Ky Thuat Dien Tu 2

8/8/2019 Ky Thuat Dien Tu 2

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K thut in t

Bin son: Nguyn Thnh Trung


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K thut in t

Chng I : Mu

1.1 Vtr mn hcK thut in t v tin hc l mt ngnh mi nhn mi pht trin. Trong mt s

khong thi gian tng i ngn, t ngy ra i tranzito ( 1948 ), n c nhng tin

b nhy vt, mang li nhiu thay i ln v su sc trong hu ht mi lnh vckhc nhau

ca i sng, dn trthnh mt trong nhng cng c quan trng nht ca cch mng k

thut trnh cao( m im trung tm l tng ho tng phn hoc hon ton, tin hc

ho, vphng php cng ngh v vt liu mi).

1.2 Cci lng, khi nim cbn khi phn tch mch in1.2.1 in p v dng in

a) in p

in p l hiu sin th gia hai im khc nhau ca mch in.Thng mt

no ca mch in c chn lm im gc ti in th bng khng, hiu in th

ca mt im bt k trong mch in so vi im c th m hoc dng v c gi

l in p ti im .

b) Dng in

Khi nim dng in l biu hin trng thi chuyn ng ca cc ht mang introng vt cht do tc ng ca trng hay do tn ti mt gradien nng ht theo khng

gian

Dng in trong mch c chiu chuyn ng t ni c in th cao n ni c

in th thp v do vy ngc chiu vi chiu chuyn ng ca in t.

Nhn xt:

-in p lun c o gia hai im khc nhau ca mch in trong khi dng in c

xc nh ch ti mt im ca mch.

- bo ton in tch tng cc gi tr dng in i vo mt im ca mch lun bngtng cc gi tr dng in i ra khi im (quy tc nt vi dng in).

-in p gia hai im A v B khc nhau ca mch nu o theo mi nhnh bt k c

in trkhc khng ni gia A v B l ging nhau.

1.2.2 Tnh cht in ca mt phn t

a) nh ngha


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K thut in tTnh cht in ca phn t bt k trong mt mch in c th hin qua mi

quan h tng h gia in p V trn hai u phn t v dng in I chy qua n v

c nh ngha l in tr(hay in trphc-trkhng) ca phn t.

-Nu mi quan h ny l t l thun

V = R.Iy R l hng s t lc gi l in trca phn t v phn t tng ng c gi

l mt in trthun.

-Nu in p trn phn t t l vi tc bin i theo thi gian ca dng in trn n,

tc l:dt

dILV= (y L l mt hng s t l)

ta c phn t l mt cun dy c in cm L.

-Nu dng in trn phn t t l vi tc bin i theo thi gian ca in p trn n,

tc l :dt

dVCI= (y C l mt hng s t l)

ta c phn t l mt tin c gi trin dung l C.

-Ngoi cc quan h nu trn trong thc t cn tn ti nhiu quan h tng ha dng v

phc tp gia in p v dng in trn mt phn t. Cc phn t ny gi chung l cc

phn t khng tuyn tnh.

c) Mt s tnh cht quan trng ca phn ttuyn tnh:

-c tuyn Vn-Ampe (th hin quan h V(I) ) l mt ng thng; in trl mt i

lng c gi tr khng i mi im-Tun theo nguyn l chng cht

-Khng pht sinh cc thnh phn tn s l khi lm vic vi tn hiu xoay chiu ( khng

gy mo phi tuyn ).

ng dng

Cc phn t tuyn tnh (R, L, C), c mt sng dng quan trng sau:

-in trlun l con sc trng cho s tiu hao nng lng (ch yu di dng nhit )

v l thng s khng qun tnh

-Mc tiu hao nng lng c nh gi bng cng sut trn n :R

VRIIVP

22. ===

-Cun dy v tin l cc phn t cbn khng tiu hao nng lng v c qun tnh

-Chng c trng cho hin tng tch lu nng lng t trng, hay in trng ca

mch khi c dng in hoc in p bin thin qua chng


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K thut in t-Gi trin trtng cng ca nhiu in trni tip nhau lun ln hn ca tng ci v

c tnh cht cng tuyn tnh.

-in dn ca nhiu in trmc song song vi nhau lun ln hn in dn ring r ca

tng ci v cng c tnh cht cng tuyn tnh

-C th thc hin chia nh mt in p (hay dng in) hay cn gi l thc hin dchmc in th ( hay mc dng in ) gia cc im khc nhau ca mch bng cch ni

ni tip hay song song cc in tr.

-Trong cch ni ni tip, in tr no ln hn s quyt nh gi tr chung ca dy.

Ngc li, trong cch ni song song, in trno nh hn s quyt nh.

-Vic ni ni tip hay song song cc cun dy dn s dn ti kt qu tng t nhi

vi cc in tr: s lm tng ( hay gim ) tr sin cm chung.

-i vi tin khi ni song song chng, in dung tng cng tng:

Css = C1 + C2 + C3+........+Cn.

-Cn khi mc ni tip th :

1/Cnt = 1/C1 + 1/C2 + 1/C3 + ......+1/Cn.

-Nu ni ni tip hay song song R vi L hoc C s dn nhn c mt kt cu mch

in c tnh cht chn lc tn s ( trkhng chung ph thuc vo tn s, gi l cc

mch lc tn s ).

-Nu ni ni tip hay song song L vi C s dn ti mt kt cu mch va c tnh cht

chn lc tn s, va c kh nng thc hin qu trnh trao i qua li gia hai dng nng

lng in-t trng, tc l kt cu c kh nng pht sinh dao ng in p hay dng

in nu ban u c mt ngun nng lng ngoi kch thch

1.2.3 Ngun in p v ngun dng in

a) Ngun sc in ng

Nu mt phn t tn hay khi chu cc tc ng khng c bn chtin t, c

kh nng to ra mtin p hay dngin mtim no ca mch in th n

c gi l mt ngun sc in ng (s. . ).

Hai thng sc trng cho mt ngun s.. l:+Gi trin p hai u lc hmch ( khi khng ni vi bt k mt phn t no khc

n hai u ca n ) gi l in p lc hmch ca ngun v k hiu l Uhm+Gi tr dng in ca ngun a ra mch ngoi lc mch ngoi dn in hon ton: gi

l gi tr dng in ngn mch ca ngun k hiu l (Ingm).


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K thut in t Vng l mt phn ca mch bao gm mt s nt v mt s nhnh lp thnh mt

ng kn m dc theo n mi nhnh v nt phi v ch gp mt ln ( tr nt c chn

lm im xut pht ).

Cy l mt phn ca mch bao gm ton b s nt v nhnh ni gia cc nt

nhng khng to nn mt vng kn no. Cc nhnh ca cy c gi l nhnh cy, ccnhnh cn li ca mch khng thuc cy c gi l b cy.

1.3 Tnh cht ca tin tc, tn hiu v phn loi tn hiu theo thi gian

1.3.1 Tin tc

-Tin tc c hiu l ni dung cha ng bn trong mt s kin, mt bin c hay mt

qu trnh no (gi l ngun tin).

-Tnh cht quan trng nht ca tin tc l n mang ngha xc sut thng k, th hin

cc mt sau:+Ni dung cha ng trong mt s kin cng c ngha ln (ta ni s kin c lng

tin tc cao) khi n xy ra cng bt ng, cng t chi. Ngha l lng tin c ln t

l vi bt nghay t l nghch vi xc sut xut hin ca s kin v c th dng xc

sut l mc o lng tin tc

+Mc o chc chn ca tin tc cng cao khi cng mt ni dung c lp i lp li (v

cbn ) nhiu ln, ta ni tin tc cn c tnh cht trung bnh thng k ph thuc vo mc

hn lon ca ngun tin, ca mi trng truyn tin v c ni nhn tin, vo tt c kh

nng gy sai nhm c th ca mt h thng thng tin.

-Tin tc khng t nhin sinh ra hoc mt i m ch l mt biu hin ca cc qu trnh

chuyn ho nng lng hay qu trnh trao i nng lng gia hai dng vt cht v

trng.

1.3.2 Tn hiu

nh ngha, phn loi

-Tn hiu l khi nim m t cc biu hin vt l ca tin tc

-Cc biu hin ny a dng v thng c phn chia lm hai nhm:

+C bn cht in t

+Khng c bn cht in t

-C th coi tn hiu ni chung l mt lng vt l bin thin theo thi gian v biu din

n di dng mt hm s hay th theo thi gian l thch hp hn c.


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K thut in t-Nu biu thc theo thi gian ca mt tn hiu l s(t) tho mn iu kin s(t) = s(t+T)

vi mi t y T l mt hng s th s(t ) c gi l mt tn hiu tun hon theo thi

gian. Gi tr nh nht trong t p tho mn iu kin s(t) = s(t+T) gi l chu k ca s(t)

V d: Tn hiu hnh sin l tn hiu tun hon:

-Cng c th chia tn hiu theo cch khc thnh hai dng cbn l bin thin lin tctheo thi gian ( tn hiu tng t ) hay bin thin khng lin tc theo thi gian ( tn hiu

xung s Digital )

Cc tnh cht ca tn hiu theo cch biu din thi gian

- di v tr trung bnh ca tn hiu

+ di ca tn hiu l khong thi gian tn ti ca n ( t lc bt u xut hin

n lc mt i ).

+Nu tn hiu s(t) xut hin lc t0 c di l ? th gi tr trung bnh ca s(t) k

hiu l: )(ts , c xc nh bi: +

=

0

0

).(1

)(t

t

dttsts

-Nng lng, cng sut, tr hiu dng

Nng lng Es ca tn hiu s(t) c xc nh bi:

=+

=0

0

).(2t

t

S dttsE +

dtts )(2

Cng sut trung bnh ca s(t) trong thi gian tn ti ca n c nh ngha bi:

S

t

t

Edttsts ==

+0

0

).(1

)( 22

Gi tr hiu dng ca s(t) c nh ngha l:

+

=

0

0

)(1 2

t

t

hd dttss =

SE

-Di ng ca tn hiu l t s gia cc gi tr ln nht v nh nht ca cng sut tc

thi ca tn hiu. Nu tnh theo n v logarit (dexibel), di ng c nh ngha l:

DdB ={ }{ }

{ }{ })(min

)(maxlg.20

)(min

)(maxlg.10

2

2

ts

ts

ts

ts=

Thng s ny c trng cho khong cng hay khong ln ca tn hiu tc

ng ln mch hoc h thng in t.

-Thnh phn mt chiu v xoay chiu ca tn hiu


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K thut in tMt tn hiu s(t) lun c th phn tch thnh thnh phn xoay chiu v thnh

phn mt chiu sao cho : =+= ssts ~)(

Vi l thnh phn bin thin theo thi gian ca s(t) v s= l thnh phn c

nh theo thi gian ( thnh phn mt chiu )

~s

-Cc thnh phn chn v l ca tn hiu

Mt tn hiu s(t) cng lun c th phn tch thnh hai thnh phn chn v lc

xc nh nh sau:

Sch(t) = Sch(-t) = 1/2 [s(t) + s(-t)]

Sl(t) = -Sl(-t) = 1/2[s(t)-s(-t)]

-Thnh phn thc v o ca tn hiu

Mt tn hiu s(t) bt k c th biu din tng qut di dng mt s phc

S(t) = Re(s(t)+j.Im(s(t))

y Re l phn thc ca cn Im l phn o ca s(t)

1.4 H thngin tin hnh

H thng in t l mt tp hp cc thit bin t nhm thc hin mt nhim v

k thut nht nh nh gia cng x l tin tc, truyn thng tin d liu, o lng thng

siu khin t chnh ......

1.4.1 H thng thng tin thu-pht

Nhim v:

H thng c nhim v truyn mt tin tc, d liu theo khng gian trn mt

khong cch nht nh t ngun tin ti ni nhn tin.

Cu trc s khi:

Ngun tin Gia cng

To dao

n caoiu ch Khuch i Phi hpk

Chn lc> Gii iu

ch

Gia cng Nhn tin

Thit b

pht

Antenpht


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K thut in t

Cc c im ch yu

+L h thng h

+Bao gm 2 qu trnh cbn: Qu trnh iu ch v qu trnh di iu ch

+Cht lng v hiu qu cng nh cc c im ca h do 3 yu t quy nh:-c im ca thit b pht

-c im ca thit b thu

-Mi trng thc hin qu trnh truyn tin

+Cc ch tiu quan trng ca h: Dng iu ch, cng sut bc x ca thit b pht,

khong cch v iu kin mi trng truyn, nhy v chn lc ca thit b thu.

1.4.2 H tiu chnh

Nhim v:H c nhim v theo di khng ch mt hoc mt vi thng s ca mt qu trnh

sao cho thng s ny phi c gi tr nm trong mt gii hn nh trc (hoc ngoi

gii hn ny) tc l c nhim vn nh thng s (tng) mt tr s hay mt di tr

s cho trc.

S cu trc

Cc c im ch yu-L h dng cu trc kn :Thng tin truyn theo 2 hng nhcc mch phn hi

-Thng s cn o v khng chc theo di lin tc v duy tr mc hoc gii

hn nh sn

- chnh xc khi iu chnh ph thuc vo

chnh xc ca qu trnh bin i t Tch thnh Uch


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K thut in t phn di ca phn t so snh ( nh ca ?U)

chnh xc ca qu trnh bin i Tx thnh Ux

Tnh cht qun tnh ca h

-C thiu chnh lin tc theo thi gian (analog) hay gin on theo thi gian min sao

t c gi tr trung bnh mong i


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K thut in t

Chng II : Cu kin in t

2.1 Khi nim v cht bn dn

2.1.1. Cu trc vng nng lng ca cht rn tinh th

Mt nguyn t bao gm c ht nhn v cc in t. Khi nguyn tng c lp

nng lng ca cc in t phn thnh cc mc ri rc. Khi a cc nguyn t li gn

nhau, do tng tc, cc mc ny b suy bin thnh nhng di gm nhiu mc st nhau

c gi l cc vng nng lng.

Ta xt dng cu trc nng lng in hnh ca vt rn tinh th:

Tu theo tnh trng cc mc nng lng trong mt vng c bin t chim ch haykhng, ngi ta phn bit 3 loi vng nng lng khc nhau:

Vng ho tr (hay cn gi l vng y), trong tt c cc mc nng lng u

b chim ch, khng cn trng thi nng lng t do.

Vng dn (vng trng), trong cc mc nng lng u cn b trng hay ch b

chim ch mt phn.

Vng cm, trong khng tn ti mc nng lng no in t c th chim

ch hay xc sut tm ht ti y bng 0.

Mi quan h gia v tr tng i cc vng nng lng v tnh cht dn in cacht rn cu trc tinh th (xt 00 K)


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K thut in t

2.1.2 Cht bn dn thun

-Hai cht bn dn thun in hnh l Gemanium (Ge) vi Eg = 0.72 eV v Silicium (Si)

vi Eg = 1.12 eV, thuc nhm bn bng tun hon Mendeleep.

-M hnh cu trc mng tinh th ( 1 chiu ) ca chng c dng sau:

Si

Si

Si

Si

Si

Si

Si

Si

Si

Si

HT

D

pi

ni


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K thut in tCc nguyn t Si lin kt vi nhau theo kiu cng ho tr bng cc i e gp

chung. 00 K Si l cht cch in.

Khi b kch thch bi 1 ngun nng lng ngoi (nhit , nh sng...) s xy ra

hin tng in ho cc nguyn t nt mng hnh thnh nn tng cp ht dn:-in t t do

-L trng

Di tc ng ca in trng ngoi cc in t t do v cc l trng chuyn

ng c hng hnh thnh nn dng in trong cht bn dn thun

Dng in ny gm hai thnh phn tng ng nhau:

+Dng chuyn ng ca cc in t t do

+Dng chuyn ng ca cc l trng v bn cht l dng dch chuyn ca cc in t

ho tr

.2.1.3 Cht bn dn tp cht

a. Cht bn dn tp cht loi n

-Tin hnh pha thm cc nguyn t thuc nhm 5 bng Mendeleep vo mng tinh th

cht bn dn nguyn cht nh cng nghc bit vi nng 1010 n 1018 nguyn

t/Cm3 ta thu c cht bn dn tp cht loi n.


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K thut in tCc nguyn t nhm 5 c 5 in tlp ngoi cng nn khi tham gia lin kt vi cc

nguyn t bn dn thun mi nguyn t tp cht s tha ra 1 in t. in t ny lin

kt yu vi ht nhn nn d dng bt ra khi ht nhn hnh thnh nn tng cp:

-in t t do-In dng tp cht

nhit phng hu ht cc nguyn t tp cht b in ho.

Cng vi qu trnh in ho cc nguyn t tp cht vn din ra qu trnh in ho

cc nguyn t bn dn thun nhng vi mc yu hn

Nh vy trong mng tinh th cht bn dn tp cht loi n tn ti hai loi ht mang

in:

+ in t t do

+L trng (mang in tch dng)

Trong in t l ht dn chim a s, l trng l ht dn thiu s(nn>>pn)

d.Cht bn dn tp cht loi p

-Nu tin hnh pha cc nguyn t thuc nhm 3 bn tun hon Mendeleep vo mng

tinh th cht bn dn thun ta c cht bn dn tp cht loi p.Cc nguyn t nhm 3 c 3 e ngoi cng nn khi tham gia lin kt vi cc

nguyn t cht bn dn thun s c mt lin kt b thiu e.


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K thut in t

Lin kt khuyt ny d dng nhn thm e hnh thnh nn cc in m tp cht vmt i s lng cc e tng ng.

Cc e b p cho lin kt b khuyt c sn sinh ra t vic in ho cc nguynt bn dn thun. (Qu trnh in ho cc nguyn t bn dn thun hnh thnh nn tngcp:in t t do v l trng)

Nh vy trong mng tinh th cht bn dn tp cht loi p tn ti hai loi ht mangin:

+Cc in t t do

+Cc l trng

Trong cc l trng l ht dn chim a s c nng ln hn nhiu cp so vinng ca cc in t t do (pp>>np)

2.2 it bn dn

2.2.1. Mt ghp p-n v tnh chnh lu ca it bn dn

a. Mt ghp p-n khi cha c in trng ngoi

Khi cho hai n tinh th cht bn dn tp cht loi p v cht bn dn tp cht loin tip xc cng ngh vi nhau ta thu c mt ghp p-n. Do c s chnh lch v nng

in t t do gia min bn dn tp cht loi n v min bn dn tp cht loi p nn

ti ni tip gip gia hai min xy hin tng chuyn ng khuch tn ca cc e t do

t min bn dn n sang min bn dn p. Qu trnh chuyn ng khuch tn ny lm hnh

thnh nn lp in m bn pha min bn dn p v lp in dng bn pha min bn dn


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K thut in tn, vng in ny nm hai bn ni ti p gip v c gi l vng ngho ( vng ny

ngho ht mang in t do v c in tr ln hn nhiu cp so vi vng cn li). Qu

trnh khuch tn tip din cho ti khi lp in m bn pha min p ln to ra lc

y ln ngn trkhng cho cc e khuch tn t min n sang.

B rng ca vng ngho khi cha c in p ngoi l lo v in p ti vng ngho

(in p gia lp in dng v lp in m) l Vtx chnh Vtx l nguyn nhn ca vic

ngn trchuyn ng khuch tn ca cc e t do t min n sang min p (ngn trdng

in chy t min p sang min n). Mun c dng in chy qua tip gip p-n cn t ti

n mt in p c chiu v ln thch hp to ra lc ln gip cc e t do vt

qua c s cn trca Vtx. iu kin tiu chun ngi ta o c Vtx = 0.7 V vi

it lm t Si v Vtx = 0.3 v vi it lm t Ge

b. Phn cc cho mt ghp p-n

Khi nim v phn cc:

Phn cc cho mt thit bc hiu l t cc in p thch hp ti cc cc ca

n xc lp ch lm vic cho n.

Vi it c 2 ch phn cc:

-Phn cc thun-Phn cc ngc

Phn cc thun cho it(tip gip pn)

Mch in di y phn cc thun cho it


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K thut in t

iu kin it c phn cc thun l:

+in p phn cc t ngc cc tnh so vi Vtx ( cc dng ca in p phn

cc t ti min bn dn p cc m ca ngun phn cc t ti min bn dn n)

+in p phn cc ln hn in p VtxKhi it c phn cc thun c dng in chy qua n theo chiu t Anode

sang cathode. Vic xut hin dng in chy qua it c gii thch nh sau:

+in p m ca VBIAS y cc e t do min n v gn vi tip gip p-n

+in p dng ca VBIAS y cc l trng min p v gn vi tip gip p-ndn ti kt qu l vng ngho hp li.

Do in p phn cc VBIAS > Vtx nn cc in t t do min n c cung cp

nng lng c th vt qua c vng ngho sang n min p. Khi sang n min

p, do phi vt qua vng ngho nn cc e t do mt i mt phn nng lng v khng

cn l e t do na m trthnh cc e tham gia lin kt. Cc e ny dch chuyn theo cc


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K thut in tl trng min p ra khi min p v trv pha cc dng ca vBIAS. Nh vy xut

hin dng in chy qua tip gip p-n trong bn pha min n l dng chuyn ng

ca cc in t t do hng v tip gip p-n, cn bn pha min p l dng chuyn ng

tng i ca cc l trng hng ra xa tip gip p-n.

Phn cc ngc cho tip gip p-n

Mch in sau phn cc ngc cho tip gip p-n

iu kin:

in p m ca VBIASt ti min p, in p dng ca VBIASt ti min n

in p phn cc ngc cho VBIAS cn nhit khi bnh thng


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K thut in tKhi it c phn cc ngc th dng in chy qua n rt nh nn c th coi

nh khng c dng in chy qua n. iu ny c gii thch nh sau:

in p dng ca ngun phn cc ko cc in t (l ht a s) min n ra xa

tip gip p-n.

in p m ca ngun phn cc ko cc l trng (l ht a s) min p ra xa tipgip p-n

C hai hin tng trn lm cho vng ngho c mrng ra. Vng ngho c

m rng ra cho n khi in p t ln vng ngho chnh bng in p phn cc. Lc

ny xut hin dng in chy qua tip gip p-n theo chiu t n sang p. y l dng in

do s chuyn ng ca cc ht thiu s nn dng in ny nh v c th b qua.

2.2.1. c tuyn Von-Ampe v cc tham s cbn ca it bn dn

Khi nim vc tuyn Von-Ampe ca it bn dn:

c tuyn V-A ca it l th th hin mi quan h gia in p V trn hai

u it v dng in I chy qua it.

thu c c tuyn V-A ca it cn phi kho st.

a. Kho st min c tuyn thun

Xt mch sau:

Mch in trn gip ta kho st tm ra c tuyn V-A ca it bn dn khi n

c phn cc thun. Cch kho st nh sau: iu chnh VBIAS v 0 V, tng dn VBIAS,


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K thut in tquan st vn k v ampe k ghi li cc cp gi tr (V,I) tng ng ri da trn s liu

thu c vc tuyn trn h trc V-I.

Kt qu thu c nh sau:

+Khi VBias = 0 thy V = 0 v I = 0.

+Tng dn VBias thy V tng v I tng chm theo V+Tip tc tng VBias cho n khi VBias >= 0.7V(vi it lm t Si) th ty tri nu

tip tc tng VBias V gn nh khng i v nhn gi tr c0.7 V trong khi th I li

tng nhanh. c tuyn thun ca it c dng nh hnh v (vng 1)

b. Kho st min c tuyn ngc v min nh thng

Xt mch sau:

Mch in trn gip ta kho st tm ra c tuyn V-A ca it bn dn khi n

c phn cc ngc. Quy trnh kho st tng t nh trong trng hp kho st nhnh

phn cc thun. Kt qu thu c nh sau:

Khi VBIAS = 0 th V = 0 v I = 0

Tng dn VBIAS thy V tng, I tng v nhanh chng t ti gi tr bo ho. Dng

qua it khi n c phn cc ngc c cng rt nh nn c th b qua. Vng ctuyn ngc ca it c th hin hnh v di y (vng 2).

Khi t ln it mt in p ngc ln s lm cho it bnh thng, dng

in ngc s tng ln t ngt, tnh cht van ca it b ph hoi.

C hai loi nh thng:

-nh thng v nhit


21/107

K thut in tnh thng v nhit do ti p xc p-n b nung nng cc b, v va chm ca ht

thiu sc gia tc trong in trng mnh. iu ny dn ti qu trnh sinh ht t

(in ho cc nguyn t cht bn dn thun, c tnh cht thc l) lm nhit ni tip

xc tip tc tng...dng in ngc tng t bin v mt ghp p-n b ph hng.

-nh thng v innh thng v in do hai hiu ng: in ho do va chm (gia ht thiu sc

gia tc trong trng mnh c105 V/cm vi nguyn t ca cht bn dn thun thng

xy ra cc mt ghp p-n rng( hiu ng Zener ) v hiu ng xuyn hm (tunel) xy ra

cc tip xc p-n hp do pha tp cht vi nng cao lin quan ti hin tng nhy

mc trc tip ca in t ho tr bn bn dn p xuyn qua ro th tip xc sang vng

bn dn n.

c tuyn c dng sau:

2.2.3. Cc tham s ca it bn dn

a. Cc tham s gii hn

-in p ngc cc i it cn th hin tnh cht van (cha bnh thng)


22/107

K thut in t-Dng cho php cc i qua it lc m: IAcf-Cng sut tiu hao cc i cho php trn van cha b hng v nhit:PAcf-Tn s gii hn ca in p (dng in) t ln van n cn th hin tnh cht van

fmax

b. Cc tham snh mc-in trmt chiu ca it

)1ln( +==s

A

A

T

A

AKd

I

I

I

V

I

VR

-in trvi phn (xoay chiu) ca it:

sA

T

A

AKd

II

V

I

Vr

+=

=

-in dung tip gip p-n: lp in tch khi l0 tng ng nh mt tin gi l in

dung ca mt ghp p-n: Cpn = Ckt + Crao2.2.4. Mt vi ng dng in hnh ca it bn dn

2.2.4.1 Cc mch chnh lu

a. Mch chnh lu mt na chu k:

-S mch

-Nguyn l hot ng

Gi sin p vo Vin l in p hnh sin c bin l Vp. na chu k dng

ca Vin(Vin>0), it c phn cc thun, dng in qua ti c chiu nh hnh v:


23/107

K thut in t

in p ra Vout c dng ging dng ca in p vo.


24/107

K thut in t na chu k m ca Vinit c phn cc ngc, dng chy qua it I = 0;

in p ra Vout = 0

-Dng ca in p ra:

-in p trung bnh trn ti

===

)(sin).(

11 2/

00

outVpdttoutV

TdtV

TV

T

p

T

outavg

( )7.0)( = pp VoutV

-in p ngc cc i t ln it(PIV)

Trong qu trnh hot ng nhng na chu k m it c phn cc ngc v

th n phi chu in p ngc. PIV l in p ngc cc i m it phi chu trong

mch chnh lu khi n c phn cc ngc.

Ta tm PIV cho it mch chnh lu na chu k trninnioutR

VVVV ==

pinR VVMaxVMaxPIV === )()(

b. Mch chnh lu hai na chu k

-S mch


25/107

K thut in t

-Nguyn l hot ng

in p cung cp ti cun s cp ca bin p l in p xoay chiu hnh sin

(110/220 V), in p gia hai u cun th cp bin p VAB cng l in p hnh sin cbin Vp. Do cu to ca bin p nn in p A (VA) v in p B (VB) c cng

bin l Vp/2nhng ngc pha nhau.

chu k dng ca VAB (VAB > 0) VA>0 cn VB


26/107

K thut in t

-Dng ca in p ra

7.02

)( = ppV

outV

-in p trung bnh trn ti


27/107

K thut in t

===2

0

2

0

)(2).(

2

2

1T

p

p

T

outavg

outVdttSinoutV

TdtV

TV

-in p ngc cc i t ln it

chu k dng ca VAB ta tm PIV cho D2.

Ta c : BoutR VVV =

. 7.02

)()2

()()()( =+==== pp

p

p

pBoutR VV

outVV

outVVVMaxVMaxPIV

chu k m, ca VAB ta tm PIV cho D1.

Ta c: AoutR VVV =

7.0

2

)()

2

()()()( =+==== pp

p

p

pAoutR VV

outVV

outVVVMaxVMaxPIV

c. Mch chnh lu cu

-S mch

-Nguyn l hot ng

in p trn cun scp bin p l in p xoay chiu hnh sin. Do , in ptrn hai u th cp bin p VAB cng l in p hnh sin, gi s bin ca in p ny

l Vp.


28/107


29/107

K thut in t-Dng ca in p ra Vout

0

T/2

T

3T/2

VAB

Vout

Vp

Vp(out)

0

T/2

T

3T/2

t

t

4.1)( = pp VoutV

-Tm in p trung bnh trn ti

===2

0

2

0

)(2).(

2

2

1T

p

p

T

outavg

outVdttSinoutV

TdtV

TV

-Tm PIV

chu k dng ca VAB ta tm PIV cho D3, D4+Tm PIV cho D3: Ta c ACAR VVVV == m 7.0= DA VV (do D1 c phn cc

thun) nn 7.07.0 +=+== outDAR VVVV t PIV = Max(VR) = Vp(out) + 0.7 = Vp

0.7.

+Tm PIV cho D4: Ta c BDR VVV = m 7.0= BC VV (do D2 c phn cc thun)

nn do 7.07.0 == CB VV 7.07.0 +=+= outDR VVV ; PIV=Max(VR) = Vp(out) + 0.7 =

Vp 0.7.

chu k m ca VAB ta tm PIV cho D1, D2. Cch tm tng t nh tm

cho D3 v D4 v kt qu tm c l PIV(D1) = PIV(D2) = Vp 0.7

2.2.4.2 Mch hn ch bin

Mch hn bin c tc dng khng ch bin ca tn hiu vo mt hoc hai

ngng in p nh trc. C mt s dng mch hn bin sau:

-Mch hn bin trn


30/107

K thut in t-Mch hn bin di

-Mch hn bin c trn ln di.

Ta xt hot ng ca mch hn bin trn:

-S mch:

Hot ng ca mch nh sau:Gi sin p vo Vin l in p hnh sin c bin l Vp vi Vp >VBIAS + 0.7.

T mch in ta thy in p phn cc cho it l: Vin - VBias.

+Khi Vin - VBias > 0.7 hay Vin > VBias + 0.7 it c phn cc thun , st p trn it

Vout - VBias = 0.7 hay Vout = VBias + 0.7.

+Khi Vin - VBias


31/107

K thut in t

Vi gi thit tn hiu vo l tn hiu in p hnh sin c bin Vp hot ng

ca mch dch mc in p c gii thch nh sau:

chu k m u tin ca Vinit c phn cc thun, t C c np in

n khi in p trn tt ti gi tr Vp 0.7. Ngay sau it c phn cc ngc, t

C ch c th phng in qua RL. Do RLc chn vi in trln nn t phng in rtchm mi chu k ca Vin v th nn t C lc ny c th coi nh mt ngun nui DC

mc ni tip vi Vin. Do , Vout = Vc + Vin = Vp 0.7 + Vin2.2.4.4 Mch nhn i in p

Gi sin p vo l in p hnh sin c bin Vp th in p ra ca mch nhn

i in p l Vout = 2(Vp 0.7). Mch nhn i in p c 2 dng

a . Dng 1(Dng chu k)

-S v hot ng ca mch c th hin trn hnh v:


32/107

K thut in t

Gi sin p trn th cp ca bin p l in p hnh sin c bin Vp.

na chu k dng ca in p trn th c p bin p, D1 c phn cc

thun, D2 c phn cc ngc t C1 c np in ti in p Vp 0.7.

na chu k dng ca in p trn th c p bin p, D1 c phn cc

ngc, D2 c phn cc thun, t C2 c np in bi in p VC1 + Vin, v thin

p trn t C2 (in p ra) t ti VC2 = 2(Vp 0.7).Trong trng hp khng ti ni ti u ra in p trn t C2 (in p ra)

khng i.

Khi c ti mc ti u ra t C2 s phng in na chu k dng k tip, v

np in na chu k m tip theo.

b. Dng 2(dng c chu k)

-S v hot ng ca mch nh sau:

Gi sin p trn th cp bin p l in p hnh sin c bin Vp.

chu k dng ca in p trn th cp bin p, D1 c phn cc thun,

D2 c phn cc ngc, t C1 c np in ti in p VC1 = Vp 0.7


33/107

K thut in t chu k m ca in p trn th cp bin p, D2 c phn cc thun, D1

c phn cc ngc, t C2 c np in ti in p VC2 = Vp 0.7.in p ra Vout = VC1 + VC2 = 2(Vp 0.7)

2.2.5 Mt s loi it c bit

2.2.5.1 it pht quang(LED)

LED l loi it c kh nng pht ra nh sng khi c phn cc thun. N

c cu to t mt mt ghp p-n trong min p v min n ri nhau v c bc trong

mt lp nha trong sut.

LED c kh nng pht ra nh sng l do s gii phng nng lng ca cc e

t do khi cc e ny ti hp vi cc l trng.


34/107

K thut in t

Bc sng ca nh sng do LED pht ra ph thuc vo vt liu ch to LED. Cng ca nh sng do LED pht ra ph thuc vo cng dng in chy qua LED.

2.2.5.2 it Zener

a.c tuyn V-A ca it Zenerit Zener thng c dng vi mc ch n nh in p. it Zener c

nhiu im rt ging vi it thng, nhng c iu c bit l it Zenerc thit k

c th hot ng c min nh thng (vi it thng khi bnh thng l bhng). c tuyn Von-ampe ca it Zener c dng sau:


35/107

K thut in t

IZK l dng ti thiu chy qua it zener khi it zener lm vic min nh thng, IZM

l dng in ln nht chy qua it zener khi it zener lm vic min nh thng.

Nu dng in ngc IZ chy qua it zener vt qu IZM s lm hng it. VZK, VZM,

VZ chnh lch nhau khng nhiu, c th coi VZK = VZT = VZM = VZ. Data sheet ca it

zener cung cp cho ta VZT, IZK, IZT, IZM.

T min nh thng ca c tuyn V-A ca it Zener c th rt ra mt vi

nhn xt sau:

+it Zener lm vic min nh thng(min n p) cn phn cc ngc cho itzener vi in p phn cc ln hn VZ+Khi it Zener lm vic trong min n p th st p trn n lun l VZ cn dng in

chy qua n c th bin thin t IZKn IZM.

b. Mt vi ng dng ca it zener

* n nh in p


36/107

K thut in tKhi lm vic min nh thng, p trn hai u it Zener gn nh khng

i trc s thay i ca dng qua it, c th li dng tnh cht ny ca it zener

thc hin vic n nh in p.

Xt mch in sau:

it 1N4740 l it zener c VZ = 10V, IZK = 0.25mA, IZM = 100 mA. T

mch in ta thy VIN = VR+ VZ = IZR + VZ;

it Zener trong mch c phn cc min nh thng nn IZ c th nhn

gi tr t 0.25mA n 100mA v th :IZKR + VZ < VIN< IZMR + VZ hay 10.055V < VIN . Khi trng thi thng bo ho th mi quan h IC = DCIB

khng cn ng na.

2.3.7 Sngng dn ca BJT


46/107

K thut in tKhi IB = 0; BJT lm vic min ngng dn (c JE v JC u c phn cc

ngc). Trong mch xut hin dng in ngc ICE0 dng ny c gi tr nh nn c th

b qua v do VCE VCC

2.3.9 ng ti mt chiu

im thng bo ho v im ngng ca BJT c c minh ho bng ngti mt chiu.

im cui ca ng ti tnh l im ngng dn l tng (IC = 0 ; VCE = VCC).

im u ca ng ti l im thng bo ho ca BJT (IC = IC(sat) ; VCE = VCE(sat)).

Tp hp cc im nm gia im ngng dn v im thng bo ho ca BJT hnh thnh

nn vng tch cc(vng khuch i) ca BJT. Tu vo dng mch m ta c th thit lp

phng trnh ng ti v v n.2.3.10. ng dng ca BJT

a. Sdng BJT khuch i tn hiu

Khuch i tn hiu c thc hiu l vc lm tng tuyn tnh bin ca tn

hiu in. BJT c thc s dng khuch i tn hiu. BJT khuch i c tn

hiu cn phn cc BJT sao cho JE c phn cc thun v JC c phn cc ngc.


47/107

K thut in tTrong mch khuch i tn ti ci lng mt chiu (DC) v i lng xoay

chiu(ac). Cc i lng mt chiu c k hiu theo quy tc ch s chnh l ch in hoa

ch s ph cng l ch in hoa (v d: IB). Cc i lng xoay chiu c k hiu theo

quy tc ch s chnh l ch in hoa ch s ph l ch in thng(v d I

B

b)

BJT c kh nng khuch i c tn hiu l do dng in colectgp ln dngin baz. Xt mch sau:

VBB, VCC phn cc cho BJT m bo JE lun c phn cc thun, JC lun c phn

cc ngc khi c cng nh khng c tn hiu vo Vin.

in p ti B l VB + Vb trong VB do VBB sinh ra, Vb do Vin sinh ra. Dng

in bazl IB + Ib, trong IB do VBB sinh ra, Ib do Vin sinh ra.

Do BJT c phn cc lm vic min tch cc nn dng in colectl: IC

+ Ic = DCIB + B ac.Ib trong IC = DCIB; Ic = ac.Ib. in p ti colect l :VCC-(IC +

Ic)RC = VCC-ICRC-IcRC = VC + Vc. Trong VC = VCC ICRC; Vc = -RcIc. Tn hiu ra ca

mch trn l Vc; tn hiu vo l Vin . Gi s Vin l tn hiu hnh sin th Vc cng l tn hiuhnh sin v c cng tn s vi Vin, ngc pha vi Vin cc in trc thc la chn

tn hiu ra Vc c bin gp Vin A ln (A>1). Nh vy ta c th s dng BJT

khuch i tn hiu. Mi quan h gia tn hiu ra v tn hiu vo ca mch c th hin

trn hnh v:


48/107

K thut in t

b. Sdng BJT lm kho ng m

BJT c thc s dng nh mt kho ng m. Khi ta phn cc cho BJT

n c th chuyn i gia trng thi thng bo ho v trng thi ngng dn. Trng thi

thng bo ho ng vi kho ng, trng thi ngng dn ng vi kho m.

2.4 Tranzito trng (FET:Field Effect Transistors)2.4.1 Gii thiu chung

-FET l loi linh kin n cc

-Dng in qua FET l dng in ca ch mt loi ht (hoc l dng ca cc in t t

do, hoc l dng ca cc l trng)

-C th chia ra lm 2 loi FET


49/107

K thut in t+JFET(Junction Field-Effect Transistor) l loi tranzito trng c cc ca tip xc

+MOSFET(Metal Oxide Semiconductor Field-Effect Transistor) l loi tranzito trng

c cc ca cch ly.

-Nu nh BJT l linh kin c iu khin bi dng in th FET l loi linh kin c

iu khin bi in p-FET c trkhng vo rt ln.

2.4.2 JFET

2.4.2.1 Cu to v hot ng ca JFET

JFET l loi FET thng hot ng vi mt l p ti p gip p-n c phn cc

ngc, chnh ti p gip p-n ny iu khin dng in chy qua knh dn ca JFET.

Knh dn JFET c th l cht bn dn tp cht loi p hoc cht bn dn tp cht loi n

Ba cc ca JFET c tn l:

+Cc mng (Drain)

+Cc ca(Gate)

+Cc ngun(Source)

Vi JFET knh n cc ca c ni vi c hai min bn dn p. Vi JFET knh pcc ca c ni vi c hai min bn dn n.

Ta xt hot ng ca JFET knh n


50/107

K thut in t

in VGG t ti cc G v S phn cc ngc cho tip gip pn. in p VDD

t ti D v S to ra dng in chy trong knh dn.

in p phn cc ngc t ti G v S lm cho vng ngho dc theo tip gip p-

n c mrng ra ch yu v pha knh dn, iu ny lm knh hp li hn do in

trknh dn tng ln v dng qua knh dn gim i. Vi cch phn cc mch trn th

in p phn cc ngc gia G v D ln hn in p phn cc ngc gia G v S lm

cho vng ngho mrng khng u.


51/107

K thut in t


52/107

K thut in t

Trong cc mch in JFET c k hiu nh sau:

2.4.2.2 Cc c tuyn ca JFETa. c tuyn ra ca JFET

JFET l loi linh kinc iu khin bi in p (khi ta phn cc cho n min

dng khng i). hiu r iu ny ta tin hnh kho st c tuyn ra ca JFET. c

tuyn ra ca JFET l th th hin mi quan h gia ID v VDS khi VGS khng i.


53/107

K thut in tTrc ht ta kho st trong trng hp phn cc cho JFET vi in p VGG = 0;

nh mch in hnh v di y:

Tng dn VDD thy VDS tng v ID cng tng tuyn tnh theo VDS. Khi tng VDD

th vng ngho c xu hng rng ra, tuy nhin khi VDD cha ln th b rng ca vng

ngho cha rng gy nh hng ti dng ID v th mi quan h gia ID v VDS l

mi quan h tuyn tnh khi VDD cn nh. Mi quan h ny c th hin trn c

tuyn ra vng t A n B. Min ny cn c gi l min ohm.

Khi VDD ln khi VDS ln lc ny b rng ca vng ngho bt u gy

nh hng n dng ID. N km hm s tng ca ID trc s tng ca VDSiu ny c

thc gii thch nh sau: VDS tng l nguyn nhn ID c th tng, nhng khi VDStng lm vng ngho rng ra y li l nguyn nhn km hm ID v th ID gn nh

khng i trc s thay i ca VDS. Mi quan h ny c th hin trn c tuyn ra

vng t B n C vng ny c gi l vng dng khng i.

Tip tc tng VDDn gi tr ln nh thng tip gip pn th ty ID

tng t ngt theo VDS min ny c gi l min nh thng khi ta phn cc JFET

lm vic min ny JFET s b hng.

c tuyn ra ca JFET trong trng hp VGS = 0 c th hin trn hnh v sau:


54/107

K thut in t

Vp v IDSS l hai i lng c trong data sheet ca JFET. IDSS l dng in ln nht m

JFET c th dn qua. Vp, IDSSc xc nh iu kin VGS = 0

Khi ta phn cc ngc cho JFET vi in p VGG khc 0. Thay i VDD kho

st mi quan h gia ID v VDS ta thu c cc ng c tuyn c dng tng t nh

trn.


55/107

K thut in t

T h cc ng c tuyn ra ca BJT ta thy ID cng gim khi VGS cng m v

im pinch-off xy ra cc Vp khc nhau vi cc gi tr khc nhau ca VGS .

Gi tr ca VGS lm cho ID gn bng 0 c gi l VGS(off) c iu c bit l

VGS(off) = -Vp. Data sheet ca JFET cung cp cho ta mt trong hai in p trn.

b. c tuyn truyn t ca JFET

Ta thy VGS (vng gi tr t 0 ti VGS(off)) iu khin dng in ID chy trong

JFET. Vi JFET knh n VGS(off)0. th th hin mi

quan h gia VGS v IDc gi l c tuyn truyn t v c dng nh trn hnh v

di y:


56/107

K thut in t

ng cong ny chnh l c tuyn truyn t ca JFET knh n n cho ta bit

gii hn hot ng ca JFET.Ta c th thu c c tuyn truyn t tc tuyn ra nh hnh di y.

ng cong c tuyn truyn t c dng parabol v c phng trnh biu din nh sau:

2))(

1(offV

VII

GS

GSDSSD = cng chnh v vy m FET cn c xp vo cc linh kin tun

theo lut bnh phng (square-law devices)


57/107

K thut in t2.4.3 MOSFET(Metal Oxide Semiconductor Field-Effect Transistor)

MOSFET l loi FET c cc ca cch ly vi knh dn. C hai loi MOSFET c

bn :

-MOSFET knh to sn (D-MOSFET)

-MOSFET knh cm ng(E-MOSFET)2.4.3.1 MOSFET knh to sn

Cu to v k hiu ca MOSFET knh to sn th hin hnh v di y:

MOSFET loi D c th hot ng mt trong hai ch :

+Ch giu (phn cc lm knh dn giu thm)

+Ch ngho(phn cc lm knh dn ngho i)

MOSFET loi D hot ng ch no l tu vo in p t ti cc ca(Gate). Vi

MOSFET loi D knh n hot ng ch ngho khi in p t ti cc ca l in pm v hot ng ch giu khi in p t ti cc ca l in p dng.

Ta xt hot ng ca MOSFET loi D knh n c hai ch l ch giu v

ch ngho

*Ch ngho


58/107


59/107

K thut in tMOSFET knh cm ng ch hot ng ch giu khng c ch ngho, khi

cha c in p phn cc thch hp th knh dn ni gia D v S cha hnh thnh. Knh

dn ch hnh thnh khi c in p thch hp t ti cc ca ca MOSFET knh cm ng.

Hnh v di y th hin cu to ca MOSFET knh cm ng loi knh n:

Trong cc mch in E-MOSFET c k hiu nh sau:

*Hot ng

Vi E-MOSFET knh n, hnh thnh knh dn cn t ti cc G in p dng

ln VGS >VGS(th) . in p dng ny lm xut hin lp mng cc in tch m

min vt liu nn dc theo lp vt liu cch in SiO2 lp in tch mng ny chnh l

knh dn ni lin D v S. Khi tng in p cc ca G ln s lm cho knh dn giu


60/107

K thut in tthm, cn nu in p t ti cc ca G nh di mc ngng hnh thnh knh dn th

knh dn khng c hnh thnh.

2.4.2.3 Cc c tuyn ca MOSFET

a. c tuyn truyn t ca E-MOSFET

Phong trnh biu din c tuyn truyn t ca E-MOSFET

: . Trong K tu thuc vo loi E-MOSFET v c thc xc

nh t data sheet ca E-MOSFET v gi tr I

2))(( thVVKI GSGSD =

D tng ng.


61/107

K thut in tChng III : K thut tng t (16 tit)

3.1 Nhng vn chung v khuch i tn hiu

3.1.1 nh ngha khuch i

Khuch i l qu trnh bin i nng lng c iu khin, nng lng ca

ngun nui cung c p 1 chiu (khng cha ng thng tin) c bin i thnh dngnng lng xoay chiu (c quy lut bin i, mang thng tin cn thit).

Theo nh ngha ny th khuch i c tn hiu cn phi c ngun nui, c

phn t lm nhim v bin i nng lng, v yu tiu khin qu trnh bin i nng

lng chnh l tn hiu vo. Thng thng phn tiu khin l BJT hoc FET hoc l

cc phn tc xy dng t BJT, FET.

3.1.2 Cu trc nguyn l xy dng mt tng khuch i, cc tham s cbn

PK

VCC

VinRTVout

B

E

C

RC

Nguyn l xy dng mt tng khuch i

Phn t cbn trong tng khuch i l PK thng thng l tranzito. Phn t

ny c in tr thay i theo siu khin ca tn hiu vo. Tuy nhin PK khuch

i c tn hiu th cn phn cc cho n .

Cc tham s cbn:

H s khuch i=

i lng u ra

i lng u vo

Vout

61H s khuch i in p AV =

B mn K thut my tnh Vin


62/107

K thut in t

H s khuch i dng in AV =

Iout

Iin


63/107

K thut in t

Tr khng vo Rin =

Vi

Iin

Tr khng ra Rout =

Vout

Iout

3.1.3 Mt s mch phn cc cho BJT

3.1.3.1 Gii thiu chung

Cc mch phn cc cho BJT xc lp ch lm vic cho BJT ta c th phn cc

cho BJT lm vic mt trong cc ch:

+Ch tch cc (JE c phn cc thun, JC c phn cc ngc) trong ch ny

BJT c kh nng khuch i tn hiu

+Ch thng bo ho (JE c phn cc thun, JC c phn cc thun)

+Ch ngng dn(JE c phn cc ngc, JC c phn cc ngc)

im Q(VCE,IC) nm trn ng ti tnh c gi l im lm vic tnh ca BJT.

Tu vo cch phn cc m c v tr tng ng ca im lm vic trn ng ti tnh. Vtr ca im lm vic c nh hng n dng ca tn hiu ra khi khuch i tn hiu

3.1.3.2 im lm vic Q

Xt mch in sau:


64/107

K thut in t

-iu chnh VBB c c IB = 200A th ta c IB C = DCIB = 200A*100 = 20mA v

VCE = VCC ICRC = 10v - 220*20mA = 5.6 V, ta c im lm vic Q tng ng

l:Q(VCE = 5.6V;IC = 20mA)

-iu chnh VBB c c IB = 300A th ta c IB

C = IC = DCIBB

= 300A*100 = 30mAv VCE = VCC - ICRC = 10v - 220*30mA = 3.4 V, ta c im lm vic Q tng ng l:

Q(VCE = 3.4V,IC = 30mA)

-iu chnh VBB c c IB = 400A th ta c IC = DCIB = 400A*100 = 40mA vB

VCE = VCC ICRC = 10v - 220*40mA = 1.2 V, ta c im lm vic Q tng ng

l:Q(VCE = 1.2V, IC = 40mA)

Vi cc VBB khc nhau ta c c cc im lm vic khc nhau nh vy ta c 3

im lm vic ba im ny cng nm trn mt ng thng v ng thng ny gi l

ng ti tnh.


65/107

K thut in t

3.1.3.3 Min hot ng tuyn tnh:

Tp hp cc im Q nm gia im ngng dn v im thng bo ho to thnh

min hot ng tuyn tnh ca BJT, t tn l min tuyn tnh bi v dc theo min ny

in p ra c mi quan h tuyn tnh vi in p vo

Xt mch sau:


66/107

K thut in tVin l in p hnh sin bin thin trn nn l in p mt chiu ti B; Vin bin

thin sinh ra dng Ib bin thin iu ho trong khong t -100A n 100 A trn nn

l dng in mt chiu IB = 300 A iu ny sinh ra dng Ic bin thin iu ho trong

khong t -10mA n 10 mA trn nn l dng mt chiu IC = 30mA; iu ny dn ti

VCE thay i trong khong(VCE(Q) 2.2V;VCE(Q) + 2.2V); (VCE(Q) l VCE ti thi im

khng c tn hiu vo). Cc kt qu thu c trn th hin hnh v di y:


67/107

K thut in t3.1.3.4 Smo dng tn hiu ra do phn cc

Khi ta phn cc im lm vic gn im ngng dn hoc gn im thng bo

ho trn ng ti tnh c th dn ti hin tng mo dng tn hiu u ra.

Khi bin tn hiu vo qu ln cng c th dn ti hin tng mo dng tnhiu u ra

Tn hiu ra c th b ct pha trn trong trng hp c khong thi gian BJT

ngng dn trong khong thi gian bin thin ca tn hiu vo; BJT c th b ct pha

di trong trng hp c khong thi gian BJT thng bo ho trong khong thi gian

bin thin ca tn hiu vo; tn hiu ra c th b ct c pha trn v pha di trong

trng hp bin tn hiu vo ln dn ti c khong thi gian BJT ngng dn v c c

khong thi gian BJT thng bo ho.

Hnh v di y th hin cc tnh hung nu trn:


68/107

K thut in t


69/107

K thut in t

3.1.3.5 Mt s mch phn cc cho BJT

C nhiu dng mch phn cc cho BJT ta ch xt mt s dng mch sau:


70/107

K thut in t+Phn cc ba z

+Phn cc bng in p colectphn hi

+Phn cc bng cu phn p

a) Phn cc baz

-S mch phn cc

-Tm im lm vic Q(VCE, IC) ca BJT

T mch in ta c:B

CCBR

VI

7.0= suy ra BDCC II = ; CCCCCE RIVV =

nh gi tnh n nh:

Q ph thuc vo DC m DC ph thuc vo nhit nn im Q ph thuc vo

nhit , cng v th m im lm vic Q i vi phng php phn cc trn km n

nh.

V d: Xc nh xem im lm vic Q ca mch di y thay i nh th no khi c

s thay i nhit . Bit vi s thay i ca nhit DC thay i t 85 n 100 v VBE

thay i t 0.7 xung 0.6 (c hai thay i ny din ra ng thi)


71/107

K thut in t

Gii:

Trc khi nhit tng ta c DC = 85; VBE = 0.7V do

AxVV

R

VVI

B

BECCB

5103.11

100000

7.012)1()1( =

=

=

mAII BDCC 61.9)1( == ; VRIVV CCCCCE 62.6)1()1( ==

Sau khi nhit tng ta c DC = 100; VBE = 0.6V do

AxVV

R

VVI

B

BECCB

5104.11100000

6.012)2()2( =

=

=

mAII BDCC 4.11)2( == ; VRIVV CCCCCE 62.5)2()2( ==

Nh vy c s thay i im lm vic khi c s thay i nhit . S thay i c

nh gi nh sau:

%6.18%10061.9

61.94.11%100

)1(

)1()2(% =

=

=

C

CCC

I

III

%1.15%10062.5

62.662.5%100

)1(

)1()2(% =

=

=

CE

CECECE

V

VVV


72/107

K thut in tb) Phn cc bng in p colectphn hi

-S mch phn cc:

-Xc nh im lm vic Q(VCE, IC)

Ta c : BEBBCCBCC VRIRIIV +++= )(

Hay: BEBBCDCBCC VRIRIV +++= )1( ty ta tm cBCDC

BECCB

RR

VVI

++

=

)1(

CR)(;)1( BCCCCEBCDC

BECCDCBDCC IIVV

RR

VVII +=

++

==

;

-nh gi tnh n nh

Nu IC tng, dn ti VC gim, dn ti IB gim, dn ti IC gim, dn ti VC tng

Nu IC gim, dn ti VC tng, dn ti IB tng, dn ti IB C tng, dn ti VC gim

Nh vy vi cch hi tip vng quanh im lm vic lun n nh.

Ta c th thy c sn nh ca im Q cn c vo biu thc ca IC, VCE. T

biu thc ca IC, VCE ta thy trn t v mu ca ICu xut hin DC nn im lm vic

t ph thuc vo DC


73/107

K thut in tc) Phn cc bng cu phn p

-S mch phn cc

Mch phn cc kiu ny c s dng rng ri trong vic phn cc BJT n

lm vic min khuch i tuyn tnh. Phng php phn cc ny s dng mt ngunin p v mt mch phn p. Khng ging nh cc phng php phn cc khc

phng php ny im lm vic gn nh khng ph thuc vo DC nn n nh ca

im lm vic rt cao.

-Xc nh im lm vic Q(VCE, IC)

xc nh im lm vic Q ta tnh ton theo trnh t sau:

+Tm VBB

+Tm VE

+Tm IE+Tm IC+Tm VCE

+Tm VB


74/107

K thut in t-Nu dng IB nh hn nhiu so vi dng I2 th ta c th b qua IB trong vic tnh

ton V

B

BB. Khi VBc tnh ton theo cng thcB 221

RRR

VV CCB +

=

-Nu IB khng nh c th b qua th vic tnh ton VB B trnn phc tp hn

khi cn xt n in trli vo baz1 chiu RIN(base) nh hnh v

T s mch trn ta d dng tm ra c ))(//()(//

221

baseRRbaseRRR

VV IN

IN

CCB

+=

Khi RIN(base) >=10R2 th ta c rh b qua RIN(base) v VBc tnh theo cng

thc: 221

RRR

VV CCB +

= .

Xc nh RIN(base)

Xem xt mch sau:


75/107

K thut in t

IN

EEBE

IN

ININ

I

RIV

I

VbaseR

+==)( ; VBE = 10R2 ta c th b qua RIN(base) trong cng thc tnh VB. Khi

221

RRR

VV CCB +

=

-Khi RIN(base) < 10R2 ta xc nh VB theo cng thc:B

))(//()(// 221baseRR

baseRRRVV IN

IN

CCB +=

+Tm VE

VE = VB VB BE = VE 0.7V

+Tm IE


76/107

K thut in t

ERE

E

VI =

+Tm IC

E

DC

C

II

1DC +=

+Tm VCE

EECC RI-RIVV CCCE =

-nh gi sn nh ca im lm vic Q

Qua cc tnh ton trn ta thy IE gn nhc lp vi DC v th IC cng c lp

vi DC dn n im lm vic Q rt n nh.

3.2 B khuch i tn hiu nh dng tranzistor lng cc-BJT

3.2.1 Phn loi cc s khuch i

C 3 loi tng khuch i tn hiu nh dng BJT tng ng vi 3 cch mc BJT

+Tng khuch i chung Emit(CE)

+Tng khuch i chung colect(CC)

+Tng khuch i chung Baz(CB)

tng khuch i E chung, tn hiu vo c a ti B-E, tn hiu ra c ly

trn C-E. tng ny cc E dng chung cho c tn hiu vo v ra v th gi l tng

khuch i emitchung,

tng khuch i C chung tn hiu vo c a ti B-C, tn hiu ra c ly

trn E-C. tng ny cc C dng chung cho c tn hiu vo v ra v th gi l tng

colectchung.

tng khuch i B chung tn hiu vo c a ti E-B, tn hiu ra c ly

trn C-B. tng ny cc B dng chung cho c tn hiuh vi v ra v th gi k tng

bazchung.

Nhng so snh, tng kt khc ca tng tng sc trnh by sau khi nghin cu

xong ba tng khuch i trn.

3.2.2 Phn tch b khuch i theo s tng ng3.2.2.1 S tng ng ca BJT trong ch khuch i tn hiu nh.

Trong cc mch khuch i tn hiu nh ta c th thay th BJT bng s tng

ng. S tng ng ny chc s dng phn tch xoay chiu ch khng s

dng phn tch phn cc.


77/107

K thut in tC hai loi s tng ng ca BJT, loi da trn tham s h, loi da trn

tham s r. y ta xem xt loi s tng ng da trn tham s r.

Trong khi phn tch xoay chiu cc tng khuch i ta c th thay th BJT bng

s tng ng sau:

Trong s trn:

+re l in trxoay chiu emit

+rb l in trxoay chiu baz

+rc l in trxoay chiu colect

+c

eac

I

I=

+b

cac

I

I=

S tng ng ca BJT hnh trn l dng y . Ngoi s tng

ng dng y cn c s tng ng dng n gin thu c t s tng

ng dng y bng cch b qua cc thng s khng thc s quan trng. S

tng ng ca BJT dng n gin nh sau:


78/107

K thut in tTrong s trn rb b b qua v nh hng ca n l nh, rc b b qua v n qu

ln (hng trm K).

Trong s tng ng ca BJT re l thng s rt quan trng. re chnh l in

trxoay chiu gia B v E khi tip gip JE c phn cc thun.

rec xc nh theo cng thc:E

eI

mVr

25' = trong IE l dng in emitmt

chiu.

Colectng vai tr nh mt ngun dng in cung cp dng in bacI


79/107

K thut in t3.2.2.2 Cc bc phn tch tng khuch i tn hiu nh dng BJT

Vic phn tch tng khuch i tn hiu nh dng BJT tri qua hai bc

Bc 1: Phn tch phn cc

Mc tiu chnh ca bc ny l xc nh im lm vic Q(VCE, IC) v mt si

lng 1 chiu khc. T, ta nh gi c gii hn ca bin tn hiu vo tnhiu ra khng b mo.

Trong bc ny cn:

+Tm s mch phn cc t s mch khuch i (s tng ng 1 chiu)

+Tm im lm vic ca tng Q(VCE, IC)

Lu :

tm c s mch phn cc t s tng khuch i ta lm nh sau:

+Tt c cc tin trong tng khuch i thay th bng hmch

+La ra phn mch c cha BJT chnh l s mch phn cc

Bc 2: phn tch xoay chiu

Mc tiu chnh ca bc ny l xc nh c kh nng khuch i tn hiu ca

tng thng qua vic tm cc thng s:

+H s khuch i in p (Av)

+H s khuch i dng in(Ai)

+H s khuch i cng sut(Ap)

+Trkhng vo ca tng(Rin)

+Trkhng ra ca tng(Rout)

Trong bc ny cn:

+Tm s tng ng xoay chiu t s tng khuch i

+Tm Av, Ai, Ap, Rin, RoutLu :

tm c s tng ng xoay chiu t s tng khuch i ta lm nh

sau:

+Tt c cc tin trong tng thay th bng ngn mch(on dy ni tt)+Ccim ni vi ngun nui mt chiu thay th bng vic ni vi im GND

ca mch.

3.2.2.3 Phn tch tng khuch i emitchung (CE)

S tng khuch i E chung:


80/107

K thut in t


-S mch phn cc

S mch phn cc trn thu c sau khi thay th cc t bng hmch v chnra phn mch cha BJT. By gita tnh cc i lung 1 chiu v tm im lm vic Q.

Tm VB

Ta c RIN(base) = DCRE;

Nu RIN(base) >=10R2 th 221

RRR

VV CCB +

=


81/107

K thut in tNu RIN(base)


82/107

K thut in t-Tm trkhng vo ca tng

)(

1111

)(////

21

21

baseRRR

baseRRRI

VR

in

in

in

inin

++=== ; trong Rin(base) c gi l in

trli vo bazxoay chiu. in trny c xc nh nh sau:

eac

b

ee

b

bin r

I

rI

I

VbaseR ')1(

')( +===

Lu :


83/107

K thut in tNu s tngng xoay chiu xut hin in trmc gia cc E vi GND

th Rin(base) = (ac + 1)(re + RE).

-Tm h s khuch i in p(Av, Avs)

in

outv

s

outvs

V

VA

V

VA == ; ; nu Rs rt nh th Avs = AV; thng thng ta tnh ton Av. T AV ta

c th tm ra Avs nu bit c Rs.

AV

b

out

in

out

V

V

V

VAv == ; vi mch trn eebLccout rIVRRIV '.);//( == nn ta c

e

Lc

ac

ac

ee

Lccv

r

RR

rI

RRIA

'

//.

1'.

)//(

+==

Avs

Ta c

))(////()(////

))(////( 2121

21 baseRRRbaseRRRR

VbaseRRRIV in

ins

sininin +

==

nn)(////

)(////

21

21

baseRRR

baseRRRRVV

in

insins

+= t suy ra v

ins

in

s

outvs A

baseRRRR

baseRRR

V

VA

)(////

)(////

21

21

+==

Lu :

H s khuch i in p trn c tnh cho trng hp c ti, mun tm h s

khuch i in p khi khng ti ta ch vic b RL-Tm h s khuch i dng in

Lc

21

in21

Lc

//RR

)(//R//

(base)//R//RR

//RR baseRRA

Vin

V

I

IA inv

out

in

outi ===

-Tm h s khuch i cng sut

ivp AAA =

-Tm trkhng ra ca tng(khi khng ti)

tm trkhng ra ca tng ta lm nh sau:

+Thay th cc ngun c lp bng 0(cc ngun ph thuc gi nguyn)

+Thay th ti bng mt ngun ginh Vtest


84/107

K thut in t+

test

testout

I

VR =

C

test

testout R

I

VR ==

Trng hp c ti Rout = RC//RL3.2.2.3 Phn tch tng khuch i colectchung (CC)

-S tng khuch i colectchung

Tng khuch i colectchung cn c gi l tng lp li Emit. Tng ny tn

hiu vo v tn hiu ra ng pha nhau.


-S mch phn cc


85/107

K thut in t

-Tm VB



RRR

VV CCB +

=

Nu RIN(base)


86/107

K thut in t

-Tm h s khuch i in p

)1(r'R)'( ee

+

=+

== e

eee

ee

in

outv

R

rRI

RI

V

VA (Re = RE//RL) trong trng h p khng ti

th Re = RE. V re


87/107


88/107

K thut in t

Tm VB



RRR

VV CCB +

=

Nu RIN(base)


89/107

K thut in t-Bc 2: Phn tch xoay chiu

-S tng ng xoay chiu

-Tm trkhng vo

eEin rRR '//=

-Tm h s khuch i in p Av

Ee

LC

ac

ac

Eee

LCc

in

outv

Rr

RR

RrI

RRI

V

VA

//'

//.

1)//'(

)//(

+===

-Tm h s khuch i dng Ai

1+===

ac

ac

e

c

in

outi

I

I

I

IA

-Tm h s khuch i cng sut Ap

ivp AAA =

-Tm trkhng ra ca tng


90/107

K thut in t

C

test

testout R

I

VR == (trong trng hp c ti RL th Rout = (RC//RL)

3.3 Khuch i c bit Darlington

Qua phn tch cc tng khuch i ta thy trkhng vo ca tng ph thuc vo

ac, ac gii hn gi tr cc i ca trkhng vo. C mt cch tng trkhng vo ca

tng l s dng cp darlington.


91/107

K thut in t xy dng c p darlington s dng 2 BJT ni vi nhau theo quy tc: cc

colectca hai BJT c ni vi nhau, cc emitca BJT th nht c ni vi cc B

ca BJT th hai nh hnh trn vi cch mc nh vy coi nh ta c BJT c h sac =

ac1.ac2

Cp Darlington thng c mc trong tng m (tng khuch i CC) nm giatng c trkhng ra cao v ti c trkhng nh.

3.4 Mch ghp ni gia cc b khuch i

3.5 Khuch i cng sut

3.5.1 nh ngha, phn loi, c im

Mch khuch i cng sut c nhim v to ra mt cng sut ln cho tn hiu kch thch ti. Cng sut ra c th t vi trm mw n vi trm watt. Nh vy mch

cng sut lm vic vi bin tn hiu ln li vo do ta khng th dng mch

tng ng tn hiu nh kho st m thng dng phng php th.

Ty theo ch lm vic ca transistor, ngi ta thng phn mch khuch i

cng sut ra thnh cc loi chnh nh sau:

- Khuch i cng sut ch A: Tn hiu c khuch i gn nh tuyn tnh, ngha l

tn hiu li ra thay i tuyn tnh trong ton b chu k ca tn hiu li vo (Transistorhot ng ch khuch i c hai na chu k ca tn hiu li vo).

- Khuch i cng sut loi AB: Transistorc phn cc gn vng ngng. Tn hiu

li ra thay i hn mt na chu k ca tn hiu vo (Transistor hot ng hn mt na

chu k - dng hoc m - ca tn hiu li vo).


92/107

K thut in t- Khuch i cng sut loi B: Transistorc phn cc ti V

BE=0 (vng ngng). Ch

mt na chu k m hoc dng - ca tn hiu li vo c khuch i.

- Khuch i cng sut loi C: Transistorc phn cc trong vng ngng ch mt

phn nh hn na chu k ca tn hiu li vo c khuch i. Mch ny thng c

dng khuch i cng sut tn s cao vi ti cng hng v trong cc ng dng c

bit.

Hnh v di y th hin dng in Ici vi cc ch khuch i

3.5.2 Khuch i cng sut kiu n ch A

a) S mch khuch i


93/107

K thut in t

b)Kho st phn cc

+S mch phn cc

+im lm vic Q(VCE, IC)

T mch in ta c:B

CCB

R

VI

7.0= suy ra BDCC II = ; CCCCCE RIVV =


94/107

K thut in t c c hiu sut ln nht ta nn phn cc sao cho im lm vic Q nm

chnh gia ng ti tnh nh hnh v trn.

c) Kho st xoay chiu

i vi tng khuch i cng sut ta kho st xoay chiu bng phng php

th.

Khi a tn hiu Viti li vo dng I

Cv in p V

CE(tn hiu ra) s thay i

quanh im lm vic Q. Vi tn hiu vo nh, th dng in baz thay i rt t nn

dng in IC

v in th VCE

li ra cng thay i t quanh im lm vic.

Khi tn hiu vo ln, in p ra s thay i rt ln quanh im lm vic Q dng

IC s thay i quanh gii hn 0 mA v VCC/RC. in p VCE thay i gia hai gii hn

0V v ngun VCC.


95/107

K thut in t

d)Kho st cng sut

*Cng sut cung cp cho tng khuch i

Cng sut a vo tng khuch i l do VCC cung cp, v th cng sut cung cp

l: CQCCi IVdcP =)(

*Cng sut ra

Dng in ra v in p ra thay i quanh in p v dng in ti im lm vic

tnh, cung cp cng sut xoay chiu trn ti RC. Cng sut ny ln hay nh tu vo tn

hiu vo ln hay nh. Cng sut xoay chiu trn ti RC c thc xc nh bng mt

s cch.

+Tnh theo gi tr hiu dng

C

Co

CCo

CCEo

R

rmsVacP

RrmsIacP

rmsIrmsVacP

)()(

)()(

)()()(

2

2

=

=

=

+Tnh theo in p nh


96/107


97/107

K thut in tS khi tng khuch i cng sut ch B

*Cng sut cung cp

3.6 Khuch i thut ton3.6.1 Khi nim chung

3.6.1.1

Danh t :khuch i thut ton(operational amplifier) thuc v b khuch i

dng mt chiu c h s khuch i ln, c hai u vo vi sai v mt u ra chung.

Tn gi ny c quan h ti vic ng dng u tin ca chng ch yu thc

hin cc php tnh cng, tr, tch phn, vvHin nay cc b khuch i thut ton ng

vai tr quan trng v c ng dng rng ri trong k thut khuch i, to tn hiu hnh

sin v xung, trong bn p v b lc tch cc v.v3.6.1.2 K hiu ca khuch i thut ton

Vn: in p u vo o

Vp: in p u vo thun

Vo:in p u ra


98/107

K thut in t-V : ngun m

+V: ngun dng

Trong mt s trng hp ta khng cc u ni ti ngun nui trong k hiu

khuch i thut ton khi ta c k hiu n gin hn nh sau:

3.6.1.3 M hnh tng ng b khuch i thut ton

A: l h s khuch i hvng,

Vo = A.Vin = A(Vp Vn)

A c gi tr ln chng vn ti hng triu, Ri ln cmega ohm, Ro nh cohm.

3.6.2 B khuch i thut ton l tng


99/107

K thut in t

Vi b khuch i thut ton l tng ta c:

+Ip = In = 0

+Rin =

+Rout = 0

+A =

Vi b khuch i thut ton l tng ta c hai quy tc quan trng l:

-Dngin vo khuch i thut ton bng khng

-in p ti li vo o bngin p ti li vo thun(Vp = Vn

3.6.2 Phn tch b khuch i sdng khuych i thut ton

3.6.2.1 B khuch i khng o

-S mch khuch i


100/107

K thut in t-S mch tng ng

Do In = Ip = 0 nn I1 = I2 => )1(R 1

2

21

+=

=

R

RVV

VV

R

Vno

onn

M

A

RR

R

R

V

VG

A

R

R

VV

A

VVVVAV

in

onn

oinnino

11

1

)

1

1()(

1

2

1

2

1

2

++

+

==

+

+=+==

Nh vy ta c (G > 0) biu thc trn chng t mch trn l mch khuch i

khng o. Nu b khuch i thut ton l l tng th khi ta c

ino VGV .=

A

inVR

RVo )1(

1

2 +=

3.6.2.2 B khuch i o

-S mch


101/107

K thut in t

-S tng ng

Do mch trn l mch tuyn tnh nn ta c th p dng nguyn l xp chng ti

nt mch 1.

Theo nguyn l xp chng th : inon VnVnV += ; vi Vn0 l in p ti nt 1 khi Vin

= 0 Vnin l in p ti nt 1 khi Vno = 0. Vic p dng nguyn l ny c th hin trn


102/107

K thut in thnh v di y:

Nh vy21

2

21

1

RR

R

RR

R

+

+

+

= inon VVV

M

in

ininino

inooo

nnnpo

V

R

R

AA

Vo

V

R

RA

AVVoV

RR

R

AV

VVA

V

A

VVAVVVAV

2

1

2

1121

2

21

2

21

1

21

2

21

1

)1

1(1

1

)1(1)

ARRR

AR(

RR

R)

1(

RR

R

RR

R)(

++=

++=

++=

+=

++

++

+====

Biu thc trn chng t mch trn l mch khuch i o. Khi A => th

ino VR

RV

1

2=

3.6.2.3 B cng o

-S mch


103/107

K thut in t

-Ta phn tch chng t mch trn l mch cng o

Ti nt N1 ta c I1 + I2 + I3 = IF m theo nh lut Ohm ta c:

3

3

3

13

32

2

2

12

21

1

1

11

1 ;; R

V

R

VV

IR

V

R

VV

IR

V

R

VV

I

inNininNininNin

=

==

==

= (v VN1 = Vp = 0)

FF

N1F RR

V-V o outVI ==

Vy inFoininin V

RRRRVo

V

R

V

R

V

R

V)

111(

R 321F3

3

2

2

1

1 ++==++ Khi R1 = R2 = R3 = RF = R th

ta c )( 321 ininin VVVVout ++=

3.6.2.4 B khuch i hiu


104/107

K thut in tMch trn l mch khuch i hiu. Tn hiu ra t l vi hiu ca Vin1 v Vin2. Ta

tm mi quan h gia Vout vi Vin1 v Vin2.

Ta c th p dng nguyn l xp chng tm ra mi quan h ny. Theo nguyn

l xp chng th: Vout = Vout1 + Vout2. Trong Vout1 l u ra ca mch khi Vin2 = 0;

Vout2 l y ra ca mch khi Vin1 = 0;

1

21

1

2

43

42

1

2

43

422

1

211

)1)((

)1)((

R

RV

R

R

RR

RVV

R

R

RR

RVV

R

RVV

ininout

inout

inout

++

=

++

=

=

Chn cc in trR1 = R2 = R3 = R4 ta c Vout = Vin2 Vin1. Biu thc trn chng

t mch trn l mch khuch i hiu.3.6.2.5 B tch phn

Mch trn c u ra Vout t l vi tch phn ca Vin v th gi l b tch phn. Ta

s phn tch chng minh iu ny.

Ta c:IR = IC (v Ip = In = 0); mdt

dVC

dt

dVcCI

VI outC

in ==== ;RR

V-V ninR nn ta

c == dtVRCVdtdV

CVin

outoutout .

1

R. Biu thc trn chng t mch trn l b tch

phn o.


105/107

K thut in t3.6.2.6 B vi phn

-S mch

-Mch trn cho in p ra Vout t l vi vi phn ca in p vo v th c tn l b

vi phn. Ta s tm biu thc th hin mi quan h gia Vout v Vin chng minh iu

ny.

Ta c Ic = IR; m

dt

dVRCVoutV

dt

dVC

VI

dt

dVC

dt

VVdC

dt

dVCI

inoutin

outinnincc

==

===

==

R

RR

Vout-Vn;

)(R

Biu thc trn chng t mch trn l b vi phn o.


106/107

K thut in t3.7 To dao ng iu ha

3.7.1 nh ngha, tham s cbn

nh ngha dao ng iu ho:dao ng iu ho

nh ngha mch to dao ng

Cc tham s cbn:3.7.2 S khi, iu kin to dao ng

-S khi

Mch to dao ng iu ho gm 2 khi chnh l khi khuch i v khi phn

hi. Khi khuch i l khi khuch i khng o c h s khuch i Av, khi phnhi c h s truyn t l .

-iu kin to dao ng iu ho

3.7.3 To dao ng LC

3.7.4 To dao ng RC

3.8 Ngun chnh lu

3.8.1 nh ngha, s khi

Ngun chnh lu hm ch b ngun c xy dng da trn vic chnh lu dng inxoay chiu thnh dng in mt chiu. S khi ca ngun chnh lu c th hin

trn hnh v:

3.8.2 Cc mch chnh lu mt pha cbn


107/107

K thut in tMch chnh lu l b phn mch khng th thiu c trong ngun chnh lu.

Cc mch chnh lu lm nhim v bin dng in xoay chiu hnh sin thnh dng na

hnh sin thc hin c iu ny l nhtnh cht van ca it

Mt s mch chnh lu mt pha c nghin cu trong chng 2 bao gm

:Mch chnh lu chu k, mch chnh lu 2 na chu k v mch chnh lu cu.3.8.3 Mch lc trong b ngun chnh lu

in p sau chnh lu cn qua mch lc gim bt thng ging. Cc mch

lc c xy dng trn cc linh kin tin, cun cm, in tr.

3.8.4 n p trong b ngun chnh lu

Mch n p l b phn mch cui cng trong b ngun chnh lu. Mch n p c

nhim vn nh in p trc s bin ng ca in p vo b ngun v s bin ng

ca ti. Tuy nhin s bin ng ny phi nm trong mt di xc nh tu thuc vo linh

kin v kt cu ca mch.

Ti liu tham kho

[1] Tp th tc gi : Xun Th, ... K thut in t, Nh xut bn Gio dc, 1999

[2] Phm Minh H : K thut mch in t, Nh xut bn Khoa hc v K thut , H

Ni, 1997

[3] Nguyn Thy Vn : K thut s, Nh xut bn Khoa hc v K thut, H Ni, 1995

[4] Phm Minh Vit, Trn Cng Nhng : K thut mch in t phi tuyn, Nh xut

bn Gio dc, H Ni, 2001

[5] Xun Th, Nguyn Vit Nguyn : Bi tp k thut in t, Nh xut bn Gio

dc, H Ni, 1999

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