Tai Lieu Thi Nghiem Mach Dien

TRNG I HC TN C THNG KHOA IN - IN T

B MN K THUT IN

TI LIU HNG DN TH NGHIM

MCH IN

Tp.H Ch Minh, thng 4 - 2010

NI QUY PHNG TH NGHIM IN-IN T

IU I. TRC KHI N PHNG TH NGHIM SINH VIN PHI:

1. Nm vng quy nh an ton ca phng th nghim. 2. Nm vng l thuyt v c k ti liu hng dn bi thc nghim. 3. Lm bi chun b trc mi bui th nghim. Sinh vin khng lm bi chun b theo ng

yu cu s khng c vo lm th nghim v xem nh vng bui th nghim . 4. n phng th nghim ng gi quy nh v gi trt t chung. Tr 15 pht khng c vo

th nghim v xem nh vng bui th nghim . 5. Mang theo th sinh vin v gn bng tn trn o. 6. Tt in thoi di dng trc khi vo phng th nghim.

IU II. VO PHNG TH NGHIM SINH VIN PHI: 1. Ct cp, ti xch vo ni quy nh, khng mang dng c nhn vo phng th nghim. 2. Khng mang thc n, ung vo phng th nghim. 3. Ngi ng ch quy nh ca nhm mnh, khng i li ln xn. 4. Khng ht thuc l, khng khc nh v vt rc ba bi. 5. Khng tho lun ln ting trong nhm. 6. Khng t di chuyn cc thit b th nghim

IU III. KHI TIN HNH TH NGHIM SINH VIN PHI: 1. Nghim tc tun theo s hng dn ca cn b ph trch. 2. K nhn thit b, dng c v ti liu km theo lm bi th nghim. 3. c k ni dung, yu cu ca th nghim trc khi thao tc. 4. Khi my c s c phi bo ngay cho cn b ph trch, khng t tin sa cha. 5. Thn trng, chu o trong mi thao tc, c thc trch nhim gi gn tt thit b. 6. Sinh vin lm h hng my mc, dng c th nghim th phi bi thng cho Nh trng v

s b tr im th nghim. 7. Sau khi hon thnh bi th nghim phi tt my, ct in v lau sch bn my, sp xp thit

b tr v v tr ban u v bn giao cho cn b ph trch. IU IV.

1. Mi sinh vin phi lm bo co th nghim bng chnh s liu ca mnh thu thp c v np cho cn b hng dn ng hn nh, cha np bo co bi trc th khng c lm bi k tip.

2. Sinh vin vng qu 01 bui th nghim hoc vng khng xin php s b cm thi. 3. Sinh vin cha hon thnh mn th nghim th phi hc li theo quy nh ca phng o to. 4. Sinh vin hon thnh ton b cc bi th nghim theo quy nh s c thi nhn im kt

thc mn hc. IU V.

1. Cc sinh vin c trch nhim nghim chnh chp hnh bn ni quy ny. 2. Sinh vin no vi phm, cn b ph trch th nghim c quyn cnh bo, tr im thi.

Trng hp vi phm lp li hoc phm li nghim trng, sinh vin s b nh ch lm th nghim v s b a ra hi ng k lut nh trng.

Tp.HCM, Ngy 20 thng 09 nm 2009 KHOA IN-IN T ( k) PGS TS. PHM HNG LIN

TRNG H TN C THNG KHOA IN IN T

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CNG HA X HI CH NGHA VIT NAM c lp - T do - Hnh phc

**************

Tai lieu hng dan th nghiem mch in

B mn K thut in, Khoa in in t 1

MC LC

Bi 1: Mch in DC, Mch Thevenin-Norton v mng mt ca 2 Bi 1.1: Mch in DC 2 Bi 1.2: Mch Thevenin-Norton v mng mt ca 11 Bi 2: Th nghim mch in AC 1 pha v 3 pha 17 Bi 2.1: Mch in xoay chiu 1 pha17 Bi 2.2: Mch in xoay chiu 3 pha24 Bi 3: Mng 2 ca tuyn tnh khng ngun v mch cng hng RLC33 Bi 3.1: Mng 2 ca tuyn tnh khng ngun 33 Bi 3.2: Mch cng hng RLC 41 Bi 4: Qu trnh qu mch tuyn tnh-mch phi tuyn49 Bi 4.1: Qu trnh qu mch tuyn tnh 49 Bi 4.2: Mch phi tuyn 57



H V TN: ................................ LP: ...................................

MSSV: ...................................

Nhn xt ca gio vin hng dn

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Bi 1 MCH IN DC, MCH THEVENIN-NORTON

V MNG MT CA Bi 1.1 MCH IN DC I/. MC CH

o dng in, o in p, o cng sut. Kho st mch in ni tip, mch in song song. Kho st nh lut K1, nh lut K2 v nguyn l xp chng. Kho st hin tng ngn mch, h mch.

II/. PHN L THUYT nh lut Ohm Quan h gia dng v p trn mt phn t in tr R:

U = R.I Cng sut trn in tr R:

P = U.I nh Lut K1 Tng i s cc dng in bt k ti mt nt bng khng.

Qui c: Cc dng in c chiu dng i vo nt th ly du (+), cn i ra khi nt th ly du (-), hoc ngc li.



nh Lut K2 Tng i s cc in p trn cc phn t dc theo cc nhnh trn mt vng bng khng.

Nguyn l xp chng p ng to bi nhiu ngun kch thch tc ng ng thi th bng tng cc p ng to bi mi ngun kch thch tc ng ring r.

III/. DNG C TH NGHIM Bng mch th nghim: bn s 1 MODULE 1A v MODULE 1B. Ngun p v ngun dng DC iu chnh c. VOM. Dy ni.

IV/. PHN TH NGHIM 1/. Kho st nh lut K1 v K2 cho mch in ni tip v song song:

a/. Mch ni tip

Mch in th nghim nh hnh 1:

R1=10 , R2=2 , R3=2 , R4=4 .

S dng mch (a), MODULE 1A Cc bc thc hin:

1. t ngun dng v tr OFF. 2. Kt ni ngun dng vo s mch. 3. Mc Ampe k o dng in chy qua in tr R1.

Hnh 1

R3

R4

A

J

AR2

R1

VVVV1111

(1)(1)(1)(1)

JJJJ



4. Bt ngun dng ON, chnh ngun dng gi tr 3A. 5. c ch s dng in qua R1. 6. Dng mt Vn K ln lt o cc hiu in th gia hai u ngun dng,

R1, R2, R3, R4. Ghi cc gi tr o c vo bng s liu. 7. Bt ngun dng v tr OFF. 8. Qu trnh lm tng t nh cc bc trn, o dng in qua cc in tr R2. 9. c v ghi cc ch s in p, dng in o c vo bng.

Phn t Ngun dng R1 R2 R3 R4 Dng Ap

Cng sut Cu hi:

1. T bng s liu, hy tnh cng sut ca mi phn t (in vo bng). 2. Hy kim chng nh lut K1 v K2. nh lut K1 ti nt 1:

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nh lut K2 cho vng V1: ...............................................................................................................................

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3. Nhn xt v mi quan h gia cng sut ca ngun dng vi tng cng sut tiu tn trn cc in tr.

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b/. Mch song song

Mch in th nghim nh hnh 2:

R1=10 , R2=22 , R3=50 , R4=50



Hnh 2

S dng mch (b), MODULE 1A Cc bc thc hin:

1. t ngun p v tr OFF.

2. Kt ni ngun p vo s mch.

3. Mc Ampe k o dng in chy qua in tr R1.

4. Bt ngun p ON, chnh n gi tr 20V.

5. c ch s dng in qua R1.

6. Dng mt Vn K ln lt o cc hiu in th gia hai u R1, R2, R3, R4. Ghi cc gi tr o c vo bng s liu.

7. Bt ngun p v tr OFF.

8. Qu trnh lm tng t, ln lt o cc gi tr dng in chy qua cc in tr R2, R3 v R4.

9. c v ghi cc ch s in p, dng in o c vo bng sau

Phn t Ngun p R1 R2 R3 R4

Dng

p Cng sut

Cu hi:

1. T bng s liu, hy tnh cng sut ca mi phn t (in vo bng trn). 2. Kim chng nh lut K1 v K2

nh lut K1 ti nt 1

V

A A

R3A

A

R2 R4

R1 (1) (1) (1) (1)

VVVV1111



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nh lut K2 cho vng V1

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3. Nhn xt v mi quan h gia cng sut ca ngun p vi tng cng sut tiu tn trn cc in tr.

2/. Nguyn l xp chng Mch in th nghim nh hnh 3:

R1=20 , R2=5 , R3=10 , R4=5 .

S dng s trong MODULE 1B.

Cc bc thc hin:

1. t ngun p v ngun dng v tr OFF.

2. Kt ni ngun p v ngun dng vo s mch.

3. Mc Ampe k o dng in chy qua in tr R2.

4. ng kho K1, h kho K2. iu ny cho php kch thch mch bng ngun p v trit tiu ngun dng (ngun p ON, ngun dng OFF),

5. Bt ngun p ON v ngun dng OFF, chnh ngun p gi tr 20V.

6. c v ghi vo bn s liu ch s dng in qua R2, gi l dng I21.

7. Dng Vn K o hiu in th gia hai u R2, gi l U21. Ghi cc gi tr o c vo bng s liu.

Hnh 3

A

V IR4R2

R1 R3

----



8. M kho K1, ng kho K2. iu ny cho php kch thch mch bng ngun dng v trit tiu ngun p (ngun p OFF, ngun dng ON), chnh ngun dng gi tr 3A.

9. c v ghi vo bn s liu ch s dng in qua R2, gi l dng I22.


11. ng c 2 kho K1 v kho K2. iu ny cho php kch thch mch bng c hai ngun dng v ngun p (c hai ngun ON).

12. c v ghi vo bn s liu ch s ca Ampe k, ch dng in qua R2, gi l dng I2.


Ngun p ON Ngun dng OFF

Ngun p OFF Ngun dng ON

Ngun p ON Ngun dng ON

I21 U21 I22 U22 I2 U2

Cu hi:

1. T bng s liu trn, so snh I2 vi I21 v I22, U2 vi U21 v U22.

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2. Nguyn l xp chng:

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3. Nhn xt v nguyn l xp chng cho mch in trn.

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4. Sinh vin hy t mc mch kho st nguyn l xp chng trn cc in tr R1, R3, R4 theo trnh t cng vic ging nh trn.



in tr R1 Ngun p ON

Ngun dng OFF Ngun p OFF

Ngun dng ON Ngun p ON

Ngun dng ON

I11 U11 I12 U12 I1 U1

Nhn xt: ............................................................................................................................

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in tr R3 Ngun p ON



Ngun dng ON

I31 U31 I32 U32 I3 U3

Nhn xt: ............................................................................................................................

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in tr R4 Ngun p ON



Ngun dng ON

I41 U41 I42 U42 I4 U4

Nhn xt: ............................................................................................................................

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3/. Hin tng ngn mch, h mch

Mch in th nghim nh hnh 4.

S dng s trong MODULE 1B.



K2

1

2

MODULE 1B

I

R1 20 R3 10

R2 5

R4 5

V

K1

Cc bc thc hin:

Ngun p cung cp cho mch o l 20V, ngun dng l 3A


2. Kt ni ngun p v ngun dng vo s mch. Kho K1 v kho K2 ng.

3. H mch gia hai im E-M bng cch h mch hai im E-F, dng dy dn ni tt cc cp im sau li vi nhau: A-B, C-D, G-H.

4. Bt ngun p v ngun dng ON.

5. o hiu in th h mch gia hai im E-M, gi l Uhm .

6. o dng in qua cc in tr R1 v R3. Gi l IR1, IR3.

7. Ghi cc gi tr o c vo bng s liu.

8. Bt ngun p v ngun dng v tr OFF.

9. Ngn mch gia hai im E-M bng cch mc gia hai im E M mt Ampe k .

10. Bt ngun p v ngun dng ON. c ch s ca Ampe k, gi l Inm .

11. Bt ngun p v ngun dng v tr OFF.

A B C D

E

F

G

H

Hnh 4

M



12. Ni tt gia E-M bng mt si dy dn, mc Ampe k gia hai im A-B o dng IR1.

13. Tng t, o dng IR3.

14. Lu , mi ln thay i v tr mc Ampe k nn bt ngun p v ngun dng v tr OFF.

H mch hai im E-M Ngn mch hai im E-M

Uhm IR1 IR3 Inm IR1 IR3

T bng s liu, hy nhn xt cc kt qu o c.

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Bi 1.2 MCH THEVENIN V MNG MT CA I/. MC CH

Minh ha cc nh l Thevenin v Norton. S dng s tng ng Thevenin v Norton kho st mng mt ca.

Kho st s phi hp tr khng gia ngun v ti truyn cng sut cc i.

II/. PHN TM TT L THUYT nh L Thevenin

C th thay th tng ng mt mng mt ca tuyn tnh bi mt ngun in p bng in p trn ca khi h mch mc ni tip vi tr khng Thevenin ca mng mt ca.

nh L Norton C th thay th tng ng mt mng mt ca tuyn tnh bi mt ngun dng

in bng dng in trn ca khi ngn mch mc song song vi tr khng Thevenin ca mng mt ca.

Trong : Uhm : in p h chng mch trn phn mng mt ca.

Inm : dng in ngn mch qua ca phn mng mt ca.

Zo : tr khng vo mng mt ca khi cho n tr nn th ng.

Mt khc, cc gi tr trong cc s tng ng trn cn c lin quan vi nhau theo cng thc sau:

+ Uhm Zo

a/ Sa/ Sa/ Sa/ S Thevenin Thevenin Thevenin Thevenin

Inm Zo

b/ Sb/ Sb/ Sb/ S Norton Norton Norton Norton



nmIhmU

oZ =

Cng thc ny gip ta xc nh c Zo thng qua vic xc nh Uhm v Inm .

Nguyn l truyn cng sut cc i: Theo nh l Thevenin, bt k mt mng mt ca no ni vi ti cng c th

a v s tng ng nh hnh sau:

Trong E v Zn c xc nh theo nh l tng ng Thevenin. Ti Zt s nhn c cng sut cc i nu:

Zt = Zn

y l iu kin phi hp tr khng gia ngun v ti

Trong bi th nghim ny, Zn = Rn v Zt = Rt u l thun tr nn iu kin phi hp tr khng s l:

Rt = Rn

III/. DNG C TH NGHIM Bng mch th nghim: bn s 1 MODULE 1C.

Ngun p v ngun dng DC iu chnh c.

VOM.

Dy ni.

IV/. PHN TH NGHIM Mch th nghim nh hnh 5:

+ E Zn

Mch tng ng

mng mt ca

Zt

c/ Mc/ Mc/ Mc/ Mng mng mng mng mt ct ct ct caaaa



R3

R4V

B

R1

RL

R5

R2

A

I

Hnh 5

Mc ch: Xy dng s tng ng Thevenin v Norton ca mch nhn t 2 cc A v B.

S dng s trong MODULE 1C.

MODULE 1C

R1

22

R3

10

R210

R45

I sourceV source

R5 47

K1 K2

A RLV

Hnh 6

1111 2222 AAAA

BBBB

CCCC

DDDD



1/. Xy dng m hnh mch tng Thevenin - Norton

Cc bc thc hin:


2. Kt ni ngun p v ngun dng vo s mch.

3. ng 2 kha K1 v kha K2.

4. H mch hai im A-B (hai u in tr R2). 5. Bt ngun p v ngun dng ON, chnh ngun p 20V v ngun dng

3A.

6. Dng Vn k o hiu in th h mch gia hai im A-B, gi l Uhm .

7. Tt ngun p v ngun dng.

8. Ngn mch A-B bng cch gn Ampe k gia A v B.


10. c gi tr trn Ampe k, gi l dng ngn mch Inm.

Uhm Inm Ro

Cu hi:

1. T bng s liu hy tnh in tr tng ng Thevenin/Norton Ro

R 0 =

2. V mch tng ng Thevenin v mch tng ng Norton. Ghi y cc gi tr ca cc phn t trong mch.



2/. Phi hp tr khng gia ngun v ti (truyn cng sut cc i) Cc bc thc hin

1. Tt ngun p v ngun dng. Bt kho K1 v tr ON, kho K2 ON.

2. Ni vo hai im A-B mt bin tr Rt bng cch ni tt hai im B-D v mc Ampe k gia hai im A-C.


4. iu chnh bin tr Rt khong 10 gi tr t khng n gi tr cc i.

5. ng vi mi gi tr ca Rt hy ghi li gi tr dng in v in p trn Rt ri in vo bng sau:

STT 1 2 3 4 5 6 7 8 9 10

Rt

U

I

P

Cu hi:

1. Tnh ton cng sut tiu tn P trn Rt ng vi tng gi tr ca Rt (in gi tr tnh ton vo bng trn).

2. T bng gi tr o c, v th cng sut tiu tn P trn Rt theo gi tr ca Rt, P = f(Rt).

1 2 3 4 5 6 7 8 9 10

P

Rt



3. T th hy xc nh gi tr ca Rt sao cho cng sut tiu tn trn Rt l ln nht. Rtmax =

4. Theo l thuyt gi tr ca Rt l bao nhiu. Rtmax (l thuyt) =

5. Nhn xt v cc gi tr ny.

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Bi 2 TH NGHIM MCH IN AC 1 PHA V 3 PHA

Bi 2.1 MCH XOAY CHIU 1 PHA

I/. MC CH Kho st cc thng s c trng ca mt on mch c kch thch bng ngun xoay chiu hnh sin nh: p, dng, tr khng, cng sut, lch pha.

V gin vect dng, p ca mt on mch.

II/. TM TT L THUYT Trong ch xc lp iu ho (ngun kch thch hnh sin), cc thng s trong mi nhnh c mi quan h vi nhau nh sau:

U = Z*I hay I = Y*U

= u - i P = U x I x Cos() = Re[Z] x I2

III/. DNG C TH NGHIM Bng mch th nghim: bn s 2 MODULE 2A.

Ngun p xoay chiu 1 pha iu chnh c, tn s 50 Hz.

VOM s.

Watt k.

Dy ni.

IV/. PHN TH NGHIM 1/. Mch thun tr, thun dung, thun cm

Mch th nghim nh hnh 7:



A

ZVac

B

AW*

*

V

Hnh 7

Da vo hnh 7, sinh vin ln lt mc mch vi cc ti Z l: R, L, C.

Vi trng hp ti R, sinh vin t v mch th nghim vi y dng c o, th nghim c tin hnh theo cc bc sau:

1. Bt cng tc ngun v tr OFF, iu chnh bin p t ngu v v tr 0V.

2. Mc mch nh hnh v, lu cc dng dng v p ca Watt k. 3. Bt cng tc ngun v tr ON, iu chnh bin p t ngu sao cho ch

s ca Vn k l 220V.

4. c cc tr s o cc gi tr in p, dng in, cng sut v ghi vo bng sau tng ng vi tng trng hp.

Qu trnh lm tng t cho cc ph ti L v C. Cc gi tr o c ghi vo bng sau

U (V) I (A) P(kW) R

L

C

T bng s liu o c, trong mi trng hp hy:

1. Xc nh gc lch pha gia hiu in th v dng in gia hai im A-B (in gi tr tnh ton c vo bng trn).

2. V gin vect dng v p ca on mch.



3. Tnh cng sut tc dng trn mi phn t mch theo l thuyt. So snh vi cng sut o c.

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4. Nhn xt.

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2/. Mch R, L, C ni tip

Mch th nghim nh hnh 8:

C

AR

V L

W*

*C

D

Vac

A

B

Hnh 8

Th nghim c tin hnh theo cc bc sau:


2. Mc mch nh hnh v 8, lu cc dng dng v p ca Watt k. 3. Bt cng tc ngun v tr ON, iu chnh bin p t ngu sao cho ch

s ca Vn k l 200V.

4. c cc tr s o tr in p, dng in, cng sut v ghi vo bng sau:

U(V) I(A) P(kW) on mch A-B

R

L

C



Lu : o cng sut trn cc phn t R, L, C ta ch vic mc cun p ca Watt k mc ti: A-C, C-D, D-B.

T bng s liu o c hy:

1. Xc nh gc lch pha gia hiu in th v dng in gia hai im A-B v tng phn t mch (gi tr tnh ton in vo bng trn).


3. Tnh biu thc UAB.I.Cos(). So snh vi cng sut o c. .............................................................................................................................

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Xc nh tr khng phc ZAB t gi tr U, I va o c trong bng s liu

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4. Nhn xt.

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3/. Mch R, L, C song song

S mch th nghim nh hnh 9:



Hnh 9

Th nghim c tin hnh theo cc bc sau


2. Mc mch nh hnh v 9, lu cc dng dng v p ca Watt k. 3. Bt cng tc ngun v tr ON, iu chnh bin p t ngu sao cho ch

s ca Vn k l 200V.


Lu : trc khi chuyn v tr cc Ampe k ti cc v tr cc nhnh o dng in nhnh th phi bt cng tc ngun v tr OFF.


R

L

C

T bng s liu o c hy:

1. Xc nh gc lch pha gia hiu in th v dng in gia hai im A-B v tng phn t mch bng gin vect (gi tr tnh ton in vo bng trn).


V Vac R L

W *

* A

C

B

A



3. Tnh biu thc UAB.I.Cos(). So snh vi cng sut o c. ..............................................................................................................................

4. Xc nh tr khng phc ZAB t gi tr U, I va o c trong bng s liu

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5. Nhn xt.

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4/. Mch R, L, C hn hp

S mch th nghim nh hnh 10:

C

A

R

V

W*

*A

L

B

VacC

Hnh 10

Th nghim c tin hnh theo cc bc sau:


2. Mc mch nh hnh v 10, lu cc dng dng v p ca Watt k. 3. Bt cng tc ngun v tr ON, iu chnh bin p t ngu sao cho

ch s ca Vn k l 200V.




Lu : trc khi chuyn v tr cc Ampe k, Volt k v Watt k vi cc v tr cc nhnh o dng in nhnh th phi bt cng tc ngun v tr OFF.


R

L

C

T bng s liu o c hy:

1. Xc nh gc lch pha gia hiu in th v dng in gia hai im A-B v tng phn t mch bng gin vect (gi tr tnh ton in vo bng trn).


3. Tnh biu thc UAB.I.Cos(). So snh vi cng sut o c. ..............................................................................................................................

4. Xc nh tr khng phc ZAB t gi tr U, I va o c trong bng s liu

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5. Nhn xt.

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Bi 2.2 MCH XOAY CHIU 3 PHA

I/. MC CH Kho st vic u dy cho ngun ti v cch mc cc dng c o trong mt mch in ba pha.

Kho st vic o cng sut trn mt mch ba pha.

II/. TM TT L THUYT Mach ba pha l h thng gm ba sc in ng, ba ti v cc dy ni chng. Mt h thng ba pha thng dng nht l h thng ba pha c ngun ba pha i xng. C hai cch mc ti l mc hnh sao v mc tam gic. Trong tng trng hp, cn lu n cc cng thc tnh ton dng, p pha v dy. o cng sut ti ba pha, ngi ta ln lt o cng sut tng pha (phng php dng ba Watt k) hay dng phng php hai Watt k. Mi phng php o c tng c im ring v phm vi kho st ring m sinh vin cn lu . Sinh vin cn n li kin thc mch ba pha trong gio trnh mch in 1 trc khi th nghim bi ny.

III/. DNG C TH NGHIM Bng mch th nghim : bn s 2 MODULE 2B, MODULE 2C

Ngun p xoay chiu 3 pha, tn s 50 Hz.

VOM s.

Watt k.

Dy ni

IV/. PHN TH NGHIM 1/. Mch 3 pha ti hnh sao - c dy trung tnh

S mch th nghim nh hnh 11. Ti Z1, Z2, Z3 l thun tr.



Hnh 11

t cc cng tt ngun V1, V2, V3 v tr OFF trc khi tin hnh mc mch.

Sinh vin t mc mch nh hnh 11 v cc ng h o lng thch hp thc hin o cc gi tr dng in pha, p pha, p dy v dng dy trung tnh ri ghi kt qu vo bng sau:

Up1 Up2 Up3 Ud1 Ud2 Ud3 Ip1 Ip2 Ip3 Itt

o cng sut trn cc ti t suy ra cng sut ca mch ba pha.

P1 P2 P3 Pt

Cu hi:

1. V th vect dng p pha v dy. Nhn xt mi quan h gia cc i lng.

ZZZZ1111

V3

V1

ZZZZ2222

ZZZZ3333

V2

0

Ip

Ip

Ip

It



...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

2. Lm li th nghim trn vi mi ti Z1, Z2, Z3 l R-C mc ni tip.



P1 P2 P3 Pt

V th vect dng p pha v dy. Nhn xt mi quan h gia cc i lng

................................................................................................................................

................................................................................................................................

................................................................................................................................

3. Lm li th nghim trn vi ti Z1, Z2 l thun tr R, Z3 l R, L, C mc ni tip (ti bt i xng).



P1 P2 P3 Pt




...........................................................................................................................................

...........................................................................................................................................

...........................................................................................................................................

2/. Mch 3 pha ti hnh sao khng dy trung tnh (i xng v bt i xng) S mch th nghim nh hnh 12. Ti Z1, Z2, Z3 l thun tr.

Hnh 12

Bt cc cng tt ngun V1, V2, V3 v tr OFF trc khi tin hnh mc mch.

Sinh vin mc mch nh hnh 12 v thc hin o dng pha, p pha, p dy ri ghi kt qu vo bng sau:

Up1 Up2 Up3 Ud1 Ud2 Ud3 Ip1 Ip2 Ip3

V1

V2

V3

ZZZZ1111

ZZZZ2222

ZZZZ3333

0

Ip1

Ip2

Ip3



o cng sut trn cc ti t suy ra cng sut ca mch ba pha:

P1 P2 P3 Pt

Cu hi:


......................................................................................................................................

......................................................................................................................................

......................................................................................................................................

2. o in p im trung tnh. Nhn xt.

Utt =

3. Lm li th nghim trn vi mi ti Z1, Z2, Z3 l R, L mc ni tip.



P1 P2 P3 Pt




..................................................................................................................................

..................................................................................................................................

..................................................................................................................................

4. Lm li th nghim trn vi ti Z1, Z2 l thun dung C, Z3 l R, C mc ni tip (ti bt i xng).



P1 P2 P3 Pt


..................................................................................................................................

..................................................................................................................................

3/. Mch 3 pha ti hnh tam gic (i xng v bt i xng) S mch th nghim nh hnh13. Ti Z1, Z2, Z3 l thun tr.

Hnh 13

V3

0

V1

V2

ZZZZ2222

ZZZZ3333

ZZZZ1111

Ip1

Ip2

Ip3 Id1

Id2

Id3




Sinh vin mc mch nh hnh v 13 v thc hin o dng pha, dng dy, p pha ( p dy) ri ghi kt qu vo bng sau:

Ud1 Ud2 Ud3 Ip1 Ip2 Ip3 Id1 Id2 Id3


P1 P2 P3 Pt

Cu hi:


..................................................................................................................................

..................................................................................................................................

..................................................................................................................................

2. Lm li th nghim trn vi mi ti Z1, Z2, Z3 l R, L mc ni tip.



P1 P2 P3 Pt




..................................................................................................................................

..................................................................................................................................

..................................................................................................................................

3. Lm li th nghim trn vi ti Z1, Z2 l thun dung C, Z3 l thun tr R (ti bt i xng).



P1 P2 P3 Pt


.........................................................................................................................

.........................................................................................................................

.........................................................................................................................



4/. Kho st hin tng mt pha

a/. Ti mc hnh sao


Sinh vin mc mch nh hnh v 11 vi Z1, Z2, Z3 l thun tr R

Bt cc cng tt ngun V1, V2, v tr ON, V3 v tr OFF.

Thc hin o dng pha (dng dy), p pha, p dy, dng dy trung tnh ri ghi kt qu vo bng sau:

So snh kt qu vi kt qu o phn 1. Nhn xt khi xy ra hin tng mt pha.

..................................................................................................................................

..................................................................................................................................

b/. Ti mc tam gic


Sinh vin mc mch nh hnh v 13 vi Z1, Z2, Z3 l thun tr R

Bt cc cng tt ngun V1, V2, v tr ON, V3 v tr OFF.

Thc hin o dng pha, dng dy, p pha, p dy ri ghi kt qu vo bng sau:

Up1 Up2 Up3 Ip1 Ip2 Ip3 Id1 Id2 Id3

Ud1 Ud2 Ud3

So snh kt qu vi kt qu o phn 3. Nhn xt khi xy ra hin tng mt pha.

.........................................................................................................................

.........................................................................................................................

.........................................................................................................................




H V TN: ................................ LP: ...................................

MSSV: ...................................


.......................................................

.......................................................

Bi 3: MNG 2 CA TUYN TNH KHNG NGUN V MCH CNG HNG RLC

Bi 3.1 MNG HAI CA TUYN TNH KHNG NGUN I/. MC CH S dng mt trong cc phng php hc tm hiu cch xc nh cc thnh phn ca ma trn c trng cho mng hai ca, kho st cc tnh cht ca mng hai ca, ng dng l thuyt mng hai ca vo vic phn tch mch. Kho st ma trn A, ma trn Z v nguyn l tng h. II/. TM TT L THUYT Ta bit h phng trnh dng A ca mng hai ca tuyn tnh khng ngun c dng:

2.

222.

21 1.

2.

122.

111.

IA UA I

IA UA U

+=

+=

Lu : Chiu dng in I2 chn cng chiu in p U2. phn th nghim ny, tin hnh xc nh cc h s Aik thng qua cc trng thi c bit ca mng hai ca l ngn mch v h mch mt trong hai ca (xem thm phn l thuyt v mng hai ca). th vect dng p ca mch s gip ta xc nh c gc pha ca cc i lng vect trn mch.

C th nh sau:

A11 = 0IU

U

2.

2.

1.

=

A12 =0UI

U

2.

2.

1.

=

A21 = 0IU

I

2.

2.

1.

=

A22 =0UI

I

2.

2.

1.

=

i vi cc dng ma trn Z:



2.

121.

11 1.

I Z I Z +=U

2.

221.

21 2.

I Z I Z +=U

Cc thng s Zik c xc nh nh sau:

Z11 = 0II

U

2.

1.

1.

=

Z12 =0II

U

1.

2.

1.

=

Z21 = 0II

U

2.

1.

2.

=

Z22 =0II

U

1.

2.

2.

=

Nguyn l tng h Dng in pht sinh ti mt nhnh di kch thch ca mt ngun p duy nht

t ti nhnh th hai, s bng dng pht sinh ti nhnh th hai khi t ti nhnh th nht cng ngun p (v l ngun duy nht).

i vi mng hai ca, kim tra nguyn l tng h, ta c th tin hnh bng th nghim nh sau: cho ngun p vo ca 1 v ngn mch ca 2, ghi nhn dng ngn mch ca 2 ny; sau cho ngun p vo ca 2 v ngn mch ca 1, ghi nhn dng ngn mch ca 1 ny. Gi tr ca dng in qua hai ca b ngn mch cho ta kim chng tnh ng n ca nguyn l tng h. III/. DNG C TH NGHIM

Bng mch th nghim : bn s 3 MODULE 3A

Ngun AC 48V.

VOM S.

Dy ni

Cc thng s mch : R1 = 30 , R2 = 20, R3= 20, C= 150F, RL= 16..

IV/. PHN TH NGHIM S mch:



Hnh 14

1/. Xc nh ma trn A khi kha K h v ng

1.1/. Kha K m

Bt cng tc ngun v tr OFF, iu chnh bin p cch ly v v tr 0V.

H kha K.

Ni ca vo (1) (1) vi ngun AC (ng ra ca bin p cch ly). H mch ca ra (2) (2). Bt cng tc ngun v tr ON, iu chnh bin p cch ly sao cho in p U1 =

48V.

Thc hin o cc gi tr dng v p trn mch ri in vo bng sau:

I1 I2 IC UR1 UR2 UC U2 Ng ra

H mch


Vn h kha K.

Ni ca vo (1) (1) vi ngun AC (ng ra ca bin p cch ly). Ngn mch ca ra (2) (2). Bt cng tc ngun v tr ON, iu chnh bin p cch ly sao cho in p U1 =

48V.

RL

R2

C

R1

R3

1111 2222

2222 1111

IIII1111 IIII2222

IIIIcccc

++++

UUUU1111

----

++++

UUUU2222

----

KKKK

A A A





Ngn mch

Cu hi:

1.1.a/. T cc s liu o c ng vi hai trng hp h mch v ngn mch ng ra, hy v gin vect ca cc i lng dng p xc nh gc pha ca cc i lng phc I1, U2 , I2 vi gi s gc pha ca U1 l bng khng.

1.1.b/. Xc nh cc h s Aik ca ma trn thng s A.

A11 =

A12 =

A21 =

A22 =

1.2/. Kha K ng

Tun t thc hin cc bc th nghim nh trong trng hp kha K m

Cc gi tr dng p o c trong trng hp ng ra h mch.


H mch

Cc gi tr dng p o c trong trng hp ng ra ngn mch.


Ngn mch



Xc nh cc h s Aik ca ma trn thng s A.

A11 =

A12 =

A21 =

A22 =

2/. Kim chng phng trnh dng A

a/. ng vi hai trng hp ca kha K, hy tnh biu thc:

T = A11A22 - A12A21

a.1/. Kha K m:

T =

Nhn xt:

.................................................................................................................................

.................................................................................................................................

a.2/. Kha K ng:

T =

Nhn xt:

.................................................................................................................................

.................................................................................................................................



b/. Mc ti RL ti ca ra (2) (2), kha K h. Thc hin o t ht cc thng s sau:

I1 I2 IC UR1 UR2 UC U2 Ng ra gn

Ti RL

T s liu o, v gin vect v xc nh gc pha ca I1, U2, I2. Gi s gc pha ca U1 l bng 0.

S dng ma trn thng s A trong trng hp kha K h, hy tnh gi tr U2 v I2 theo U1, I1 bit. Hy so snh vi kt qu o c, nhn xt.

2.

U =

2.

I =

Nhn xt :

...............................................................................................................................

...............................................................................................................................

3/. Xc nh ma trn Z khi kha K h v ng

3.1/. Kho K h


H kha K.

Ni ca vo (1) (1) vi ngun AC (ng ra ca bin p cch ly). H mch ca ra (2) (2). Bt cng tc ngun v tr ON, iu chnh bin p cch ly sao cho in p U1 =

48V.





H mch


Vn h kha K.

Ni ca ra (2) (2) vi ngun AC (ng ra ca bin p cch ly). H mch ca vo (1) (1). Bt cng tc ngun v tr ON, iu chnh bin p cch ly sao cho in p U2 =

48V.


I1 I2 IC UR1 UR2 UC U1 Ng vo

H mch

Cu hi:

3.1.a/. T cc s liu o c ng vi hai trng hp h mch ng vo v ng ra, hy v gin vect ca cc i lng dng p xc nh gc pha ca cc i lng phc I1, U2 , I2 vi gi s gc pha ca U1 l bng khng.

3.1.b/. Xc nh cc h s Zik ca ma trn thng s Z.

Z11 =

Z12 =

Z21 =

Z22 =

3.1.c/. Nhn xt ma trn Z thu c:



.................................................................................................................................

3.2/. Kho K ng

Tun t thc hin cc bc th nghim nh trong trng hp kha K m

Cc gi tr dng p o c trong trng hp ng ra h mch.


H mch

Cc gi tr dng p o c trong trng hp ng vo h mch.

I1 I2 IC UR1 UR2 UC U1 Ng vo

H mch

Xc nh cc h s Zik ca ma trn thng s Z.

Z11 =

Z12 =

Z21 =

Z22 =

Nhn xt ma trn Z thu c:

.................................................................................................................................

.................................................................................................................................



4/. Kim chng nguyn l tng h v tuyn tnh a/. Nguyn l tng h: Mch th nghim nh hnh v, ln lt cho kha K h v ng hy o cc dng in I1 v I2 ng vi hai trng hp :

Ni ca vo vi ngun 48V, ngn mch ca ra.

Ni ca ra vi ngun 48V, ngn mch ca vo.

Ni ca vo vi ngun 48V Ngn mch ca ra.

Ni ca ra vi ngun 48V Ngn mch ca vo.

K h I1 = I2 = I1 = I2 =

K ng I1 = I2 = I1 = I2 =

Hy nhn xt kt qu o c.

...............................................................................................................................

b/. Tnh cht tuyn tnh :

Mc ti RL vo ca ra (2) (2) ca mng hai ca, kha K h. Chnh in p vo ca vo l U1 = 48V, o xc nh gi tr in p ng ra

U2 =

Da vo tnh cht t l hy tnh in p ng ra khi in p ng vo l 24V

U2 (U1 = 24V) = Hy mc mch v o c kim chng (khi U1 = 24V)

U2 =

Nhn xt :

...............................................................................................................................



H V TN: ................................ LP: ...................................

MSSV: ...................................


.......................................................

.......................................................

Bi 3.2 MCH CNG HNG RLC I/. MC CH Bi th nghim gip sinh vin hiu c mt s c tnh ca mch cng hng R L C, cch xc nh tn s cng hng ca nhnh, xc nh bng thng ca mch cng hng v kho st hnh dng cc tn hiu trong mch khi hin tng cng hng xy ra.

II/. TM TT L THUYT mch RLC ni tip, tr hiu dng in p trn cc phn t khng gn cng

hng s rt ln so vi in p vo mch. mch RLC song song th dng in qua mc li LC gn cng hng s rt ln so vi dng in cp cho nhnh.

Ti tn s cng hng, bin tn hiu ng ra s l cc i. V khong tn s, m bin hm truyn t p gim 2 bin cc i, c gi l bng thng ca mch cng hng (k hiu l B). H s phm cht Q ca mch cng hng c th c tnh bng cng thc :

Q = f0 / B f0 : Tn s cng hng. o lch pha hai tn hiu bng dao ng k

Gi s 2 tn hiu a vo dao ng k ng x v y c biu thc :

x = asin(t) y = bsin(t + )

Dao ng k s biu din trn mn hnh tn hiu y qut theo tn hiu x nh hnh 15.

2a Hnh 15

y

x

2b Y0 r

2R

r



xc nh gc pha bng dao ng k, ngi ta c hai cch n gin:

Cch 1 : (Dng th) Ta c th thy rng, ti t = 0 th x = 0 v y = Y0

=> bsin () = Y0 => sin() = Y0 / b y b l gi tr cc i ca tn hiu y, c th xc nh trn mn hnh v Y0 cng

tng t. Nh vy gc lch pha s c tm t :

Sin() = Y0 /b Phng php ny r rng n gin, nhng thc ra n ch hu hiu cho cc gi tr

gc pha nh. Khi sin() thay i nhanh theo ( nh hn hay bng 450). Cn vi cc gi tr gn 900 th tr sin() thay i rt chm, v chnh xc s gim.

Cch 2 :(Dng dao ng k so pha) R rng chng ta c th cho cng lc hai tn hiu vo dao ng k (chn MODE ca

tng qut dc l DUAL hay CHOP) v so pha da vo cc thng s dc c trn mn hnh. Cch ny cn cho ta thy c s nhanh chm pha ca hai tn hiu.

III/. DNG C TH NGHIM Bng mch th nghim : bn s 3 MODULE 3B

My pht sng

Dao ng k

Dy ni

Cc thng s mch : C = 1 f ; L=10 mH ; Rs = 100. ; RL = 10.

IV/. PHN TH NGHIM 1/. Mch R-L-C ni tip

Mc mch th nghim nh hnh 16 :



iu chnh my pht sng ch pht tn hiu hnh sin.

Mc dao ng k vo nh hnh v 16 quan st tn hiu Vout nh hnh v. iu chnh bin tn hiu ng ra my pht sng khong 5V. Cho tn s ca my

pht sng thay i t 100Hz n 10KHz, t xc nh tn s cng hng ca mch, f0.

Lu : ti tn s cng hng tn hiu ra Vout(t) s c bin ln nht. So snh kt qu t c vi cng thc l thuyt.

f0 =

f0 (l thuyt) = Hy xc nh bin tn hiu ng ra ng vi cc tn s tn hiu ng vo theo bng

sau :

f(Hz) 100 200 400 800 1000 1500 2000 3000 6000 10000 Vout

Lu : ng vi mi tn s ca Vin, ta phi iu chnh li bin tn hiu vo gi c nh l 5V, bi v khi thay i tn s my pht bin ca tn hiu ra cng i theo.

Cu hi :

1.a/. Vi 10 gi tr o c v mt gi tr ti tn s cng hng, sinh vin hy v p ng tn s ca mch cng hng trn : th Vout theo tn s ng vo f, Vout(f).

My My My My Pht Pht Pht Pht SngSngSngSng R

CLR+

VVVVinininin

-

+

VVVVoutoutoutout

-

Hnh 16

CH 1CH 1CH 1CH 1

DAO DAO DAO DAO NGNGNGNG KKKK

GNDGNDGNDGND



100 200 400 800 1000 1500 2000 3000 6000f(Hz)

Vou

t

1.b/. Tnh ton bng thng mch cng hng theo tn s f1 v f2. Vi f1 v f2 l cc tn s m in p trn Rs gim i 2 ln so vi gi tr ti tn s cng hng.

f1 =

f2 =

Bng thng cng hng B :

B = f2 f1

Xc nh h s phm cht Q Q = 1.c/. o tr hiu dng in p trn t khi tn s f thay i t f0/3 n 3f0, vi

khong 10 gi tr. V th Uc(f) ?

f(Hz) UC(V)



f(Hz)

Uc

(V

)

2. Mch R v L - C song song


My My My My Pht Pht Pht Pht SngSngSngSng

RS RL

C L

+

VVVVinininin

-

DAO DAO DAO DAO NGNGNGNG KKKK

CH 1CH 1CH 1CH 1

Hnh 17

Vout

GNDGNDGNDGND



iu chnh my pht sng ch pht tn hiu hnh sin.

Mc dao ng k vo nh hnh v 17 quan st dng sng trn in tr Rs. iu chnh bin tn hiu ng ra mt pht sng khong 5V. Cho tn s ca my

pht sng thay i t 100Hz n 10KHz, t xc nh tn s cng hng ca mch, f0. Lu : ti tn s cng hng tn hiu ra in p trn in tr Rs s c bin nh nht. So snh kt qu t c vi cng thc l thuyt.

f0 =

f0 (l thuyt) = Hy xc nh bin tn hiu ng ra ng vi cc tn s tn hiu ng vo theo bng

sau :

f(Hz) 100 200 400 800 1000 1500 2000 3000 6000 10000 Vout

Lu : ng vi mi tn s ca Vin, ta phi iu chnh li bin tn hiu vo gi c nh l 5v, bi v khi thay i tn s my pht bin ca tn hiu ra cng i theo.

Cu hi :

2.a/. Vi 10 gi tr o c v mt gi tr ti tn s cng hng, sinh vin hy v p ng tn s ca mch cng hng trn : th Vout theo tn s ng vo f, Vout(f).

100 200 400 800 1000 1500 2000 3000 6000f(Hz)

Vou

t



2.b/. Tnh ton bng thng mch cng hng theo tn s f1 v f2. Vi f1 v f2 l cc tn s m in p trn Rs bng 2 ln so vi gi tr ti tn s cng hng.

f1 =

f2 =

Bng thng cng hng B :

B = f2 f1

Xc nh h s phm cht Q Q =

2.c/. o tr hiu dng in p trn t khi tn s f thay i t f0/3 n 3f0, vi khong 10 gi tr. V th Uc(f) ?

f(Hz) UC(V)

f(Hz)

Uc

(V

)



H V TN: ................................ LP: ...................................

MSSV: ...................................


.......................................................

.......................................................

Bi 4 QU TRNH QU MCH TUYN TNH - MCH PHI TUYN

Bi 4.1 QU TRNH QU MCH TUYN TNH - I/. MC CH Bi th nghim ny gip sinh vin hiu c mt s c tnh qu mch tuyn tnh, bao gm cc mch: R-C, R-L v mch R-L-C. Thng qua cc c tnh ny, sinh vin c th kim nghim c cc phng php phn tch mch qu hc phn l thuyt v hiu thm c mt s qu trnh vt l xy ra trong cc phn t mch thc t.

II/. TM TT L THUYT Qu trnh qu l qu trnh xut hin khi mch chuyn t mt ch xc lp ny sang ch xc lp khc (xem thm l thuyt v qu trnh qu ). Thng thng thi gian qu rt ngn nn quan st ngi ta s dng ngun kch thch chu k c bin bin thin t ngt (ng m theo chu k). II/. DNG C TH NGHIM

Bng mch th nghim: bn s 4 MODULE 4A My pht sng. Dao ng k, VOM. Dy ni. Cc thng s mch: VR = 10 K; R =100 ; C = 50 nF ; L = 1 mH.

III/. PHN TH NGHIM : 1/. Mch qu cp mt R-C:

t cng tc ngun my pht sng v tr OFF.




Cc bc thc hin

iu chnh bin tr VR = 1K.

Mc u o ca dao ng k vo hai m E G quan st tn hiu ng ra ca my pht sng.

Cu hi:

1.1. Bt ngun my pht sng v iu chnh my pht tn hiu xung vung c bin l 5V v tn s 1 KHz.

V tn hiu ng ra ca my pht sng.

T

Uou

t

VR R

C My Pht Sng

E F

G

H

Hnh 18

DAO NG K

CH 1

GND



1.2. Mc u o ca dao ng k vo hai m H G quan st tn hiu in p trn t C, uC(t).

V dng sng ca tn hiu uC(t).

T

Uc (t)

T dng sng uC(t), hy tnh thi hng ca mch ny. .................................................................................................................................

So snh vi l thuyt. Nhn xt ..............................................................................................................................

..............................................................................................................................

..............................................................................................................................

1.3. Mc u o ca dao ng k vo hai m F H quan st tn hiu in p trn in tr R, uR(t).

V dng sng ca tn hiu uR(t).

T

uR(t)



T dng sng uR(t) tnh thi hng ca mch ny. ...............................................................................................................................

...............................................................................................................................

...............................................................................................................................

So snh vi l thuyt. Nhn xt. .................................................................................................................................

.................................................................................................................................

1.4. Thay i gi tr ca VR. Nhn xt v s thay i thi hng.

.................................................................................................................................

.................................................................................................................................

2/. Mch qu cp mt R-L t cng tc ngun my pht sng v tr OFF. Mc mch th nghim nh hnh 19:

Cc bc thc hin iu chnh bin tr VR= 1k. Mc u o ca dao ng k vo hai m E G quan st tn hiu ng ra

ca my pht sng. Cu hi

2.1. Bt ngun my pht sng v iu chnh my pht sng pht tn hiu xung vung c bin l 5V v tn s 1 Khz.


E F H

G

My Pht Sng

GN

DAO NG K

CH

Hnh 19

VR R

L



T

Uou

t

2.2. Mc u o ca dao ng k vo hai m H G quan st tn hiu in p trn cun dy L, uL(t).

V dng sng ca tn hiu uL(t).

T

uL(t

)

T dng sng ca uL(t) hy tnh thi hng ca mch. ...............................................................................................................................

...............................................................................................................................

So snh vi l thuyt. Nhn xt. .................................................................................................................................

.................................................................................................................................





T

uR(t)

T dng sng uR(t) tnh thi hng ca mch . So snh vi l thuyt. Nhn xt.

.................................................................................................................................

.................................................................................................................................

2.4. Thay i gi tr ca VR. Nhn xt v s thay i thi hng.

...................................................................................................................................................

...................................................................................................................................................

3/. Mch qu cp hai R-L-C Bt cng tc ngun my pht sng v tr OFF. Mc mch th nghim nh hnh 20 :

Hnh 20

GND

DAO NG K

CH1

E F H

K G

My Pht Sng

VR

C

R

L



Cc bc thc hin

iu chnh bin tr VR = 0.

Mc u o ca dao ng k vo hai m E G quan st tn hiu ng ra ca my pht sng.

Cu hi

3.1. Bt ngun my pht sng v iu chnh my pht sng pht tn hiu xung vung c bin l 5V v tn s 1Khz.


T

Uou

t



T

uR(t)

3.3. Mc u o ca dao ng k vo hai m H K quan st tn hiu in p trn t C, uC(t).



V dng sng ca tn hiu uC(t).

T

uC(t)

3.4. Nhn xt v cc dng sng o c. .................................................................................................................................

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3.5. Bng l thuyt hy, xc nh in tr ti hn ca mch Rth. Cho bit khi no th mch trng thi dao ng v khng dao ng.

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3.6. Thay i gi tr ca VR, hy nhn xt v s thay i dng sng uC(t). .................................................................................................................................

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3.7. V dng sng ca tn hiu uC(t) ng vi VR = 2k.

T

uC(t)



H V TN: ................................ LP: ...................................

MSSV: ...................................


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Bi 4.2 MCH PHI TUYN I/. MC CH

Bi th nghim ny gip sinh vin n tp cc l thuyt c bn v phn t phi tuyn v mt s hiu ng c bit gy ra bi tnh phi tuyn ca phn t mch. II/. TM TT L THUYT

Mt trong cc m hnh in hnh cho cc phn t mch phi tuyn ph bin trong thc tin l m hnh in tr phi tuyn v in cm phi tuyn. Hin nay c rt nhiu phng php kho st mch phi tuyn nh : phng php gii tch, phng php th, phng php s . . . . Mc d c khuyt im l chnh chnh xc khng cao, nhng phng php th cho php ngi kho st tip cn v xem xt nh tnh mt cch nhanh chng cc mch c cha phn t phi tuyn. Bi th nghim ny ch yu i vo phng php th, ng dng n trong vic kho st c tuyn mch, hay cc qu trnh xy ra trong mch khi c mt phn t tr phi tuyn.

Trong bi th nghim ny, in tr phi tuyn c kho st l Diode. xy dng c tuyn ca in tr phi tuyn ta s dng phng php Volt-Amper : Kch thch cho Diode bng mt ngun tn hiu sin, tn hiu in p v tn hiu dng in tng ng c a vo ng qut ngang v qut dc ca dao ng k. th hin ra trn dao ng k chnh l c tuyn V-A ca phn t phi tuyn i = f(u). Tuy nhin, chng ta phi chuyn th Volt Volt trn mn hnh sang th Volt - Amper bng cch ly i xng qua trc tung(thay u bng u).

Phn tch mch phi tuyn bng phng php th: Phn tch mch ang xem xt thnh 2 phn nh hnh bn di:

Phn th nht l phn mch tuyn tnh. Phn mch ny c biu din di dng mch tng ng Thevenin, gm ngun UT mc ni tip vi in tr RT. Mi quan h gia dng v p ti hai cc ca phn mch tuyn tnh c c trng bi phng trnh ng ti:

+ UUUUTTTT RRRRTTTT

PhPhPhPhn mn mn mn mch tuych tuych tuych tuyn tnhn tnhn tnhn tnh

PhPhPhPhn mn mn mn mch phi tuych phi tuych phi tuych phi tuyn n n n

++++

uuuu

----

iiii iiii



T

T

RuUi =

(1)

Lu : ng ti (1) ct trc honh ti im c gi tr bng in p ngun UT v c h s gc l -1/RT . Nu in p ngun UT thay i th ng ti s thay i nhng chng lun song song vi nhau.

Phn th hai l phn mch phi tuyn. Phn t phi tuyn ny c c trng bi c tuyn V-A: i = f(u) (2).

Nu UT l ngun in p mt chiu: ng ti (1) s khng thay i theo thi gian. Giao im ca hai th (1) v (2) cho ta bit gi tr dng v p trn phn t phi tuyn.

Nu UT l ngun in p xoay chiu: ng ti (1) s thay i lin tc theo thi gian, t th ta s thu c dng sng ca dng in i(t) nh hnh v:

uuuu

iiii

U

I

(1)(1)(1)(1)

(2)(2)(2)(2)

UT

T

T

R

U

uuuu

iiii

UT

(1)(1)(1)(1)

(2)(2)(2)(2)

tttt

iiii

Um

Um/RT

tttt



Nu UT bao gm ngun in p mt chiu v xoay chiu: ng ti (1) s thay i lin tc theo thi gian xung quanh ng ti tnh, t th ta s thu c dng sng ca dng in i(t) nh hnh v:

T th chng ta nhn thy rng nu bin ca thnh phn xoay chiu ca ngun UT tng ln th dng in qua phn t phi tuyn s khng l hnh Sin na.

III/. DNG C TH NGHIM Bng mch th nghim: bn s 4 MODULE 4B. My pht sng. Ngun p DC iu chnh c. Dao ng k. VOM. B dy ni.

IV/. PHN TH NGHIM 1/. Xy dng c tuyn V-A ca phn t tr phi tuyn: Diode chnh lu v Diode Zener a/. c tuyn V-A ca Diode chnh lu:

Bc cng tc ngun my pht sng v tr OFF. Mc mch th nghim nh hnh 21.

uuuu

iiii

UT

(1)(1)(1)(1)

(2)(2)(2)(2)

tttt

iiii

UDC

UDC/RT

tttt



Cc bc thc hin

Mc dao ng k nh hnh v, in p trn in tr c a vo knh Y, v in p ngc trn Diode c a vo knh X.

Bc ngun my pht sng v iu chnh my pht sng pht tn hiu hnh sin c bin l 5V v tn s khong 50Hz hoc nh hn.

Chn H-Mode ca dao ng k l X-Y. Cu hi:

1a.1. Hy v th nhn c t dao ng k.

1a.2. T th nhn c hy suy ra c tuyn V-A ca Diode vi chiu dng v p trn Diode nh hnh 22.

My pht sng

CH Y

GND

CH X

D

R

Hnh 21

iiii

+ u + u + u + u ----

D

Hnh 22



b/. c tuyn V-A ca Diode Zener:

Sinh vin mc mch th nghim nh hnh 23.

Thc hin cc bc th nghim ging nh phn trn.

Cu hi:

1b.1. Hy v th nhn c t dao ng k.

1b.2. Xc nh c tuyn V-A ca Diode Zener, vi qui c dng v p trn Diode Zener nh trong hnh 24.

R

D

CH Y

GND

CH X

My

pht

sng

Hnh 24

Z + u + u + u + u

iiii Z D

Hnh 23



2/. Mch tr phi tuyn ngun DC:

Cc bc thc hin

t cng tc ngun DC v tr OFF.

Sinh vin mc mch th nghim nh hnh 25.

Bc cng tc ngun DC ON.

Ln lt iu chnh gi tr ngun DC l 1V, 2V, 5V, 10V. Dng VOM s o cc gi tr Uout tng ng ri in vo bng sau:

Vdc 1V 2V 5V 10V

Uout

Nhn xt

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Hy lm li cc bc th nghim trn ng vi mch hnh 26.

Vdc 1V 2V 5V 10V

Uout

D

1K

1KVdc

+

Uout

Hnh 25

D

1KVdc

1K+

Uout

-

Hnh 26



3/. Mch tr phi tuyn ngun iu ha:

Cc bc thc hin


Mc mch th nghim nh hnh 27.

Bc ngun my pht sng v iu chnh my pht sng pht tn hiu hnh Sin c bin l 5Vv tn s 1Khz.

Cu hi:

3.1. Dng dao ng k xem v v li dng sng Uout(t).

T

Uou

t

3.2. Xc nh li dng sng Uout(t) theo phng php th.

D

1KMy

Pht

Sng

+

Uout

- Hnh 27



3.3. Dng Volt k in t o in p ra Uout. Hy kim chng gi tr o c bng khai trin Fourier.

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4/. Mch tr phi tuyn ngun AC v DC:

Cc bc thc hin



Chnh ngun Vdc = 5V.

Bt ngun my pht sng v iu chnh my pht sng pht tn hiu hnh Sin c bin l 1V v tn s 1Khz.

Cu hi:


T

Uou

t

4.2. Kim tra li theo phng php th.

4.3. Thay i bin my pht sng l 2V, v li dng sng Uout(t).

Vdc

D

1KVac

My

Pht Sng

+

Uout

- Hnh28



4.4. Tng bin ng ra my pht sng, nhn xt v s thay i Uout(t). Gii thch.

5/. Mch chnh lu bn k:

Cc bc thc hin:



Bc ngun my pht sng v iu chnh my pht sng pht tn hiu hnh Sin c bin l 5V v tn s 50Hz.

Cu hi:


T

Uou

t

5.2. Lp li th nghim trn ng vi cc trng hp sau:

a. C t C1;

C2C1

D

C31KMy

Pht

Sng

+

Uout

-

Hnh 29



T

Uou

t

b. C t C1 v C2;

T

Uou

t

c. C t C1, C2 v C3.

T

Uou

t

5.3. Nhn xt v gii thch.

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Tai Lieu Thi Nghiem Mach Dien