Kỹ Thuật điện tử

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GIO TRNH

K THUT IN T

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1

LI NI UGio trnh K thut in t c bin son da theo nhiu ti liu ca nhng tc

gi c xut bn, cp nht thng tin trn mng sau chn lc, tng hp m c bitl bi ging mn K thutin t v kinh nghim thc t ging dy ca ti.

Mn K thut in t c th gii thiu ngi c thy c hnh nh thu nh

ca lnh vc in t v cn thit cho nhng ai mun tm hiu tng qut v in t. Tuy

nhin do chng trnh hc cc khoa ngoi ngnh in t c nhiu mn tm hiu

in t, mn K thut in t c yu cu ging 15 tit l thuyt v 30 tit thc hnh.

Gio trnh K thutin t nhm lm ti liu dy hc mn k thut in t (l thuyt).

Hc sinh sinh vin cn c chun b trc,t tr li cu hi v bi tp sau mi chng,

chn p n cho cc cu trc nghim, h thng li kin thc hc v kin thc cn tm

hiu thm. Trong gio trnh ti trnh by 6 chngv phn ph lc:

Chng 1:C s in hc.

Chng 2: Linh kin th ng.

Chng 3: Cht bn dn diode.

Chng 4: Transistor mi ni lng cc.Chng 5: Transistor hiu ng trng.

Chng 6: Linh kin c vng in tr m.

Ph lc: Cu hi trc nghim, phn ny ti son ring cho mi chng kt hp

vi cu hi bi tp sau mi chng gip hc sinh sinh vin t kim tra v cng c kin

thc ca mnh.

Tuy c nhiu c gng nhng v thi gian v trnh ca bn thn c gii hn nn

ti liu kh trnh sai st. Ti mong nhn c s gp chn thnh ca bn c.

Tp.HCM nm 2009

GV bin son

L Th Hng Thm

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Chng 1: C s in hc

3

Chng 1

C S IN HC

1.1. Ngun gc ca dng in1.1.1. Cu to vt cht

Khi nghin cu v th gii xung quanh, cc nh khoa hc cho rng mi vt u ccu to t cc phn t nh nht khng th chia ct. Theo thuyt nguyn t th nguyn tl phn t nh nht ca vt cht.

Cui th k 19, nhng cuc tm ti v kho st khoa hc chng t nguyn t khngphi l phn t nh nht. Bng thc nghim cc nh khoa hc khng nh s tn ti

ca electron trong nguyn t, electron mang in tch m.

Nm 1911, t kt qu th nghim, nh Vt l ngi Anh Rutherford a ra munguyn t Rutherford nhng cn nhng hn ch trong vic din t, gii thch cc qutrnh thuc lnh vc vi m. Nm 1913, nh Vt l an mch Niel Bohr a ra munguyn t mi trn c s tha nhn nhng thnh cng ca Rutherford v a ra hai tin

: Tin 1 (tin v cc trng thi dng) Tin 2 ( tin v tn s bc x)n nay, mi ngi tha nhn mi nguyn t c cu to gm ht nhn, quanh n

l cc electron chuyn ng trnnhng qu o xcnh. Cc electron sp xp trnnhng lp v k tipnhau. K t ht nhn ra, cc lp v c k hiu: K,L, M, N, O, P, Q; s lng t tng ng l 1, 2, 3,,7; mi lp c s electron gii hn. Ht nhn mangin tch dng gm c neutron l ht khng mangin, proton l ht mang in tch dng.

V d:Cu to ca nguyn t Henh hnh 1.1.

Hnh 1.1. Cu to ca nguyn t He.

Bnh thng, nguyn t trng thi trung ha in, ngha l nguyn t c s lngproton bng s lng electron.

1.1.2. in tchin l mt thuc tnh ca ht, lng mang tnh cht in gi l in tch.

n v o in tch c tnh bng Coulomb (C).in tch nguyn t: e = 1,6.10-19 C.

+ +

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Chng 1: C s in hc

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T nghin cu thcnghim dn n qui c gi loi in ging nh loi in xuthintrn thanh thy tinh sau khi cxt vo la l in dng, loi in ging loi inxut hin trn la l in m. Mi vt cht u c th tr thnh nhim in ngha l cmang mt in tch.

Mt vt hay mt phn t ca vt cha n1e in tch dng, -n2e in tch m th intch tonphn ca n l: q = (n1 - n2)e. (1.1)

Bnh thng, c n1 = n2 nn tng i s nhng in tch trong mt th tch ca vtbng 0. Khi n1 n2, vt c gi l vt mang in tch.

Ngoi cc ht c bn electron, proton, neutron, ngi ta cn pht hin nhiu ht cbn khc: positron (e+), ht pi (+, 0, - ).

Tng qut, tng in tch ca mt h c lp khng i.

Ngoi ra, ln ca mt in tch khng thay itrong cc h qui chiu qun tnh

khc nhau. Do , ln ca mt in tch khng ph thuc vo trng thi ng yn hayang chuyn ng ca in tch.

Cc ht mang in tng tc nhau: cc ht tri du ht nhau, cc ht cng du ynhau.

Khi kho st cc lc tng tc gia nhng ht tch in, nm 1785, nh vt l ngiPhp Coulomb pht hin ra nh lut sau v c gi nh lut Coulomb:

Lc tng tc gia hai in tch im q1, q2 trng thi ng yn, cch nhau mtkhong r c:

- Phng lng thng ni gia hai in tch im.- ln t l thun vi tch ln cc in tch v t l nghch vi bnh phng

khong cch gia chng v ph thuc vo mi trng.- Chiu l chiu ca lc y nu hai in tch cng du, lc ht nu hai in tch

tri du. ln lc tng tc gia hai in tch im q1, q2 trng thi ng yn, cch nhau

mt khong r c xc nh theo nh lut Coulomb:

2

21

r

qq

KF (1.2a)F: lc tng tc (N)q1, q2: in tch (C)r: khong cchgia hai in tch im (m)Hng s t l Kty thuc h thng n v.H thng n v SI:

r04

1K (1.2b)

K = 9.109

Nm2

/C2

H thng n v CGSE: K = 1

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Chng 1: C s in hc

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Mt nguyn t trung ha in khi s lng proton bng s lng electron. Mtnguyn t c s lngproton khc s lng electron th tr thnh ion:

- ion dng khi s lng proton ln hn s lng electron.- ion m khi s lng proton nh hn s lng electron.

V d:- Mt in t thot li khi nguyn t th in t ny c gi l in t t do,

nguyn t cn li l ion dng.- Mt nguyn t khi mt in t tr thnh ion dng cn nu nguyn t nhn

thm in t th tr thnh ion m.

1.1.3. in trngNng lng phn b lin kt vi in tch cho chng ta mt hnh nh v in trng.

Trong khng gian xut hin mt in tch q th n to ra xung quanh c mt in trng

lan truyn trong khng gian.Tnh cht c bn ca in trng l khi c mt in tch qtt trong in trng th

in tch chu tc dng ca lc in.

in trng l dng vt cht tn ti xung quanh in tch v tc dng ln in tch

khc t trong n.

Hnh 1.2. Biu din chiu ca ng sc.Chiu ca ng sc l chiu t in tch dng sang in tch m.

Ngi ta biu din in trng bng cc ng sc, mt cc ng sc dng ch cng in trng.

tq

FE (1.3)

+ -

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Chng 1: C s in hc

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E: cng in trng (V/m)F: lc in trng (N)

qt: in tch (C)V in t mang in tch m nn lc tc ng ln in t ngc chiu vi in

trng hay ni cch khc, mt in t t do s di chuyn ngc chiu vi in trng.

1.1.4. in th - hiu in thTrong trng thca mt in tch q, mt in tch im qtt cch q mt khong r,

s c th nng:

r

qq.

4

1W t

r0

p (1.4)

Do , th nng ca mt in tch im q tti mt im bng cng ca lc tnh in

khi dch chuyn in tchim qtt im ra xa v cc.

Th nng ny chnh l th nng tng tc ca hai in tch q v qt.Nu q, qt cng du th WP > 0.Nu q, qt tri du th WP < 0.Khi r th WP 0

Ti cng mt im A ca tnh in trng nhng in tch im khc nhau q t1, qt2,

qt3, s c th nng WP1, WP2, WP3, , nhng t s:

r

q

4

1...

q

W

q

W

q

W

r03t

3p

2t

2p

1t

1p

A (1.5)

Ac gi l in th ca in trng ti im A. Al mt i lng c trng chotnh in trng do in tch im q to ra ti im A ang xt.

in th ti mt im c tr s bng cng ca lc in trng tc dng vo n vin tch dng khi in tch ny di chuyn t im ra xa v cc.

q

AA

(1.6a)

hay

A

A SdE (1.6b)

Tng t nh nc ch chy thnh dng gia hai ni c a th khc nhau, bng thcnghim cc nh vt l chng t rng: cc htmang in tch ch chuyn ng c hng tothnh dng in gia hai im c in th khc

nhau.

mch in hnh 1.3, ti A c in th VA,ti B c in th VB. dch chuyn in lngq t v tr A sang v tr B tc to dng in tA sang B th ngun in phi to ra mt nng

BA

+ -

Ngun in

Hnh 1.3. Mch in kn.

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Chng 1: C s in hc

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lng l UAB > 0. (UAB< 0 th dng in c chiu t B v A).UAB = VAVB = - UBA (1.7)

UAB, UBAgi l hiu in th gia A v B. Ngoi ra, hiu in th gia A v B c th k hiu l U, U1.im ni chung ca

mch inc chn lm im gc(im t, im mass). im ny c in th bng 0.

Khi cho im A ni trc tip xung mass th im A c in th VA = 0.K hiu ni mass, ni t (Ground GND)

Hnh 1.4. K hiu mass, GND.

n v o in th, hiu in th: Volt (V)1 kV (kilovolt) = 10

3V = 1000 V

1 mV (milivolt) = 10-3

V = 0,001 V

1.1.5. Dng in mch hnh 1.3, nu c chnh lch in th gia A v B th c s dch chuyn ca

cc ht mang in theo mthng xc nh. Khi hnh thnh dng in chy trongmch. Ngc li, khng c chnh lch in th gia A v B th khng c s dch chuynca cc ht mang in nn khng c dng trong mch.

Dng in l dng chuyn di c hng ca cc ht mang in.

dt

dqI (1.8)

I: cng dng in (A)dq: in lng (C)dt: khong thi gian ngn (s)

Theo qui c dng in c chiu t dng sang m.n v o cng dng in: Ampere (A)

1 mA (miliampere) = 10-3

A

1 A (microampere) = 10-6

A

1.2. Dng in mt chiu

Khi dng in v in th phn b trong mt h mch khng thay i theo thi gianth mch c xem nh trng thi tnh hay trng thi DC (Direct Current state).

1.2.1. nh ngha

Dng in mt chiu l dng inc chiu v cng dng in khng i theothi gian.

1.2.2. Cng dng in

Cng dng in o bng lng in tch ca cc ht mang in chuyn ng chng qua tit din dydn trong mt n v thi gian.

GN

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Chng 1: C s in hc

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dt

dqI (1.9)

I: cng dng in (A)dq: in lng (C)dt: khong thi gian ngn (s)

Dng in khng i:

t

QI (1.10)

Q l tng cc in tch i qua tit din dy dn trong khong thi gian t.

1.2.3. Chiu ca dng in

Dng introng mch c chiu qui c hngt ni c in th cao sang ni c inth thp. Chiu ca dng in ngc vi chiu chuyn ng ca in t(ngc vi chiu

dch chuyn ca in tch m). Chiu ca dng incng chiu dch chuyn ca intch dng.Theo qui c: chiu ca dng in l t dng sang m.

1.2.4. Ngun in mt chiu

Cc loi ngun mt chiu:

- Pin, acquy.

- My pht in mt chiu.Khi s dng ngun mt chiu, cn bit hai thng s quan trng ca ngun l in p

lm vic v in lng.

in lng Q c n v Ampere gi (Ah). in lng Q ch lng in c npv cha trong ngun. Thi gian s dng s ty thuc cng dng in tiu th vc tnh theo cng thc:

I

Qt (1.11a)

Q: in lng (Ah)

I: cng dng in (A)

t: thi gian (h)V d:

Ngun in mt chiu c in lng 50Ah, nu dng in tiu th l I = 1 A th thigian s dng ti a l:

1

50

I

Qt = 50 (h) (1.11b)

Theo l thuyt nu dng tiu th l 10 A th thi gian s dng l 5h hay nu dngin tiu th l 50A th thi gian s dng l 1 h.

Thc t th khi dng in tiu th ln qua ni tr ca ngun s sinh ra nhit ln lmh ngun trc khi t thi gian s dng theo cng thc trn. trnh h ngun th phi gii hn dng in tiu th mc:

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Chng 1: C s in hc

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(1.11c)

Q: in lng (Ah)I: cng dng in (A)t: thi gian (h)

K hiu:

Ngun c nh:

E: sc in ng.r: in tr trong (in tr ni).

Ngun iu chnh tr s c:

Hnh 1.5. K hiu ca ngun mt chiu.

-Ngun mt chiu: V, U, VCC, VBB, E,-Ngun dng: +VCC

-Ngun m: - VCC-Ngun i xng: VCC

1.2.5. Cch mc ngun in mt chiu

- Mc ni tip.- Mc song song.- Mc hn hp.

V d:Mi ngun c E = 1,5 V, Q = 4,5 Ah, r = 1 .- Mc ni tip.

Hnh 1.6.on mch c ngun mc ni tip.

Ta c: Et = 3 V, Qt = 4,5 Ah, rt = 2 .

- Mc song song.

Hnh 1.7. on mch c ngun mc song song.

Ta c: Et = 1,5 V, Qt = 9 Ah, rt = 0,5 .- Mc hn hp.

E , r

VCC

+ -

VCC

+ -

E, r

+ -

E, r

+ -

Et, rt

+ -

E, r

+ -

E, r + -

Et, rt

10

QI

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Chng 1: C s in hc

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Hnh 1.8. on mch c ngun mc hn hp.

Ta c: Et = 3 V, Qt = 9 Ah, rt = 1 .

1.2.6. Cngcng sut

Dng in chy qua bng n lm bng n chy sng, chy qua bp in, bn isinh ra nhit, chy qua ng c lm ng c quay. iu ny c ngha l nng lng in

c th chuyn i thnh cc dng nng lng khc: quang nng, nhit nng, cnng,.Nh vy dng in thc hin c mt cng:

A = U.I.t = R.I2.t (1.12)

A: cng ca dng in c gi l in nng (J) (Joule)U: in p (V)I: cng dng in (A)t: thi gian dng in chy (s)

R: in tr ()1 J = 1 Ws nhng thc t thng dng Wh hay KWh.

1 KWh = 1000 Wh = 3600000 Ws.

Cng sut ca dng in l cng ca dng in sinh ra trong mt n v thi gian.

K hiu: P, n v: Watt (W).

P = U.I = RI2

(1.13)

1.3. Dng in xoay chiu

Khi dng in v in th phn b trong mt h mch thay i theo thi gian thmch c xem nh trng thi ng hay trng thi AC (Alternative Current state).

1.3.1. nh ngha

Dng in xoay chiu hnh sinl dng in c chiu v cng dng in bin itheo thi gian mt cch tun hon vi qui lut hnh sin.

1.3.2. Cci lng c trng cho dng in xoay chiu hnh sinCci lng c trng cho dng in xoay chiu hnh singm c: gi trnh (gi

tr cc i), gi tr trung bnh, gi tr hiu dng, gi tr tc thi, chu k, tn s, tn s gc,gc pha, pha ban u.

Dng in xoay chiu: i = I0 sint (A) c:- Gi tr nh (gi tr cc i) l I0.

+ -

Et, rt+ -

E, r

+ -

E, r

+ -

E, r

+ -

E, r

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Chng 1: C s in hc

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- Gi tr hiu dng2

II 0 . (1.14a)

- Tn s gc = 2f. (1.14b)

- Tn s lT

1f . (1.14c)

- Chu k lf

1T . (1.14d)

- Gc pha l 100t rad.

- Pha ban u bng 0.- Gi tr tc thi ti thi im t l i.

V d:

* Dng in xoay chiu: i = 14,14sin100t (A) c:

- Gi tr nh (gi tr cc i) l 14,41 A.- Gi tr hiu dng 10 A.

- Tn s gc100 rad/s.

- Tn s l 50 Hz.- Chu k l 0,02 s.


- Pha ban u bng 0. in p xoay chiu: u = U0 sint (V) c:

- Gi tr nh (gi tr cc i) l U0.

- Gi tr hiu dng2

UU 0 . (1.15a)

- Tn s gc = 2f. (1.15b)

- Tn s lT

1f . (1.15c)

- Chu k lf

1T . (1.15d)


- Pha ban u bng 0.- Gi tr tc thi ti thi im t l u.

V d:

* in p xoay chiu: u = 311,1sin100t (V) c:

- Gi tr nh(gi tr cc i) l 311,1 V.- Gi tr hiu dng 220 V.

- Tn s gc100 rad/s.

- Tn s l 50 Hz.

- Chu k l 0,02 s.


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Chng 1: C s in hc

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- Pha ban u bng 0.Dng in xoay chiu i = I0 sint (A) chy qua on mch ch c thun in tr R th

hiu in th gia hai u in tr l:u = U0 sint (V) (1.16)Dng in xoay chiu i = I0 sint (A) chy qua on mch ch c t C th hiu in

th gia hai u t l:u = U0 sin(t - /2)(V) (1.17)

Dng in xoay chiu i = I0 sint (A) chy qua on mch ch c cun cm L th hiu

in th gia hai u cun cm L l:

u = U0sin(t + /2) (V) (1.18)

Tm li:

- Hiu in th gia hai u in tr thun R cng pha vi dng in chy qua

in trR.- Hiu in th gia hai u t in chm pha hn dng in chy qua t in mt

gc l /2.

- Hiu in th gia hai u cun cm nhanh phahn dng in chy qua cun

cm mt gc l /2.

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Chng 1: C s in hc

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CU HI V BI TP

1. Nu cu to ca mt nguyn t trng thi bnh thng. Khi mt nguyn t khng trng thi trung ha in th n tr thnh ion g?

2. in tch l g? Cho bit n v o in tch.Xc nh lc tng tc gia cc intch.3. in trng l g? Xc nh vect cng in trng.4. in th l g? Phn bit khi nim in th, hiu in th, mass (GND), k hiu

ca n.5. Dng in l g?Dng in mt chiu l g? Dng in xoay chiu l g? Xc nh

chiu ca dng in trn mch in. Nu cng thc tnh cng dng in. 6. So snh pha ca hiu in th gia hai u ti vi pha ca dng in chy qua ti,

nu ti l:

a. in tr thun.b. t in.c. cun cm.

7. Mi ngun c sc in ng E, in lng Q, in tr ni r. Nu cng thc tnhEt, Qt, rtca onmch gm haingun mc:

a. ni tip.b. song song.

8. Cho mch nh hnh1.6. Vi mi ngun c E = 1,5 V, Q = 4,5 Ah, r = 1 .

Xc nh Et, Qt, rtca on mch.9. Cho mch nh hnh 1.7. Vi mi ngun c E = 1,5 V, Q = 4,5 Ah, r = 1 .

Xc nh Et, Qt, rtca on mch.10. Cho mch nh hnh 1.8. Vi mi ngun c E = 1,5 V, Q = 4,5 Ah, r = 1 .

Xc nh Et, Qt, rtca on mch.11. Nu biu thc lin quan gia ba i lng: tn s gc, tn s, chu k.12. Cho bit gi tr cc i, hiu dng, trung bnh, nh, tn s gc, tn s, chu k dao

ng ca dng in xoay chiu: i = 1,414sin100t (A).

13. Cho bit gi tr cc i, hiu dng, trung bnh, nh, tns gc, tn s, chu k daong ca in p xoay chiu: : u = 31,11sin100t (V)

14. Ta ni in p xoay chiu 220 V ch gi tr hiu dng hay gi tr cc i cain p ny?

15. Ti sao ta phi tnh gi tr trung bnh ng vi mt bn k ca in p xoay chiu?Nu cng thc tnh gi tr trung bnh ng vi mt bn k ca in p xoay chiu.

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Chng 2: Linh kin th ng

14

Chng 2

LINH KIN TH NG

2.1. in tr

2.1.1. Khi nim

in tr(resistor) l mt linh kin c tnh cn tr dng in v lm mt s chc nngkhc ty vo v tr ca in tr trong mch in.

2.1.2. K hiu - n v

Hnh 2.1. K hiu in tr.

n v : Ohm ()

1 k = 103

1 M = 103 k = 106

2.1.3. in tr ca dy dn

in tr ca dy dn l i lng c trng cho tnh cn tr dng in ca dy dn.

K hiu: R; n v: (Ohm)

in dn l i lng c trng cho tnh dn in ca dy n. in dn l nghcho ca in tr.

K hiu: G ; n v: S (siemens)

R

1G (2.1a)

T thc nghim ta rt ra kt lun: mt nhit nht nh, in tr ca mt dy dn

ty thuc vo cht ca dy, t l thun vi chiu di ca dy v t l nghch vi tit dinca dy.

SlR (2.1b)

R: in tr ca dy dn ()l : chiu di ca dy dn (m)

S: tit din ca dy dn (m2)

: in tr sut (m)

in tr sut:

S o in tr ca dy dn lm bng mt cht no v c chiu di 1m, tit din

thng 1 m2 c gi l in tr sut ca cht .

R

R

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15

Vi nhng cht khc nhau th in tr sut ca n cng khc nhau. in tr sut bin i theo nhit v s bin i ny c xc nh theo cng thc sau:

= 0(1+at) (2.1c)

0: in tr sut o 00C.

a: h s nhit t: nhit (0C)

: in tr sut nhit t.Bng 2.1 a ra tr s trung bnh ca in tr sut ca mt s cht dn in thng

gp:

Cht (.m) Cht (.m)

Bc 0,016.106 Km 0,06.106

ng 0,017.106 Thp 0,1. 106

Nhm 0,026.106 Photpho 0,11.106

Vonfarm 0,055.106 Ch 0,21.106

Bng 2.1. in tr sut ca mt s cht dn in thng gp.

2.1.4. nh lut Ohm

a. nh lut Ohm cho on mch thun in tr

Nm 1926, nh vt l ngi c George Simon Ohm thit lp bng thc nghimnh lut sau: cng dng in trong mt onmch t l thun vi hiu in thgia hai u onmch v t l nghch vi in tr ca onmch.

R

UI (2.2)

I: cng dng in (A)U: hiu in th gia hai u onmch (V)

R: in tr ()

b.nh lut Ohm tng qut i vi on mch

Hnh 2.2. on mch AB.

Dng in chy trong on mch c tnh bi cng thc:

A BV1,r1 V2,r2

R

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16

tR

VI

BA

(2.3)

A: in th ti A.

B: in th ti B.

Rt: in tr ca on mch AB.Rt = R + r1 + r2

Qui c ngun in ty theo chiu dng in:

Ngun pht (cp in), qui c V > 0

Ngun thu (tiu th in), qui c V < 0

c. nh lut Ohm tng qut cho mch knDng in chy trong mt mch kn c tnh bi cng thc:

tR

VI

(2.4a)

I: cng dng in chy trong mch kn.

V: tng in th c trong mch kn.Rt: in tr ca ton mch.

Thcra, vi on mch AB (hnh 2.2) nu hai u A,B ca on mch trng nhau, ta

c mt mch kn. Khi A = Bv cng thc tnh dng in tr thnh:

21

21

t rrR

VV

R

VI

(2.4b)

V d khc:Ta c:

21

21

t RR

VV

R

VI

(2.4c)

Hnh 2.3. Mch in kn.

2.1.5. nh lut Kirchhoff

Thc t, ta thng gp cc mng in phn nhnh phc tp gm nhiu nt v vngmng.

Mt nt in l ch ni cc nhnh in v phi c t nht ba nhnh in tr ln.

Vng mng l vng kn do cc onmch to thnh.a. nh lut Kirchhoff th nht (nh lut nt)

V1R1

R2

V2

I

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Tng i s cc cng dng in ti mt nt bng khng.

0)I(n

1k

k

(2.5a)

Ti nt c nnhnh in. Qui c: cng dng in ti nt mng du +, cng dng in i khi nt mng du -.

Hay ni cch khc: Tng cc cng dng in ti nt bng tng cc cng dng in i khi nt .

Ivo = I ra (2.5b)

V d:Ti nt A ta c:I 1 - I 2 - I 3 + I 4 + I 5 = 0 (2.5c)

Hay I 1 + I 4 + I 5 = I 2 + I 3 (2.5d)

Hnh 2.4. Nt A c 5 nhnh in.

b. nh lut Kirchhoff th hai (nh lut vng mng)

Trong mt vng mng, tng ca tng i s cc sc in ng v tng i s cc

gim in th trn ccphn t khc bng khng.

0)RI()V(n

1k

n

1kkkk

,

,

,,

(2.6a)

Qui c:Sc in ng mang du + nu chiu i chn trn vng mng xuyn vo ccdng ca ngun in. Sc in ng mang du - nu chiu i chn trn vng mngxuyn vo cc m ca ngun in.

Cng dng in mang du + nu n cng chiu vi chiu chn v mang du -nu n ngc chiu vi chiu chn.

V d: Xt mch nh hnh 2.5 ta c:Vng I:

- V1 + I1(r1 + R1)I2(r2 + R2) + V2 = 0 (2.6b)

Vng II:

- V2 + I2(r2 + R2)I3(r3 + R3) + V3 = 0 (2.6c)

2.1.6. Phn loi

in tr c th phn loi da vo cu to hay da vo mc ch s dng m n c

nhiu loi khc nhau.2.1.6.1. Phn loi theo cu to

I2

I1

I3

I4

I5 A

V1,r

I

I1R1

R2

R3II

V2,r

V3,r

I2

I3

Hnh 2.5. Mch in gm hai vng

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a.in tr than (carbon resistor)

Ngi ta trn bt than v bt t st theo mt t l nht nh cho ra nhng tr s

khc nhau. Sau , ngi ta p li v cho vo mt ng bng Bakelite. Kim loi p st hai u v hai dy ra c hn vo kim loi, bc kim loi bn ngoi gi cu trc bn

trong ng thi chng c xt v m. Ngoi cng ngi ta sn cc vng mu cho bittr s in tr. Loi in tr ny d ch to, tin cy kh tt nnn rtin v rt thngdng. in tr than c tr s t vi n vi chc M. Cng sut danh nh t 0,125 Wn vi W.

b.in tr mng kim loi (metal film resistor)

Loi in tr ny c ch to theo qui trnh kt lng mng Ni Cr trn thn gm cx rnh xon, sau ph bi mt lp sn. in tr mng kim loi c tr s in tr nnh, khong in tr t 10 n 5 M. Loi ny thng dng trong cc mch dao

ng v n c chnh xc v tui th cao, t ph thuc vo nhit . Tuy nhin, trongmt s ng dng khng th x l cng sut ln v n c cng sut danh nh t 0,05 Wn 0,5 W.Ngi ta ch to loi in tr c khong cng sut danh nhln t 7 W n1000 W vi khong in tr t20 n 2 M. Nhm ny cn c tn khc l in trcng sut.

c.in tr oxit kim loi (metal oxide resistor)

in tr ny ch to theo qui trnh kt lng lp oxit thic trn thanh SiO2. Loi ny c n nh nhit cao, chng m tt, cng sut danh nh t 0,25W n 2 W.

d.in tr dy qun (wire wound resistor)

Lm bng hp kim Ni Cr qun trn mt li cch in snh, s. Bn ngoi c phbi lp nha cng v mt lp sn cch in. gim ti thiu h s t cm L ca dyqun, ngi ta qun s vng theo chiu thun v s vng theo chiu nghch.

in tr chnh xc dng dy qun c tr s t 0,1 n 1,2 M, cng sut danh nhthp t 0,125W n 0,75 W.

in tr dy qun c cng sut danh nh cao cn c gi in tr cng sut. Loi

ny gm hai dng:- ng c tr s 0,1 n 180 k, cng sut danh nh t 1 W n 210 W.- khung c tr s 1 n 38 k, cng sut danh nh t 5 W n 30 W.

2.1.6.2.V mc ch s dng

a. in tr c nh

in tr c nh l loi in tr c tr s c nh khng thay i c. Tr s nyc nh sn xut n nh c sai s trong phm vi cho php.

Nhm in tr c nh chia ra cc loi:

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in tr chnh xc: c th l dng mng kim loi hoc dy qun, c thit k dng trong cc mch i hi sai s trong phm vi hp, n nh ln, ting n thp vh s nhit thp. Loi dy qun tng i ln v ch c mt khong in tr t 0,1 n 1,2 Mnhng n c n nh cao nht. Cc hiu ng ca in cm L v indung C ca in tr dy qun khin n khng thch hp dng tn s ln hn 50 kHz

ngay c khi qun c bit gim in cm v in dung lin kt. in tr mng kimloi khng bn nh in tr dy qun song c in cm nh hn. in tr mng kim loithng c v hoc hn kn hoc c nha phenol. N c khong in tr t 10 n

5M.

in tr bn chnh xc: c thit k cho cc mch i hi n nh nhit

lu di. in tr thng nh hn in tr chnh xc v r hn, ch yu lm chc nnghn dng v gim p trong cc mch.

Loi in tr Khong in tr Khong cng sut danh nh

Oxit kim loiKim loi gmThan kt ta

10 n 1,5 M

10 n 1,5 M.10 n 5 M.

0,25 W n 2 W0,05 W n 0,5 W0,125 W n 1 W

in tr a dng: loi ny nh, r tin, thng hay dng trong mch in t mdung sai ban u l khng quan trng (v d: 5% hoc ln hn), n nh di hn l

khng quan trng. Khng c dng nhng in tr ni cn h s nhit ca intr thp v mc n thp. Khong in tr t 2,7 n 100 M. Tr s in tr trn0,3Mbt u b gim tn s xp x 100 kHz, trn tn s 1 MHz tt c cc tr s u

b gim. Khong cng sut danh nh t 0,125 W n 2 W.

in tr cng sut: c dng dy qun hoc dng mng, l loi c khong cngsut danh nh cao, c dng trong cc b ngun cng sut, ccb chia p...

b. in tr c tr s thay i c:

Bin tr(VR = Variable Resistor): l loi in tr c tr s thay i cBin tr dy qun: dng dy dn c in tr sut cao, ng knh nh, qun trn li

cch in bng s hay nha tng hp hnh vng cung 2700. Hai u hn hai cc dn in

A, B. Tt c c t trong mt v bc kim loi c np y. Trc trn vng cung cqun dy l mt con chy c trc iu khin a ra ngoi np hp. Con chy c hnvi cc dn in C.

Bin tr dy qun thng c tr s nh t vi n vi chc . Cng sut kh ln,c th ti vi chc W.

Bin tr than: ngi ta trng mt lp than mng ln hnh vng cung bng bakelit. Haiu lp than ni vi cc dn in A v B. gia l cc C ca bin tr v chnh l con

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chy bng kim loi tip xc vi lp than. Trc xoay c gn lin vi con chy, khi xoaytrc (chnh bin tr) con chy di ng trn lp than lm cho tr s bin tr thay i. Bintr than cn chia lm hai loi: bin tr tuyn tnh, bin tr phi tuyn.

Bin tr than c tr s t vi trm n vi Mnhng c cng sut nh.

Hnh 2.6. Hnh dng v k hiu ca bin tr.

Nhit intrl loi in tr m tr s ca n thay i theo nhit (thermistor).

Nhit tr dng ( PTC = Positive Temperature Coefficient) l loi nhit tr c h snhit dng.

Nhit tr m ( NTC = Negative Temperature Coefficient) l loinhit tr c h snhit m.

VDR(Voltage Dependent Resistor) l loi in tr m tr s ca n ph thuc inp t vo n. Thng thVDR c tr s in tr gim khi in p tng.

in tr quang (photoresistor) l mt linh kin bn dn th ng khng c mini P N. Vt liu dng ch to in tr quang l CdS (Cadmium Sulfid), CdSe

(Cadmium Selenid), ZnS (st Sulfid) hoc cc tinhth hn hp khc.

Hnh 2.7. Cu to ca in tr quang.in tr quang cn gi l in tr ty thuc nh sng (LDR Light Dependent

Resistor) c tr s in tr thay i ty thuc cng nh sng chiu vo n.

Hnh 2.8.Hnh dng v k hiu ca in tr quang.K hiuv hnh dngca in tr quangnh hnh 2.8.

CdS

nh sng

LDRCdS

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Khi b che ti th in tr quang c tr s rt ln, khi c chiu sng th dn inca cht bn dn tng do cc cp in t t do v l trng hnh thnh nhiu tc l in

tr gim nh. in tr quang c tr s in tr thay i khng tuyn tnh theo sngchiu vo n. Khi trong bng tiintr quang c tr s khong vi megaohm, tr s cain tr quang trong bng ti vi nhiu trng hp ng dng cn phi bit. N cho ta

dng in r ln nht vi mt in th trn in trquang. Dng r qu ln s dn ns sai lch khi thit k mch in. Khi c chiu sng in tr quang c tr s rt nhkhong vi chc n vi trm Ohm.

H s nhit ca in tr quang t l nghch vi cng chiu sng. Do gimbt s thay i ca in tr quang theo nhit , in tr quang cn c cho hot ng

vi mc chiu sng ti a. mc chiu sng thp v tr s in tr quang cao cho ta ssai bit kh ln so vi tr s chun.

in tr quang c ng dng lm b phn cm bin quang trong cc mch t ngiu khin bi nh sng; mch o nh sng; mch chnh hi t ca mt s thit b; mchtr chi in t,

c.Mt s in tr khc:

in tr cu ch.in tr xi mng.

in trchip.in tr dn

Hnh 2.9. Hnh dng ca mt s loi in tr.

2.1.7. Cch mc in tr

a.Mc ni tip

Hnh 2.10.Mch in tr mc ni tip.

Xt mch nh hnh 2.10, vi:

I1: cng dng in chy qua R1I2: cng dng in chy qua R2

U1: hiu in th gia hai u R1U2: hiu in th gia hai u R2

R2

I

+

U

I2

I1

R1 Rt

+

U

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Ta c: I1 = I2 = I (2.7)

U = U1 + U2 (2.8)

Rt = R1 + R2 (2.9a)

Nu c nhiu in tr mc ni tip th

Rt = R1 + R2 + + Rn (2.9b)

b. Mc song song

Hnh 2.11.Mch in tr mc song song.

Xt mch nh hnh 2.11, vi:

I1: cng dng in chy qua R1I2: cng dng in chy qua R2

U1: hiu in th gia hai u R1U2: hiu in th gia hai u R2Ta c: U1 = U2 = U (2.10)

I = I1 + I2 (2.11)

21t R

1

R

1

R

1 hay

21

21

t RR

RR

R

1

(2.12a)

Nu c nhiu in tr mc song song vi nhau th:

n21t R

1...

R

1

R

1

R

1 (2.12b)

2.1.8. Cch c tr s in tr

a. c tr s in tr theo qui c vng mu:

in tr 4 vng mu

- Vng A, B ch tr s tng ng vi mu. - Vng C ch h s nhn.

- Vng D ch sai s.

Hnh 2.12. in tr4 vng mu.

A B C D

R1

R2

Rt

I1R

I2

II

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V d:

tm bc = 2,7 k 10%

tm vng nh = 2,7 k 5%

vng nh = 2,2 k 5%

Nulc vng nh = 1,5 k 5%Camcam vng nh vng nh = 3,3 5%

Mu Vng A, B Vng C Vng D

en

Nu

Cam

Vng

LcLam

Tm

Xm

TrngVng nhBc

Mu thnin tr

0

1

2

3

4

5

6

7

8

9

-------------

-------------

-------------

x10 = x1

x101 = x10

x102 = x100

x103 = x1000

x104 = x10000

x105 = x100000

x106= x1000000

x107 = x10000000

x108 = x100000000

x109= x1000000000

x10-1 = x0,1

x10-2 = x0,01

------------------------

----------

1%

2%

3%--------------------------

-------------

-------------

-------------

-------------

5%

10%

20%

Bng 2.2. Bng qui c mu in tr.

in tr 3 vng mu:

Ln lt c k hiu A, B, C. ngha ca cc vng mu tng t loi in tr 4vng mu: vng A, B ch tr s tng ng vi mu. Vng C ch h s nhn. Sai s xem

nh mu ca thn in tr.V d:

tm = 2,7 k 20%

in tr 5 vng mu:

Loi in tr 5 vng mu c k hiu l vng A, B, C, D, E: 3 vng A, B, C ch trs tng ng vi mu, vng D ch h s nhn, vng E ch sai s.

V d:

Nuen en en nu = 100 1%

b. c tr s in tr theo qui c chm mu

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Trn thn in tr, mt u in tr c mu B khc vi mu ca thn in tr (A),gia thn c chm mu (C). ngha cc mu v cch c tr s in tr nh trn.

V d:Mt in tr c thn mu xanh l cy, mt umu , gia thn c chm vng, tr

s ca n l 520 k.

c.in tr c ghi s trn thn

i vi in tr c ghi s trn thn th hai s u l s c ngha, s th ba ch snhn.

V d:

Trn thn in tr c ghi 103 th tr s in tr l 10 k.Ngoi ra trn thn in tr c ghi con s v ch th con s ch tr s in tr, ch ch

bi s: R= x1; K = x103; M = x106.

V d: 5R = 5 .4K7 = 4,7 k.

V l thuyt, linh kin in tr c th c gi tr bt k t thp nht n cao nht.

Trong thc t, cc linh kin in tr c khong in tr t 0,1n 100 M.

Cc gi tr tiu chun: 1.0; 1.2; 1.5; 1.8; 2.2; 2.7; 3.3; 3.9; 4.3; 4.7; 5.1; 5.6; 6.8; 7.5;8.2; 9.1. Cc linh kin in tr thng c ch to vi gi tr l cc gi tr tiu chun

nhn vi bi s ca 10.

V d: in tr: 10; 100 ; 1,5 k; 2,7 k; 5,6 k.

2.1.9. Cng sut ca in tr

Cng sut ca in tr l tr s ch cng sut tiu tn ti a ca n. Cng sut chung ny do nh sn xut cho bit di dng ghi sn trn thn hoc kch thc ca intr. Kch thc in tr ln th cng sut ca n ln. Cng sut ca in tr thay itheo kch thc vi tr s gn ng nh bng 2.3.

Bng 2.3. Cng sut ca in tr thay i theo kch thc.

Nn chn cng sut chu ng ln hn hay bng 2 ln cng sut tnh ton.

Cng sut Chiu di ng knh2 W 1,6 cm 10 mm

1 W 1,2 cm 6 mm

0,5 W 1 cm 4 mm

0.25 W 0,7 cm 3 mm

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2.1.10. ng dng

in tr c nhiu ng dng trong lnh vc in v in t:- Ta nhit: bp in, bn i.- Thp sng: bng n dy tc.

- B cm bin nhit, cm bin quang.- Hn dng, chia dng.

- Gim p, chia p,.

IRR

RI

21

21

(2.13a)

IRR

RI

21

12

(2.13b)

CC

21

11 V

RR

RV

(2.14)

Mch chia dng nh hnh 2.14cn c gi l mch phn dng. Mch chia p nhhnh 2.15 cn c gi l mch phn p hay cu phn p (mch chia th / mch phn th/ cu phn th).

Hnh 2.14. Mch chia dng.

Hnh 2.15. Mch chia p.

+VCC

R1

R2

V1

Hnh 2.13. Mch dng R hn dng, gim p.

R

9V/3WVCC

12V

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2.2. T in

2.2.1. Khi nim

T in (capacitor) l linh kin c tnh tch tr nng lng in di dng in trng.

2.2.2. Cu to k hiu

T in c cu to gm hai bn cc bngcht dn in (kim loi) t song song gn nhau

nhng cch in bi lp in mi gia.K hiuca t in:

Hnh 2.17. K hiu ca t khng phn cc (a), t c phn cc (b), t bin i (c).

2.2.3. S dn in ca t

Xt mch nh hnh 2.14.

Khi kha K h th n tt.ng kha K, ta thy n lesng ln ri tt. Nu i ngunVDCbng ngun VAC th khi K h n tt, K ng ta thy

n sng lin tc.

2.2.4. in dung

in dung (capacitance) l i lng c trng kh nng tch in ca t.

K hiu: C, nv: Farad (F)Thng dng cc c s ca Farad:

Microfarad: 1 F = 10-6 F

Nanofarad: 1 nF = 10-9 F

Picofarad: 1 pF = 10-12 F

Femptofarad: 1 fF = 10-15 F

in dung ph thuc cht in mi, t l thun vi tit din ca bn t v t l nghchvi khongcch gia hai bn t (b dy ca lp in mi).

d

SC (2.15a)

(a) (b) (c)

Cht in mi

Bn cc

Dy ni ra

Hnh 2.16. Cu to ca t in.

+ ++ ++ +

-- -- - -- -

K

VDC

Hnh 2.18. Mch th nghim s dn in ca t.

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Vi:C: in dung (F)S: tit din ca bn t (m2)d: khong cch gia hai bn t (m)

= r.0 (2.15b)

r: hng s in mi tng i.0: hng s in mi khng kh; 0 = 8,85.10

-12 F/m.

Mt s cht in mi thng dng lm t: Khng kh kh, giy tm du, gm, oxitnhm, mica.

Cht in mi Hng s in mir

Khng kh kh

Giy

Gm

Mica

1

3,6

5,5

4 5

Bng 2.4. Hng s in mi ca mt s cht.

in dung c th o bng t s in tch ca t trn hiu in th gia hai bn t in.

U

QC (2.16)

C: in dung ca t (F)Q: in tch (C)U: hiu in th gia 2 bn t (V)

Nng lng tch tr t in l:

2CU2

1W (2.17)

W: nng lng (J)

C: in dung (F)U: hiu in th gia 2 bn t (V)

2.2.5. in th lm vic

i vi mi t in, ch c th t vo n mt in pln nht no , ty theo kt

cu ca lp in mi. Nu in p t vo qu ln in mi s b nh thng v tr nndn in, lm t in b hng khng dng c na.

in th lm vic (Working Volt = WV) chnh l in th ln nht cho php p vohai u t m t chu ng c. Thng in th ny c ghi trn t.

2.2.6. Mch tng ng ca t in

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Ngoi in dung, mt t in thc t cn c in tr v in cm nh trong mchtng ng hnh 2.15.

Hnh 2.19. Mch tng ng ca t in.

Vi RSl in tr ni tip do cc dy dn, cc u tip xc v cc in cc.RP: in tr sun do in tr sut ca cht in mi v vt liu lm v, do hao in

mi.

L: h s t cm (in cm) tp do dy dn v cc in cc.

in tr tng ng ni tip ESR(Equivalent Series Resistance) l in tr AC cat in phn nh c in tr ni tip RSv in tr song song RPti tn s cho hao ca cc phn t ny c th biu th bng hao ca mt in tr R trong mch tngng.

Dung khng: l i lng c trng cho sc cn in ca t.

K hiu: XChoc ZC, n v: Ohm ().

Biu thc:fC2

1

C

1XC

(2.18)

XC: dung khng (): tn s gc (rad/s)C: in dung (F)f: tn s (Hz)Tng tr ( tr khng): khi lm vic tn s cao th phi tnh thm in cm u ra:

2

CL

2 XXRZ (2.19a)

Z: tng tr()

R: ESR()

XC: dung khng ()

XL: cm khng (), XL= L = 2fL (2.19b)L: h s t cm (H)

H s cng sut PF (Power Factor). Thut ng PF ch hao in trong t khi lmvic vi in p AC. t in l tng, dng s sm pha hn in p gia hai u t l900. T thc t, do tn hao cht in mi, in cc v cc u tip xcnn gc pha nhhn 900. PF c nh ngha l t s in tr tng ng ni tip R v tng tr Z. PF c

n v: %.

R L C

RP

RS L

C

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H s tiu tn DF (Dissipation Factor) l t s in tr tng ng ni tip R vdung khng XC. DF c n v: %. DF xp x bng PF khi PF 10%.

H s phm cht Q (Quality Factor) l nghch o ca h s tiu tn. N thng pdng cho t trong cc mch iu hng.

Dng in r(r) DC l dng chy qua t khi c t in p DC vo t. in tr cch in: l t s ca in p t vo t trn dng in r (r) v thng

biu th bng M.

2.2.7. Cch mc t in

a. Mc ni tip

Hnh 2.20. Mch t in mc ni tip.

in tch np vo t c tnh theo cng thc:

21QQQ (2.20)

22

112211 C

Q

U;C

Q

UUCUCQ

Mt khc:t

tC

QU.UCQ

m : 21 UUU (2.21)

21t C

1

C

1

C

1 (2.22a)

Nu c nhiu t ghp ni tip th:

(2.22b)

b. Mc song song

n21t C1...

C1

C1

C1

+

U

+

U

C2

+

C1

+

C t

+

+

U

C1+

+

U

C2+

C t+

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Hnh 2.21.Mcht in mc song song.

Hiu in th gia hai u t C1, C2: U = U1 = U2 (2.23)in tch np vo t C1: Q1 = C1Uin tch np vo t C2: Q2 = C2.U

in tch np vo t Ct: Q = Ct.Uin tch np vo t C1, C2bng in tch np vo t Ct nn:

Q = Q1 + Q2 (2.24)

Ct.U = C1U + C2U = (C1 + C2) U

Ct = C1 + C2 (2.25a)Nu c nhiu tmc song song th:

(2.25b)

2.2.8. Hin tng np x ca t

Xt mch nh hnh 2.18,gi s t cha tch in, ta

bt kha K sang v tr s 1

th t bt u np in, lngin tch c tch trn hai

bn t tng dn n khi hiu

in th gia hai u t gnbng ngun VDC (99%VDC)th qu trnh np in ca t c chm dt, t c xem nh np y, nu khng ctc ng khc th hin tng vn khng i.

Khi cha tch in th hiu in th gia hai u t bng khng. Trong qu trnh npin th hiu in th gia hai u t thayi theo dng hm s m:

)e(1V(t)v

t

DCC

(2.26a)e = 2,71828

= R.C (2.26b)

R: in tr ()C: in dung (F)

: thi hng np x ca t (s)t: thi gian t npin (s)Trong qu trnh np in, t c dng

innp thay i theo dng hm s m:

Ct = C1 + C2+ Cn

+ +C

K

21

R

VDC

Hnh 2.22. Mch kho st hin tng np -x ca t.

0 2 3 4 5 t

V C (t)/IC (t)

63

8695

98 99

37

145

2 1

Hnh 2.23. c tuyn np ca t.

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t

DCC e

R

V(t)i

(2.27)

Theo l thuyt, thi gian t np y l v hn (vc(t) = VDC). Trn thc t, sau thi

gian 5t np c 99%VDC, lc ngi ta xem nh t np y.

Khi t np y, ta bt K qua v tr s 2, t C x in qua R, hiu in th gia haiu t thay i theo biu thc:

t

DCC eV(t)v

(2.28a)

e = 2,71828

= R.C (2.28b)

R: in tr ()C: in dung (F)

: thi hng np x ca t (s)t: thi gian t x (s)

Dng x ca t thay i theo biu thc:

t

DCC e

R

V(t)i

(2.29)

tc np x nhanh trong thi gian lc u t 0 n , sau chmli trongthi gian sau.

2.2.9. Phn loia. Da theo mc ch s dng

T c nh: l t c tr s in dung c nh. Tr s ny c nh sn xut n nhc sai s trong phm vi cho php.N c chia lm hai dng:

- T c cc (polar): t c phn cc tnh dng v m. - T khng phn cc(nonpolar): t c hai cc nh nhau.

T bin i: l loi t c tr s in dung c iu chnh thay i theo yu cu s

dng.

Hnh 2.24. Hnh dng t bin i.

b. Da theo cht in mi

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- T ha: l loi t c phn cc tnh. T ha c bn cc l nhng l nhm, in mil lp oxit nhm rt mng c to bng phng php in phn. in dungca t hakh ln.

Khi dng phi rp ng cc tnh dng v m. in th lm vic thng nh hn500V.

- T ha tantalum (Ta): l t c phn cc tnh, c cu to tng t t ha nhng dngtantalum thay v dng nhm. T Tantalum c kch thc nh nhng in dung ln. inth lm vic ch vi chc volt.

- T giy: l loi t khng phn cc tnh. T giy c hai bn cc l nhng l nhmhoc thic, gia c lp cch in l giy tm du v cun li thnh ng.

- T mng: l t khng phn cc tnh.T mng ccht in mi l mng cht donh: polypropylene, polystyrene, polycarbonate, polyethelene. C hai loi t mng

chnh: loi foil v loi c kim loi ha.Loi foil dng cc ming kim loi nhm hay thic to cc bn cc dn in. Loi

c kim loi ha c ch to bng cch phun mng mng kim loi nh nhm hay kmtrn mng cht do, kim loi c phun ln ng vai tr bn cc. Vi cng gi tr in

dung v nh mc in p nh thng th t loi kim loi ha c kch thc nh hn loifoil. u im th hai ca loi kim loi ha l n t phc hi c.iu ny c ngha l

nu in mi b nh thng do qu in p nh thng th t khng b h lun m n tphc hi li. T foil khng c tnh nng ny.

- T gm (ceramic): l loi t khng phn cc tnh. T gm c ch to gm chtin mi l gm, trng trn b mt n lp bc lm bn cc.

- T mica: l loi t khng phn cc tnh. T mica c ch to gm nhiu mingmica mng, trng bc, t chng ln nhauhoc ming mica mng c xp xen k vicc ming thic. Cc ming thicl ni vi nhau to thnh mt bn cc, Cc ming thic chn ni vi nhau to thnh mt bn cc. Sau bao ph bi lp chng m bng sp

hoc nha cng. Thng t mica c dng hnh khi ch nht.

Ngoi ra, cn c t dn b mt c ch to bng cch t vt liu in mi gm giahai mng dn in (kim loi), kch thc ca n rt nh. Mng t in (thanh t in) ldng t c nh sn xut tch hp nhiu t in bn trong mt thanh (v) tit kim

din tch. Ngi ta k hu chn chung v gi tr ca cc t.

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33

Hnh 2.25. Hnh dng ca mt s loi t.

2.2.10. Cch c tr s in dung

- T c ghi s trn thn: .1 c ngha l t c in dung C = 0,1 F; .01 c ngha l t

c in dung C = 0,01 F.- T c ghi s trn thn:103K c ngha l t c in dung:C = 10000 pF 10%.

Hai s u l s c ngha, s th ba ch s nhn. Ch ch sai s: J = 5%.K = 10%, M = 20%.

- T c ghi hai ch s trn thn, v d: 47/50 th s u l in dung, n v l pF, sth hai l tr s in p lm vic, n v l volt.

- T ha: cc tnh c ghi bng du + hoc du -. n v in dung l microfarad,

in p lm vic n v l volt.V d: trn thn t ha ghi 2200F25V c ngha l t c: C = 2200 F, WV = 25 V

Qui c mu i vi t in tng t qui c mu i vi in tr.

T in gm dng hnh ng c 5 vng mu nh nhau nhng c vng th nm cch xahn. ngha cc vng mu: vng th nht, vng th hai ch s tng ng vi mu, vngth ba ch s nhn, vng th t ch sai s, vng th nm ch c im ring ca n.

T in gm dng hnh ng c 4 vng mu nh nhau nhng c vng th nm rng

hn. ngha cc vng mu: vng th nht, vng th hai ch s tng ng vi mu, vngth ba ch s nhn, vng th t ch sai s, vng th nm ch h s nhit .

c bit i vi t dn b mt c ba cch m ha thng dng, c ba u dng n v

pF.

H 33 k hiu ch in hoa v thng: trn thn t ghi mt k hiu v theo sau l s(0 9) ch s nhn.

K hiu S nhn

A1.0

B1.1

C1.2

D1.3

E1.5

F1.6

G1.8

H2.0

J2.2

K2.4

a2.5

L2.7

M3.0

N3.3

b3.5

P3.6

Q3.9

d4.0

R4.3

e4.5

S4.7

f5. 0

T5.1

U5.6

m6.0

V6.2

W6.8

n7.0

X7.5

t8.0

Y8.2

y9.0

Z9.1

0 = x1

1 = x10

2 = x100

3 = x1000

4 = x10000

5 = x100000

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34

V d:J3 = 2.2 x 1000 = 2200 pF

P2 = 3.6 x 100 = 360 pF

S1 = 4.7 x 10 = 47 pF

H 24k hiu ch in hoa: trn thn t ghi mt k hiu v theo sau l s (1 9) chs nhn.

V d:05 = 5 pF

82 = 82 pF

A1 = 10 x 10 = 100 pF

N3 = 33 x 1000 = 33000 pF

K hiu S nhn

A10

B11

C12

D13

E15

F16

G18

H20

J22

K24

L27

M30

N33

P36

Q39

R43

S47

T51

U56

V62

W68

X75

Y82

y90

Z91

1 = x10

2 = x100

3 = x1000

4 = x10000

5 = x100000

Lu : vi h ny th cc gi tr nh hn 100pF s c ghi trc tip, cc gi tr ln

hn 100pF c ghi bng mt ch vi mt s.

H 24 k hiu ch in hoa v s: trn thn t ghi mt k hiu v s nhn c quinh bi mu ca k hiu .

K hiu S nhn (mu)

A1.0

B1.1

C1.2

D1.3

E1.5

H1.6

I1.8

J2.0

K2.2

L2.4

N2.7

O3.0

R3.3

S3.6

T3.9

V4.3

W4.7

X5.1

Y5.6

Z6.2

36.8

47.5

78.2

99.1

cam = x1.0

en = x10

lc = x100

lam = x1000

tm = x10000

= x100000

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35

V d:W mu cam = 4.7 x 1.0 = 4.7 pF

2.11. ng dng

T thng c dng lm t lc trong cc mch lc ngun, lc chn tn s hay cho

qua tn s no .T c mt trong mch lc th ng, mch lc tch cc, .T lin lc ni gia cc tng khuch i. T kt hp vi mt s linh kin khc tao nhng mch

dao ng,.Ngy nay cn c t nano tng dung lng b nh nhm p ng nhu cucng cao ca con ngi.

2.3. Cun cm

2.3.1. Cu to k hiu

Cun cm (inductor) / cun dy (coil)l dy dn qun nhiu vng lin tip trn 1 ci

li. Li ca cun cm c th l mt ng rng (li khng kh), st bi hay st l.Ty theo loi li, cun cm c cc khiu khc nhau:

li khng kh li st bi li st l

Hnh 2.26. K hiu ca cun cm.

2.3.2. H s t cm

H s tcm l i lng c trng cho kh nng tch tr nng lng t trng cacun cm.

K hiu: Ln v o: Henri (H)Milihenri: 1 mH = 10-3 H

Microhenri: 1 H = 10-6 HH s tcm ph thuc vo s vng dy, tit din, chiu di v vt liu lm li ca

cun cm.

L = 0rn2

l.S = 0r

d2

4(2.30)

0 = 4 .10-7 H/m

r: h s t thm tng i ca vt liu lm li i vi chn khng.n: s vng dyS: tit din li (m2)L: chiu di li (m)

d: ng knh ca li (m)

Mt khc , h s tcm cn tnh bi cngthc sau:

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36

I

nL

(2.31)

I: bin thin dng in (A)

: bin thin t thng (Wb)

Nng lng np vo cun cm

Dng in chy qua cun cm to ra nng lng tr di dng t trng.

WL =12

LI2 (2.32)

2.3.3. Mch tng ng ca cun cm

Ngoi h s tcm L, mt cun cm thc t cn c in tr tn hao (in tr nitip) RS, c khi k n in dung k sinh C nh mch tng ng hnh 2.23.

Hnh 2.27. Mch tng ng cha k in dung k sinh (a), k n in dung k sinh (b).

H s phm cht Q (Quality Factor):

S

L

R

X

Q (2.33)

RS: in tr ni tip ()

Cm khng l i lng c trng cho sc cn in ca cun cm.

XL: cm khng ()XL= L =2fL (2.34)

: tn s gc (rad/s)L: h s t cm (H)

f: tn s (Hz)2.3.4. Hin tng t cm

Nu dng in I chy trong mt cun cm thay i theo thi gian, th cun cm s tcm ng v sinh ra mt scin ng cm ng.

t

IL

t

ne

(2.35)

I: bin thin dng in (A)

: bin thin t thng (wb)t: khong thi gian bin thin (s)

RS L

(a)

RS L

C

(b)

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37

L: h s t cm (H)e: sc in ng cm ng (V)n: s vng dy qun ca cun cm.Sc in ng cm ng sinh ra dng in gi l dng in cm ng.

2.3.5. HcmKhi c hai hay nhiu cun cm th s thay i dng in trong mt cun cm s lm

t thng thay i, cc cun cm cn li phn ng bng cch sinh ra cc sc in ngcm ng. Khi ngi ta gi l c hin tng h cm gia cc cun cm.

K hiu: Mn v o: Henri (H)

V d: c hai cun cm L1, L2t gn nhau. Khi dng qua L1thay i, t trng sinhra t cun L1 lm nh hng n cun L2v ngc li. Nh vy c hin tng h

cm gia hai cun cm L1, L2. H s h cm:

21LLKM (2.36)

L1, L2: h s t cm (H)M: h s h cm (H)K: h s lin kt (h s ghp), 0 K 1

H s K ty thuc cch ghp. Nu hai cun dy cng qun trn mt li t th K = 1;hai cun dy t xa nhau, khng nh hng ln nhauhay c chn t gia hay t thng

gc vi nhau th K = 0.2.3.6. Cch mc cun cm

a. Mc ni tip

(2.37)

Hnh 2.28. Cun cm mc ni tipb. Mc song song

(2.38)

L2 Lt

L1

Lt = L1 + L2

21tL

1

L

1

L

1

L2

Lt

L1

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38

Hnh 2.29.Cun cm mc song song

2.3.7. Hin tng np x ca cuncm

Xt mch nh hnh 2.26, gi s cun cm cha tch tr nng lng in. Bt kha Ksang v tr s 1 cun cm pht sinh sc in ng cm ng bng ngun VDCnhng ngc

du chng li dngin do ngun VDC cungcp, do lc u dngin chy qua cun cm

bng khng. Sau dngin qua cun cm tng

ln theobiu thc sau:

)e-(1R

V

(t)i

t

DC

L

(2.39a)

R

L (2.39b)

l thi hng npxca cun cm.Ngc vi dng in, hiu in th

gia hai u cun cm lc u bngngun VDCnhng sau gim dn theo

biu thc:

t

DCL eV(t)v

(2.40)

Sau thi gian 5 th cun cm xemnh c np y, nu khng c tcng khc th hin tng vn khng thay

i.Khi cun cm np y ta bt kha K

sang v tr s 2. Dng in x c thay

i theo hm s m:

t

DCL e

R

V(t)i

(2.41)

Trong qu trnh x nng lng in th hiu in th gia hai u cun cm thay itheo biu thc:

t

DCL eV(t)v

(2.42)

Sau thi gian 5th cun cm s x ht nng lng in tch tr ca n.

2.3.8. Phn loi ng dngC nhiu cch phn loi cun cm:

1

K

L

2

VDC

+

R

Hnh 2.30. Mch kho st hin tng np -x ca cun cm.

0 2 3 4 5 t

VL (t)/ IL (t)

63

86

9598 99

37

145

2 1

Hnh 2.31. c tuyn np ca cun cm.

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39

Phn loi theo kt cu: cun cm 1 lp, cun cm nhiu lp, cun cm c li khngkh, cun cm c list bi, cun cm c li st l

Phn loi theo tn s lm vic: cun cm m tn,cun cm cao tn.

Cun cm 1 lp li khng kh: gm mt s vng dy qun vng n st vng kia hoc

cch nhau vi ln ng knh si dy. Dy c th cun trn khung bng vt liu cchin cao tn (gm; thy tinh; nha)hay nu cun cm cng th c th khng cn

khung m ch cn hai np gi hai bn.Loi dy s dng: dy ng thng (f > 50MHz) hay dy Litz (f < 2 MHz).

Cun cm nhiu lp li khng kh: khi tr s cun cm ln, cn c s vng dy nhiu,nu qun mtlp th chiu di cun cm qu div in dung k sinh qu ln. kchthc hp lv gim c in dung k sinh, ngi ta qun cc vng ca cun cm thnhnhiu lp chng ln nhau theo kiu t ong (kiu ton dng tin).

Cun cm c li st bi (bt st t): rt ngn kch thc ca hai loi trn bngcch lng vo gia n mt li ferit. Thn li c rng xon c. Hai u c kha 2 rnh.

Ngi ta dng mtci quay vt nha iu chnh li ln xung trong lng cun cm tng hay gim tr s incm ca cun cm.

Hnh dng li c dng hnh tr hay hnh xuyn. Tn s lm vic:100 kHz100 MHz.

Cun cm c li st ming (st l): dng dy ng trng men cch in, c qunthnh tng lp u n, vng n st vng kia, lp n st lp kia bng mt lt giy bngcch in. Li t l cc l thp Si , thp Si ht nh hng. Hnhdng li: dng ch E, I,

U, T, .Mi l thp c cch in bi lp ph rt mng oxit st, thp Si hoc varnis.Vt liu cch in lm tng in tr trong phn ct ngang ca li gim dng in xoy

nhng vn cho php mt t thng cao qua li.

Hnh 2.32. Mt s dng cun cm.

Hin nay, nh sn xut ch to nhiu loi cun cm c sn di dng linh kin dtr tac th dng ngay hoc p ng nhu cu ring ta thit k qun dy, chn licho cun cm.Sau y l ba cch qun dy tham kho, cch qun hnh (c) c indung k sinh nh hn hai cch kia.

Cun cm c ng dng lm micro in ng, loa in ng, rle, bin p, cun

dy trong u c a,.Trong mch in t, cun cm c th mch lc ngun, mchlc tn s, mch dao ng cng hng, mch to (chnh sa) dng sng, dng xung,

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40

Cuon so cap Cuon thu capI 2I 1

2.4. Bin th

2.4.1. Khi nim

Bin th (transformer) l dng c dng bin i in p hay dng in xoay chiunhng vn gi nguyn tn s.

2.4.2. Cu tok hiu

Cu to v hnh dng ca bin th nh hnh 2.29. Bin th gm 2 cun dy ng

trng men cch inqun trn mt li thp t khp kn: cun nhn in p vo gi lcun s cp, cun cho ly in p ra l cun th cp.Li t khng phi l mt khi stm gm nhiu l st mng ghp song song cch in nhau trnh dng in xoy(Foucoult) lm nng bin th.

Ngoi ra, li ca bin th c th l st bi hay khng kh.

Hnh 2.33. Cu toca bin th.

K hiu ca bin th nh hnh 2.30.

Hnh 2.34. K hiu bin th li khng kh (a), li st bi (b), li st l (c).

2.4.3. Nguyn l hot ng

Hnh 2.35. Cu to ca bin th.

Khi cho dng in xoay chiu c in th V1vo cun s cp, dng in I1s to rat trng bin thin chy trong mch t v sang cun dy th cp, cun th cp nhn

(a) (b) (c)

V1V2

I1 I2

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c t trng bin thin s lm t thng qua cun dy thay i, cun th cp cm ngcho ra dng in xoay chiu c in th V2.

V1 = - N1t

(2.43)

V2 = -N2t (2.44)

N1: s vng dy ca cun s cp.N2: s vng dy ca cun th cp.

V1: in p vo hai u cun s cp.V2: in p ly ra hai u cun th cp.

: bin thin t thng (wb)

t: khong thi gian bin thin (s)

2.4.4. Cc cng thc ca bin th

T l v in th V2V1

=N2N1

(2.45)

T l dng in: I2I1

=N1N2

(2.46)

T l v cng sut: P1 = V1 I1 ; P2 = V2I2

L tng ta c: P1 = P2 (2.47a)

V1.I1 = V2 . I2

Thc t: P2 < P1 (2.47b)

Hiu sut: = P2P1

. 100% (2.47c)

T l v tng tr: R2 =V2I2

; R1 =V1R1

R1R2

=

N1

N2

2(2.48)

2.4.5. Phn loi - ng dng

Da theo tn s lm vic: bin th m tn, bin th trung tn,bin th cao tn.

Da theo cu to: bin th c li st l, bin th c li st bi, bin th c li khng

kh,

Da theo mc ch s dng: bin th ngun, bin th loa, bin thxut m, bin thxung, bin th o pha,

ng dng ch yu ca bin th l lm thay i in th, dng in theo yu cu thct.

Bin th cng hng l bin th cao tn, cun s cp hoc cun th cp c mc

song song vi mt t in, hnh thnh mch cng hng. Nu c hai cun u c mc t

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in th ta c bin th cng hng kp. Li ca bin th cng hng lm bng ferrite cth iu chnh c.Mt s bin th cng hng dng tn scao hn c li khng kh.

Hnh 2.36. Hnh dng ca bin th.

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CU HI N TP

1. in tr l g? Hy k tn mt s loi in tr v ni vi ng dng ca n. Nu vicch c tr s in tr.

2. in tr ca dy dn l g? Nu cng thc tnh v cho bit tn, n v ca cc ilng trong cng thc.in tr ca dy dn ph thuc vo nhng yu t no ca dy?

3. Tm hiu nh lut Ohm, nh lut Kirchhoff, ng dng ca n. 4. in tr c my cch mc c bn? Hy k tn v v on mch tng ng gm hai

in tr.Vit biu thc quan h gia cc i lng I, U, R trong on mch.Nu nhnxt.

5. T in l g? Hy k tn mt s loi t in v ni vi ng dng ca n. Nu vicch c tr s in dung.

6. in dung l g? Nu cng thc tnh v cho bit tn, n v ca cc i lng trong

cng thc.in dung ph thuc vo nhng yu t no ca t in?7. Dung khng l g? Nu cng thc tnh v cho bit tn, n v ca cc i lng

trong cng thc.Dung khng ph thuc vo nhng yu t no?8. T in c my cch mc c bn? Hy k tn v v on mch tng ng gm hai

t in. Vit biu thc quan h gia cc i lng Q, U, C trong on mch. Nu nhn

xt.

9. Trnh by hin tng np x ca t. Vit biu thc tnh dng in, hiu in th

gia hai u t trong qu trnh np, x ca t in. Nu nhn xt.

10. Vit cng thc tnh nng lng tch tr vo t. cho bit tn, n v ca cc ilng trong cng thc.

11. Cun cm l g? Hy k tn mt s loi Cun cm v ni vi ng dng ca n.Nu vi cch c tr s in cm.

12. H s t cm l g? Nu cng thc tnh v cho bit tn, n v ca cc i lngtrong cng thc.H s t cm ph thuc vo nhng yu t no cacun cm?

13. Cm khng l g? Nu cng thc tnh v cho bit tn, n v ca cc i lng

trong cng thc.Cmkhng ph thuc vo nhng yu t no?

14. Trnh by hin tng np x ca cun cm. Vit biu thc tnh dng in, hiuin th gia hai u cun cm trong qu trnh np, x ca cun cm. Nu nhn xt.15. Vit cng thc tnh nng lng tch tr vo cun cm. cho bit tn, n v ca cc

i lng trong cng thc.16. So snh qu trnh np x ca t vi qu trnhnp x ca cun cm.

17.Bin th l g? Hy k tn mt s loi bin th v ni vi ng dng ca n.18. Nu nguyn l hat ng ca bin th.19. Cho bit cc cng thc ca bin th. 20. Hy k tn v v k hiu ca nhng linh kin hc.

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Chng 3: Cht bn dn - diode

44

Chng 3

CHT BN DN DIODE

3.1. Cht bn dn

3.1.1. Khi nim

S dn in ca mt cht ty thuc vo s in t (electron) nm lp v ngoi cngca nguyn t. Da trn c s ny ngi ta xc nh s dn in ca mtcht nh sau:

- Cht dn in (conductor) l mt cht c s in t lp ngoi cng t hn rtnhiu so vi s in t bo ha ca lp .

- Cht cch in (insulator) l mtcht c s in t lp ngoi cng bng hoc gn

bng s in tbo ha ca lp .- Cht bn dn (semiconductor) l mt cht c s in t lp ngoi cng nm

khong gia hai loi trn.

Ngoi ra, ngi ta c th phn bit chtdnin, cht cch in, cht bn dn, datheo khi nimin tr sut, in dn sut,. C th ni cht bn dn c dn innm khong gia kim loi v cht cch in.Ta c th iu chnh, thay i dn in

ca cht bn dn.

Cht bn dn dng nguyn t c tm thy trong nhm IV ca bng h thng tunhon. Loi tiu biu ca ngnh in t: Silicium (Si), Germanium (Ge).

Cht bn dn dng hp cht c to thnh bng cch kt hp cc nguyn t nhm

III v V, II v VI, c loihp cht gm ba hay bnnguyn t. V d: AlGaAs, GaAsP,AlGaAsSb, GaInAsP. Trng hp c bit dng hp cht nhm IV: SiC, SiGe.

- Hp cht gm hai nguyn t III v V: AlAs, AlP, AlSb, GaAs, GaP, GaSb, InAs,InP, InSb.

- Hp cht gm hainguyn t II v VI : CdSi, CdTe, HgS, ZnS, ZnTe.

3.1.2. Bn dn thun

- Khi nim: Bn dn thun l bn dn duy nhtkhng pha thm cht khc vo.

- S dn in ca bn dn thun :

Xt bn dn tinh khit Si, Si c 4 in t lp

ngoi cng, 4 in t ny s lin kt vi 4 in t cabnnguyn t k cn n, hnh thnh mi lin kt gi

l lin kt cng ha tr. nhit thp cc lin kt bn vng nn tt

Si

Si

SiSi Si

Hnh 3.1. Cu trc tinh th Si.

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c cc in t b rng buc trong mng tinh th, do Si khng dn in.

nhit tng i cao hoc c cung cp nng lng di dng khc: chiu nh

sng, mt trong nhng mi lin kt b ph v, in t thot ra tr thnh in t t do,li trong mng tinh th mt chtrng thiu in t gi l l trng, l trng mang in

tch dng. Nhit cng cao th s in t t do v l trng hnh thnh cng nhiunhng mt ca chng (nng trong mt n v th tch) l bng nhau v thng khiu ni = pi (3.1)

Khi khng c in trng th in t t do v l trng chuyn ng nhit hn lonkhng u tin theo phng no nn khng c dng in.

Khi c in trng t vo tinh th bn dn, di tc dng ca lc in trng int v l trng chuyn ng c hng:in t chuyn ng ngc chiu in trng, ltrng chuyn ngcng chiu in trng lm xut hin dng in trong bn dn.

Nh vy, dng in trong bn dn thun l dng chuyn di c hng ca in t tdo v l trng di tc dng ca in trng.

3.1.3. Bn dn tp cht

Bn dn tp cht l bn dn c pha thm cht khc vo. Ty vo cht khc l cht no

m c hai loi bn dntp cht:bn dn loi N vbn dn loi P.

a.Bn dn loi N

Pha thm mt lng rt t phosphore (P) vo

cht bn dn Si theo t l8

10

1, s dn in ca Si

tng ln 10 ln. P l cht nhm V, c 5 in t lp ngoi cng. Bnin t ca nguyn t P linkt vi 4 in t ca bn nguyn t Si khc nhaunm cn n. Nh vy, P cn tha li mt in tkhng nm trong lin kt ha tr. in t tha nyrt d dng tr thnh in t t do, nguyn t tp

cht P khi b ion ha v tr thnh mt iondng. Nu c in trng p vo, cc ht dn tdo s chuyn ng c hng, to nn dng in. Nu pha cht P cng nhiu th dn

in ca bn dn Si cng tng ln.

Tp cht nhm V cung cp in t cho cht bn dn cbn nn c gi l tp cht

cho (donor). Cht bn dn c pha thm tp cht nhm V gi l bn dn loi N(Negative).

Nu gi Ndl nng tp cht cha trong mt n v th tch cht bn dn c bn thkhi c cung cp nng lng y , ton b cc nguyn t tp cht b ion ha.

Nng in t t do do tp cht cung cp l:

Si

Si

SiSi

P

Hnh 3.2. Bn dn loiN.

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nd = Nd (3.2)

Ngoi s in t t do nh tp cht cung cp, cht bn dn c bn vn c qu trnhsinh ra cc cp in t - l trng do tc ng ca nhit (hoc nh sng,)ging nh

bn dn thun. Vy tng nng in t t do trong cht bn dn loi N l:

nn = Nd + pn (3.3)pnl nng l trng trong bn dn loi N. nn > pnnn bn dn loiN c ht ti dn

in a s l in t, ht ti dn in thiu s l l trng. C trng hp ngi ta b quavai tr ca ht ti dn in thiu s, ly gn ng i vi bn dn loi N l:

nn Nd (3.4)

b. Bn dn loi P

Pha thm mt lng rt t Bore (B) vo cht bn

dn Si theo t l 810

1

, s dn in ca Si tng ln

hn10 ln. Bl cht nhm III, c 3 in t lpngoi cng. Ba in t ca nguyn t B lin kt vi3 in t ca ba nguyn t Si k cn n. Nh vy,B cn thiu mt in t cho lin kt cui cng. Nd dng nhn thm mt in t ca nguyn t gnn c ngha l ch cn mt kch thch n (nhit ,

nh sng) l mt trong nhng in t ca cc mi

lin kt hon chnh bn cnh s n th vo milin kt th t (mi lin kt thiu in t trn). Nguyn t tp cht lc tr thnh ionm, iu ny lm pht sinh mt l trng.Nh vy, c c mt nguyn t tp cht th cthm mt l trng, nng tp cht cng cao th s l trng cng nhiu. Nu c in

trng p voth cc l trng ny s tham gia dn in.

Tp cht nhm IIItip nhn in t tcht bn dn c bn sinh ra cc l trngnn c gi l tp chtnhn (acceptor). Cht bn dn c pha thm tp cht nhm IIIgi l bn dn loi P (Positive).

Nu gi Nal nng tp cht cha trong mt n v th tch cht bn dn c bn thkhi c cung cp nng lng y , ton b cc nguyn t tp cht b ion ha.

Nng in t t do do tp cht cung cp l:

pa = Na (3.5)

Ngoi s l trng dotp chtto ra, trong cht bn dn c bn cng c qu trnh sinhra cc cp in t - l trng do tc ng ca nhit (hoc nh sng,) ging nh bn

dn thun. Vypp l tng nng l trng trong cht bn dn loi P; npl nng in

t trong bn dn loi P. Ta c:

pp = Na + np (3.6)

Si

Si

SiSi

In o

Hnh 3.3. Bn dn loi P.

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Ta thy pp > npnn bn dn loi P c ht ti dn in a s ll trng, ht ti dnin thiu s l in t. C trng hp ngi ta b qua vai tr ca ht ti dn in thius, ly gn ng i vi bn dnloi P l:

pp Na (3.7)

3.1.4. Mi ni PN

*Chuyn ng biu kin ca l trng.

Gi s in t ti v tr s 1, l trng v tr s 2, in t dch chuyn t 1 sang 2 li bn 2 in t v bn 1 l trng.Nh vy, in t dch chuyn t 1 sang 2 cn ltrng c xemnh dch chuyn t 2 sang 1. Sdch chuyn ca l trng gi l chuynng biu kin ca l trng. iu ny cho tathy in t v l trng chuyn ng ngcchiu nhau, in t di chuyn t m sang dng, ngc li l trng di chuyn t dngsang m.

Sau khi hnh thnh mu bn dn loi P, N; cho hai mu bn dn ny tip xc vinhau. Ta c mt lp tip xc P N (mi ni P - N). Ti ni tip xc P - N c hintng trao i in tch. in t t vng N khuch tn sang vng P v ngc li l trngt vng P khuch tn sang vng N. S dch chuyn ny to ra dng thun (dng khuchtn) iFc chiu t P N.

Hnh 3.4. Mi ni P N.

Ti ni tip xc in t v l trng ti hp nhau, bn vng P s tn ti in tch m(ion m), bn vng N s tn ti in tch dng (ion dng) tn ti mt in trngtrong (in trng ni ti) to ra dng in nghch(dng in tri) iN. iNngc chiu viiF. Khi iN = iFth s khuch tn ca cc ht ti a s ngng li.

Vng cn mt tip xc gi l vng him (vng khim khuyt). trng thi cn bng,

hiu in th tip xc gia bn dn P v bn dn N c mt gi tr nht nh V. Hiu thny ngn cn, khng cho ht ti (ht dn) tip tc di chuyn qua mt ranh gii , duy trtrng thi cn bng, nn c gi l hng ro in th.

Bn dn chnh (bn dn c bn) loi Si c V = 0,6 VGe c V = 0,2 V

3.2. Diode bn dn

3.2.1. Cu to k hiu

Diode bn dn (semiconductor diode) l dng c bn dn c mt mi ni P-N. T mubn dn lai P tip xc kim loi a chn ra (cc ra) anode (A: cc dng). Mu bn dn

P NP N

+++

---

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lai N tip xc kim loi a chn ra cathode (K: cc m). Bn ngoi c bc bi lpplastic.

C nhiu cng ngh ch to: cy ion, khuch tn cht kch tp vo bn dn c tpcht loi ngc li, ko lp epitaxy,.

V d: Mt diode c th to ra bng cch bt u vi mu bn dn loi N c pha tpcht Ndv chuyn i c chn lc mt phn ca mu bn dn thnh loi P bng cch

thm cc tp cht nhn in t c N a > Nd. im m vt liu thay i t loi P sang loiN c gi l tip xc luyn kim (mi ni luyn kim) (metallurgical junction). Mu bndn loi P tip xc kim loi a ra cc anode (A). Mu bn dn loi N tip xc kim loia ra cc Cathode (K).

A: Anode: cc dngK: Cathode: cc m

Hnh 3.5. Cu to (a), k hiu (b) ca diode.

3.2.2. Nguyn l hat ng

Ta c th cp in diode mt trong nhng trng thi sau:VA > VK: VAK> 0: diode phn cc thun.VA = VK: VAK= 0: diode khng phn cc.VA < VK: VAK< 0: diode phn cc nghch.

a.Phn ccthun

Phn cc thun diode: ta ni A vi cc dng ca ngun, K vi cc m ca ngun.

in tch mca ngun y in t trong N v lp tip xc. in tch dng ca

ngun y l trng trong P v lp tip xc, lm cho vng khim khuyt cng hp li. Khilc y ln th in t t vng N qua lp tip xc, sang vng P v n cc dng cangun.Lc y ln l lc diode c VAKt gi tr V, lc ny diode c dng in

chy theo chiu t A sang K.

V c gi l in th ngng (in ththm, in th m).

i vi loi Si c V= 0,6 V (0,7 V); Gec V= 0,2 V.

b. Phn cc nghch

P NCathodeAnode

(a)

A K

(b)

+ -

ooo

ooo

+++

---

VDC

Hnh 3.6. Mch phn cc thun diode.

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Phn cc thun diode: ta ni A vi cc m ca ngun, K vi cc dng ca ngun.

in tch m ca ngun s ht l trng ca vng P,in tch dng ca ngun s ht

in t ca vng N, lm cho in t v l trng cng xa nhau hn. Vng khim khuytcng rng ra nn hin tng ti hp gia in t v l trng cng kh khn hn. Nh

vy, s khng c dng qua diode. Tuy nhin, mi vng bn dn cn c ht ti thiu snn mt s rt t in t v l trng c ti hp to nn dng in nh i t N qua P gil dng nghch (dng r, dng r). Dng nyrt nh cvi nA. Nhiu trng hp coi nhdiode khng dn in khi phn cc nghch.Tng in p phn cc nghch ln th dng

xem nh khng i, tng qu mc th diodeh (b nh thng). Nu xt dng in r thdiode c dng nh chy theo chiu t K v A

khi phn cc nghch.

c.Khng phn cc:

Khi ta dng ngun VDCiu chnh c v chnh v 0, lc mch c VA = VK=0 hay VAK= 0 hoc trng hp khc VA = VK 0 nhng VAKvn bng 0. Lc ny diodekhng c phn cc. V khng c s chnh lch in th nn khng c s dch chuynca cc ht ti nn khng c dng in.

3.2.3. c tuyn Volt Ampe

IS: dng nghchbo ha.V: in th ngng.

VB: in th nh thng.k: hng s Boltzman, k= 1,38.10-23 J/0KT: nhit tuyt i ca cht bn dn,

nhit thng T = 3000K.

q

kTT = 0,025 V 0,026 V = 26 mV (3.8)

1eII 0,026V

SD

D

(3.9a)

Phn cc thun: VD >0 0,026VD

e 1 0,026V

SD

D

eII (3.9b)

Khng phncc: VD = 0 0,026VD

e =1 ID = IS (11) = 0 (3.9c)

Phn cc nghch: VD < 0 0,026VD

e 1 ID = IS (1) = -IS (3.9d)

Du (-) ch chiu dng in qua diode khi phn cc nghch ngc vi chiu dngin qua diode khi phn cc thun.

+-

o +++

---

VDC

oo

oo

o

Hnh 3.7. Mch phn cc nghch diode.

VB

ID

0

V VDIS

Hnh 3.8. c tuyn Volt Ampe.

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3.2.4. in tr diode

C hai loi in tr lin quan n diode:- in tr tnh: in tr i vi dng in mt chiu.

D

DD

I

VR (3.10)

Khi diode c phn cc thun c dng ln chy qua diode nn in tr thun nh.Khi diode c phn cc nghch c dng r nh chy qua diode nn in tr thun

ln.Ngi ta li dng c tnh ny o kim tra diode bng my o V.O.M.

in tr thun v in tr nghch ca diode ph thuc vo cht bn dn lm diode lGe hay Si theo bng sau:

in tr thun in tr nghch

Diode Si Vi vi trm k

Diode Ge Vi vi M

Bng 3.1. in tr ca diode.

Kt qu:

in tr thun = in tr nghch = 0 th diode b nh thng.in tr thun = in tr nghch = th diode b t.in tr thunngnhng in tr nghch gim xung kh nhiu th diode b r,r

khng dng c.in tr thun, in tr nghch ng nh bng trn th diode tt.in tr ng: in tr i vi tn hiu xoay chiu.

DD

Dd

I

0,026

i

vr (3.11)

Ngoi ra, i vi diode l tng: nu n c phn cc thun th khng c in tr vnu n c phn cc nghch th c in tr v cc. Vy diode l tng c xem nhcng tc (ON hay OFF) ph thuc vo cc tnh ca in p t vo diode.

Mch tng ng ca diode i vi tn hiu xoay chiu nh hnh 3.15.

Hnh 3.9. Mch tng ng ca diode i vi tn hiu xoay chiu.

rd

r1

Ct

(a)

rd

Ct

(b)

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r1: in tr ca hai cht bn dn (ngoi vng him), thng b qua. rd: in tr ng(in tr vi phn): in tr i vitn hiu xoay chiu.

DD

Dd

I

0,026

i

vr (3.12)

Ct: in dung tng ng ca diode gm in dung mi ni Cjv in dung khuchtn Cd.Ct = Cj + Cd (3.13)

Tr s Ct thay i ph thuc in p t vo diode. Vi tn hiu tn s thp, nhhng ca Ctc th b qua. Nhng vi tn hiu tn s cao th nh hng ca Ctl ng

k. Chnh in dung ny lm gim tr khng theo chiu nghch tn s cao, lm xu ctnh chnh lu ca diode v lm chm tc ng m khi dng diode nh kha in t.

3.2.5. Phn loi

Nh bit diode c bn l mt mi ni P N nhng c th da theo kt cu, datheo cng dng m ta phn bit cc loi diode nh sau:

Da theo kt cu lp tip xc P N

C hai loi: diode tip im v diode tip mt.Diode tip im: l diode c mt tip xc gia hai lp bn dn P N rt nh gn nh

mt im (th tch rt nh) c bc bi lp v thy tinh.Dng in nh mc rt b(khong vi chc miliampe), in p ngc khng vt qu vi chc volt.

Diode tip mt: l diode c mt tip xc gia hai lp bn dn P N l mt mtphng, lp v bn ngoi l nha. Dng in nh mc kh ln (khong vi trm

miliampe n vi trm ampe), in p ngc t n vi trm volt.

Da vo cng dng

Diode chnh lu: Hnh dng to, thucloi tip mt, hat ng tn s thp. Diodechnh lu dng i in xoay chiu sangin mt chiu. y l loi diode rt thng

dng, thng c bc nha mu en, cvch trng nh hnh 3.10.

Khi dng cn quan tm hai thng s: in p ngc cc i v dng thun ti a cadiode, c th mc ni tip tng in p ngc, mc song song tng dng chu ng.

Diode tch sng: hnh dng nh thuc loi tip im, hotng tn s cao. Cnglm nhim v nh diode chnh lu nhng ch yu l vi tn hiu nh v tn s cao.Diode ny chu dng t vi mA n vi chc mA. Thng l loi Ge.

Diode xung l diode dng trong cc mch c tc chuyn trng thi rt nhanh vn c tn s hat ng cao hn nhiu so vi diode thng.

Hnh 3.10.Hnh dng diode chnh lu.

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Cc my in t hin i thng dng b ngun cung cp in theo kiu ngt m(switching), to ra dng in xoay chiu dng xung c tn s kh cao, ti vi chc ngnHz. Sau dng diode xung chnh lu thnh in DC cung cp cho my. Trong int s, ta c th dng diode xung lm cc chuyn mch in t hai trng thi: dn khi

phn cc thun, ngng (tt) khi phn cc nghch.

Hnh dng diode xung cng tng t diode thng, mun phn bit ta phi dng schtra cu tra.

Cc thit b xung cn dng loi khc gi l diode Schottky. Loi ny c cu to hikhc so vi diode thng, tc chuyn trng thi ca n rt cao.

Diode zener: c cuto ging diode thng nhng cht bn dn c pha tp chtvi t l cao hn v c tit din ln hn diode thng, thng dng bn dn chnh l Si.

Hnh 3.11. K hiu ca diode zener.

c tuyn volt ampe trong qu trnhnh thng gn nh song song vi trcdng in, ngha l in p gia A v Kgn nh khng i. Ta li dng u imny dng zener lm phn t n nhin p.

Hnh 3.12. c tuyn volt ampe ca diode zener.

Lu : Diode zener dng n pkhi c phn cc nghch. Khi phncc thun diode zener ging diodethng.

Cc nh ch to thay i nng

tp cht to ra cc loi diodezener c gi tr n p Vz khc nhau,v d: 5 V; 6 V; 6,8 V; 7,5 V;

Hnh 3.13 l mch n p n gin c in p ra trn ti Vt = Vzl mt tr s khngi trong khi in th ngun cung cp VDCthay i. Tuy nhin cn khi VDC < Vz thmch cha n p, VDC = Vzth zener mi bt u ghim p.

Diode quang - diode cm quang (photodiode)c cu to bn dn ging nh diodethng nhng t trong v cch in c mt mt lnha hay thy tinh trong sut

nhn nh sng bn ngoi chiu vo mi ni P-N ca diode, c loi dng thu knh hi t tp trung nh sng.

VZ

ID

0

V VDIS

+VDC

V z = V t

Z

R

Rt

Hnh 3.13. Mch n p n gin.

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Hnh 3.14. Cu to ca diode quang.

K hiuca diode quang nh hnh 3.15:

Hnh 3.15.K hiu ca diode quang

Qua th nghim cho thy khi photodiode c phn cc thun th hai trng hp mini PN c chiu sng hay che ti dng in thun qua diode thay i t. Ngc lidiode b phn cc nghch, mi ni P N c chiu sng th dng in nghch tng lnln hn nhiu ln so vi khi b che ti. Do nguyn l trn nn diode quang c s dng

trng thi phn cc ngc trong cc mch iu khin nh sng.Diode pht quang: LED (Light Emitting Diode)

Hnh 3.26. K hiu (a), hnh dng (b) ca LED.

Diode pht quang c cu to gm mtmi ni PN, tip xc kim loi a ra cc A(Anode), K (cathode). Diode pht quang c lm t cc cht GaAs, GaP, GaAsP,SiCDiode pht quang l diode pht sng khi c dng chy qua n. Diode ny c th

pht ra nhiumu sc khc nhau.

- Diode GaAs cho ra nh sng hng ngoi m mt nhn khng thy c, n c s tihp vng dn vng ha tr l trc tip. Bc x pht sinh ch yu l qua s ti hp.

Nng lng photon khong 1,4eV.

A K

(a) (b)

PHOTODIODE

KA

A K

nh sng

chiu vo

P

SiO2 Vng him

N

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- Diode Ga AsP vi s ti hp trc tip v nng lng ln hn 1,7eV cho ra nh sngkh kin, khi thay i hm lng photpho s cho ra nh sng khc nhau nh , cam,

vng.

- Diode GaP pha thm tp cht (Nit v ZnO) s cbc x cho ra nh sng. Ty loi

tp cht m diode c th cho ra cc mu t , cam, vng, xanh l cy.- Diode SiC khi pha thm tp cht s cho ra nh sng mu xanh da tri. LED mu

xanh da tri cha ph bin v gi thnh cao.

Do khc nhau v vt liu ch to nn in p ngng ca cc loi LED cng khcnhau.

LED c V = 1,6 V 2 V

LED cam c V = 2,2 V 3 V

LED xanh l c V = 2,7 V 3,2 V

LED vng c V = 2,4 V 3,2 V

LED xanh da tri c V = 3 V 5 V

LED hng ngoi c V = 1,8 V 5 V

Tng t diode thng, LED cng c batrng thi:

VAK> 0: LED c phn cc thun.VAK= 0: LED khng c phn cc.VAK< 0: LED c phn cc nghch.

LED ch pht sng trong trng hp dn in (cho dng chy qua) khi n c phncc thun v VAKnm trong khong mc ngng cho php caLED. Nhng trng hp

cn li LED tt.

Lu :c tuyn volt ampe ca LED tng t c tuyn volt ampe ca diodethng nhng khong mc ngng cho php ca LED ty loi LED v mc ngng nyln hn mc ngng ca diode thng. in p nghch ti aca LED tng i thp.

Khi dng thng mc in tr ni tip vi LED hn dng qua LED.

LED hai mu

LED hai mu l loi LED i gm hai LED nm song song v ngc chiu nhau,trong c mt LED v mt LED xanh l cy

hay mt LED vng v mt LED xanh l cy.

Loi LED hai mu thng ch cc tnh cangun hay chiu quay ca ng c.

K hiu LED i loi hai munh hnh 3.17.

Nu chn A1c in p sao cho v nmtrong khong mc ngng cho php th LED1

A2A1

LED1

LED2

Hnh 3.17. K hiu LED hai mu.V > 0A1A2

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sng v ngc li nu chn A2 c in p sao cho v nm trong khong mcngng cho php th LED2 sng.

Tng qut:

- Khi ch c dng qua LED1th LED sng mu ca LED1.

- Khi ch c dng qua LED2th LED sng mu ca LED2.- Khi khng c dng qua hai LED th LED tt.

LED ba mu

LED ba mu cng l loi LED i nhng khng ghp song song m hai LED ch cchung cc cathode, trong mt LED ra chnngn, mt LED mu xanh l cy ra chn di, chngia l cathode chung.

K hiu LED i loi ba mu nh hnh 3.18.Nuchn A1 c in p dng th LED sng, nu chnA2c in p dng th LED xanh sng, nu chn A1

v A2c in p dng th haiLED u sng v chora nh sng mu vng.

Tng qut:

- Khi ch c dng qua LED1th LED sng mu ca LED1.- Khi ch c dng qua LED2th LED sng mu ca LED2.

- Khi c dng qua hai LED th LED sng mu pha ca mu LED 1 v mu LED2.- Khi khng c dng qua hai LED th LED tt.

Mt s mch ng dng ca LED

Mch bo ngun DC

Khi s dng LED iu quan trng l phi tnh in tr ni tip vi LED c tr s

thch hp trnh dng in qua LED qu ln s lm h LED.

in tr trong mch bo ngun DC c tnh theo cng thc:

R =LED

LEDDC

I

VV

Hnh 3.19.Mch bo ngun DC.

A1

A2

LED1

LED2

Hnh 3.18.K hiu LED ba mu.

D1

Rt

LED

C

D2

VDC

VAC

3 6

5

1 4

R

V < 0A1A2

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Mch bo ngun AC

Hnh 3.20.Mch bo ngun AC.

Trong mch bo ngun AC, LED ch sng khic phn cc thun bng bn k thchhp, khi LED b phn cc nghch th diode D c phn cc thun nn dn in gicho mc in p ngc trn LED l VD = 0,7V trnh h LED.

in tr trong mch bo ngun AC ctnh theo cng thc:

R =LED

LEDAC

I

VV (3.14)

LED c ng dng nhiu trong cc mch in t: mch bo v thit b, mch quangbo, mch n trang tr, mch chi, mch kim sot in p cho xe hi,.c bitLED c tch hp thnh nhiu dngn rt p v tin li. Hnh 3.21 l mt dngbngn ng dng LED.Tui th ca LED cao hn bng n thng, ty loi LED m ta cctrng chiu sng khc nhau.

Hnh 3.21. Dng bng n ng dng LED.

Hnh 3.22. Ma trn LED.

Ngoi ra, LED pht ra tia hng ngoi (IRED) dng truyn tn hiu trong cc bghp quang, c tn hiu, mch iu khin t xa,

LED by an

LED byon c loi anode chung v loi cathode chung. Hin nay LED by onc dng nhiu trong cc thit b hin th s.

R

VACLED

D

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Hnh 3.23Mch tng ng vi cu to ca LED loi K chung (a), A chung (b).

Hnh 3.24. Hnh dng ca LED by on.

LED by on l tp hp tm LED c ch to dng thanh di sp xp nh hnh3.23 v c k hiu bng tm ch ci l a, b, c, d, e, f, g, p. Phn ph ca LED by onl mt chm sng p ch du phy thp phn. Du chm ny l mt LED p tng ngc pht sng. Khi cho cc thanh sng vi cc s lng v v tr thch hp ta c nhng

ch s t 0 n 9v nhng ch ci t A n F.

Diode bin dung (Varicap)

Hnh 3.25. K hiu diode bin dung.

Diode bin dung (Varicap) l loi diode c in dung k sinh thay i theo in p

phn cc.Cu to diode ti mi ni P-N c hng ro in th lm cho in t ca vng N

khng sang c vng P. Khong cch ny coi nh mt lp cch in c tc dng nhin mi trong t in v hnh thnh t in k sinh, k hiu CD. in dung CDc tr scng c tnh theo cng thc :

d

SC

D (3.15)

Trong : : hng s in mi.

S: tit din mi ni.d: b dy lp in mi thay i theo hiu in th VD.

A K

ca

GND

b d e g p

(a)

db e

+Vcc

a gc p

(b)

d

g

f b

e

P

c

a

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Diode bin dung c dng ch yu trong cc mch cng hng vi vai tr l mt tin bin i theo in p iu chnh tn s cng hng ca mch. V d: trong cc

b tuner ca TV, b iu hng ca my radio,.

Thc t, khi dng ta cn lu :

- Loi diode.- Dng thun ti a ca diode.

- in p ngc ti a m diode chu c.- c bit vi loi diode zener ta cn xem in p ghim Vz.

Trn thn diode thng c ghi mt s k hiu di dng ch s hay vng mu. Ta cth c trc tip hoc tra cu bit c vi thng s ca diode trc khi s dng n.

V d:DZ5.6 VZ = 5,6 VDZ9.1 VZ =9,1 V

3.2.6. Mch chnh lu

a. Mch chnh lu bn k

Xt mch nh hnh 3.26, bin thdng gim in p xoay chiu xung

tr s thch hp.

Gi s bn k u ti A l bn kdng, D c phn cc thun nn dn

in, c dng ILqua ti vi chiu t trn hng xung, v cho ra in th trn ti VDCdng bn k dng gn bng VA. Bn k k tip ti A l bn k m, D phn cc nghchnn khng c dng hay dng qua ti bng khng v VDC = 0.

Gi tr trung bnh ca in p ra:

)(sin2

12

0

0 ttdUV

(3.16)

Hnh 3.27. Dng sng vo, ra ca mch chnh lu bn k.

b. Mch chnh lu ton k

-

t

+ + +

- -t

+ + +

VA

VDC C t lc

DA

RLVAC

Hnh 3.26. Mch chnh lu bn k.

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Dng hai diode

Xt mch nh hnh 3.28. Mchdng bin p o pha, cun thcp c ba u ra, im gia chiacun th thnh hai na cun bng

nhau. iu ny gip cho diode D1v D2 lun phin dn in trongmi bn k, c th l: gi s bn k

u ti A l bn kdng, tngng ti B l bn k m. Ta c D1dn in, D2ngng dn, cp dng qua ti c chiu t

trn hng xung to hiu in th VDCgia 2 u ti. Bn k k tip A l bn k m,tng ng ti B l bn k dng. Ta c D1ngng dn, D2 dn in, cp dng qua ti cchiut trn hng xung, to ra VDC.

Hnh 3.29. Dng sng vo, ra ca mch chnh lu ton k.

Gi tr trung bnh ca in p ra:

)(sin2

2

00

ttdUV

(3.17) Dng cu diode

Xt mch nh hnh 3.30. Gi sbn k u ti A l bn k dng thta c D1 v D3dn in, cp dngqua ti c chiu t trn hngxung. D2 v D4ngng dn. Bn kk tip ti A l bn k m th ta cD1 v D3ngng dn, D2 v D4dnin, cp dng qua ti c chiu t

-

t

+ + +

- -

t

VA

VDC

+ + + + + +

C t lc C

D1A

VAC

D2

RL

Hnh 3.28. Mch chnh lu ton k dng hai diode.

D2

A

RL

VAC

D1

D3

D4

Hnh 3.30. Mch chnh lu ton k dng cu diode.

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trn hng xung.

Dng sng vo, ra ca mch nh hnh 3.29.Nh vy, nhng mch trn c in p ra trn ti l in p mt chiu cn b nhp

nhy. gim bt nhp nhy, nng cao cht lngra ta mc thm t lc C song song vi

ti.c. Chnh lu m dng

Mch dng bin p o pha v cudiode.

C1 v C2l 2 t lc ngun.Ng ra l hai ngun in p mt

chiu i xng VCC.

d. Mch nhn p

Mch c tc dng chnh lu v nngcao c in pra ln 2, 3, n ln in p nh ca ngun xoay chiu.

Mch chnh lu tng i in thkiu Schenbel

Hnh 3.32.Mch chnh lu nhn i in p kiu Schenbel.

Gi s bn ku ti A l bn km, tng ng ti B l bn k dng, D1dn in,D2ngng dn, dng in chy t dng qua D1np vo t C1mt hiuin th VDC ccc tnh nh hnh v bn k k tip ti A l bn k dng, ti B l bn k m, D1ngngdn, D2dn in vi in th p vo D2gm: in th t C1ni tip vi in th xoay

chiu bn kdng. Nh vy D2dn np vo t C2mt hiu in th l 2VDCcp incho ti.

Mch chnh lu tng i in thkiu Latour

Gi s ti A l bn kdng, D1dn in, D2ngng dn, dng in qua D1np vot C1mt hiuin th l U2. Bn k k tip ti A l bn k m, D1ngng dn, D2dnin, dng in qua D2np vo t C2mt lng in th VDC. Nh vy c chu k inxoay chiu vo, in th mt chiu ng ra gm hiu in th gia hai u t C1cngvi hiu in th gia hai ut C2c np t C3. N chnh l 2VDC cp in cho ti.

C1

C2

D2

D1

V0 = 2VDCVAC

Hnh3.31.Mchchnh lu m dng-V

CC

D2

A

VAC

D1

D3

D4

+VC

C1

C2

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Hnh 3.33.Mch chnh lu nhn i in p kiu Latour.

VACV0 = 2VDC

D1

D2

C1

C2

C3

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CU HI N TP

1. Hy phn bit cht cch in, cht bn dn, cht dn in. Cho v d.2. Bn dn thun l g? Nu s dn in ca bn dn thun.

3.

Bn dn tp cht l g? C my loi? K tn v nu c trng ca n.4. Hy gii thch c ch dn in ca cht dn in, cht cch in, cht bn dn,bn dn loi N, bn dn loi P, mi ni P N theo l thuyt vng nng lng.

5. Diode bn dn l g? Nu nguyn l hotng ca n. Cho bit iu kin n

dn in, iu kin n ngng dn. Hy v v gii thch c tuyn volt ampe cadiode.

6. Khi no cn dng diode mc ni tip, diode mc song song?7. Nu cch o th diode.

8. Hy k tn v v k hiu ca mt s loi diode bn dn v cho bit vi ng dngca n.

9. Diode zener cn c gi l diode g? Ti sao? 10.Diode quang l g? Nu nguyn l hot ng ca diode quang.11.Cho bit vi mch ng dng ca diode quang.12.LED l g? Nu nguyn l hot ng ca LED.13.LED by on l g? V tr cc LED a, b, c, d, e, f, g, p trn LED by on l c

nh hay thay i c? Ti sao?14.Hy v nhng on sng tng ng trn LED by on hin th cc ch s 0, 1,

2,, 9.15.Hy k tn LED sng, LED tt trong LED by on khi dng n hin th cc ch

s 0, 1, 2, ., 9.16.Hy k tn mt s loi LED v v k hiu tng ng, cho bit vi ng dng ca

n.

17.Hy k tn nhng linh kin quang in t hc v chia n ra hai nhm linh kinbin i tn hiu quang in, in quang.

18.Hy v v gii thch nguyn l hotng ca mt s mch ng dng c trnh

by trn.

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Chng 4: Transistor mi ni lng cc

63

Chng 4

TRANSISTOR MI NI LNG CC

Transistor mi ni lng cc (BJT)c pht minh vo nm 1948 bi John Bardeenv Walter Brittain ti phng th nghim Bell (M). Mt nm sau nguyn l hotng

ca n c William Shockley gii thch. Nhng pht minh ra BJT c trao giithng Nobel Vt l nm 1956. S ra i ca BJT nh hng rt ln n s pht trinin t hc.

BJT Bipolar Junction Transistor Transistor mi ni lng cc Transistor tip

xc lng cc Transistor tip gip hai ccTransistor lng ni Transistor lngcc.

4.1. Cu to k hiu

Hnh 4.1. Cu to (a) mch tng ng vi cu to (b) k hiu (c) ca BJT loi NPN.

Hnh 4.2. Cu to (a) mch tng ng vi cu to (b) k hiu (c) ca BJT loi PNP.

BJT l mt linh kin bn dn c to thnh t hai mi ni P N, nhng c mt

vng chung gi l vng nn.Ty theo s sp xp cc vng bn dn m ta c hai loi BJT: NPN, PNP.

C

E

B

N

P

N

C

E

B

C

E

B

(a) (b) (c)

C

E

B

P

N

P

E

C

B

E

C

VBB PNP

(b)(a) (c)

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Ba vng bn dn c tip xc kim loi ni dy ra thnh ba cc:

- Cc nn: B (Base)- Cc thu: C (Collector)- Cc pht: E (Emitter)

Trong thc t, vng nn rt hp so vi hai vng kia. Vng thu v vng pht tuy ccng cht bn dn nhng khc nhau v kch thc v nng tp cht nn ta khng th

hon i v tr cho nhau.

4.2. Nguyn l hotng

Khi cha c ngun cp in VCC, VEEth BJT c hai mi ni P N trng thi cnbng v hng ro in th mi mi ni duy tr trng thi cn bng ny.

Vi hnh 4.3, ta chn ngun VCC VEEv tr s in tr sao cho tha iu kin:

- Mi ni P N gia B v E (lp tip gip, lp tip xc JE) c phn ccthun.

- Mi ni P N gia B v C (lp tip gip, lp tip xc JC) c phn ccnghch.

- VBEt th ngng ty loi BJT.

in t t cc m ca ngun VEE di chuyn vo vng pht qua vng nn, ng ltrv cc dng ca ngun VEEnhng v: vng nn rt hp so vi hai vng kia v ngun VCC VEEnn a s in t t vng nn vo vng thu, ti cc dng ca ngun VCC,mt t in t cn li vcc dng ca ngun VEE. S dch chuyn ca in t to thnhdng in:

- Dng vo cc nn gi l dng IB.

- Dng vo cc thu gi l dng IC.

- Dng t cc pht ra gi l dng IE.

Ngoi ra, mi ni P N gia B v C c phn cc nghch cn c dng r (r) rtnh gi l ICBO.

+- -

IC

Rc

e

IE

+VEE

e

IB

Vcc

e

-

RE

Hnh 4.3. Mch kho st gii thch nguyn l hotng ca BJT.

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4.3. H thc lin h gia cc dng in

Hnh 4.4. Mch tng ng vi hnh 4.3

S dch chuyn ca cc in t nh trn cho thy:

IE = IB + IC (4.1)

IC= IE (4.2)

= (Tng s in t dch chuyn n vng thu) / (Tng s in t dch chuyn t vngpht)

H s gn bng 1.

T (4.2) ta c:

II CE (4.3)

Th (4.3) vo (4.1) ta c:

BC

BC

CB

C

I1

I

I1)

1(I

II

I

(4.4)

t1

(4.5)

c gi l h s khuch i dng.

IC= IB (4.6)

Kt hp (1) v (4) ta c h thc thng dng:

IE = IB + IC IC= IB (4.7)

Mi ni gia nn v thu phn cc nghch cn c dng in r (dng r nh diodephn cc nghch) gi l ICBOrt nh (c A). Vy nu xt dng r ta c:

IC= IE + ICBO (4.8)

III CBOCE

(4.9)

Th (4.9) vo (4.1) ta c:

ICIB

IE

RC

VEE

RE

VCC

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1

III

1

II

1

I

II1)

1(I

II

II

CBOBC

CBOBC

CBOBC

CBCBOC

(4.10)

1

IIIIII CBOBCCBE

(4.11)

Khi b qua dng in r ICBO th phng trnh (4.11) tr thnh phng trnh (4.7),phng trnh (4.10) tr thnh phng trnh (4.6).

4.4. Cc cch mc c bn

4.4.1. BJT mc kiu cc pht chungMch dng BJT mc kiu cc pht chung (Common Emitter CE)nh hnh 4.5.

Hnh 4.5. BJT mc kiu cc pht chung.

4.4.2. BJT mc kiu cc nn chung

Mch dng BJT mc kiu cc nn chung (Common Base CB)nh hnh 4.6.

Hnh 4.6. BJT mc kiu cc nn chung.

RCRB1

RB2

C2

C1

RE

Vi

VO

+VCC

RB1

RB2

C1

C2RC

RECB

+ VCC

VO

Vi

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4.4.3. BJT mc kiu cc thu chung

Mch dngBJT mc kiu cc thu chung (Common Collector CC)nh hnh 4.7.

Hnh 4.7. BJT mc kiu cc thu chung.

CE:

-Tn hiu vo B so vi E, tn hiu ra C so vi E.- Pha gia tn hiu vo v ra: o pha.- H s khuch i Ai, Avln.

CB:

-Tn hiu vo E so vi B, tn hiu ra C so vi B.

- Pha gia tn hiu vo v ra: cng pha.- H s khuch i Avln, Ai 1.

CC:

- Tn hiu vo B so vi C, tn hiu ra E so vi C.- Pha gia tn hiu vo v ra: cng pha.- Hs khuch i Ailn, Av 1.

4.5. c tuyn ca BJT

Hnh 4.8. Mch kho st c tuyn ca BJT.

+VCC

RE

C1

C2

RB1

RB2

VO

Vi

VBB

VCC

RC

RB

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Xt mch nh hnh 4.8. Vi VBEl hiu inth gia cc nn B v cc pht E. VCEl hiu inth gia cc thu C v cc pht E.

4.5.1. c tuyn ng vo IB(VBE) ng vi VCE = const

Chn ngun VCC dng xc nh c VCE =

const. Chnh ngun VBB thay i VBEt 0 tng lnn gi tr nh hn in th ngng Vth o dng IB 0. Tip tc tng ngun VBB c VBE = V th btu c dng IB v IBcng tng theo dng hm s mnh dng IDca diode phn cc thun.

Hnh 4.9.c tuyn ng vo ca BJT

4.5.2. c tuyn truyn dn IC(VBE) ng vi VCE = const

kho st c tuyn ny, ta o, chnh ngun tng t c tuyn ngvo nhngdng th

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