(490700575) Bao Cao TH Cau Kien Dien Tu

Bo co th nghim cu kin in t GVHD: Nguyn Thanh Sn

MC LCMC LC...................................................................................................................................1BI 1:..........................................................................................................................................2SEMICONDUCTOR FUNDAMENTAL...................................................................................2

Ch 1: Gii thiu v cht bn dn.....................................................................................2Ch 2 : Diode v chnh lu bn k.....................................................................................2Ch 3 : Chnh lu ton k v mch lc...............................................................................6Ch 4 : Tip gip ca Transistor v s phn cc Dc cho Transistor PNP........................11Ch 5 : ng ti v h s khuych i ca Transistor...................................................12

BI S 2:..................................................................................................................................16CC NGUYN L C BN CA F.E.T...............................................................................16

I. THIT B.......................................................................................................................16II. NI DUNG:.................................................................................................................16

CH 1: LM QUEN VI BNG MCH.....................................................................16CH 2: FET C CNG TIP GIP JFET................................................................16CH 3: MCH KHUCH I BNG JFET................................................................19CH 4: NGUN DNG BNG JFET..........................................................................20CH 5: MOSFET CNG I......................................................................................22CH 6: TRANSISTOR N NI - UJT.......................................................................25

BI S 3:..................................................................................................................................27THYRISTOR V CC MCH IU KHIN CNG SUT................................................27

Tm hiu thyristor v cc mch iu khin cng sut...........................................................27II. DNG C:...................................................................................................................27III. NI DUNG:.................................................................................................................27

CH 1: LAM QUEN VI BN MCH........................................................................27CH 2: SCR....................................................................................................................29CH 3: IU KHIN SCR BNG TN HIU DC......................................................31CH 4 : IU KHIN SCR BNG TN HIU AC....................................................32Ch 5:................................................................................................................................34IU KHIN CNG SUT KHI SCR C KCH DN BNG MCH UJT.............34CH 6: TRIAC....................................................................................................37CH 7: IU KHIN CNG SUT AC BNG TRIAC...........................................38

BI S 4:..................................................................................................................................40

Nhm 4B1 Lp 06DT11


BI 1:SEMICONDUCTOR FUNDAMENTAL

I.Thit b:- My hin sng 2 chiu- My to sng sin- Bng mch

II.Ni dung th nghim: Ch 1: Gii thiu v cht bn dn

a. Mc ch:Hiu nh ngha gii thch v vt liu bn dn v chc nng ca n trong cc dng bn

dn.Nhn dng v kim chng nguyn l hot ng ca mt s dng c bn dn. b. Kt lun:

- Diode v transistor c cu trc t vt liu bn dn, thng l silicon vGermanium.

- Diode c mt tip gip PN, Transistor c 2 tip gip PN.- Diode,zener, LED khc nhau v k hiu.- K hiu ca PNP transistor c chiu mi tn cc BASE,cn NPN th chiu mi tn

ch vo cc Emiter- Diode c 2 cc Anode v Cathode.Transistor c 3 cc Emiter,Base,collector.

Ch 2 : Diode v chnh lu bn k a. Mc ch :

Hiu, gii thch v m t cc nguyn l v cc c tnh hot ng ca Diode bn dn.Mch th nghim :

Cc thng s o c trn mnh l : VA = -10VDC VR1 = -9.37VR2 = -0.1mV

Vi cc gi tr nh th CR1 c phn cc thun, CR2 c phn cc nghch.

Nhm 4B1 Lp 06DT12


Cc thng s o c trn mnh l : VA = 10VDC VR1 = 0.1mVVR2 = 9.39V

Vi cc gi tr nh th CR1 c phn cc nghch, CR2 c phn cc thun.Da vo nh lut Ohm ta xc nh c dng in qua in tr R2 :

Da v gi tr ca IR2 ta xc nh c dng qua CR2 : ICR2 = 2.85mA

VA (V) VR2 (V) ICR2 = VR2 /3.3K (mA) VD = VA VR2 (V)0.75 0.29 0.088 0.46

5 4.41 1.34 0.5910 9.38 2.84 0.62

Kt lun : - c tuyn dng mt chiu ca diode m t dng v in p thun v ngc.- Khi in p phn cc thun tng, vt qua in p chn th dng tng nhanh chng,

vi st p trn diode nh.- Khi diode phn cc ngc, c dng r nh chy qua. Cho n khi t c in p

nh thng, dng ngc tng nhanh chng, diode b nh thng.- Diode c in tr thun thp, in tr nghch cao.

Nhm 4B1 Lp 06DT13

mA85.2RVI

2

2R2R


Th nghim 2 : Hnh minh ha mi quan h gia dng sng vo v sng ra ca chnh lu bnk dng.

Khi Vi (pk) = 4V Vo(pk) = Vi(pk) VF = 4V

Khi Vo (pk) = 5V Vo (avg) = 0.318 x Vo (pk) = 1.59V

Quan h gia sng vo v sng ra ca chnh lu bn k dng.Mch chnh lu diode bn k:

Nhm 4B1 Lp 06DT14


* VI (pk) = 1V Vo (pk) = VI (pk) 0.6 = 0.4VVi 0.6V l st p trn CR1

* VI (pk) = 2V Vo (pk) = VI (pk) 0.6 = 1.4V* Vo(pk) = 3V Vo(avg) = 0.318 x Vo(pk) = 0.954V

Kt lun : - Chnh lu bn k l qu trnh bin i tn hiu xoay chiu thnh xung mt chiu sut

na chu k.- Diode v in tr ti hnh thnh mt chnh lu bn k.- Khi anode ca diode c kt ni vi tn hiu xoay chiu, th tn hiu ra l xung

dng trong bn k dng.- Khi cathode ca diode c kt ni vi tn hiu xoay chiu, th tn hiu ra l xung m

trong bn k m.- St p trn diode gim in th ra ca chnh lu ton k.

Ch 3 : Chnh lu ton k v mch lc.Mc ch : Hiu, gii thch v kim chng mch chnh lu ton k, mch lc v mch

nhn i in p.Th nghim 1 : Chnh lu ton k

Vo(pk) = 10V Vo (avg) = 0.636 x Vo(pk) = 6.36V Chnh lu ton k:

Nhm 4B1 Lp 06DT15


Ta c : VI (pk-pk) = 20V, fI = 100Hz fo = 2x fI = 200Hz Vo(pk) = VI(pk) 2x0.6 = 8.8V Vo(avg) = 0.636 x Vo(pk) = 5.59V

Kt lun : - Mch chnh lu cu ton k chuyn i c 2 pha ca tn hiu xoay chiu vo thnh

xung ra 1 chiu- Tn s ca tn hiu ra gp 2 ln tn s vo- 4 diode to thnh mch chnh lu cu. v vy dng ch chy qua 2 diode trong cng 1

thi gian- 2 diode dn s hng in p lm cho dng ti lun chy theo 1 hng- in p nh ra nh hn din p nh vo v do st p trn 2 diode.- Vo(avg)=0.636xVo(pk)

Th nghim 2 : Mch lc

Vo(pk) = 10 V Vo(pk) = 12V

Thm t C1 vo hnh v:

Nhm 4B1 Lp 06DT16


Vi VOM, dng in DC ra = 12V

Kt ni thm R2 vo mch nh sau :

Da vo my hin sng, in th gn sng : Vr(pk) = 50mV

in th DC ra : Vo = 9.2V

Ni thm C2 song song vi C1 v R2 nh sau : Da vo my hin sng, in th gn sng :

Vr(pk) = 60mVin th DC ra :

Vo = 8.7VKt lun :

- T lc ra c gi l t ha khi kt ni ngang vi tn hiu chnh lu ra.- T x nhanh chng ti in th chnh lu nh.- in th ra gia 2 xung bng khng, t np c x v cung cp dng qua ti.- Trc khi in p t ri chm, mt xung ra khc t b chnh lu s c np li cho

n khi t t n in p nh.- Thi gian x trn t ln hn thi gian np.- gn sng tn ti trong di Volt c th c gim xung n di mV.

Th nghim 3 :

Mch nhn i in pCho mch nh hnh v :

Nhm 4B1 Lp 06DT17


VI(pk) = 10V Vo(pk) = 2x VI(pk) = 20VVC1 = VC2 = 10V

gn r = 20mV(pk-pk)* Thay i RL = 39K

gn r = 54mV(pk-pk) Vo = 12V

Kt lun : - 2 diode v 2 t lc c th to thnh mch nhn i in p.- Mch nhn i in p chnh lu tn hiu vo v lc tn hiu ra c in p bng 2 ln

in p vo.- Ti mi na chu k ca tn hiu xoay chiu, gm mt diode dn v 1 t np.

- Bi v : + 2 t mc ni tip ngang qua in p ra nn tn hiu ra mt chiu bng tng in p

trn 2 t.+ 1 t np trong sut na bn k ca tn hiu AC, tn s gn sng ra ca mch nhn

i in p bng 2 ln tn s tn hiu vo.

Exercise 2 : Zener Diode

VCR1 = -0.6V

Ni mch nh hnh v :

Nhm 4B1 Lp 06DT18


c tuyn ca diode Zener

Kt lun : - Diode c thit k hot ng an ton trong min nh thng.- Ti phn cc thun,diode Zener ng vai tr nh mt diode chnh lu.- Da vo c tuyn phn cc ngc ca diode Zener ch rng diode Zener s ngng

dn khi VZ t ti im nh thng.- Ti VZ, dng ngc tng nhanh, in p ngc tng rt chm.- Trong mch diode Zener, in tr c mc ni tip vi in tr Zener gii hn

dng c gi tr bng dng IZT.Kt lun :

- Diode Zener c s dnh trong cc mch iu ha in p do mc in p Zenerhu nh khng thay i.

- Diode Zener s gi mc in p ra bng mc in p Zener, bt chp cc bin thinca in p ngun cung cp v in tr ti.

- Dng in tng trong mch n nh in p bng diode Zener l tng dng chy quadiode Zener v dng chy qua ti.

- Cc Diode Zener thc hin tt s iu chnh in p bi v IZ c th thay i ng ktheo s thay i nh ca in p t vo khi lm vic vng nh thng.

- Lng tng ln ca dng ti s b bng lng gim ca dng Zener, c tnh nys cho kh nng iu ha ti ca b n nh in p.

- n nh ca ti theo phn trm c o bng thay i in p trn ti do sthay i ti.

Ch 4 : Tip gip ca Transistor v s phn cc Dc cho Transistor PNP.

Mc ch : Xc nh v gii thch cc c tnh v nguyn l hot ng ca Transistor, pdng bng cch o th cc transistor, v kho st chuyn mch bng transistor.

Exercise 2 :

Nhm 4B1 Lp 06DT19


VE = -1.5VVBE = 0mVVCE = -13.6V

(R1 = 10K)VBE = -0.74VVCE = -0.04VVR2 = 13V

K/V13RV

I2

2RC

R1 = 1M)VR2 = 3.48V

K/V10x48.3RV

I 62

2RC

IC = 0.00348mAVBE = -0.65VVCE = -9.7V

Kt lun : - Transistor lng cc c th lm vic nh mt chuyn mch bng s thay i dng

Base t mc 0 n gi tr ln nht.- tip gip JE phn cc thun, th base ca transitor PNP sillicon phi c mc

khong 0.5Vdc n 0.8Vdc, m hn so vi cc Emiter.- in tr ca tip gip JC ty thuc vo dng IB.- Tip gip JE c phn cc thun s lm cho in tr Collector Emitter rt thp,

cho php dng chy trong mch tng t nh mt chuyn mch kn.- Khi tip gip JE c phn cc ngc, dng IB = 0 gy ra in tr Collector - Emittor

rt cao, chn dng chy trong mch Collector, tng t nh chuyn mch h.- Khi dng Collector ln nht, transistor dn bo ha nn VCE gn bng khng.- Khi dng IB = 0 th khng c dng IC,transistor vng ngt, VCE gn bng in p

ngun cung cpIE = IB + IC

V mc dng IB khng ng k, nn IC v IE gn nh bng nhau.

Ch 5 : ng ti v h s khuych i ca Transistor

Mc ch : Hiu, gii thch v kim chng cc trng thi hot ng v nh hng ca hs khuch i ca Transistor nn cc dng in ca transistor bng cch s dng ng ti DC

Th nghim 1 :

VBEO = -2.499V

Nhm 4B1 Lp 06DT110


c tnh quan h VBEO v IBEO

Kt lun : + Tip gip PE ca transistor c phn cc thun hay ngc ph thuc vo in th

ca ngun cung cp.+ c tnh phn cc thun DC ca tip gip BE transistor tng t nh cc diode khc+ Dng in phn cc gia Base v Emittor t 0 n di microamp cho n khi VBEO

t 0.5V+ VBE nm trong khong 0.5V n 0.8V, dng IBEO tng nhanh chng vi VBEO tng rt

nh.+ Sau khi dng Base Emitter t 2mA, in p phn cc thun gn nh khng i

Kt lun : - Dng IC ln ca transistor c iu khin bi dng IB nh- T s dng IC v IB c gi l h s khuych i dng transistor.- Dng IE = IC + IB- IB = 5%IE

Kt lun : - H c tuyn dng Collector l th dng ca Collector theo in p Collector-

Emitter khi dng base l thng s.- Do hu nh khng i vng tch cc ca transistor nn th theo in p

Collector-Emitter khi dng base khng i l c tuyn tng rt t.- im Q hay im lm vic l im giao cho gia ng ti v IB v s c xc

nh bng iu kin phn cc DC ca transistor.- Cc mch transitor c s dng khuch i tn hiu nh, thng c thit k

c im nm trung tm ng ti. iu ny s cho khong hot ng trong vng tch cc ivi tn hiu AC t vo.

- Vic xc nh ng ti s b nh hng theo cc thay i ngun cung cp collectorhay tr s ca in tr collector.

Nhm 4B1 Lp 06DT111


BI S 2:CC NGUYN L C BN CA F.E.T

I. THIT B- FACET Base Unit.- Bng mch FET FUNDAMENTAL CIRCUIT.- Ngun +15V, -15V.- ng h vn nng.- My to sng Sin.- My hin th sng.

II. NI DUNG: CH 1: LM QUEN VI BNG MCH.

1. Mc ch:Nhn dng cc dng c bn dn chnh v cc khi mch trn bng mch FET.

2. Ni dung th nghim:Th nghim 1.1: Nhn dng v xc nh v tr ca mch.Th nghim 1.2: Kho st nguyn l hot ng ca mch dao ng bng UJT.Mc mch nh hnh v:

Dng my hin th sng xc nh dng sng ti cc E ca UJT. Dng sng nh hnhv:

CH 2: FET C CNG TIP GIP JFET.1. Mc ch: Kho st v hiu c cc c tnh v nguyn l hot ng ca transistor JFET.2. Ni dung th nghim:

Th nghim 2.1: c tuyn lm vic ca JFET.Mc mch nh hnh v:o IDS

Nhm 4B1 Lp 06DT112


Dng my hin sng iu chnh VDS = 10VdcDng Multimeter o IDSS :IDSS = 7.96 mAGim VDS n 8Vdc: IDSS khng i.Gim VDS n 0Vdc, tng VDS t 0 n 2Vdc: Dng IDS tng khi VDS tng.

Mc mch li nh hnh v: o IDS khi thay i VGS

Chnh VDS =10Vdc. VGS = 0Vdc. o IDSS.IDSS = 8mA.Gim VGS n VGS = -1Vdc, ta thy IDS gim.Gim VGS n VGS = -10Vdc, dng IDS gim n 0mA.

Kt lun:- VGS phn cc ngc iu khin dng IDS- VGS phn cc thun khng dng cho JFET.- VGS = 0Vdc, IDSS cc i.- IDSS gim n 0mA ti 1 gi tr xc nh ca VGS phn cc ngc.- C 2 vng hot ng ca JFET: vng thun tr, vng bo ha (hay vng thtknh).

Nhm 4B1 Lp 06DT113


Th nghim 2.2: H c tuyn ca JFET.Mc ch: Quan st h c tuyn ID VDS xc nh c tnh ca JFET.Ni dung:

Mc mch nh hnh v:

Dng my hin sng chnh GEN 15Vpk-pk tn s f = 1000Hz.Dng ng h o chnh VGS = 0VdcKt ni my hin sng nh hnh v, chnh my hin hin sng ch XY,

hin th c tuyn ID VDS .

Chnh VGS = -1Vdc.Quan st dng sng.T t gim VGS n -10Vdc.Quan st dng sng.Ta c h c tuyn:

Nhm 4B1 Lp 06DT114


Kt lun:- VGS phn cc xc nh c tuyn hot ng ca JFET.- IDS c th gim n 0mA ti 1 gi tr xc nh ca VGS.- VG phi m i vi JFET knh n.- H c tuyn ca JFET c to ra bng cch thay i VDS v VGS.

CH 3: MCH KHUCH I BNG JFET.1. Mc ch:

Kim chng nguyn l hot ng vi tn hiu mt chiu v tn hiuj xoay chiu mch khuch i JFET.2. Ni dung th nghim:

Th nghim 3.1: Hot ng DC ca mch khuch i bng JFET.

Mc ch: Kho st v o cc in p mt chiu ca mch khuch i bng JFET.Ni dung:

Mc mch nh hnh v:Dng ng h o xc nh VR2: VR2 = 1.47Vdc.Xc nh VGS: VGS = -1.47Vdc.Xc nh VR3: VR3 = 1.45VdcTheo nh lut Ohm ta c: IDS =VR3/ R3 =1.45/2.7=0.537mATa thy kt qu IDS bng kt qu ti R2 :

IR2=VR2/R2 =1.47/2.7=0.544mADng ng h o vn nng o VD: VD=13.40VdcKt qu ny gn bng vi kt qu o theo l thuyt

VD=VDD-VR3=15V-1.45V=13.55V

Kt lun:- in p ri trn in tr ngun to ra in p phn cc- Trong mch phn cc ngun VDD = VD + VR3.- Phn cc ngun hay t phn cc gim tc ng ca s thay i IDSS.- Dng mng v dng ngun xp x bng nhau.

Kch hot CM17 thay i R3.o VD = 13.03 VdcKhi khng kch hot CM17, VD = 13,41 Vdc.Ta c: VD = VDD ID.R3.Do ID, VDD khng i, VD gim => CM17 kch hot lm R3 gim.

Th nghim 3.2: Hot ng AC ca mch khuch i bng JFET.

Mc ch: o h s khuch i AC ca mch khuch i bng JFET.Ni dung:

Mc mch nh hnh v:

Nhm 4B1 Lp 06DT115


Dng my hin sng iu chnh GEN 100mVpk-pk, 1000Hz.Dng knh 2 ca my hin sng xc nh dng sng ra ti cc D => sng c bin

ln hn. Vpk-pk = 0,44 V.Kch hot CM18 R3 gim t 2,7K=> 2.1K.Vpk-pk gim cn 0.34V.B C2 th in p ra Vpk-pk gim.Dng sng u vo v u ra ngc pha nhau.

Kt lun:- T ni song song vi RS lm gim tc ng ca RS n h s khuch i.- Vi mch cc ngun chung, tn hiu u ra ngc pha vi tn hiu u vo.- H s khuch i ca JFET khi khng c hi tip l hm s ca RL v gm.

Av = RL.gm.

CH 4: NGUN DNG BNG JFET.

1. Mc ch:Kho st nguyn l hot ng ca mch ngun dng bng JFET.

2. Ni dung th nghim:

Th nghim 4.1: Nguyn l hot ng ca ngun dng hng bng JFETMc ch: Quan st v o dng ra ca mt ngun dng hng bng JFET

Ni dung:

Mc mch nh hnh v:

Nhm 4B1 Lp 06DT116


iu chnh theo chiu kim ng h lm ngn mch R2 mch ch cn R1 =100.o IDSS : IDSS=14.3mA

iu chnh R2 tng tr (R1+R2)=400 o IDSS : IDSS=14.25mA

Kt lun: - Mt JFET c th cung cp mt dng khng i khi hot ng trong vng ngt hoc

vng bo ha - Gi tr ca ngun dng khng i khi thay i gi tr in tr- Ngun dng ca mt JFET c th dng trong phn cc 0 ,cc cng v ngun ngn

mch vi nhau - Gi tr ln nht ca ngun dng khng i xy ra khi cc cng v cc ngun c

ngn mch vi nhau

Th nghim 4.2: S phn b cng sut v in p ca JFET

Mc ch: Xc nh s phn b in p v cng sut ca ngun dng bng JFET.Ni dung:

Mc mch nh hnh trn (th nghim 4.1) iu chnh theo chiu kim ng h lm ngn mch R2 mch ch cn R1 =100.o IDSS :

IDSS=14.3mA

Kt ni mch nh hnh v:

Ngn mch R2 o VR1VR1=1.412Vdc

o in p ri trn D&S ca Q1 :VDS=13.44Vdc

Cng sut tiu tn trn JFET vi in tr l 100 c tnh theo cng thc :

P=IDSS.VDS=14.3mA.13.44Vdc=192.2mWiu chnh R2 tng tr (R1+R2)=400 .o VDS:

VDS=9.22VdcCng sut tiu tn trn JFET vi in tr l 400 c tnh theo cng thc :

P=IDSS.VDS=14.3mA.9.22Vdc=131.85mWKt lun:

- in p cung cp c phn phi qua mi thnh phn ca ngun dng khng imc ni tip.

- S phn b in p gia ngun dng (JFET) v ti ph thuc vo gi tr ca in trti.

- Cng sut tiu tn ca ngun dng khng i tng khi in tr ti gim.

Nhm 4B1 Lp 06DT117


CH 5: MOSFET CNG I1. Mc ch:

Kho st hot ng DC v AC ca MOSFET.2. Trnh t th nghim:

Th nghim 5.1: Cc ch hot ng ca MOSFET.Mc ch: Xc nh nh hng ca s phn cc cc ch hot ng ca MOSFET

bng cch dng cc mch o thng dng.

Ni dung:

Mc mch nh hnh v:

Dng my hin sng chnh GEN 10Vpk-pk, 100Hz.iu chnh R1 VGS = 0Vdc.t my hin th sng ch XY, knh 1( X) ni vo cc D, knh 2 (Y) ni vo cc S.Quan st dng sng, sau ln lt iu chnh R1 VGS n +0,3V, -0,3V. Ta c

dng sng nh hnh v:

Kt lun:- i vi MOSFET ch tng cng knh, in p phn cc dng c t, ch

ngho knh, in p phn cc m c t vo.- MOSFET knh N kiu tng cng/ngho knh c th hot ng vi in p G -

dng hoc m, v khng to nn dng cng.- in p G dng lm tng dng IDS i vi MOSFET knh N kiu tng cng/ngho

knh.

Nhm 4B1 Lp 06DT118


- in p G m lm gim dng IDS i vi MOSFET knh N kiu tng cng/nghoknh.

- IDSmax khng xut hin ti VGS = 0Vdc.

Th nghim 5.2: B khuch i in p bng MOSFET.

Mc ch: Xc nh cc c tnh hot ng ca b khuch i in p bng MOSFETknh N.

Ni dung:Mc mch nh hnh v:

Mc ng h o in th ti cc D, iu chnh R1 VD = 7.5Vdc.Dng my hin sng chnh GEN 200mVpk-pk, 1000 Hz.

Nhm 4B1 Lp 06DT119


Xc nh dng sng ti cc D qua knh 2 ca my hin sng. Ta thy tn hiu ra c bin ln hn tn hiu u vo.

Mc thm C4 song song vi R6, bin tn hiu u ra tng so vi khi khng c t.B t C4, tn hiu u ra ngc pha so vi tn hiu u vo.

Mc s mch li nh hnh v:

Chnh R1 ht sang chiu kim ng h.iu chnh GEN 200mVpk-pk, 1000Hz.iu chnh R1 ngc chiu kim ng h, sau cng chiu kim ng h thay i Vdc ti G2. Ta thy bin sng ra ti D thay i.

Kt lun:- Trong ch phn cc bnh thng cho MOSFET, VD xp x VDD.- Trong b khuch i S chung, tn hiu ra ngc pha tn hiu vo.- Dng t ni song song vi RS lm tng h s khuch i.- MOSFET l thit b iu khin bng in p.- Trong MOSFET cng i c 2 cng u iu khin dng ID.

CH 6: TRANSISTOR N NI - UJT.1. Mc ch:

Kho st nguyn l hot ng ca mch dao ng tch thot da trn mch th nghims dng UJT.2. Trnh t th nghim:

Th nghim 6.1: Cc c tnh lm vic ca UJT. Mc ch:

Kim chng cc c tnh lm vic ca UJT. Ni dung:Mc mch nh hnh v:

Nhm 4B1 Lp 06DT120


Dng ng h o in tr xc nh in tr gia B1 - B2, B2 - B1, E-B1, E-B2 ta cRBB(B1 - B2 ) = 5,84K, RBB(B2 B1 ) = 5,87K, suy ra B1 - B2 ging nh mt in tr; EB1,EB2 ging nh tip gip PN.Mc mch nh hnh v:

o VC1, ta c VC1 = 7,07 V.VJ =VC1 = 7,07V.=> = VJ/ VBB = 0.707.

Dng my hin sng xc nh dng sng ti cc E. Ta thy dng IE nh xut hin tigi tr VV.

Kt lun:- Mt Ohm k c th dng kim tra UJT.- in tr ni xut hin ging nh 1 gi tr in tr.- UJT c 1 tip gip gia PN gia E v B.- in p nh Emitter c quan h vi t s dng ni ti.- Vp = VJ VD.- c tuyn lm vic ca UJT c 3 vng: vng ngt, vng in tr m v vng boha.- UJT cho thy c tuyn in tr m khi dn.

Th nghim 6.2: Mch to dao ng bng UJT.

Mc ch:Kho st hot ng ca mch dao ng tch thot bng UJT.

Ni dung:

Ni mch nh hnh v:

Nhm 4B1 Lp 06DT121


Ni knh 1 ca my hin sng ti cc E, sng c dng rng ca.Xc nh chu k ca sng: T =1,1ms => f = 1/T = 909Hz.Tnh f theo cng thc: f = 1/(R1xC2) = 1/(10.103.0,1.10-6) = 1000HzKhi kch hot CM6 thay i gi tr ca C2, tn s ca tn hiu ti E gim, => CM6

lm C2 tng.B kch hot CM6, ni knh 2 ca my hin sng n cc B1, dng sng ti B1 l xung

dng, c tn s bng tn s ca sng ti E.Ni knh 1 vo B1, knh 2 vo B2, dng sng trn B2 l xung m. Dng sng trn B1 v

trn B2 ngc pha nhau, c tn s bng nhau.

Kt lun:- Dng sng ti E l dng rng ca.- Dng sng ti B1 v B2 ln lt l xung dng v xung m.- Xung ti cc Base xut hin khi UJT dn.- Thi gian dn ca UJT bng thi gian x ca t.- Tn s hot ng ca mch ph thuc vo t nh thi, in tr np (R1) v in tr

m rng ti B1.- Tip gip lm x t qua B1 v in tr ti B1.

- in tr m rng ti B1 cn to xung ti B1. BI S 3:

THYRISTOR V CC MCH IU KHIN CNG SUT

I. MC CH:Tm hiu thyristor v cc mch iu khin cng sut.

II. DNG C:- FACET Base Unit.- Bng mch THYRISTOR & POWER CONTROL CIRCUITS.- Ngun +15V, -15V.- ng h vn nng.- My to sng Sin.- My hin th sng.

III. NI DUNG:

CH 1: LAM QUEN VI BN MCH.

1. Mc ch:Nhn dng bng mch v cc thyristor trn bng mch THYRISTOR & POWER

CONTROL CIRCUITS.2. Ni dung th nghim:

Th nghim 1.1: Nhn dng cc thyristor v cc cu kin in t tren bng mch.

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Th nghim 1.2: Lin kt cc mch thyristor trn bng mch THYRISTOR & POWERCONTROL CIRCUITS.

THYRISTOR & POWER CONTROL CIRCUITS.

Lp bng mch SILICON CONTROLED RECTIFIER (SCR)iu chnh VA=6 vdco VR4 =0.1mv dcNhn S1 v o li VR4 =5.22 Vdc

MC BNG MCH: SCR DC GATE HALF-WAVE AND FULL-WAVE

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Bng mch: SCR AC GATE AND UJT HALF-WAVE AND FULL- WAVE

Kt ni knh 1 v knh 2 ca my hin sng ngang qua hai du in tr R8 nh hnh v

Kt lun:

- SCR c s dng ch yu chuyn mch trong cc mch iu khin AC v DC.SCR l cu kin 3 cc

- Triac c 3 cc, l chuyn mch c iu khin bng cng. G . Triac l dng chai chiu v hot ng nh hai SCR mc sng song

- Transistor in ni l mt dng c 3 cc c mt tip gip PN , c th c dng lmtr tn hiu ti cng SCR

CH 2: SCR1. Mc ch:

Kim chng nguyn l hot ng c bn ca SCR

2. Ni dung th nghim:

* Th nghim 2.1 :

Mc ch:

- o th 1 SCR bng ng h vn nng- Cn phi c ng h c chc nng o in tr hay chc nng o diod- Tip gip G-K ca SCR ch l mt tip gip PN nn c th o nh 1 diod

* Th nghim 3

1.Mc ch th nghimo in p kch khi cng v dng gi ca 1 SCR2.

2.Trnh t th nghim

Mc mch nh hnh v

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- Lc u R3 c gi tr ln nht (CCW), SCR khng dn- Xoay nh R3 v cho ti khi SCR dn (c dng qua R4), VAK=1V v o dc VGT=0.643V-Tt SCR bng cch cho h mch A v R4

Mc mch nh hnh o dng gi

-Khi SCR dn ta o dc VR4=5.23V suy ra dng chy qua R4 l IR4 = VR4/R4 =5.23/220 = 0.024A

-Xoay R3 ngc chiu kim ng h (CCW) cho ti khi SCR tt , ti thi im SCRgn tt ta c VR4=0.33Vdc, suy ra dng gi IH= VR4/R4= 0.33/220=0.0015A

Kt lun : - in p kch khi cng VGT l mc in p cng nh nht cn thit m SCR - xcs nh gi tr cua in p kch khi cng , ta quan st SCR khi tng in p ti

cng cho n khi SCR chuyn sang dn- Dng gi ca SCR IH L dng anode thun nh nht cn thit gi SCR trang thi

dn - xc nh gi tr dng gi ta quan st dng anode thun khi gim dng anode cho

ti khi in p ti cng tng ln t ngt SCR chuyn sang tt CH 3: IU KHINSCR BNG TN HIU DC

I. MC CH Kim chng s iu khin DC ca cc mch chnh lu dung SCR

II. TRNH T TH NGHIM

* Th nghim 3.11. Mc ch: Kim chng hot ng ca SCR nh mt b chnh lu bn k c iu khin

2 .Trnh t th nghim

Mc mach nh hnh v :

-Cha nhn S1 th SCR cha dn

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-Nhn S1 th SCR dn , tn hiu ra trn R4 c dng chnh lu na chu k -Tn hiu trn SCR c dng nh hnh v

* Th nghim 3.2

1.Mc dch th nghim -Kim chng vic iu khin bng SCR vi b chnh lu bn k 2.Trnh t th nghim

Mc mch nh hnh :

- Khi cha nhn S1 , SCR khng dn khng c in th trn R4- Nhn S1 SCR dn, dng dn c duy tr bi R3- Tn hiu ra trn R4 nh hnh v -Tt SCR bng cch nh cng kt ni gia R3 v A

* Th nghim 3.3 1.Mc ch th nghim

Kim chng vic iu khin bng SCR i vi b chnh lu ton k

2. Trnh t th nghim Mc mch nh hnh v

- Mt SCR c th iu khin b chnh lu ton k

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- Khi SCR khng dn th s khng c dng chy qua R4 hay SCR , Q1- Khi S1 ng s chuyn SCR sang dn , dng in lc ny chy qua R4 v SCR - SCR vn trng thi dn bi v dng gi c cung cp bi R3- S chnh lu ton k ca tn hiuu vo xy ra ti CR2 sau kh kch dn

CH 4 : IU KHIN SCR BNG TN HIU AC1.Mc ch :kim chng s iu khin bng tn hiu ac ca mt SCR khi s dng ccmch iu khin pha bn k v ton k.2. Ni dung th nghim:

Th nghim 4.1 IU KHIN PHA BN K BNG SCRMc mch nh hnh v :(mch iu khin pha bn k)

u vo 18Vpk-pk ,60Hz sineDng sng hin th 2 u R8:

Kt lun :

-in tr cng thay i c th iu khin gc dn ca SCR xp x t 180 n 90 -Khi mc thm t n cng l nguyn nhn lm dch chuyn pha tn hiu cng , lm

tng khong iu khin ca gc dn SCR t 180 n 0 -Mt mch iu khin pha SCR c th iu khin tc ca motor mt chiu-Khi mt mch iu khin pha SCR c dng iu khin tc c ca motor mt

chiu th motor s pht ra mt in p trong thi gian SCR ngng dn

* Th nghim 4.2 IU KHIN PHA TON K BNG SCR

Mch nh hnh v:

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Thay i R2 theo CW quan st dng sng trn R8:

Dng sng trn cc cng ca SCR khi R2 t CW:

iu chnh R2 xem s thay i ca gc dn SCR t 90 n 180 :

Kt lun : - Mt SRC c th iu khin ti ca my chnh lu ton k.- Mch iu khin pha c th iu chnh gc dn cu SCR.- Cng sut cc i ti ca mch chnh lu ton k c th gp i.

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Ch 5:

IU KHIN CNG SUT KHI SCR C KCH DN BNG MCH UJT

1. Mc ch:Hiu r ng dng ca UJT lm cu kin kch khi cho SCR

2. Trnh t th nghim:Th nghim 5.1: c tuyn V-A ca UJT.Mc ch th nghim 5.1: kim chng h c tuyn ca UJTMc mch nh hnh v;

iu chnh my pht n 18Vpp v 60Hz siniu chnh R2 n CW :Quan st trn my hin th sng ta thy dng sng:sng vo sng ra :hnh :1

Hnh:1 hnh :2iu chnh tng R2 chm ta thy dng sng: Th hin hnh 2

Tng R2 n cc i ta c dng sng sau: Tn hiu nh m emiter l: Ve=-7,9 VKt lun :

- UJT c kch m khi t ti in th VP( firing voltage)- Chng ta c th iu khin xung ra B1 t 0 n 180 .- UJT tt khi in th emitter gim xung di in th Vv

* Th nghim 5.2 IU KHIN PHA BN K / TON K BNG UJT

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Mc ch : Kim chng mch iu khin pha bn k v ton k vi UJT

Mch iu khin pha bn k bngUJT :

iu chnh R2 nh nht ta thy dng sng trn R8 l:

Mch chnh lu ton k iu khin bng UJT:

iu chnh R2 n CW ta thy dng sng ra trn R8 l:

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KT LUN: - Bng cch s dng mch iu khin pha bn k dng UJT ,SCR ta c th iukhin cng sut ca ti t gc dn 0o n 180o- Xung c to ra trn UJT c th c dng kch m SCR - Bng cch s dng mch iu khin pha ton k dng UJT ,SCR ta c th iukhin cng sut ca ti t gc dn 0o n 360o.

CH 6: TRIAC

1. Mc ch: Kim chng hot ng ca cc TRIAC thng dng2. Trnh t th nghim:

Th nghim 6-1 :S DN IN THEO 2 HNG TRIAC

Mc ch : Kim chng v gii thch h c tuyn dn 2 chiu ca triac.

Mc mch nh hnh v bng mch TRIAC AC POWER CONTROL nh hinh v.

iu chnh VA =6Vdc .ta o c in th MT2 l 6Vdc.( v triac khng dn).n S1 triac s dn (VMT2=0.8vdc)thay VAbng ngun -6Vdc: => o c VMT2= -6Vdc.n S1 triac s dn (VMT2= -0.8vdc)

KT LUN :-Triac l mt thit b dn 2 chiu-Triac ging nh CSR yu cu tn hiu kch dn cc cng m- tt triac, ngt dng MT2

Th nghim 6-2: CC CH KCH DN CHO TRIAC

Mc ch : Kim chng v nhn bit 4 ch kch dn ca triac.

Mc mch nh hnh v : bng mch TRIAC AC POWER CONTROL

Nhm 4B1 Lp 06DT131


iu chnh VA= 6Vdc ln lt thay i cc tnh ca cc ngun VA, VG kimchng 4 kiu dn ca triac.

KT LUN : -Triac c th c kch dn bng 1 trong 4 kiu .-Triac c th dn khi MT2 dng hoc m hn so vi MT1-in th kch dn cng ca triac c th dng hoc m.-St p trn triac ang dn gn 0,8V dc

CH 7: IU KHIN CNG SUT AC BNG TRIAC

1 Mc ch : Kim chng vic s dng triac iu khin cng sut ac.2 Trnh t th ngim :

Th nghim 7.1

Mc mch nh hnh v :

Mch triac bn k

iu chnh my pht 18Vpk-pk, 60Hz sine. iu chnh R1 n CW ta quan st c dng sng trn MT2 :

iu chnh R1 theo CCW, gc dn thay i nh sau:

Quan st dng sng ra trn cc cng( gate) khi R1 CW nh sau:

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KT LUN:

- Mt mch triac bn k dn khi n t n in th kch khi cng.- Hot ng ca mt mch triac bn k ging nh mt SCR.- Chng ta c th s dng mch iu khin pha cc cng ca triac diu khingc dn ca triac.

Th nghim 7.2 Mch triac ton k :

iu chnh R1 n CW ta quan st dng sng ra

in th trn cng khi triac m l 1V

KT LUN :

- Mt mch triac chnh lu ton k dn theo 2 chiu.- Chng ta c th iu khin gc dn ca triac trn c 2 k.- Chng ta c th kch m triac bng in p cng dng hoc m

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BI S 4:

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I. THIT BII. NI DUNG:II. DNG C:III. NI DUNG:

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