các dạng bài tập Đtcs 2013
DESCRIPTION
tai lieu chuyen nganh dien tu cong suatTRANSCRIPT
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CC DNG BI TP IN T CNG SUT V THIT B CHUYN I TCS
Bi tp 1. Cho mch chnh lu tia 1 pha khng iu khin, bit in p ngun xoay chiu u2 = 12Sin314t [V], RT = 2 .
Hy xc nh cc thng s la chn diode cho mch chnh lu trn, v s mch, dng sng in p trc, sau chnh lu v dng sng dng in trn ti (bit diode chnh lu c ch to t Si).
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, 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).
Bi tp 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.
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?
Bi tp 3. 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.
Bi tp 4. Cho mch chnh lu cu 1 pha, bit cc thng s tng t nh bi 3 nhng ti l RL, dng lin tc gn phng. V dng sng in p v dng in trn ti.
Bi tp 5. Ngi ta dng thit b chnh lu cu 1 pha np in cho c quy, c E = 110V, dng np Id = 50A, in p ngun xoay chiu l u = 311Sin314t.
a. Tnh t1 ti thi im thit b chnh lu bt u dng np cho c quy trong tng na chu k, thi gian dn dng ca mi diode trong mt chu k.
b. in tr R phi bng bao nhiu m bo dng np yu cu.
Bi tp 6. 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;
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2
+
SCR1
SCR2
E L R U1
U2
ud
id
Hnh 1
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.
Bi tp 7. 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 8. 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.
Bi tp 9. 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 10. Cho mch chnh lu tia 2 pha nh hnh v 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 = 225
0;
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 = 450;
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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;
g. Gi s ti c Lng = 10mH, LT = , R = 2, E = 0V, hy tnh dng in trung bnh trn ti khi cc SCR c khch gc = 300?
Bi tp 11. 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 = 20
0 (R, L, E
khng thay i gi tr); c. Nu b E, ti ch cn R, L, hy tnh Id khi = 60
0 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 = 30
0;
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.
Bi tp 12. Cho mch in nh hnh v, bit u2 = 70Sin100t [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?
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
F
E
u2 u2
A B
1 2 3 4 5
C
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4
Bi tp 13. Cho mch in nh hnh 13.1; 13.2, bit in p vo l hnh Sin, f = 50Hz. a. Hy v dng sng in p ti cc im A; B; E; B1. b. 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? c. Nu tn hiu ng ra ti B1 qu nh khng kch cho SCR th cn gii quyt nh
th no?
Hnh 13.1
ui
RS
4k7
100k
12V C
10k
10k G1
G2
B
A
E
B1
BAX
12V
C
VR
R4
R3 R2 UAC E
UJT B1
A R1
Hnh 13.2.
B
OUT
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Bi tp 14. 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 = 0
0;
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 ti 49,72A; 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.
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.
Bi tp 15. 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 = 60
0 ;
c. Xc nh gc kch ca cc SCR khi ti c cng sut P = 2,65 kW.
Bi tp 16. 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 = 0
0 ; 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 v 2
=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.
Bi tp 17. 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; 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.
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HNG DN
I. Cc bi tp chng 2 (t bi 1-9) p dng ch yu cc cng thc tng hp vo bng 3.3, phn s v dng sng U, I coi trong gio trnh. - Nu cho vt liu ch to diode l Si th tn hao trn diode l 0.7V, nu b qua
tn hao th coi diode l l tng. - Nu in p ngun xoay chiu cho di dng gi tr tc thi th bit c tn
s f v phi xc nh gi tr hiu dng ca in p ngun AC.
II. Cc bi tp chng 3, ch yu p dng cc cng thc trong bng tng hp 3.3, 3.4 v 3.5
Ch cc vn sau: 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 ( =
+);
- 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, RLE khi 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. 9. Cc cu hi trong bi 13, coi ni dung ca bi 3.3 t trang 88 91 v bi
3.14 trang 127.
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Bi tp 18. Cho mch iu chnh in p xoay chiu 1 pha
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 Cos ca 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 TLV = 120mS, thi gian ngh TN = 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: 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.
Bi tp 19: Cho mch in nh hnh v, 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.
Bi tp 20: Hy v nhng kiu mch c th iu chnh c in p xoay chiu 1 pha?
G1
X
G2
XL = L
Y
SCR2
SCR1
RT
~ u
D2
SCR
D1
D4 D3
Ti
~ u
A B
C
D
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Bi tp 21: Cho mch in nh hnh v, 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.
HNG DN: - 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);
- Mc c; d, Ui cc thng s ca mch chnh lu xc nh theo bng 3.3; - Mc a; b theo ph lc bng hng dn xc nh dng sng v cc thng s ca
cc mch bin i in p ca TI cui gio trnh ta c:
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) 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 in ng ra v qua diode D0: ID0 = I0 = U0/R
C
D0
PWM ID0 R
Ui
U0
L
D3 D1
D4 D6
D5
D2
~ u
IQ IT
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9
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.
Bi tp 22: Cho th dng sng in p nh hnh v. 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.
HNG DN: - Dng sng in p ng ra trn ch c th c cp t mch nghch lu p qua bin
p tng p; - Ngoi ra nu cho in p xoay chiu ng ra c gi tr nh (di 40V) th c th
dng cc mch nghch kiu cu hoc na cu trc tip t ngun DC; - Trng hp khc c th cho s mch nghch lu, bit tn s, dng xung iu
khin linh kin bn dn, cn v dng sng in p trn ti?
Bi tp 23: Cho mch nghch lu p 3 pha nh hnh v 23.1. 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 1800, lch pha nhau 600. b. Hy lp bng trng thi in p pha v in p dy (bng 23.1) 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.
R C
PW
M
Ui IQ
Q
D0
L
ID0
t
Ut
1mS
0
+220V
-220V 1mS
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10
Bng 23.1
0 60o 60o 120o 120o 180o 180o 240o 240o 300o 300o 360o
UA0
UB0
UC0
UAB
UBC
UCA
Bi tp 24.Cho mch iu ch SPWM nh hnh v (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.
U +
-
S1
S4
S3
S6
S5
S2
D1
D1
A
D3
D2
D2
Hnh 23.1. Mch nghch lu p 3 pha
0 ZA
D3
B
C ZB
ZC
ien ap
o chia
0
URC (tam giac) Uk (sin chuan)
U
0
U0
0
-U
+U