dng ch ph thuc vo thng lng nh sng m n

nhn c. in trtrong ca t bo quang in rt ln v c th xc nh t dc

ca c tuyn vng bo ho:

=

aka

dV

dI1

nhy ph ca t bo quang in c biu din thng qua gi tr ca dng anottrong vng bo ho, thng vo c10 - 100 mA/W.

2.3.3. T bo quang in dng kh

T bo quang in dng kh c cu to tng t t bo quang in chn khng,

ch khc ch th tch bn trong ca n c in y bng kh, thng l kh acgon,

di p sut c10-1 - 10-2 mmHg.

K

A Ia

E

Rm

Ia (A)

Vak (V)

4,75 mW

2,37 mW

0,95 mW

4

3

2

1

0 20 40 60 80 100 120

Hnh 2.25 S t ng ng v c tr ng I - v c a t bo quang i n chn khng

a) b)


45/179

Khi in p thp hn 20V, c tuyn I - V c dng ging nh t bo quang in

chn khng. Khi in p cao, in t chuyn ng vi tc ln lm ion ho cc

nguyn t kh, kt qu l dng anot tng ln t 5 - 10 ln.

2.3.4. Thit b nhn quang

Khi b mt vt rn b bn ph bi cc in t c nng lng cao, n c th pht

x cc in t (gi l pht x th cp). Nu sin t pht x th cp ln hn sin

t ti th c kh nng khuch i tn hiu. S khuch i c thc hin bng cc

thit b nhn quang (hnh 2.27).

Cc in t ti (in t scp) c pht x t mt photocatot t trong chn

khng v b chiu sng. Sau chng c tiu t trn c cc th nht ca dy cc

in cc (dynode) ni tip. B mt cc in cc ni tip ph bng vt liu c kh nng

pht xin t th cp. Theo chiu i tin cc th nht n cc in cc tip theo,

in th ca cc in cc tng dn sao cho cc in t sinh ra tin cc th k s b

ht bi in cc th (k+1). Kt quin cc sau sin t ln hn in cc trc

.

0,5.10-2 lm

Hnh 2.26 c tr ng v nh y c a t bo quang i n d ng kh

Ia (A)

Vak (V)

2

1

0 20 40 60 80 100 120

10-2 lm

1,5.10-2 lm

2.10-2 lm

Vak (V)

6

2

0 20 40 60 80

4

8

nh

yt

ng

i

Hnh 2.27 Thi t b nhn quang1)b Photocatot 2) Dynode ( i n c c th c p)

K A

1

2


46/179

H s khuch i ca thit b nhn quang xc nh theo cng thc:

( )ntcM =

c - h s thu nhn in t hu hiu ca cc cc.t- h s chuyn ti hu hiu tin cc ny sang in cc khc.

- h s pht x th cp (sin t th cp pht ra khi c mt in tp vo in

cc).

Vi sin cc n = 5 - 15, h s pht x th cp = 5 - 10 v t > 90%, th M

~106 - 108.

Chng III

Cm bin o nhit

3.1. Khi nim cbn

Nhit l mt trong s nhng i lng c nh hng rt ln n tnh cht vtcht. Bi vy trong nghin cu khoa hc, trong cng nghip cng nh trong i sng

hng ngy vic o nhit l rt cn thit. Tuy nhin vic xc nh chnh xc mt

nhit l mt vn khng n gin. a s cc i lng vt l u c th xc nh

trc tip nhso snh chng vi mt i lng cng bn cht. Nhit l i lng ch

c tho gin tip da vo s ph thuc ca tnh cht vt liu vo nhit .


47/179

3.1.1. Thang o nhit

o nhit trc ht phi thit lp thang nhit . Thang nhit tuyt i

c thit lp da vo tnh cht ca kh l tng.

Theo nh l Carnot: hiu sut ca mt ng cnhit thun nghch hot ng

gia hai ngun c nhit 1 v 2 trong mt thang o bt k ch ph thuc vo 1 v2:

( )( )

2

1

F

F

= (3.1)

Dng ca hm F ph thuc vo thang o nhit . Ngc li vic chn dng hm F s

quyt nh thang o nhit . t F() = T, khi hiu sut nhit ca ng cnhit

thun nghch c vit nh sau:

2

1

TT1= (3.2)

Trong T1 v T2 l nhit ng hc tuyt i ca hai ngun.

i vi cht kh l tng, ni nng U ch ph thuc vo nhit ca cht kh v

phng trnh c trng lin h gia p sut p, th tch v v nhit c dng:

p.v=G()

C th chng minh c rng:

G()=RT

Trong R l hng s kh l tng, T l nhit ng hc tuyt i.

c th gn mt gi tr s cho T, cn phi xc nh n v cho nhit . Mun

vy ch cn gn gi tr cho nhit tng ng vi mt hin tng no vi iu kin

hin tng ny hon ton xc nh v c tnh lp li.

Thang Kelvin (Thomson Kelvin - 1852): Thang nhit ng hc tuyt i,

n v nhit l K. Trong thang o ny ngi ta gn cho nhit ca im cn bng

ba trng thi nc - nc - hi mt gi tr s bng 273,15 K.

Thang Celsius (Andreas Celsius - 1742): Thang nhit bch phn, n v

nhit l oC v mt Celsius bng mt Kelvin.

Nhit Celsius xc nh qua nhit Kelvin theo biu thc:

T(oC)= T(K) - 273,15 (3.3)

Thang Fahrenheit (Fahrenheit - 1706): n v nhit l oF. Trong thang o

ny, nhit ca im nc tan l 32oF v im nc si l 212oF.


48/179

Quan h gia nhit Fahrenheit v nhit Celssius:

( ) ( ){ }32FT9

5CT oo = (3.4)

( ) ( ) 32CT5

9FT oo += (3.5)

Bng 3.1 Cho cc gi tr tng ng ca mt s nhit quan trng theo cc thang o

khc nhau.

Bng 3.1

Nhi Kelvin (K) Celsius (oC) Fahrenheit (oF)

im 0 tuyt i 0 -273,15 -459,67

Hn hp nc - nc 273,15 0 32

Cn bngnc - nc - hi 273,16 0,01 32,018

Nc si 373,15 100 212

3.1.2. Nhit o c v nhit cn o

Gi s mi trng o c nhit thc bng Tx, nhng khi o ta ch nhn c

nhit Tc l nhit ca phn t cm nhn ca cm bin. Nhit Tx gi l nhit

cn o, nhit Tc gi l nhit o c. iu kin o ng nhit l phi c

s cn bng nhit gia mi trng o v cm bin. Tuy nhin, do nhiu nguyn nhn,

nhit cm bin khng bao git ti nhit mi trng Tx, do tn ti mtchnh lch nhit Tx - Tc nht nh. chnh xc ca php o ph thuc vo hiu s

Tx - Tc , hiu s ny cng b, chnh xc ca php o cng cao. Mun vy khi o cn

phi:

- Tng cnng s trao i nhit gia b cm bin v mi trng o.- Gim s trao i nhit gia b cm bin v mi trng bn ngoi.Chng ta hy kho st trng hp o bng cm bin tip xc. Lng nhit truyn

t mi trng vo b cm bin xc nh theo cng thc:( )dtTTAdQ cx =

Vi: - h s dn nhit.

A - din tch b mt trao i nhit.

T - thi gian trao i nhit.

Lng nhit cm bin hp th:


49/179

cmCdTdQ =

Vi: m - khi lng cm bin.

C - nhit dung ca cm bin.

Nu b qua tn tht nhit ca cm bin ra mi trng ngoi v gi , ta c:

( ) ccx mCdTdtTTA =

tng cng trao i nhit gia mi trng c nhit cn o v cm bin ta

phi dng cm bin c phn t cm nhn c t nhit thp, h s dn nhit cao, hn

ch tn tht nhit t cm bin ra ngoi th cc tip im dn t phn t cm nhn ra

mch o bn ngoi phi c h s dn nhit thp.

3.1.3. Phn loi cm bin o nhit

Cc cm bin o nhit c chia lm hai nhm:

- Cm bin tip xc: cm bin tip xc vi mi trng o, gm:+ Cm bin gin n(nhit k gin n).

+ Cm bin in tr(nhit in tr).

+ Cp nhit ngu.

- Cm bin khng tip xc: ho k.Di y nghin cu mt s loi cm bin cbn.

3.2. Nhit k gin n

Nguyn l hot ng ca nhit k gin nda vo s gin nca vt liu khi

tng nhit . Nhit k loi ny c u im kt cu n gin, d ch to.

3.2.1. Nhit k gin ndng cht rn

Thng c hai loi: gm v kim loi, kim loi v kim loi.

t =AmC

, g i l h ng s th i

gian nhi t, ta c:

=

dt

TT

dT

cx

c

Nghi m c a ph ng trnh c d ng:

=t

xc keTT Hnh 3.1. Trao i nhi t c a c m

Tx

T1

0,63Tx

t

Hnh 3.2 Nhi t k gin na) Nhi t k m - kim lo i b) Nhi t k kim lo i - kim lo i

2

1a)

1

2

b)

A


50/179

- Nhit k gm - kim loi(Dilatomet): gm mt thanh gm (1) t trong ngkim loi (2), mt u thanh gm lin kt vi ng kim loi, cn u A ni vi h thng

truyn ng ti b phn ch th. H s gin nnhit ca kim loi v ca gm l k v

g. Do k > g, khi nhit tng mt lng dt, thanh kim loi gin thm mt lng

dlk, thanh gm gin thm dlg vi dlk>dlg, lm cho thanh gm dch sang phi. Dch

chuyn ca thanh gm ph thuc dlk- dlg do ph thuc nhit .

- Nhit k kim loi - kim loi: gm hai thanh kim loi (1) v (2) c h s ginnnhit khc nhau lin kt vi nhau theo chiu dc. Gi s1 > 2 , khi gin nnhit

hai thanh kim loi cong v pha thanh (2). Da vo cong ca thanh kim loi xc

nh nhit .

Nhit k gin ndng cht rn thng dng o nhit di 700oC.

3.2.2. Nhit k gin ndng cht lng

Nhit k gm bnh nhit (1), ng mao dn (2) v cht lng (3). Cht lng s dng

thng dng l thu ngn c h s gin nnhit =18.10-5/oC, v nhit k bng thu

tinh c =2.10-5/oC.

Khi o nhit , bnh nhit c t tip xc vi mi trng o. Khi nhit

tng, cht lng gin nv dng ln trong ng mao dn. Thang o c chia trn

v theo dc ng mao dn.

Di nhit lm vic t - 50 600oC tu theo vt liu ch to v bc.

3.3. Nhit kin tr

3.3.1. Nguyn l

Nguyn l chung o nhi t b ng cc i n tr

l d a vo s ph thu c i n tr su t c a v t li u theo

nhi t .

Trong tr ng h p t ng qut, s thay i i n tr

theo nhi t c d ng:

( ) ( )00 TTF.RTR =

R0 l i n tr nhi t T 0, F l hm c tr ng cho

v t li u v F = 1 khi T = T0.

1

2

3

Hnh 3.3 Nhi t k gi n


51/179

Hin nay thng s dng ba loi in tro nhit l: in trkim loi,

in trsilic v in trch to bng hn hp cc oxyt bn dn.

Trng hp in trkim loi, hm trn c dng:

( )32

0 CTBTAT1R)T(R +++= (3.6)Trong nhit T o bng oC, T0=0

oC v A, B, C l cc h s thc nghim.

Trng hp in trl hn hp cc oxyt bn dn:

=

0

0T

1

T

1Bexp.R)T(R (3.7)

T l nhit tuyt i, B l h s thc nghim.

Cc h sc xc nh chnh xc bng thc nghim khi o nhng nhit

bit trc. Khi bit gi tr cc h s, t gi tr ca R ngi ta xc nh c nhit cn o.

Khi bin thin ca nhit T (xung quanh gi tr T) nh, in trc th coi

nh thay i theo hm tuyn tnh:

( ) ( )( )T1TRTTR R +=+ (3.8)

Trong :

( ) dT

dR

TR

1R = (3.9)

c gi h s nhit ca in trhay cn gi l nhy nhit nhit T. nhy

nhit ph thuc vo vt liu v nhit , v d0oC platin (Pt) c R=3,9.10-3/oC.

Cht lng thit bo xc nh gi tr nh nht m n c tho cmin0

R

R, do

cng xc nh s thay i nh nht ca nhit c th pht hin c:

minoR

min

R

R1T

=

V d nu 6

min0

10R

R =

v vi nhng php o quanh im 0oC, vt liu l platin th

4min 10.6,2T

= oC.


52/179

Thc ra, in trkhng ch thay i khi nhit thay i do s thay i in tr

sut m cn chu tc ng ca s thay i kch thc hnh hc ca n. Bi vy i vi

mt in trdy c chiu di l v tit din s, h s nhit c dng:

dT

ds

s

1

dT

dl

l

1

dT

d1

dT

dR

R

1R +

==

t:dT

d1

= ;dT

dl

l

1l = ;

dT

ds

s

1s =

slR +=

Vi ls 2= ta c: lR =

Trn thc t thng l>> nn c th coi = R .

3.3.2. Nhit kin trkim loi

a) Vt liuYu cu chung i vi vt liu lm in tr:

-C in trsut ln in trban u R0 ln m kch thc nhit k vnnh.

-H s nhit in trca n tt nht l lun lun khng i du, khng trit tiu.-C bn c, ho nhit lm vic.-D gia cng v c kh nng thay ln.

Cc cm bin nhit thng c ch to bng Pt v Ni. Ngoi ra cn dng Cu, W.- Platin :

+ C th ch to vi tinh khit rt cao (99,999%) do tng chnh xc

ca cc tnh cht in.

+ C tnh trv mt ho hc v tnh n nh cu trc tinh th cao do m bo

tnh n nh cao v cc c tnh dn in trong qu trnh s dng.

+ H s nhit in tr0oC bng 3,9.10-3/oC.

+ in tr100oC ln gp 1,385 ln so vi 0oC.

+ Di nhit lm vic kh rng t -200oC 1000oC.

- Nikel:+ C nhy nhit cao, bng 4,7.10-3/oC.

+ in tr100oC ln gp 1,617 ln so vi 0oC.

+ D b oxy ho khi nhit cao lm gim tnh n nh.

+ Di nhit lm vic thp hn 250oC.


53/179

ng c s dng trong mt s trng hp nh tuyn tnh cao ca in tr

theo nhit . Tuy nhin, hot tnh ho hc ca ng cao nn nhit lm vic thng

khng vt qu 180oC. in trsut ca ng nh, do ch to in trc in

trln phi tng chiu di dy lm tng kch thc in tr.

Wonfram c nhy nhit v tuyn tnh cao hn platin, c th lm vic

nhit cao hn. Wonfram c th ch to dng si rt mnh nn c th ch to c

cc in trcao vi kch thc nh. Tuy nhin, ng sut d sau khi ko si kh b trit

tiu hon ton bng cch do gim tnh n nh ca in tr.

Bng 3.2

Thng s Cu Ni Pt W

Tf(oC) 1083 1453 1769 3380

c (JoC-1kg-1) 400 450 135 125

(WoC-1m-1) 400 90 73 120

l x106 (oC) 16,7 12,8 8,9 6

x108 (m) 1,72 10 10,6 5,52

x103 (oC-1) 3,9 4,7 3,9 4,5

b) Cu to nhit kin tr

trnh s lm nng u o dng in chy qua in trthng gii hn gi

tr mt vi mA v in tr c nhy nhit cao th in trphi c gi tr ln.

Mun vy phi gim tit din dy hoc tng chiu di dy. Tuy nhin khi gim tit

din dy bn li thp, dy in trd bt, vic tng chiu di dy li lm tng

kch thc in tr. hp l ngi ta thng chn in trR0oC c gi tr vo

khong 100, khi vi in trplatin s c ng knh dy cvi m v chiu di

khong 10cm, sau khi qun li s nhn c nhit k c chiu di c1cm. Cc sn

phm thng mi thng c in tr0oC l 50, 500 v 1000, cc in trln

thng c dng o di nhit thp.

-Nhit kcng nghip: s dng cho mc ch cng nghip, cc nhit k phi c v

bc tt chng c va chm mnh v rung ng, in trkim loi c cun v bao

bc trong thu tinh hoc gm v t trong v bo v bng thp. Trn hnh 3.4 l cc

nhit k dng trong cng nghip bng in trkim loi platin.

81 2 3

45

1

7

6


54/179

- Nhit kbmt:

Nhit k b mt dng o nhit trn b mt ca vt rn. Chng thng c

ch to bng phng php quang ho v s dng vt liu lm in tr l Ni, Fe-Ni

hoc Pt. Cu trc ca mt nhit k b mt c dng nh hnh v 3.5. Chiu dy lp kim

loi cvi m v kch thc nhit k c1cm2.

c trng chnh ca nhit k b mt:

- nhy nhit : ~5.10-3/oC i vi trng hp Ni v Fe-Ni

~4.10-3/oC i vi trng hp Pt.

- Di nhit s dng: -195oC 260 oC i vi Ni v Fe-Ni.

-260oC 1400 oC i vi Pt.

Khi s dng nhit k b mt cn c bit lu n nh hng bin dng ca bmt o.

3.3.3. Nhit kin trsilic

Silic tinh khit hoc n tinh th silic c h s nhit in trm, tuy nhin khi

c kch tp loi n th trong khong nhit thp chng li c h s nhit in tr

dng, h s nhit in tr~0,7%/oC 25oC. Phn t cm nhn nhit ca cm bin

2200

2400R()

Hnh 3.5 Nhi t k b m t


55/179

silic c ch to c kch thc 500x500x240 m c m kim loi mt pha cn

pha kia l b mt tip xc.

3.3.4. Nhit kin troxyt bn dn

a) Vt liu ch to

Nhit in tr c ch to t hn hp oxyt bn dn a tinh th nh: MgO,

MgAl2O4, Mn2O3, Fe3O4, Co2O3, NiO, ZnTiO4.

S ph thuc ca in trca nhit in trtheo nhit cho bi biu thc:

=

0

2

0

0T

1

T

1exp

T

TR)T(R (3.11)

Trong R0() l in trnhit T0(K).

nhy nhit c dng:

2R T

b+=

V nh hng ca hm mn in trchim u th nn biu thc (3.11) c th vit

li:

= 00 T

1

T

1

BexpR)T(R (3.12)

V nhy nhit:

2R T

B=

Vi B c gi tr trong khong 3.000 - 5.000K.

b) Cu to

Trong d i nhi t lm vi c ( -55

200oC) c th l y g n ng gi tr i n tr

c a c m bi n theo nhi t theo cng th c:

( ) ( )2000T TTBTTA1RR ++=

Trong R0 v T0 l i n tr v nhi t

tuy t i i m chu n.

S thay i nhi t c a i n tr t ng

i nh nn c th tuy n tnh ho b ng

cch m c thm m t i n tr ph .


56/179


57/179

Tng t ti mt tip xc u t do (nhit t0) cng xut hin mt hiu in

th eAB(t0).

Gia hai u ca mt dy dn cng c chnh lch nng in t t do, do

cng c s khuch tn in t v hnh thnh hiu in th tng ng trong A l

eA(t,t0) v trong B l eB(t,t0).

Sc in ng tng sinh ra do hiu ng nhit in xc nh bi cng thc sau:

)t,t(e)t,t(e)t(e)t(eE 0B0A0BAABAB +++= (3.13)

V eA(t0,t) v eB(t,t0) nh v ngc chiu nhau c th b qua, nn ta c:

)t(e)t(eE 0BAABAB +=

Nu nhit hai mi hn bng nhau, chng hn bng t0 khi sc in ng tng:

0)t(e)t(eE 0BA0ABAB =+=

Hay:)t(e)t(e 0AB0BA = (3.14)

Nh vy:

)t(e)t(eE 0ABABAB = (3.15)

Phng trnh (3.15) gi l phng trnh cbn ca cp nhit ngu. T phng trnh

(3.15) nhn thy nu gi nhit t0 = const th:

)t(fC)t(eE ABAB =+= (3.16)

Chn nhit mt mi hn t0 = const bit trc lm nhit so snh v o sc in

ng sinh ra trong mch ta c th xc nh c nhit t mi hn th hai.

Sc in ng ca cp nhit khng thay i nu chng ta ni thm vo mch mt

dy dn th ba (hnh 3.9) nu nhit hai u ni ca dy th ba ging nhau.

Tht vy:

- Trong trng hp a:)t(e)t(e)t(e)t,t(E 0CA0BCAB0ABC ++=

V:

0)t(e)t(e)t(e 0CA0BC0AB =++

Nn:

)t(e)t(e)t,t(E 0ABAB0ABC =

2 3t0t0

A B

C

2

3

t0

t1

C4

B


58/179

- Trng hp b:)t(e)t(e)t(e)t(e)t,t,t(E 1CB1BC0ABAB01ABC ++=

V:

)t(e)t(e1CB1BC

=

Nn:

)t(e)t(e)t,t(E 0ABAB0ABC =

Nu nhit hai u ni khc nhau s lm xut hin sc in ng k sinh.

3.4.2. Cu to cp nhit

a) Vt liu ch to

ch to cc nhit in c th dng nhiu kim loi v hp kim khc nhau.

Tuy nhin chng phi m bo cc yu cu sau:

- Sc in ng ln ( d dng ch to dng co th cp).- C bn chc v ho hc nhit lm vic.- D ko si.- C kh nng thay ln.- Gi thnh r.

Hnh 3.10 biu din quan h gia sc in ng v nhit ca cc vt liu dng

ch to in cc so vi in cc chun platin.

21

3

8

910

54

6

7

11

T

Ed


59/179


60/179

b) Cu to

Cu to in hnh ca mt cp nhit cng nghip trnh by trn hnh 3.12.

u lm vic ca cc in cc (3) c hn ni vi nhau bng hn vy, hn kh

hoc hn bng tia in t. u t do ni vi dy ni (7) ti dng co nhcc vt ni

(6) dy t trong u ni dy (8). cch ly cc in cc ngi ta dng cc ng s

cch in (4), s cch in phi trv ho hc v bn cv nhit nhit lm

vic. bo v cc in cc, cc cp nhit c v bo v (1) lm bng s chu nhit

hoc thp chu nhit. H thng v bo v phi c nhit dung nh gim bt quntnh nhit v vt liu ch to v phi c dn nhit khng qu nh nhng cng khng

c qu ln. Trng hp v bng thp mi hn u lm vic c th tip xc vi v

gim thi gian hi p.

3.4.3. Mch o v dng c thcp

Hnh 3.12 C u t o c p nhi t1) V b o v 2) M i hn 3) Dy i n c c 4) S cch i n

5) B ph n l p t 6) Vt n i dy 7) Dy n i 8) u n i dy

1

324 5 6

7

8

Hnh 3.11 S c i n ng c a m t s c p nhi t ng uE-Chromel/Constantan R- Platin-Rodi (13%)/PlatinJ- S t/Constantan S- Platin-Rodi (10%)/PlatinK- Chromel/Alumel B-Platin-rodi 30% / Platin-rodi 6%

E

J

K

R

S

B

toC

E


61/179

Nhit cn o c xc nh thng qua vic o sc in ng sinh ra hai

u dy ca cp nhit ngu. chnh xc ca php o sc in ng ca cp nhit

ngu ph thuc nhiu yu t. Mun nng cao chnh xc cn phi:

- Gim thiu nh hng ca tc ng ca mi trng o ln nhit u t do.- Gim thiu s st p do c dng in chy qua cc phn t ca cm bin v mcho.

a) S mch o dng milivn k

- S:

Trn hnh 3.13 biu din so thng dng s dng milivn k tin.

Khi nhit hai u t do (2) v (3) bng nhau th sc in ng trong mch

chnh l sc in ng ca cp nhit, nu chng khc nhau th trong mch xut hin

sut in ng k sinh cc mi ni v lm sai lch kt quo.

o trc tip hiu nhit gia hai im ngi ta dng so vi sai nh

hnh 3.14.

Trong s ny, c hai u 1 v 2 ca cp nhit l u lm vic tng ng vi

nhit t1 v t2. Kt quo cho php ta xc nh trc tip gi tr ca hiu s hai nhit

t1- t2.

1

2 3

t

t0 t0

A B

C

Hnh 3.13 S m ch o

m

1 2t1 t2

3 4t0 t0

Hnh3.14 S o vi sai

A AB

mV

)t,t(E '0AB

)t,t(E 0'0AB

Hnh 3.16 Hi u ch nhnhi t u t do

A A A AB B B B

Hnh 3.15 S m c n i ti p

mV

toC

E


62/179

Trng hp nhit mi trng o khng khc nhiu nhit u t do, tng

nhy php o c th mc theo s ni tip n cp nhit nh hnh 3.15. Sc in

ng tng ca b mc ni tip bng )t,t(nE 0AB .

- B nhit u tdo:

Thng thng cp nhit ngu c chun vi t0 = 0oC ng vi:

)t(e)t(e)t,t(E 0ABAB0AB =

Nu nhit u t do bng 0t '0 th gi tr sc in ng o c:

)t(e)t(e)t,t(E '0ABAB'0AB =

Rt ra:

[ ])t(e)t(e)t,t(E)t,t(E 0AB'0AB'0AB0AB +=

Hay:

)t,t(E)t,t(E)t,t(E 0'0Ab

'0AB0AB +=

Gi tr )t,t(E '0AB l lng hiu chnh xc nh t thang chia ca cp nhit ngu

dng theo gi tro nhit u t do t

0.Di y trnh by mt s phng php b nhit u t do.

- Dng dy b: loi trnh hng ca nhit i tng o ln u t do c th mc dng

co theo s hnh 3.17.

)t(e)t(e)t(e)t(eE 0CD'0BD

'0CAAB +=

Chn dy dn C v D sao cho )t(e)t(e '0DB'0CA = ( 3412 = ), khi :

)t(e)t(eE 0CDAB = V e(t0) = 0, nn:

)t(e)t(eE 0ABAB =

C

2 3t0 t0

t0

t0

D1

C

A

E


63/179

- Dng cu b:Trn hnh 3.18 gii thiu s dng cu b tng nhit u t do.

Cu b gm in trR1, R2, R3 lm bng manganin (hp kim cha 99,4%Cu,

0,6%Ni) c h s nhit in trbng khng, cn R lm bng ng c h s nhit in

tr4,25 4,28.10-3oC-1 . Khi nhit u t do t0 = 0, cu cn bng UAB=0. Gi s

nhit u t do tng ln t0, khi R tng ln lm xut hin mt in p Uc.

Ngi ta tnh ton sao cho in p ny b vo sc in ng nhit mt lng ng

bng lng cn hiu chnh, ngha l Uc= EAB(t0,t0). Nh vy trn ca vo ca dng

co c in p:

)t,t(EU)t,t(E 0ABcd'0AB =+

Sai s b ca cu tiu chun khi nhit t0 thay i trong khong 0 - 50oC l 3oC.

- nh hng ca in trmch o:

Xt mch o dng milivn kin t (hnh 3.19).

t0

B ngu n

Hnh 3.18 C u b nhi t u t do

Rd R1

R2 R3

F

D

C

mV

t1t1

t0

A Bt

Hnh 3.17 B nhi t u t do b ng dy b

1t

A B 2

3

4

D

B

t0

Rd2 3

t1 t1

A B

t0

t0

Rt

RV


64/179

Gi: Rt l in trca cp nhit.

Rd l in trdy ni.

Rv l in trtrong ca milivn k.

Khi in p gi hai u milivn k xc nh bi cng thc:

Vdt

V0ABm

RRR

R)t,t(EV

++=

Rt ra:

++=

V

dtm0AB

R

RR1V)t,t(E (3.17)

Theo biu thc (3.17) khi Rv >> Rt+Rd th:

m0AB V)t,t(E

- nh hng ca Rt: i vi cp cromen/alumen hoc cp cromen/coben c in trRt kh nh nn s thay i ca n t nh hng ti kt quo. i vi cp PtRd - Pt c

in trRt kh ln (~ 15) nn s thay i ca n nh hng ng k ti kt quo.

- nh hng ca Rd: thng thng Rd kh nh nn t nh hng ti kt quo.- nh hng ca RV: Rv = Rkd + Rf.

in tr ph Rf ca milivn k thng ch to bng vt liu c R = 0 nn

khng nh hng, s thay i Rv khi nhit tng ch yu do s thay i ca in tr

khung dy Rkd (ch to bng ng R = 4,2.10-3/oC). gim sai s nn chn RP/Rkd

ln.

b) S mch o xung i dng in th k

Trn hnh 3.20 trnh by so bng phng php xung i, da theo nguyn

tc so snh in p cn o vi mt in p ri trn mt on in tr.

Hnh 3.19 nh h ng c a i n tr m ch o

1t

G

K D

EX

E

EMP

R

GEX Rd

RGiP

i0iC A B

R c

E

Rx

RR


65/179

Theo s hnh (3.20a) ta c:

)RRR(IRIE GxdPABCX +++=

P0C III +=

)RRR(IR)II(E GxdPABP0X ++++=

GXdAB

AB0XP

RRRR

RIEI

+++

=

Nu EX = I0RAB th IP = 0, tc l in th k ch khng, khi in p ri trn AB

bng gi tr EX cn o.

Ta c:

RL

lR AB =

RL

lIE 0X =

Nu cnh c I0, L, R ta c Ex ph thuc n tr vo l tc l ph thuc vo v tr

con chy ca ng ho.

Trn s hnh (3.20b), EM l mt pin mu, RM l mt in tr mu bng

manganin. Khi ng P vo K th in p ri trn RMc so snh vi pin mu. Nu

kim in k ch khng th khng cn iu chnh dng I0, nu kim in k lch khi

khng th dch chuyn Rc kim in k v khng. Khi o ng P vo D v x dch

bin trR kim in k ch khng, khi Ex = UAB.

3.5. Ho k

Cc cm bin quang thuc loi cm bin o nhit khng tip xc, gm: ho k

bc x ton phn, ho k quang hc.

3.5.1. Ho k bc x ton phn

Nguyn l da trn nh lut: Nng lng bc x ton phn ca vt en tuyt i

t l vi lu tha bc 4 ca nhit tuyt i ca vt.

t

- +


66/179

4TE = (3.18)

Trong : l hng s, T l nhit tuyt i ca vt en tuyt i (K).

Thng thng c hai loi: ho k bc x c ng knh hi t, ho k bc x c knh

phn x.

Trong s hnh (3.21a): nh sng t ngun bc x (1) qua thu knh hi t (2)

p ti b phn thu nng lng tia bc x (4), b phn ny c ni vi dng co

th cp (5).

Trong s hnh (3.21b): nh sng t ngun bc x (1) p ti gng phn x

(3) v hi t ti b phn thu nng lng tia bc x (4), b phn ny c ni vi dng

co th cp (5).

B phn thu nng lngc th l mt vi nhit kin trhoc l mt t hp cp

nhit, chng phi tho mn cc yu cu:

+ C th lm vic bnh thng trong khong nhit 100 - 150oC.

+ Phi c qun tnh nhit nh v n nh sau 3 - 5 giy.

+ Kch thc nh tp trung nng lng bc x vo o.

Trn hnh 3.22 trnh by c u

t o c a m t b thu l t h p c p nhi t.

Cc c p nhi t (1) th ng dng c p

crmen/cben m c n i ti p v i nhau.

Cc v t en (2) ph b ng b t platin.

Ho k dng g ng ph n x t n

th t n ng l ng th p ( ~ 10%), ho k

dng th u knh h i t c th t n th t Hnh 3.22 B thu n ng l ng1) C p nhi t 2)L p ph platin

1

2

5 5

Hnh 3.21 Ho k b c x ton ph na) Lo i c ng knh h i t b) Lo i c knh ph n x

1) Ngu n b c x 2) Th u knh h i t 3) G ng ph n x4) B phn thu n ng l ng 5) D ng c o th c p

12

4

a) b)

1

34


67/179

30 - 40%. Tuy nhin loi th nht li

c nhc im l khi mi trng nhiu bi, gng b bn, phn x gim do tng

sai s.

Khi o nhit bng ho k bc x sai s thng khng vt qu 27oC, trong

iu kin:

+ Vt o phi c den xp x bng 1.

+ T l gia ng knh vt bc x v khong cch o (D/L) khng nh hn

1/16.

+ Nhit mi trng 20 2oC.

Khong cch o tt nht l 1 0,2 mt.

3.5.2. Ho k quang in

Ho k quang in ch to da trn nh lut Plng:

=

1e

CI

RT

C

5

1

2T

(3.19)

Trong l bc sng, C1, C2 l cc hng s.

Trong th c t en c a v t o


68/179

Khi o, hng ho k vo vt cn o, nh sng t vt bc x cn o nhit (1)

qua vt knh (2), knh lc (3), v cc vch ngn (4), (6), knh lc nh snh (7) ti th

knh (8) v mt. Bt cng tc K cp in nung nng dy tc bng n mu (5), iu

chnh bin trRb sng ca dy tc bng n trng vi sng ca vt cn o.

Sai s khi o:

Sai s do en ca vt o < 1. Khi To xc nh bi cng thc:

=1

lnCT

1

2do

Cng thc hiu chnh: To = Tc + T

Gi tr ca T cho theo th.

Ngoi ra sai s ca php o cn do nh hng ca khong cch o, tuy nhin sai

s ny thng nh. Khi mi trng c bi lm bn ng knh, kt quo cng bnh

hng.

mA

1 2 34

5 78

Rb

Hnh 3.25 S ho k quang h c1) Ngu n b c x 2)V t knh 3) Knh l c 4&6) Thnh ng n

5) Bng n m u 7) Knh l c nh sng 8) Th knh

6

K


69/179

Chng IV

Cm bin o v tr v dch chuyn

4.1. Nguyn l o v tr v dch chuyn

Vic xc nh v tr v dch chuyn ng vai tr rt quan trng trong k thut.

Hin nay c hai phng php cbn xc nh v tr v dch chuyn.Trong phng php th nht, b cm bin cung cp tn hiu l hm ph thuc

vo v tr ca mt trong cc phn t ca cm bin, ng thi phn t ny c lin quan

n vt cn xc nh dch chuyn.

Trong phng php th hai, ng vi mt dch chuyn cbn, cm bin pht ra

mt xung. Vic xc nh v tr v dch chuyn c tin hnh bng cch m s xung

pht ra.

Mt s cm bin khng i hi lin kt chc gia cm bin v vt cn o v tr

hoc dch chuyn. Mi lin h gia vt dch chuyn v cm bin c thc hin thng

qua vai tr trung gian ca in trng, t trng hoc in t trng, nh sng.

Trong chng ny trnh by cc loi cm bin thng dng dng xc nh v tr

v dch chuyn ca vt nhin th kin tr, cm bin in cm, cm bin in

dung, cm bin quang, cm bin dng sng n hi.

4.2. in th kin tr

Loi cm bin ny c cu to n gin, tn hiu o ln v khng i hi mch

in c bit x l tn hiu. Tuy nhin vi cc in th kin trc con chy c

hc c s c xt gy n v mn, s ln s dng thp v chu nh hng ln ca mi

trng khi c bi v m.

4.2.1. in th k dng con chy chc

a) Cu to v nguyn l lm vic

Cm bin gm mt in trcnh Rn, trn c mt tip xc in c th di

chuyn c gi l con chy. Con chy c lin kt chc vi vt chuyn ng cn

kho st. Gi tr ca in trRx gia con chy v mt u ca in trRn l hm phthuc vo v tr con chy, cng chnh l v tr ca vt chuyn ng.

- i vi in th k chuyn ng thng (hnh 4.1a):nx R

L

lR =

(4.1)

- Trng hp in th k dch chuyn trn hoc xon:


70/179

n

M

x RR

=

(4.2)

Trong M < 360o khi dch chuyn trn (hnh 4.1b) v M > 360

o khi dch chuyn

xon. (hnh 4.1c)

Cc in trc ch to c dng cun dy hoc bng dn.

Cc in trdng cun dy thng c ch to t cc hp kim Ni - Cr, Ni - Cu, Ni - Cr - Fe, Ag - Pd qun thnh vng xon dng l xo trn li cch in (bng thu

tinh, gm hoc nha), gia cc vng dy cch in bng emay hoc lp oxyt b mt.

Cc in trdng bng dn c ch to bng cht do trn bt dn in l

cacbon hoc kim loi cht ~10-2m.

Cc in trc ch to vi cc gi tr Rn nm trong khong 1kn 100k,

i khi t ti M.

Cc con chy phi m bo tip xc in tt, in tr tip xc phi nh v nnh.

b) Cc c trng

- Khong chy c ch ca con chy:

Rn, , L

Rx, l

RRn

Rn

R

Hnh 4.1 Cc d ng i n th k1) i n tr 2) Con ch y

a)

b) c)

12

1

1

21

2


71/179

Thng thng u hoc cui ng chy ca con chy t s Rx/Rn khng n

nh. Khong chy c ch l khong thay i ca x m trong khong Rx l hm

tuyn tnh ca dch chuyn.

- Nng sut phn gii:

i vi in trdy cun, phn gii xc nh bi lng dch chuyn cc i

cn thit a con chy t v tr tip xc hin ti sang v tr tip xc ln cn tip

theo. Gi s cun dy c n vng dy, c th phn bit 2n-2 v tr khc nhau vin ca

con chy:

+ n v tr tip xc vi mt vng dy.

+ n - 2 v tr tip xc vi hai vng dy.

phn gii ca in trdng dy ph thuc vo hnh dng v ng knh ca

dy in trv vo khong ~10m.

phn gii ca cc in tr kiu bng dn ph thuc vo kch thc ht,

thng vo c~ 0,1 m.

- Thi gian sng:

Thi gian sng ca in k l s ln s dng ca in th k. Nguyn nhn gy

ra h hng v hn ch thi gian sng ca in th k l s mi mn con chy v dyin trtrong qu trnh lm vic. Thng thi gian sng ca in th k dng dy dn

vo c106 ln, in k dng bng dn vo c5.107 - 108 ln.

4.2.2. in th k khng dng con chy chc

khc phc nhc im ca in th k dng con chy c hc, ngi ta s

dng in th k lin kt quang hoc t.

Kho ng ch y c

Cu i ng ch y

u ng ch y x

Hnh 4.2 S ph thu c c a i n

tr i n th k vo v tr con ch y

Rx

Hnh 4.3 phn gi i c a i n th

k d ng dy


72/179

a) in th k dng con tr quang

Hnh 4.4 trnh by s nguyn l ca mt in th k dng con tr quang.

in th k trn dng con tr quang gm iot pht quang (1), bng o (2), bng

tip xc (3) v bng quang dn (4). Bng in tro c phn cch vi bng tip xc

bi mt bng quang dn rt mnh lm bng CdSe trn c con tr quang dch

chuyn khi trc ca in th k quay. in trca vng quang dn gim ng k trong

vng c chiu sng to nn s lin kt gia bng o v bng tip xc.

Thi gian hi p ca vt liu quang dn cvi chc ms.

b) in th k dng con tr t

Hnh 4.5 trnh by s nguyn l mt in th k t gm hai tin trR1 v

R2 mc ni tip v mt nam chm vnh cu (gn vi trc quay ca in th k) bao

ph ln mt phn ca in trR1 v R2, v tr phn b bao ph ph thuc gc quay catrc.

in p ngun ESc t gia hai im (1) v (3), in p o Vm ly tim

chung (2) v mt trong hai u (1) hoc (3).

Khi in p o c xc nh bi cng thc:

Hnh 4.4 i n th k quay dng con tr quang1) iot pht quang 2) B ng o 3) B ng ti p xc 4) B ng quang d n

i n tr

Th i gian

1

2

3 4

~20 ms


73/179

S1

S

21

1m E

R

RE

RR

RV =

+=

(4.3)

Trong R1 l hm ph thuc v tr ca trc quay, v tr ny xc nh phn ca R1 chu

nh hng ca t trng cn R = R1

+ R2

= const.

T hnh 4.5b ta nhn thy in p o ch tuyn tnh trong mt khong ~90oi

vi in k quay. i vi in k dch chuyn thng khong tuyn tnh ch cvi

mm.

4.3. Cm bin in cm

Cm bin in cm l nhm cc cm bin lm vic da trn nguyn l cm ng

in t. Vt cn o v tr hoc dch chuyn c gn vo mt phn t ca mch t gy

nn s bin thin t thng qua cun o. Cm bin in cm c chia ra: cm bin t

cm v h cm.

4.3.1. Cm bin tcm

a) Cm bin tcm c khe tbin thin

- Cm bin t cm n: trn hnh 4.6 trnh by s nguyn l cu to ca mts loi cm bin t cm n.

Hnh 4.6 C m bi n t c m1) Li s t t 2) Cu n dy 3) Ph n ng

XV XV

a) b) c)

1

2

3 12

31

2

R

R2R1

1

2

3

Hnh 4.5 i n th k i n t

30%

50%

70%

0O 180O 360O

Vm/ES

a) b)


74/179

Cm bin t cm n gm mt cun dy qun trn li thp c nh

(phn tnh) v mt li thp c th di ng di tc ng ca i lng o (phn ng),

gia phn tnh v phn ng c khe hkhng kh to nn mt mch t h.

S hnh 4.6a: di tc ng ca i lng o XV, phn ng ca cm

bin di chuyn, khe hkhng kh trong mch t thay i, lm cho t trca mch t

bin thin, do h s t cm v tng trca cun dy thay i theo.

S hnh 4.6b: khi phn ng quay, tit din khe hkhng kh thay i,

lm cho t trca mch t bin thin, do h s t cm v tng trca cun dy

thay i theo.

H s t cm ca cun dy cng c th thay i do thay i tn hao sinh

ra bi dng in xoy khi tm st t dch chuyn di tc ng ca i lng o Xv

(hnh 4.6c).

Nu b qua in trca cun dy v t trca li thp ta c:

==

sW

R

WL 0

22

Trong :

W- s vng dy.

sR

0

= - t trca khe hkhng kh.

- chiu di khe hkhng kh.

s - tit din thc ca khe hkhng kh.

Trng hp W = const ta c:

+

= dL

dss

LdL

Vi lng thay i hu hn v s ta c:


75/179

( )

+

=

20

002

0

02 sW

sW

L

(4.4)

nhy ca cm bin t cm khi khe hkhng kh thay i (s=const):

2

0

0

0

1

LLS

+

==

(4.5)

nhy ca cm bin t cm khi thay i tit din khng kh ( = const):

0

0s

s

L

s

LS =

=

(4.6)

Tng trca cm bin:

==sW

LZ 02

(4.7)

T cng thc (4.7) ta thy tng trZ ca cm bin l hm tuyn tnh vi tit din khe

hkhng kh s v phi tuyn vi chiu di khe hkhng kh .

c tnh ca cm bin t cm n Z = f() l hm phi tuyn v ph thuc tn s

ngun kch thch, tn s ngun kch thch cng cao th nhy ca cm bin cng cao

(hnh 4.7).

L = f()

Z = f()

Z500Hz = f()

Hnh 4.7 S ph thu c gi a L, Z v i chi u dy khe h khng kh

Z, L


76/179

- Cm bin t cm kp lp theo kiu vi sai: tng nhy ca cm bin v tng

on c tnh tuyn tnh ngi ta thng dng cm bin t cm kp mc theo kiu vi

sai (hnh 4.8).

c tnh ca cm bin t cm kp vi sai c dng nh hnh 4.9.

b) Cm bin tcm c li tdi ng

Cm bin gm mt cun dy bn trong c li t di ng c (hnh 4.10).

XV

XV

XV

a) b) c)

Hnh 4.8 C m bi n t c m kp m c theo ki u vi sai

L = f

Hnh 4.9 c tnh c a c m bi n t c m kp l p vi sai

L1 - L2 = f()

L2 = f()

L

Hnh 4.10 S nguyn l c m bi n t c m c li t1) Cu n dy 2) Li t

l0 lf

l

XV

12


77/179

Di tc ng ca i lng o XV, li t dch chuyn lm cho di lf

ca li t nm trong cun dy thay i, ko theo s thay i h s t cm L ca cun

dy. S ph thuc ca L vo lf l hm khng tuyn tnh, tuy nhin c th ci thin

bng cch ghp hai cun dy ng dng vo hai nhnh k st nhau ca mt cu in

trc chung mt li st.

4.3.2. Cm bin h cm

Cu to ca cm bin h cm tng t cm bin t cm ch khc ch c thm

mt cun dy o (hnh 4.11).

Trong cc cm bin n khi chiu di khe hkhng kh (hnh 4.11a) hoc tit

din khe khng kh thay i (hnh 4.11b) hoc tn hao do dng in xoy thay i

(hnh 4.11c) s lm cho t thng ca mch t bin thin ko theo sut in ng e

trong cun o thay i.

- Cm bin n c khe hkhng kh:

T thng tc thi:

==

siW

R

iW 011t

i - gi tr dng in tc thi trong cun dy kch thch W1.

Hnh 4.11 C m bi n h c m1) Cu n s c p 2) Gng t 3) li t di ng 4) Cu n th c p (cu n o)

a)

3

XV

1

2

~

b)

1 2

3

c)

XV

3

4

~

XV

4

~

1 2

d)

XV

~~

e)

~

XV

~

)

44

1


78/179

Sc in ng cm ng trong cun dy o W2:

dt

di.

sWW

dt

dWe 012t2

=

=

W2 - s vng dy ca cun dy o.

Khi lm vic vi dng xoay chiu tsinIi m = , ta c:

tcosIsWW

e m012

=

v gi tr hiu dng ca sut in ng:

=

=s

kIsWW

E 012

I - gi tr hiu dng ca dng in, IWWk 012 = .

Vi cc gi tr W2, W1, 0, v I l hng s, ta c:

+

= dE

dss

EdE

Hay( )200

kss

kE+

=

(4.8)

nhy ca cm bin vi s thay i ca chiu di khe hkhng kh (s = const):

2

0

0

0

2

0

20 1

E

1

ksES

+

=

+

=

=

(4.9)

Cn nhy khi tit din khe hkhng kh s thay i ( = const):

0

0

0

Ss

Ek

s

ES =

=

=

(4.10)

0

00

ksE

= - sc in ng h cm ban u trong cun o W2 khi XV = 0.

Ta nhn thy cng thc xc nh nhy ca cm bin h cm c dng tng t

nh cm bin t cm ch khc nhau gi tr ca E0 v L0. nhy ca cm bin h

cm S v SS cng tng khi tn s ngun cung cp tng.


79/179

- Cm bin vi sai: tng nhy v tuyn tnh ca c tnh cm bin ngi ta

mc cm bin theo s vi sai (hnh 4.11d,,e). Khi mc vi sai nhy ca cm bin

tng gp i v phm vi lm vic tuyn tnh mrng ng k.

- Bin th vi sai c li t: gm bn cun dy ghp ng trc to thnh hai cm bin

n i xng, bn trong c li t di ng c (hnh 4.12). Cc cun th cp c ni

ngc vi nhau sao cho sut in ng trong chng trit tiu ln nhau.

V nguyn tc, khi li tv tr trung gian, in p o Vmu ra hai cun th

cp bng khng. Khi li t dch chuyn, lm thay i mi quan h gia cun scp

vi cc cun th cp, tc l lm thay i h s h cm gia cun scp vi cc cun

th cp. Khi in trca thit bo ln, in p o Vm gn nh tuyn tnh vi hiu

s cc h s h cm ca hai cun th cp.4.4. Cm bin in dung

4.4.1. Cm bin tin n

Cc cm bin tin n l mt tin phng hoc hnh tr c mt bn cc gn

cnh (bn cc tnh) v mt bn cc di chuyn (bn cc ng) lin kt vi vt cn

o. Khi bn cc ng di chuyn s ko theo s thay i in dung ca tin.

- i vi cm bin h

Documents

GT Cam Bien Dien Cong Nghiep