bÀi tẬp Điện tử công suất

Bi tp in t cng sut

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CC KHI NIM V CC LINH KIN

IN T CNG SUT C BN

1.1. Gii thch gi tr trung bnh v gi tr hiu dng ca mt i lng in? Tnh in p

trung bnh v in p hiu dng ca ti c dng sng sau:

t [ms]

50

vi(t) [V]

20 50 600 30 80

Hnh 1.1

t [ms]

50

20 50 600 30 80

-30

vi(t) [V]

Hnh 1.2

x = wt [rad]

50

p 3p 5p-p 0 2p

vi(t) [V]

Hnh 1.3

CHNG 1

Bi tp in t cng sut

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t [ms]

40

20 60 800 30 50

vi(t) [V]

Hnh 1.4

100

p 2p 3p0

x = wt [rad]

vi(t) [V]

Hnh 1.5

100

p 2p 3p0

-100

4p

x = wt [rad]

vi(t) [V]

Hnh 1.6

100

p 2p 3p0

x = wt [rad]

vi(t) [V]

Hnh 1.7

Bi tp in t cng sut

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p 2p 3p0

100

100 2

x = wt [rad]

vi(t) [V]

Hnh 1.8

x = wt [rad]

vi(t) [V]

p 2p 3p0 p/2 3p/2 5p/2

100

Hnh 1.9

Hnh 1.10

1.2. in p trn ti cm (R + L) c dng sng nh hnh 1.7, do in cm ca ti rt ln (cm

khng ca ti rt ln so vi in tr ca ti XL >> R) nn dng in ca ti xem nh

c nn thng v c gi tr 10 [A]. Hy tnh cng sut trung bnh trn ti?

1.3. in p trn ti cm (R + L) c dng sng nh hnh 1.9, do in cm ca ti rt ln (cm

khng ca ti rt ln so vi in tr ca ti XL >> R) nn dng in ca ti xem nh

c nn thng v c gi tr 10 [A]. Hy tnh cng sut trung bnh trn ti?

1.4. in p t trn ti in tr 10 c hm biu din u = 220sin(100pt) [V]. Hy xc nh:

a. Hm cng sut tc thi ca ti

b. Cng sut tc thi ln nht

c. Cng sut trung bnh ca ti

1.5. in p v dng in trn ti l nhng hm tun hon theo thi gian vi chu k T =

100ms nh sau:

10V; 0 t 60msu(t)

0V; 60ms t 100ms

=

;

0; 0 t 50msi(t)

4A; 50ms t 100 ms

=

T/ 2 T

t

U 2U/ 3

U/ 3

0

-U/ 3

- 2U/ 3 T/ 6 T/ 3

2T/ 3 5T/ 6

Bi tp in t cng sut

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Xc nh in p hiu dng v dng in hiu dng, cng sut tc thi, cng sut trung

bnh v nng lng tiu th ca ti trong mi chu k.

1.6. Xc nh cng sut trung bnh trn ti. Cho bit in p ti khng i u(t) = U= 24VDC v

dng in qua ti tun hon c hm biu din trong mi chu k T = 100ms nh sau:

0; 0 t 50msi

4A; 50ms t 100 ms

=

1.7. Dng in qua phn t hai cc c dng i = 20sin(100pt) [A]. Hy xc nh cng sut tiu

th trung bnh trn phn t trn nu phn t hai cc l:

a. in tr 5;

b. Cun dy c cm khng 10mH;

c. Sc in ng E = 6V.

1.8. Dng in i = 2 + 20sin100pt [A] i qua mch RLE mc ni tip. Xc nh cng sut tiu

th trung bnh trn mi phn t R, L v E, cho bit R = 3 , L = 10mH v E = 12V.

1.9. Mt l in tr cng sut 1.500W khi s dng ngun u 220 2 sin(100 t)= p [V]. Nu

iu khin cng sut l in theo chu k 12 pht vi trnh t ng in 5 pht v ngt

in 7 pht. Hy xc nh:

a. in p hiu dng trn ti.

b. Cng sut tc thi cc i

c. Cng sut tiu th trung bnh

d. Nng lng tiu th di dng nhit trong mi chu k.

1.10. Hy xc nh tr hiu dng in p, dng in v cng sut tiu th trung bnh bi ti khi

cho bit qu trnh in p v dng in ca n c dng:

u 2,5 10cos(100 t) 3 2 cos(200 t / 3)= p p p [V]

i 1,5 2cos(100 t) 1,1cos(200 t / 3)= p p p [A]

1.11. Cho dng in i 1,5 2cos(100 t) 1,1cos(200 t / 3)= p p p [A] i qua ti gm R-C mc

song song vi R = 100 v C = 50F. Xc nh cng sut tiu th trn mi phn t ca ti.

1.12. Cho in p u 2,5 10cos(100 t) 3 2 cos(200 t / 3)= p p p [V] t trn ti RLE mc ni

tip vi R = 4, L = 10mH v E = 12V. Xc nh cng sut tiu th trn mi phn t.

1.13. in p v dng in qua ti biu din bi hm sau:

20u 20 cos(n t)[V];

nn 1

= p

=

5i 5 cos(n t)[A]

2nn 1

= p

=

Xc nh cng sut trung bnh trn ti (chnh xc n n = 4).

1.14. Cho ngun 20

u 20 sin(100n t)nn 1

= p

=[V] cung cp ti RLE ni tip vi R = 20, L =

250mH v E = 36V. Xc nh cng sut trung bnh trn cc phn t ti.

Bi tp in t cng sut

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1.15. Nu iu kin dn ca SCR? So snh SCR v Diode; SCR v Triac; SCR v GTO v cu

to, nguyn l hot ng v ng dng? Khi tnh chn SCR, cn ch n cc thng s

no?

1.16. Ti sao gi SCR l linh kin ch iu khin kch ng c? Khi SCR ang dn, nu cc

bin php ngt SCR?

1.17. Gii thch hin tng t kch ca SCR l g? Hin tng ny c nh hng nh th no

n vic iu khin SCR?

1.18. Da vo s chuyn mch ca linh kin, hy phn nhm cc loi linh kin in t cng

sut v nu ng ng ca mi loi tng ng?

1.19. So snh BJT v FET v cu to, nguyn l hot ng, ch lm vic, u nhc im,

xung kch v ng dng?

1.20. Nu v gii thch cc phng php bo v linh kin v mch in (mch iu khin,

mch cng sut ti, mch ngun)?

1.21. Ti sao phi hn ch tc tng dng v tng p trn SCR?

1.22. Nu mt s hng sn xut linh kin in t cng sut trn th gii?

Bi tp in t cng sut

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CHNH LU KHNG IU KHIN

Ch : Cc bi tp c b qua cc tn tht trong mch bao gm tn tht cng sut v in

p trn ngun, linh kin v dy dn.

2.1. Cho mch chnh lu 1 pha, na chu k ti thun tr R = 100 () nh hnh 2.1 vi in p

ngun vo u 220 2 sin t 220 2 sinx(V)i = w = , tn s ngun f = 50 (Hz). Hy:

a) V dng sng in p ngun, in p ti, dng in ti v in p trn diode D?

b) S dng phn mm Matlab hoc PSIM m phng v v cc dng sng nh trn?

c) Tnh in p trung bnh, in p hiu dng v dng in trung bnh trn ti?

d) Tnh xc nh cc thng s la chn diode v my bin p ngun? (bit diode

chnh lu c ch to t Si).

ui D

u0

M

N

R

Hnh 2.1

HNG DN: c cc thng s la chn diode chnh lu, trc tin cn phi tnh

in p trung bnh trn ti Ud, tnh dng qua ti Id v dng qua diode IDtt, tm in p

ngc ln nht trn diode theo in p xoay chiu UPIVDtt (Bng cch v dng sng trn

diode, xt gi tr in p ln nht ri trn diode khi diode khng dn), sau chn theo

tiu chun:

- IDst (1.25 1.3)IDtt;

- UPIVDst (1.6)UPIVDtt .

Trong IDst; UPIVDst l cc thng s dng v p lm vic nh mc cho trong s tay tra

cu ca nh sn xut (Datasheet).

2.2. Cho mch chnh lu tia 2 pha, bit in p xoay chiu trn mi cun th cp my bin p

u2 = 21,2 Sin314t [V], ti R=1 (b qua tn hao trn diode).

a) Tnh dng in trung bnh qua ti v qua mi diode, in p ngc ln nht trn mi

diode;

b) Gi s ti c gn thm ngun E=12V ni tip th cc thng s trn thay i nh th

no?

c) V s mch chnh lu, dng sng in p trc, sau chnh lu v dng sng dng

in trn ti.

CHNG 2

Bi tp in t cng sut

Trang 7

d) Gi s ti l RL (h s t cm rt ln), hy v dng sng in p v dng in trn ti

trn cng 1 th?

2.3. Cho mch chnh lu 1 pha, na chu k ti thun tr R = 100 (), E = 220 (V) nh hnh

2.2 vi in p ngun vo u 220 2 sin t 220 2 sinx(V)i = w = , tn s ngun f = 50 (Hz).

Hy:

a) V dng sng in p ngun, in p ti, dng in ti v in p trn diode D?

b) S dng phn mm Matlab hoc PSIM m phng v v cc dng sng nh trn?

c) Thit lp cng thc v tnh in p trung bnh, in p hiu dng v dng in trung

bnh trn ti?

d) Tnh chn diode v my bin p ngun?

ui D

u0

M

N

R

E

Hnh 2.2

2.4. Cho mch chnh lu 1 pha, na chu k ti cm c R = 100 (), L = 0,1 (H) nh hnh 2.3

vi in p ngun vo u 220 2 sin t 220 2 sinx(V)i = w = , tn s ngun f = 50 (Hz),

gc tt dng = 4,625 (rad) = 2650. Hy:

a) V dng sng in p ngun, in p ti, dng in ti v in p trn diode D?

b) Thit lp cng thc Tnh in p trung bnh, in p hiu dng v dng in trung

bnh trn ti?

c) Tnh chn diode v my bin p ngun?

ui D

u0

M

N

R

L

Hnh 2.3

2.5. Cho mch chnh lu cu 1 pha dng diode ch to t Si. Bit gi tr hiu dng ca in

p ngun xoay chiu l U = 24V.Ti l R, c dng in trung bnh Id = 12A.

a) Hy tnh cng sut tiu trn ti v cc thng s la chn diode.

b) V s mch, dng sng in p trc sau chnh lu v dng sng dng in trn

ti.

2.6. Cho mch chnh lu cu 1 pha,bit cc thng s tng t nh bi 2.5 nhng ti l RL,

dng lin tc gn phng. V dng sng in p v dng in trn ti.

2.7. Cho thit b chnh lu cu 1 pha np in cho c quy, c sc in ng E = 120V,

dng np Id = 40A. Tr hiu dng ca in p ngun l 220V, tn s 50Hz.

Bi tp in t cng sut

Trang 8

a) Tnh t1 thi im thit b chnh lu bt u cung cp dng np cho c quy trong tng

na chu k v - thi gian dn dng ca mi diode.

b) Xc nh gi tr in tr R m bo dng np yu cu.

c) Tnh tr hiu dng ca dng ti.

d) Tnh hiu sut ca thit b.

ui

M

N

u0 R

E

D3

D2D4

D1

Hnh 2.4

2.8. Cho thit b chnh lu khng iu khin ba pha hnh tia, ba diode cp dng cho mt mch

ti gm sut in ng E = 120V, in tr R = 5 . Tr hiu dng ca in p pha U =

220V, tn s ngun xoay chiu f = 50 Hz.

a) V dng sng dng in qua ti v qua mt diode khi E = 120V.

b) V dng sng dng in qua ti v qua mt diode khi E = 220V. Nhn xt cc dng

sng trong hai trng hp trn.

c) Tnh tr trung bnh in p trn ti Ud, dng in qua ti Id, dng qua mt diode ID khi

E = 120V.

d) Tnh tr hiu dng dng chy qua mi cun dy th cp my bin p ngun khi E =

120V.

p s: c) Id = 27,48A, ID = 9,16A; d) I2 = 15,6A.

2.9. Cho mch chnh lu tia 3 pha khng iu khin, cp dng cho mt mch ti gm b c

quy c E = 120V, R = 2 , gi tr hiu dng ca in p pha l U = 220V, tn s ngun

in xoay chiu l f = 50 Hz.

a) Tnh dng in trung bnh qua ti v qua mi diode;

b) V s mch, dng sng in p v dng in trn ti;

c) Tnh in p ngc ln nht trn mi diode;

d) Tnh dng in trung bnh qua ti v qua mi diode khi c quy np ti tr s E = 170V,

v dng sng in p v dng in trn ti trong trng hp ny.

2.10. Cho mch chnh lu cu 3 pha khng iu khin c cp dng t my bin p 3 pha ni

/ , bit in p ngun cun dy th cp l 400V, ti R = 10

a) Tnh dng in trung bnh qua ti, qua mi diode v in p ngc ln nht m mi

diode phi chu;

b) Gi s khi c thm ti E = 100V th cc thng s trn thay i nh th no?

c) V mch chnh lu v dng sng in p trn ti trng hp a.

Hng dn: Khi bin p ni kiu tam gic th in p ng ra l in p dy.

Bi tp in t cng sut

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2.11. Cho mch chnh lu tia 6 pha khng dng cun khng cn bng, lm ngun cp dng

cho my hn c in tr R = 0.15 , bit in p dy hiu dng cun s cp my bin p

ni Y/YY l 380V, t s bin p l Kba = 6,3.

a) Tnh dng in hn, dng trung bnh qua mi diode v in p ngc ln nht trn

mi diode;

b) Gi s khi c gn thm cun khng cn bng th cc thng s trn thay i nh th

no, cho bit tc dng ca cun khng cn bng?

c) V s chnh lu v dng sng in p trn ti trong cc trng hp trn?

d) Mch chnh lu tia 6 pha thng c s dng cho nhng loi ti no, ti sao, nu tn

mt v loi ti?

Ghi ch: Mi pha bn th cp c 2 cun dy v vy c c in p U2 cp cho

mch chnh lu ta cn chia Kba cho 2.

2.12. Hy tnh dng in trung bnh qua ti R=10 , qua mi diode v in p ngc ln nht

trn mi diode trong cc s sau khi chng cho ra cng mt in p Ud = 200V khi

khng dng t lc v c t in lc phng in p trn ti:

a) S tia 1 pha, tia 2 pha, cu 1 pha;

b) S tia 3 pha, cu 3 pha;

c) S tia 6 pha khng dng cun khng v c cun khng cn bng;

d) V s mch v dng sng in p trn ti trong trng hp c t in lc phng

in p.

Bi tp in t cng sut

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+

SCR1

SCR2

E L R U1

U2

ud

id

Hnh 2.1

CHNH LU C IU KHIN

3.1. Nu iu kin dn ca SCR? So snh SCR v Diode v cu to, nguyn l hot ng v

ng dng? Khi tnh chn SCR, cn ch n cc thng s no?

3.2. Nu iu kin dn ca SCR? So snh SCR v Triac v cu to, nguyn l hot ng v

ng dng? Khi tnh chn SCR, cn ch n cc thng s no?

3.3. Ti sao gi SCR l linh kin ch iu khin kch ng c? Khi SCR ang dn, nu cc

bin php ngt SCR?

3.4. Cho b chnh lu cu 1 pha iu khin hon ton vi tn s ngun f = 50Hz, in p pha

ngun xoay chiu u(t) = 220 2 sin100 tp (V), gc kch = 450.

a) Vi ti thun tr R = 100 , tnh in p trung bnh v dng in trung bnh trn ti?

Thit lp cng thc tng qut tnh in p hiu dng trn ti?

b) Vi ti R = 10 , E = 200 V mc ni tip, v dng sng in p v dng in trn ti?

Thit lp cng thc tng qut tnh in p trung bnh trn ti?

c) Vi ti R = 10 , E = 100 V mc ni tip, v dng sng in p v dng in trn ti?

Thit lp cng thc tng qut tnh in p trung bnh trn ti?

d) Ti cm R = 10 , L = 100 mH mc ni tip, xc nh ch dng in ti v in p

trung bnh trn ti?

e) Tnh lin tc v gin on ca dng in trn ti ph thuc vo cc yu t no? Vi ti

cm, trng hp no dng in lin tc, trng hp no dng in b gin on?

3.5. Cho mch chnh lu tia 2 pha nh hnh v 2.1 bit t

s bin p Kba = U1/U2 = 2, gi tr hiu dng ca U1 =

380V, f = 50Hz cp dng cho ti R = 1.5; L c gi

tr xc nh, E = 50V (b qua in tr thun ca cun

cm v st p trn cc SCR, Lng = 0, RLE khng thay

i gi tr).

a) Tnh dng in trung bnh trn ti v qua mi SCR

khi gc kch cho cc SCR = 600, gc tt dng

= 2250;

b) Tnh dng in trung bnh trn ti khi = 300;

c) Gi s b E, ti ch cn R, L, tnh dng in trung bnh trn ti khi cc SCR c kch

vi gc = 750 v = 45

0;

d) Tnh dng in trung bnh trn ti khi cc SCR c kch vi = 900 trong trng hp

b L, E ti ch cn R;

e) Gi s ngi ta thay ti L bng ti LT = , hy tnh Id khi = 900;

f) V dng sng in p trn ti trong cc trng hp trn;

CHNG 3

Bi tp in t cng sut

Trang 11

g) Gi s ti c Lng = 10mH, LT = , R = 2, E = 0V, hy tnh dng in trung bnh trn

ti khi cc SCR c kch gc = 300?

3.6. Cho b chnh lu cu 1 pha iu khin hon ton vi cc tham s sau: p pha ngun AC

120V, f = 50Hz. Ti R - L mc ni tip R = 10, L = 100 mH. Gc kch = p/3. Xc nh

ch dng in ti v tr trung bnh ca n.

3.7. Cho mch chnh lu cu 1 pha iu khin ton phn, bit in p hiu dng ngun xoay

chiu hnh sin l 200V, tn s in p ngun 50Hz, cung cp dng cho ti R = 10 , E =

40V, L c gi tr xc nh (b qua in tr thun ca cun cm RL v st p trn SCR,

Lng =0).

a) Tnh dng in trung bnh trn ti Id khi gc kch cho cc SCR = 450, gc tt dng

= 2100;

b) Tnh dng in trung bnh trn ti Id khi gc kch cho cc SCR = 200 (R, L, E khng

thay i gi tr);

c) Nu b E, ti ch cn R, L, hy tnh Id khi = 600 v = 10

0;

d) Nu b L, E ti ch cn R hy tnh dng trung bnh trn ti Id v dng in trung bnh

qua SCR, khi = 300;

e) Nu thay 2 SCR chung anode bng 2 diode, b E ti ch cn R, L, tnh dng in trung

bnh qua mi SCR v diode khi cc SCR c kch vi = 450 (dng lin lc);

f) V dng sng in p v dng in trn ti trong cc trng hp trn.

3.8. Tnh tr trung bnh p v dng chnh lu, cng sut ti tiu th ca b chnh lu mch tia

ba pha iu khin. Ti c R= 10 [], E=50 [V] v L=0. p ngun U=220 [V]; gc iu

khin = p/3 [rad].

3.9. Cho b chnh lu mch hnh tia 3 pha iu khin mc vo ti cha R = 10 v in cm

L ca ti rt ln lm dng ti lin tc v phng. in p pha ca ngun xoay chiu 3 pha

c tr hiu dng U = 220 V. Mch trng thi xc lp, gc kch = 600.

a) Tnh tr trung bnh ca in p chnh lu v dng chnh lu?

b) Tnh cng sut trung bnh ca ti?

c) Tnh tr trung bnh dng qua mi linh kin?

3.10. Mt mch chnh lu tia 3 pha c iu khin c cp ngun t my bin p ni tam

gic/sao (/Y), bit in p hiu dng cun s cp my bin p l 660V, t s bin p Kba

= 1,73, ti c in tr thun l R = 5 (b qua st p trn cc linh kin bn dn).

a) Tnh dng in trung bnh qua ti v qua mi diode khi cc SCR c kch vi gc

= 00;

b) Tnh dng in hiu dng cun th cp I2.

c) Tnh cng sut tiu th trn ti khi gc kch cho cc SCR = 450;

d) Tnh dng in trung bnh trn ti khi cc SCR c kch vi = 300;

e) Xc nh gc kch cho cc SCR khi dng in trung bnh trn ti49,72A;

f) Gi s c gn thm LT = ni tip vi R, hy tnh in p trung bnh trn ti khi cc

SCR c kch vi gc vi = 600;

g) Gi s LT c gi tr xc nh = 2100, hy tnh dng in trung bnh trn ti trong cc

trng hp khi 1 = 900, 2 = 60

0, v 3 = 45

0.

Bi tp in t cng sut

Trang 12

h) Tnh in p ngc ln nht trn mi SCR.

Hng dn: T s bin p Kba = U1/U2 ,v s cp ni tam gic nn U1 l in p dy

ng ra, c cng t l th U2 cng phi in p dy ng ra.

3.11. Cho mch chnh lu tia 3 pha khng iu khin, bit in p dy hiu dng cun th cp

bin p U2 = 220V, cung cp dng cho ti tr c cng sut tiu th P = 3kW (b qua st

p trn cc linh kin bn dn).

a) Tnh dng in trung bnh trn ti v trn mi diode;

b) Tnh cng sut tiu th trn ti khi thay cc diode bng cc SCR vi gc kch = 600 ;

c) Xc nh gc kch ca cc SCR khi ti c cng sut P = 2,65 kW.

3.12. Cho b chnh lu mch hnh cu 3 pha iu khin mc vo ti cha R = 10 v in

cm L ca ti rt ln lm dng ti lin tc v phng. in p pha ca ngun xoay chiu 3

pha tn s 50 Hz c tr hiu dng U = 220 V. Mch trng thi xc lp, gc kch = 600.

a) Tnh tr trung bnh ca in p chnh lu v dng chnh lu?

b) Tnh cng sut trung bnh ca ti?

c) Tnh tr trung bnh dng qua mi linh kin?

3.13. Cho b chnh lu mch tia 3 pha iu khin mc vo ti cha R = 10 , E = 50 V v ti

rt ln lm dng ti lin tc v phng. p ngun xoay chiu 3 pha c tr hiu dng U =

220 V. Mch trng thi xc lp.

a) Tnh tr trung bnh ca in p chnh lu v dng chnh lu khi gc iu khin = p/3

[rad].

b) Tnh cng sut trung bnh ca ti.

c) Tnh tr trung bnh dng qua mi linh kin.

d) Tnh tr hiu dng dng qua mi pha ngun.

e) Tnh h s cng sut ngun .

3.14. Cho b chnh lu cu 3 pha iu khin hon ton vi cc tham s sau: p dy ngun AC

480V, f = 50Hz. Ti R = 10, L = 50 mH. Xc nh gc kch dng ti trung bnh bng

50A.

3.15. Cho mch chnh lu cu 3 pha iu khin ton phn c ti R =10, in p dy hiu

dng cun th cp bin p U2 = 380V, f = 50Hz.

a) Tnh dng in trung bnh trn ti v qua mi SCR khi chng c kch vi gc =

00; Tnh cng sut tiu th trn ti;

b) Tnh in p ngc cc i trn mi SCR;

c) Tnh dng in trung bnh trn ti khi cc SCR c kch vi gc 1 =450 v2 =75;

d) Nu thay 3 SCR chung anode bng 3 diode, tnh dng in trung bnh trn ti khi cc

SCR c kch vi gc = 450.

3.16. Cho mch chnh lu tia 6 pha khng iu khin, khng dng cun khng cn bng bit

in p dy hiu dng cun th cp U2 = 200V, cung cp dng cho ti thun tr c cng

sut tiu th P = 10kW (b qua st p trn cc linh kin bn dn).

a) Tnh dng in trung bnh trn ti v trn mi diode;

Bi tp in t cng sut

Trang 13

b) Gi s thay cc diode bng cc SCR, tnh cng sut tiu th trn ti khi cc SCR c

kch vi gc = 300 ;

c) Tnh dng in trung bnh trn ti khi cc SCR c kch vi = 900 ;

d) Xc nh gc kch ca cc SCR khi ti tiu th ht cng sut P = 5kW;

e) Tnh in p ngc ln nht trn mi SCR.

3.17. Cho mch in nh hnh 2.2, bit u2 = 70Sin100pt [V], RT = 1.5 , LT = .

a) Nu tn gi v chc nng ca cc khi 1; 2; 3; 4; 5 trong s ;

b) Tnh dng in trung bnh trn ti khi gc kch cho cc SCR = 600;

c) Hy v gin xung (dng sng in p) ti cc im A; B; C; D; E; F v trn bin tr

VR v dng sng in p trn ti khi = 450 trong na chu k u (A l dng B l

m);

d) Ti sao ng vo IN- ca 2 b so snh trn hnh v li ly in p trn cng mt bin

tr VR?

3.18. Cho mch in nh hnh 2.3; 2.4, bit in p vo l hnh Sin, f = 50Hz.

a) Xung ng ra trn cun th cp BAX (hnh 13.1) c th iu khin cho cc SCR trong

mch chnh lu cu iu khin bn phn khng?

b) Hy v dng sng in p ti cc im A; B; E; B1.

c) Nu tn hiu ng ra ti B1 qu nh khng kch cho SCR th cn gii quyt nh th

no?

Vcc

Vcc

XX

Y

Y

LOAD

~U1

Vcc

Mach ieu khien ong bo ien ap mot chieu 1 pha tia dung SCR

50k

100k

4007

4007

T2

T1

50k

40074007

BAX

NPN

4007

103

+

LM324

224

+

LM324

9V

4007

40074007

BAX

NPN

4007

103

+

LM324

224

+

LM324

9V

4007

1k

47k

330

1M

100k

47k

1M

2k2

1k

47k

330

1M

100k

47k

1M

2k2

VR

A

D

B

12V

12V

RT

SCR1

SCR2

LT

E

D

B

F

E

D

B u2 u2

A B

1 2 3 4 5

C

12

V

C

V

R R

4

R

3 R

2 UAC E UJ

T B1

A R1

Hnh 2.4.

B

OU

T

Hnh 2.2

Hnh 2.3.

Bi tp in t cng sut

Trang 14

MCH BIN I V NG CT

IN P XOAY CHIU

4.1. Mt l in tr cng sut 1.500W khi s dng ngun u 220 2 sin(100 t)= p [V]. Nu iu

khin cng sut l in theo chu k 12 pht vi trnh t ng in 5 pht v ngt in 7

pht. Hy xc nh:

a) in p hiu dng trn ti.

b) Cng sut tc thi cc i

c) Cng sut tiu th trung bnh

d) Nng lng tiu th di dng nhit trong mi chu k

4.2. Cho b bin i p xoay chiu mt pha cp ngun cho ti thun tr R = 10 . Ngun xoay

chiu c tr hiu dng bng 220V, tn s ngun 50Hz, gc kch = p/3[rad] .

a) Tnh tr hiu dng p ti?

b) Tnh cng sut tiu th ca ti?

c) t c cng sut ti bng 4 kW, tnh ln gc kch?

4.3. Cho mch iu chnh in p xoay chiu 1 pha nh hnh 4.1 bit RT = 20, XL= 0, in p

hiu dng ngun xoay chiu U = 380V, f = 50Hz.

a) Tnh cng sut tiu th ca ti trong trng hp in

p trn ti l ln nht;

b) Tnh cng sut tiu th trn ti khi cc SCR c

iu khin vi gc kch = 300, tnh h s cng sut

Cosca mch v v dng sng in p trn ti

c) Tnh cng sut tiu th trn ti khi cc SCR c

iu khin theo t l thi gian ng ngt, bit thi

gian lm vic Ton = 120mS, thi gian ngh Toff =

40mS;

d) Cn phi khng ch gc kch TH bng bao nhiu in p trn ti khng tr

thnh DC khi XL = 10 (xung iu khin l xung ngn);

e) V dng sng dng in, in p trn ti khi gc kch cho SCR1 = 900, SCR2 c

thay th bng 1 diode (ti c c R v XL = 10, gi s na chu k u Y c in th

dng).

f) Mch to xung iu khin bi 12 c th iu khin cho cc SCR trong mch ny

c khng, nu cn khng ch gc kch TH nh bi ny th thc hin nh th no?

HNG DN:

CHNG 4

G1

X

G2

XL = Lw

Y

SCR2

SCR1

RT

~ u

Hnh 4.1

Bi tp in t cng sut

Trang 15

Mch trn l mch iu chnh in p xoay chiu 1 pha dng 2 SCR (cng c th dng

TRIAC). Cc yu cu ca bi tp u c th xc nh theo cc cng thc trong gio trnh.

C th cho trc dng ti hoc cng sut xc nh cc thng s khc ca mch.

4.4. Cho mch in nh hnh 4.2 , bit in p xoay chiu u = 311Sin314t [V], (b qua st p

trn cc linh kin bn dn).

a) Tnh cng sut tiu th trn ti l thit b gia

nhit c R= 5 t ti hai im AB khi gc

kch cho SCR = 450, v dng sng in p

trn ti;

b) Tnh dng in trung bnh qua SCR khi ni tt

AB, ti l R = 2, LT = , dng lin tc phng

ch xc lp t ti CD khi gc kch cho

SCR = 300, v dng sng dng in v in p trn ti;

c) Thc hin tng t nh trng hp b nhng LT = 0, = 600;

d) Cho bit tc dng ca cc diode trong hai trng hp trn?

HNG DN:

- Mch trn l mch iu chnh in p xoay chiu khi ti t AB, iu khin theo

pha (Phase control);

- Trng hp b ti RL dng lin tc t CD nh mch chnh lu cu 1 pha;

- Trng hp c, mch thun tr.

4.5. Hy v nhng kiu mch c th iu chnh c in p xoay chiu 1 pha?

D2

SCR

D1

D4 D3

Ti

~ u

A B

C

D

Hnh 4.1

Bi tp in t cng sut

Trang 16

MCH BIN I IN P MT CHIU

5.1. Cho b DC DC gim p l tng nh hnh 5.1 vi in p vo Vin = 20V:

Hnh 5.1. B DC - DC gim p.

Dng sng dng in qua cun dy nh sau:

Hnh 5.2. Dng sng dng in qua cun dy.

Hy xc nh:

a) H s xung D (Duty cycle) v tn s xung iu khin;

b) in p trung bnh trn ti;

c) in cm ca cun dy;

d) in tr ca ti;

e) in tr ca ti khi khi dng in qua cun dy bin gii gia gin on v lin tc

(dng ti hn).

5.2. Cho b DC DC gim p l tng nh hnh 5.1 vi in p vo Vin = 24V, Vout = 12V,

cng sut cc i ca ti 100W, tn s ng ct 40kHz. Tm in tr ln nht ca ti, in

cm ti hn ca cun dy, lch dng in qua cun dy khi cng sut ti hn ca ti:

a) Pcrit = 10W;

b) Pcrit = 20W.

5.3. Cho b DC DC gim p l tng nh hnh 5.1 vi in p ra Vout = 12V, tm gi tr in

dung khi gn sng ca in p u ra ln nht bng 1% so vi in p trung bnh trn

ti ( VCpp 0.01 x 12 V = 120mV)?

CHNG 5

Bi tp in t cng sut

Trang 17

5.4. Cho b DC DC gim p l tng nh hnh 5.1 vi in p vo Vin = 20V, L = 10 mH, C

= 20F, R = 20, tn s ng ct 20kHz, h s xung D = 0.6. Mch ch xc lp.

Tnh:

a) in p trung bnh trn ti;

b) gn sng.

c) Xc nh ch lm vic ca mch dn lin tc CCM hay gin on DCM?

5.5. Cho mch DC - DC kiu tng p c Vin = 20V, L = 10 mH, C = 20F, R = 20, tn s

ng ct f = 50 kHz, h s xung D = 0,6. Tnh:

d) in p trung bnh trn ti;

e) gn sng.

f) Xc nh ch lm vic ca mch dn lin tc CCM hay gin on DCM?

5.6. Cho b DC DC tng/gim p (Buck Boost converter) l tng nh hnh 5.3 vi in p

vo V1 = 20V, L = 10 mH, C = 20F, R = 20, tn s ng ct 50kHz, h s xung D =

0.6. Mch ch xc lp. Tnh:

a) in p trung bnh trn ti; b) gn sng. c) Xc nh ch lm vic ca mch dn lin tc CCM hay gin on DCM?

Hnh 5.3. B DC - DC tng/gim p.

5.7. Cho mch in nh hnh 5.4, bit in p dy ngun xoay chiu 3 pha u = 660Sin314t [V],

(b qua st p trn cc linh kin bn dn), MOSFET c iu khin vi xung iu ch

c rng ton= 60mS, toff = 30mS, R= 5.

a) Tnh cc thng s la chn MOSFET (IQ, UDS); b) Tnh cc thng s la chn diode ( ID0, UD0); c) Tnh dng in trung bnh qua mi diode chnh lu D1-D6; d) Tnh in p ngc ln nht trn mi diode D1 D6.

Hnh 5.4. B DC - DC gim p.

C

D0

PWM ID0 R

Ui

U0

L

D3 D1

D4 D6

D5

D2

~ u

IQ IT

Bi tp in t cng sut

Trang 18

5.8. Cho cc mch in nh hnh 5.5, hy nu tn mch v gii thch nguyn l hot ng ca

mch.

Hnh 5.5a. Mch DC - DC .

Hnh 5.5b. Mch DC - DC .

5.9. Cho cc mch in nh hnh 5.6, gii thch nguyn l hot ng ca mch.

Hnh 5.6a. Mch DC - DC Forward

Hnh 5.6b. Mch DC - DC Fly-back

Bi tp in t cng sut

Trang 19

Hnh 5.6c. Mch DC - DC Half-Brigde

Hnh 5.6d. Mch DC - DC Brigde

Bi tp in t cng sut

Trang 20

MCH NGHCH LU V BIN TN

6.1 Cho th dng sng in p nh hnh 6.1.

a) V mch nghch lu c th to ra c in p xoay chiu trn khi in p ca ngun

DC l 36V?

b) Tnh tn s in p xoay chiu, xc nh t s bin p c in p UAC = 220V.

Hnh 6.1. Dng sng nghch lu.

6.2 Cho mch nghch lu p 3 pha nh hnh v 6.2.

a) Hy v s ni dy ca ti t bc 1 n 6 khi cc IGBT c iu khin bng xung

vung vi thi gian dn l 1200 v 180

0, lch pha nhau 60

0.

b) Hy lp bng trng thiin p pha v in p dy (bng 6.1) trn ti tng ng vi cc

xung iu khin trn.

c) Hy v dng sng cc in p pha UA0; UB0; UC0v cc in p dy UAB; UBC; UCA

trn ti tng ng vi cc xung kch nh trn.

CHNG 6

t

Ut

1mS

0

+220V

-220V 1mS

U +

-

S1

S4

S3

S6

S5

S2

D1

D1

A

D3

D2

D2

Hnh 6.2. Mch nghch lu p 3 pha

0 ZA

D3

B

C ZB ZC

Bi tp in t cng sut

Trang 21

Bng 6.1

0 60o

60o

120o

120o

180o

180o

240o

240o

300o

300o

360o

UA0

UB0

UC0

UAB

UBC

UCA

6.3 Cho mch iu ch SPWM nh hnh 6.3 (fURC =14fUk). Hy v dng xung iu ch ng ra

khi sng Sin c a vo IN+, cn sng tam gic c a vo IN- ca b so snh trong

cc trng hp khi OP-AMP s dng ngun n +12V v ngun i 12V.

Hnh 6.3. Dng sng iu ch SPWM

in p

chia

0

URC (tam giac) Uk(sin chuan)

U

0

U0

0

-U

+U

Bi tp in t cng sut

Trang 22

HNG DN

Chng 1.

Hng dn cch tnh cng sut trong cc mch in t cng sut:

Phng php tng qut tnh cng sut trung bnh:

Cng sut trung bnh c tnh: t T01

P p(t)dtT

t0

= [W]

Trong :

- p(t) = u(t). i(t): Cng sut tc thi ca mt ti c xc nh bng tch in p

tc thi u(t) v dng in tc thi i(t) qua ti tng ng.

- t0: thi im bt u mt chu k;

- T: Chu k ca p(t).

1. in p v dng in mt chiu (DC).

a) u(t) v i(t) khng i, u(t) = U v i(t) = I:

P = U.I

Hoc P = R.I2

Hoc P = U2/R

b) u(t) v i(t) c dng xung: tnh nh phng php tng qut.

c) Khi dng in qua ti khng i , i(t)= I = const:

t T t T0 01 1P I.u(t)dt I. u(t)dt I.Ud

T Tt t0 0

= = =

Trong : Ud l in p trung bnh.

d) Khi dng in qua ti khng i, u(t)= U = const:

t T t T0 01 1P U.i(t)dt U. i(t)dt U.Id

T Tt t0 0

= = =

Trong : Id l dng in trung bnh.

2. in p v dng in xoay chiu hnh sin (AC).

2.1. Mt pha

Biu thc in p: u(t) U sin( t )m u= w [V];

Biu thc dng in: i(t) I sin( t )m i= w [V];

Biu thc cng sut: p(t) U sin( t )xI sin( t )m u m i= w w [V];

1 1p(t) U .I cos( ) U .I cos(2 t )m m u i m m u i

2 2= - w

Bi tp in t cng sut

Trang 23

t T01 1

P p(t)dt U .I cos( )m m u iT 2

t0

= = -

P U .I cosrms rms=

Trong : Urms, Irms: ln lt l in p hiu dng [V] v dng in hiu [A];

= u - i: lch pha gia p v dng.

2.2. Ba pha cn bng

Biu thc in p pha: u(t) U sin( t )m u= w [V];

Biu thc dng in pha: i(t) I sin( t )m i= w [V];

Biu thc cng sut mt pha: p (t) U sin( t )xI sin( t )m u m i1f= w w [V];

Cng sut 3 pha: p (t) p (t)3f 1f

= [W].

t T01 3

P p (t)dt U .I cos( )3f m m u iT 2

t0

= = - [W]

P 3U .I cosrms rms= [W].

Hoc P 3U .I cosdrms drms= [W].

Trong :

Urms, Irms: ln lt l in p hiu dng pha[V] v dng in hiu dng pha

[A];

Udrms, Idrms: ln lt l in p hiu dng pha[V] v dng in hiu dng

pha [A];

= u - i: lch pha gia p v dng.

Nu bit in tr ca ti (hoc phn t cn tnh cng sut): P = R.I2 RMS = U

2 RMS/R trn

mi pha.

3. in p v dng in bao gm nhiu thnh phn hi.

Biu thc in p: u(t) = U0 + u1(t) + u2(t) ++ un(t) [V];

Biu thc dng in: i(t) = I0 + i1(t) + i2(t) ++ in(t) [V];

Trong :

U0, I0: thnh phn in p v dng in mt chiu;

un(t), in(t): thnh phn sng hi hnh sin bc n.

un(t) = Um(n)sin(wt + u(n)) = 2 Urmssin(wt + u(n)),

in(t) = im(n)sin(wt + u(n)) = 2 Irmssin(wt + i(n)), ,

- Cch 1: p(t) = u0.i0 + u1(t). i1(t) + u2(t). i2(t) ++ un(t). in(t) [W];

Hay p(t) U .I u (t).i (t) U .I p (t)0 0 n n 0 0 n1 1

= = [W];

Bi tp in t cng sut

Trang 24

t Ton n1

P U .I p (t)dt0 0 nTn1 ton

= [W];

- Cch 2: n 1

P U .I U .I .cos( )0 0 m(n) m(n) u(n) i(n)21

= -

Hay: n

P U .I U .I .cos( )0 0 rms(n) rms(n) u(n) i(n)1

= -

Nu bit in tr ca ti (hoc ca phn t cn tnh cng sut): P = R.I2

rms = U2 rms/R ,

Vi:

2 2 2U U ... Um(1) m(2) m(n)2 2 2 2 2U U U U U ... Urms 0 0 rms(1) rms(2) rms(n)2

= =

Hay: n n1 12 2 2 2U U U U Urms 0 0m(n) rms(n)2 21 1

= =

2 2 2I I ... Im(1) m(2) m(n)2 2 2 2 2I I I I I ... Irms 0 0 rms(1) rms(2) rms(n)2

= =

Hay: n n1 12 2 2 2U U U U Urms 0 0m(n) rms(n)2 21 1

= =

Chng 2

Chn la linh kin

c cc thng s la chn diode chnh lu, trc tin cn phi tnh in p trung bnh

trn ti Ud, tnh dng qua ti Id v dng qua diode IDtt, tm in p ngc ln nht trn diode

theo in p xoay chiu UPIVDtt, sau chn theo tiu chun:

- IDst (1.25 1.3)IDtt;

- UPIVDst (1.6)UPIVDtt .

Trong IDst; UPIVDst l cc thng s dng v p lm vic nh mc cho trong s tay tra cu

ca nh sn xut (Datasheet).

Chng 3

1. i vi cc mch 1 pha ti RL hoc RLE, p dng ng dng cng thc cn dng php

th nh sau:

Khi bit gc tt dng th ly X = - bit phn ko di ca sc in ng t cm

eL v pha bn k m sau l bao nhiu, t suy ra:

- Nu > X, ta c dng in gin on (< +);

- Nu = X, ta c gii hn ca dng in lin tc v gin on (= +);

Bi tp in t cng sut

Trang 25

- Nu < X, ta c dng in lin tc (= +).

i vi mch tia 3 pha khi cho LT l mt gi tr xc nh, cn xc nh theo iu kin:

- Khi < 5/6 +, ta c dng gin on;

- Khi = 5/6 +, ta c dng in lin tc hoc gii hn ca lin tc vi gin on.

2. i vi cc mch 1 pha hoc 3 pha ti RL, RLEkhi cho L hoc LT = , th lun c dng

lin tc phng.

3. Khi cho Lng = 0, th khng c hin tng trng dn, khi Lng 0, th mch c hin tng

trng dn, cn phi tnh dng in trong trng hp c trng dn.

4. i vi cc mch tia 3 pha, cu 3 pha v tia 6 pha ti R, cn xc nh gc kch trong

phm vi dng lin tc hay gin on p dng cng thc tnh Ud.

5. i vi cc dng bi tp cho trc dng ti hoc cng sut ti, yu cu phi xc nh gc

kch . xc nh ng dng cng thc, cn tnh Ud, Id hoc Pd gc gii hn ca

dng lin tc v gin on v so snh vi gi tr cho rt ra kt lun p dng dng

cng thc no.

6. Trong cc mch chnh lu 3 pha, 6 pha, khi cho in p dy, p dng c cc cng

thc trong bng 3.3; 3.4; 3.5, cn phi i t in p dy sang in p pha.

7. Trong mch chnh lu tia 2 pha c 2 cun th cp, mi cun c in p l U2.

8. Cc dng s , dng sng v cc cng thc tnh U, I coi trong gio trnh l thuyt.

Chng 5

i vi cc mch c kt gia mch chnh lu cu 3 pha

- Mch trn l mch kt hp gia mch chnh lu cu 3 pha (c th l cc mch chnh lu

khng iu khin khc) v mch iu chnh in p mt chiu kiu gim p (c th l

mch tng p). in p ng ra ca mch chnh lu cu 3 pha l in p ng vo ca

mch DC DC (Ui);

Mch gim p:

- in p ng ra: Uo = Ui.D

- Dng in ng ra v qua MOSFET: IQ = I0 = U0/R

- in p ngc trn diode D0: UD0 = Ui

- in p trn MOSFET: UDS = Ui

- Dng in qua diode D0: ID0 = I0(1-D)

- iu kin dng qua ti bin gii gia gin on v lin tc (dng ti hn):

Bi tp in t cng sut

Trang 26

Trong : Lcrit: in cm ti hn

Rbig: in tr ti

fSW: Tn s ng ct

Mch tng p:

- in p ng ra: U0 = Ui

1

1 D-

- Dng qua MOSFET: IQ = I0

1

1 D-

- in p ngc trn diode D0: UD0 = U0

- in p trn MOSFET: UDS = U0

- Dng inng ra v qua diode D0: ID0 = I0=U0/R

Trong D = ton/ton+toff = ton/T : Duty cycle

Ngoi ra ta c U0 = Ui.ton.f (f = 1/T), khi cho bit ton; toff; hoc ton; f ta u c th tnh c U0. Cng c th ngi ta cho trc dng in hoc cng sut tiu th trn ti, cn xc nh tn s

ng ct f hoc rng xung iu ch ton hoc t s D.

Nu ti khng phi l thun tr m RE hc RLE (ng c DC) th ta phi p dng cc cng thc thch hp tnh I0.

R C

PW

M

Ui IQ

Q

D0

L

ID0