giao trinh mach dien tu - chuong 1

8/14/2019 giao trinh mach dien tu - chuong 1

1/23

Chng 1: Mch Diode

CHNG I

MCH DIODE

Trong chng ny, chng ta kho st mt s mch ng dng cn bn ca diode bndn (gii hn diode chnh lu v diode zener - Cc diode c bit khc sc bn nlc cn thit). Ty theo nhu cu ng dng, cc m hnh l tng, gn ng hay thc sca vo trong cng vic tnh ton mch.

1.1 NG THNG LY IN (LOAD LINE):

Xem mch hnh 1.1a

Ngun in mt chiu E mc trong mch lm cho diode phn cc thun. Gi ID ldng in thun chy qua diode v VD l hiu th 2 u diode, ta c:

Trong : I0 l dng in r nghch

=1 khi ID ln (vi mA trln)

=1 Khi ID nh v diode cu to bng Ge

=2 Khi ID nh v diode cu to bng Si

Ngoi ra, t mch in ta cn c:

E - VD - VR= 0

Tc E = VD + RID (1.2)

Trng Vn Tm I-1 Mch in T


2/23

Chng 1: Mch Diode

Phng trnh ny xc nh im lm vic ca diode tc im iu hnh Q, c gi lphng trnh ng thng ly in. Giao im ca ng thng ny vi c tuyn ca diodeID = f(VD) l im iu hnh Q.

1.2. DIODE TRONG MCH IN MT CHIU

- Ngc li khi E < VK, mch c xem nh h, nn:ID = IR= 0mA ; VR= R.IR= 0V ; VD = E - VR= E

1.3. DIODE TRONG MCH IN XOAY CHIU - MCH CHNH LU

Mch chnh lu l ng dng thng dng v quan trng nht ca diode bn dn, c mcch i tin xoay chiu (m thng l dng Sin hoc vung) thnh in mt chiu.

1.3.1. Khi nim v tr trung bnh v tr hiu dng

1.3.1.1. Tr trung bnh: Hay cn gi l tr mt chiu

Tr trung bnh ca mt sng tun hon c nh ngha bng tng i s trongmt chu k ca din tch nm trn trc 0 (dng) v din tch nm di trc 0 (m)chia cho chu k.

Mt cch tng qut, tng i s din tch trong mt chu k T ca mt sng tunhon v(t) c tnh bng cng thc:

Mt vi v d:



3/23

Chng 1: Mch Diode

Dng sng Tr trung bnh

1.3.1.2. Tr hiu dng:hiu dng ca mt sng tun hon( th d dng in) l tr

s tng

Ngi ta nh ngha trng ca dng in mt chiu IDC m khi chy qua mt in trR trong mt chu

k s c nng lng ta nhit bng nhau.



4/23

Chng 1: Mch Diode

Vi th d:Dng sng Tr trung bnh v hiu dng



5/23

Chng 1: Mch Diode

Hnh 1.6



6/23

Chng 1: Mch Diode

1.3.2. Mch chnh lu na sng (mt bn k)

Trong mch ny ta dng kiu mu l tng hoc gn ng ca diode trong vic phn tchmch.

Dng mch cn bn cng cc dng sng (th d hnh sin) ng vo v ng ra nh hnh 1.7

Diode ch dn in khi bn k dng ca vi(t) a vo mchTa c:

- Bin nh ca vo(t)Vdcm = Vm - 0.7V (1.6)

- in th trung bnh ng ra:

- in thnh phn cc nghch ca diode l:VRM=Vm (1.8)

Ta cng c th chnh lu ly bn k m bng cch i u diode.



7/23

Chng 1: Mch Diode

1.3.3. Chnh lu ton sng vi bin th c im gia

Mch cbn nh hnh 1.8a; Dng sng 2 cun th cp nh hnh 1.8b

- bn k dng, diode D1 phn cc thun v dn in trong lc diode D2 phn ccnghch nn xem nh hmch (hnh 1.9)

- bn k m, diode D2 phn cc thun v dn in trong lc diode D1 phn cc nghchnn xem nh hmch (Hnh 1.10)

l trong 2 trng hp, ILu chy qua RL theo chiu t trn xung v dng inu c mt hai bn k. in thnh 2 u RL l:

Vdcm=Vm-0,7V (1.9)

V in thnh phn cc nghch mi diode khi ngng dn l:

VRM=Vdcm+Vm=2Vm-0,7V (1.10)

- Dng sng thng trc 2 u RL c din thnh 1.11



8/23

Chng 1: Mch Diode

Ngi ta cng c th chnh lu to ra in th m 2 u RL bng cch i cc ca2 diode li.

1.3.4. Chnh lu ton sng dng cu diode

Mch cbn

- bn k dng ca ngun in, D2 v D4 phn cc thun v dn in trong lc D1

v D2 phn cc nghch xem nh hmch. Dng kiu mu in th ngng, mch in cv li nh hnh 1.13



9/23

Chng 1: Mch Diode

- bn k m ca ngun in, D1 v D3 phn cc thun v dn in trong lc D2, D4phn cc nghch xem nh hmch (Hnh 1.14)

T cc mch tng ng trn ta thy:- in thnh Vdcm ngang qua hai u RL l:

Vdcm =Vm-2VD=Vm-1.4V (1.12)

- in thnh phn cc nghch VRMmi diode l:

VRM=Vdcm+VD=Vm-VD

VRM =Vm-0,7V (1.13)

l dng in trung bnh chy qua mi cp diode khi dn in ch bng 1/2dng in trung bnh qua ti.



10/23

Chng 1: Mch Diode

1.3.5. Chnh lu vi t lc

Ta xem li mch chnh lu ton sng vi bin th c im gia. Nh kt qa phntrn:

- in thnh 2 u RL l: Vdcm=Vm-0,7V- in th trung bnh 2 u RL l: VDC=0,637Vdcm

Nu ta thay RL bng 1 tin c in dung C. Trong thi im t t=0 n t=T/4, t Cs np nhanh n in thnh Vdcm. Nu dng r ca tin khng ng k, t C s khngphng in v in th 2 u tc gi khng i l Vdcm. y l trng hp l tng.Thc t, in th trung bnh thay i t 0,637Vdcm n Vdcm. Thc ra ngun in phi cungcp cho ti, th d RL mc song song vi t C. bn k dng t C np in n tr Vdcm.Khi ngun in bt u gim, t C phng in qua RL cho n khi gp bn k k tip t Cmi np in li n Vdcm v chu k ny c lp i lp li. Hnh 1.16 m t chi tit dng sng2 u t C (tc RL). Hiu th sng dnh i nh c k hiu l Vr(p-p).

Do in thnh ti a l Vdcm nn in th trung bnh ti thiu lVdcmin=Vdcm-Vr(p-p)



11/23


12/23

Chng 1: Mch Diode

H s sng d quyt nh cht lng ca mch chnh lu.

* Phng trnh in thsng d

Nu gi ic l dng phng in ca tin c in dung C v VC l in th 2 u t in th:

Nu s thay i in th 2 u t l tuyn tnh th dng in ic l dng in mtchiu.

Nu coi sng d c dng tam gic th dng phng ca t l hng s v k hiu l IDC. chnh l dng in qua ti

Vi f l tn s ca ngun in chnh lu.Nu gi frl tn s sng d, ta c

Nh vy sng d t l thun vi dng in trung bnh qua ti v t l nghch vi indung C. Sng d s tng gp i khi chnh lu na sng v lc f=fr

1.4. MCH CT (Clippers)

Mch ny dng ct mt phn tn hiu xoay chiu. Mch chnh lu na sng l mtth dn gin v mch ct.



13/23

Chng 1: Mch Diode

1.4.1. Mch ct ni tip

Dng cn bn nh hnh 1.20. Hnh 1.21 cho thy p ng ca mch ct cn bn ivi cc dng sng thng dng khi coi diode l l tng.

By ginu ta mc thm mt ngun in th mt chiu V ni tip vi diode nh hnh1.22b. Nu tn hiu vo vi(t) c dng hnh sin vi in thnh l Vm th ng ra s c dngnh hnh v 1.22c vi in thnh Vm-V tc V0=Vi-V (coi diode l tng)

1.4.2. Mch ct song song

* Mch cn bn c dng



14/23

Chng 1: Mch Diode

Hnh 1.24 l p ng ca mch ct song song cn bn vi cc dng sng thng dng(diode l tng)

* Mch c phn ccTa cng c th mc thm mt ngun in th 1 chiu V ni tip vi diode. Dngsng ng ra s ty thuc vo cc tnh ca ngun in mt chiu v diode.

Th d: ta xc nh v0 ca mch in hnh 1.25 khi vi c dng tam gic v diode xem nh ltng

- Khi diode dn in: v0=V=4V- Khi vi=V=4V, Diode i trng thi t ngng dn sang dn in hoc ngc li- Khi viV=4V, di

Hnh

ode ngng dn Vo= vi1.26 l dng v bin

ca ng ra v0



15/23

Chng 1: Mch Diode

1.5. MCH GHIM P (Mch kp - clampers)

u. Mch phi c mt tin, mtdiode

n tr s V v v0=0Vnn C xin

khn

y l mch i mc DC (mt chiu) ca tn hiv mt in tr. Nhng mch cng c th c mt ngun in thc lp. Tr s ca

in trR v tin C phi c la chn sao cho thi hng =RC ln hiu th 2 ut gim khng ng k khi t phng in (trong sut thi gian diode khng dn in). Mchghim p cn bn nh hnh 1.27

D ng kiu mu diode l tng ta thy:- Khi t: 0 T/2 diode dn in,t C np nhanh- Khi t: T/2 T, diode ngng, t phng in qua R. Do =RC ln

g ng k, (thng ngi ta chn T10).Lc ny ta c: v0=-2V

im cn ch l trong mch ghim p bin nh i nh ca vi v vo lun bngnhau.

inh vin th xc nh v0 ca mch in hnh 1.29S



16/23

Chng 1: Mch Diode

1.6. MCH DNG DIODE ZENER:Cng tng t nh diode chnh lu, vi diode zener ta cng dng kiu mu gn ng

trong vic phn gii mch: Khi dn in diode zener tng ng vi mt ngun in thmt chiu vz (in th zener) v khi ngng n tng ng vi mt mch h.

1.6.1. Diode zener vi in th ng vo vi v ti RL cnh

Mch cn bn dng diode zener c dng nh hnh 1.30

Khi vi v RL cnh, s phn tch mch c th theo 2 bc:- Xc nh trng thi ca diode zener bng cch tho ri diode zener ra khi mch vtnh hiu th V hai u ca mch h



17/23

Chng 1: Mch Diode

Cng sut tiu tn bi diode zenerc xc nh bi

Pz=Vz.Iz (1.23)Cng sut ny phi nh hn cng sut ti a PZM=VZIZM ca diode zener (IZM: dng

in ti a qua zener m khng lm hng)Diode zener thng c dng trong cc mch iu ha in th to in th

chun. Mch hnh 1.30 l 1 mch iu ha in thn gin to ra in th khng i 2 u RL. Khi dng to in th chun, in th zener nh l mt mc chun so snh vimt mc in th khc. Ngoi ra diode zener cn c s dng rng ri trong cc mchiu khin, bo v...

1.6.2. Ngun Vi cnh v RL thay i

Khi Vi cnh, trng thi ngng hoc dn ca diode zener ty thuc vo in trti RL

Do R cnh, khi Diode zener dn in, in th VR ngang qua in trR s cnh:VR=Vi - Vz

Do dng IRcng cnh:

Dng IZ s nh nht khi IL ln nht. Dng IZc gii hn bi IZM do nh sn xut chobit, do dng in nh nht qua RL l ILmin phi tha mn:



18/23


19/23

Chng 1: Mch Diode

- Ta cng c th mc mch chnh lu tng i in th theo chiu dng

- bn k dng ca ngun in D1 dn, C1 np in VC1=Vm trong lc D2 ngng.- bn k m D2 dn, C2 np in VC2=Vm trong lc D1 ngng.- in th ng ra V0=VC1+VC2=2Vm

1.7.2. Mch chnh lu tng ba, tng bn



20/23

Chng 1: Mch Diode

u tin C1 np in n VC1=Vm khi D1 dn in bn k dng. Bn k m D2 dnin, C2 np in n VC2=2Vm (tng in thnh ca cun th c p v t C1). Bn kdng k tip D2 dn, C3 np in n VC3=2Vm (D1 v D2 dn, D2 ngng nn in th2Vm ca C2 np vo C3). Bn k m k tip D2, D4 dn, in th 2Vm ca C3 np vo C4 ...

in th 2 u C2 l 2Vm2 u C1+C= l 3Vm2 u C2+C4 l 4Vm

BI TP CUI CHNG 1

******Dng kiu mu in th ngng gii cc bi tp t 1 n 8Bi 1: Xc nh VD, VRv ID trong mch in hnh 1.36

Bi 2: Xc nh VD2 v ID trong mch in hnh 1.37

Bi 3: Xc nh V0, v ID trong mch in hnh 1.38

Bi 4: Xc nh I, V1, V2 v V0 trong mch hnh 1.39

Bi 5: Xc nh V0, V1, ID1 v ID2 trong mch hnh 1.40



21/23

Chng 1: Mch Diode

Bi 6: Xc nh V0 trong mch hnh 1.41

Bi 7: Xc nh I1, I2, ID2 trong mch hnh 1.42

Bi 8: Xc nh dng in I trong mch hnh 1.43

Bi 9: Dng kiu mu diode l tng, xc nh V0 trong 2 mch hnh 1.44a v 1.44b

Bi 10: Dng kiu mu in th ngng, xc nh v0 trong mch hnh 1.45

Bi 11: Thit k mch ghip p c c tnh nh hnh 1.46 v hnh 1.47



22/23

Chng 1: Mch Diode

Bi 12: Cho mch in hnh 1.48

a. Xc nh VL, IL, IZ v IRnu RL=180

b. Xc nh gi tr ca RL sao cho diode zener hot ng khng qa cng sutc. Xc nh gi tr ti thiu ca RL zener c th hot ng c.

Bi 13: a. Thit k h thng mch c dng hnh 1.49 bit rng VL=12V khi IL thay i t0 n 200mA. Xc nh RS v VZb. Xc nh PZM ca zener.

Bi 14: Trong mch in hnh 1.50, xc nh khong thay i ca vi sao cho VL=8V vdiode zener hot ng khng qa cng sut.



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Chng 1: Mch Diode

giao trinh mach dien tu - chuong 1

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