BÀI TẬP Điện tử công suất
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Bài tập Điện tử công suất Trang 1 CÁC KHÁI NIỆM VÀ CÁC LINH KIỆN ĐIỆN TỬ CÔNG SUẤT CƠ BẢN 1.1. Giải thích giá trị trung bình và giá trị hiệu dụng của một đại lƣợng điện? Tính điện áp trung bình và điện áp hiệu dụng của tải có dạng sóng sau: t [ms] 50 v i (t) [V] 20 50 60 0 30 80 Hình 1.1 t [ms] 50 20 50 60 0 30 80 -30 v i (t) [V] Hình 1.2 x = wt [rad] 50 p 3p 5p -p 0 2p v i (t) [V] Hình 1.3 CHƢƠNG 1
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BÀI TẬP Điện tử công suất
Transcript of BÀI TẬP Điện tử công suất
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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
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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
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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
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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.
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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?
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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
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Bi tp in t cng sut
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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.
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Bi tp in t cng sut
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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.
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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.
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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
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Bi tp in t cng sut
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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.
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Bi tp in t cng sut
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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
