thiết kế môn học điện tử công suất lọc bụi tĩnh điện

Chng 1: Tng quan v cng ngh lc bi1, Gii thiu chung v cng ngh lc bi : Nn kinh t ngy cng pht trin khng ngng dn p ng c nhu cu ca con ngi v vt cht v vn ho nhng mt tri ca n l ko theo tnh trng nhim mi trng ngy cng trm trng. Vit Nam ti nhng vng tp trung nhiu cng nghip tnh trng khi bi ,kh c hi thi ra mi trng gy nhim l rt ng lo ngi.Do vic trang b cc h thng x l bi cho cc nh my x nghip l thc s cn thit v c vai tr ngy cng quan trng. Khi thit k h thng lc bi vn t ra i vi cc nh my l chn h thng lc bi no cho ph hp vi nh my ca mnh trong s rt nhiu phng php lc bi hin nay .Cc phng php lc bi thng dc s dng hin nay l: a. Lc bi s dng bung lng bi. b. Lc bi kiu li tm-xiclon c. Lc bi kiu qun tnh d.Lc bi bng li lc vi,thp,giy,.. e. Lc bi tnh in Trong cc phng php loc bi trn th phng php lc bi tnh in ang c s dng ph bin bi v phng php ny c u im nh hiu sut thu bi cao,chi ph nng lng thp,c th lm vic vi p sut chn khng hoc p sut cao,v c bit l c th iu khin v t ng ho hon ton.Sinh vin: Phm c Thnh Lp T 50 H2 1

2, Phm vi ng dng : Phng php lc tnh in l phng php tng i hiu qu i vi cc nh my cng nghip c mt lng bi ln nh nh my xi mng , nh my phn bn luyn kim,nghin ,cng nghip gm..v..v .. 3, Phn tch nguyn l lm vic v yu cu cng ngh ca thit b lc bi tnh in : Hiu sut ca thit b ny ph thuc vo tnh cht ca khng kh, bn bi,tnh cht cc in,thng s in ca thit b,tc chuyn ng v s phn b ng u lng kh trong trng in t. S lc bi tnh in cu to t cc linh kin c bn hnh thnh bi 2 tm in cc ng song song.Gia 2 in cc ny c thit lp mt hiu in th mt chiu tng i cao,cng in trng chng gy ra tng i ln dn n cc ht bi s b ion ha mnh lit.Di tc dng ca lc in trng gia hai bn cc,cc ion b ht v pha bn cc tri:ion m v cc dng, ion dng v cc m. Cc dng ca thit b lc bi thng c ni t.Cc ht bi sau khi dch chuyn v cc in cc s lng li trn b mt in cc.Theo mc tch t bi trn b mt in cc m nh k rung lc in cc hoc xi nc ra in cc loi b bi. t c cc yu cu trn, ta phi thit k mt mch iu khin cho hai bn cc v chnh lu in p cho mch. Vi mch iu khin cng ngh lc bi tnh in,cn to ra mt hiu in th 1 chiu tng i cao,c th iu khin c.Ngoi ra,s ny khng lm lch pha li in,thit k b lc n gin v in p ngc trn cc van c th bng 0. Tuy nhin ngoi vic thit k mch iu khin,cng cn phi ti cc bin php bo v an ton cho mch v cc van bn dn v trong qu trnh lc bi lng kh gia 2 bn cc b ion ha nn d to ra dng in gia 2 bn cc gy ngn mch.Do ta phi thit k mt h thng chng ngn mch v t ng ng mch vo in p lm vic sau khi kt thc phng in. in p ca thit b lc bi phi c tng dn n nh m bo cho lng bi c ht n nh v trnh s phng in khng kim sot c gia cc bn cc. Thit b lc bi tnh in hin c trn th trng quc t .

Sinh vin: Phm c Thnh Lp T 50 H2

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4, Yu cu v iu khin : Ap dng nguyn l c bn trn ta s thit k mt mch iu khin cho hai bn cc p ng cc yu cu t ra. Vi cng ngh lc bi ny khi thit k ta gp phi mt s vn sau: - Th nht l in p trn cao p lc rt cao, vo c 40KV n 100KV. Vi in p cao ny ta s rt kh chn van,c th phi v gi thnh ca h thng s cao. - Th hai l trong qu trnh lc do lng kh gia hai bn cc khi ion ho to thnh dng in nn h thng rt hay b ngn mch.V vy ta phi thit k mt h thng chng ngn mch v t ng ng mch vo in p lm vic sau khi kt thc phng in. in p ca thit b lc bi phi c tng dn n nh m bo cho lng bi c ht n nh v trnh s phng in khng kim sot c gi cc bn cc.


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Chng 2 : Thit k mch cng sut2.1, Cc s chnh lu :La chn phng n : Do chnh lu cu 3 pha u im hn cc mch chnh lu khc v h s dng my bin p, in p ngc t ln rt ph hp vi c im cu ti v cht lng in p 1 chiu u ra. Nh ni cng s b, do p ng y cc yu cu ca ti, ta s la chn mch chnh lu cu 3 pha i xng . Tuy nhin, mch chnh lu cu 3 pha i s c 2 phng n ph, l : chnh lu cu 3 pha iu khin c v chnh lu cu 3 pha khng iu khin c. Nn ta xt 2 phng n: a, Phng n th 1 : Dng mt b chnh lu cu 3 pha c th iu chnh c gc m dng s cc thysistor t sau my bin p.1-

S mch chnh lu cu 3 pha i xng :


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S cu to: mch gm 1 ngun xoay chiu 3 pha l tng mc vo b chnh lu gm 6 thysistor ( chia lm 2 nhm: V1,V3,V5 mc chung anot; V2,V4,V6 mc chung catot)

Ta c th chia s trn lm 2 phn. Phn th nht mc chung anot,s in p pha anot s l

Phn th 2 mc chung catot th s in p phia catot s l

Theo php cng in p ta s c: Ud = UdA - UdK Nn s in p qua ti s l:


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Ta c tr trung bnh p chnh lu qua ti l:

Vi U l gi tr hiu dng ca in p pha U= in p ngc ln nht trn van l : Un max= 6 U (i) Dng trung bnh qua ti l id = (ii) Dng trung bnh chy qua van l : iV = id/3 (iii) Nhn xt: (iv) (v) u im: c th iu khin cc thysistor ng m d dng thng qua gc m thay i in p qua ti v cng sut s kh ln. Hn ch nh cht lng in p ra xu, ph thuc vo gc m .S ny ch nn dng cho mch c cng sut ln v khng quan tm nhiu ti cht lng in p u ra.

Hot ng ca s : Ban u ngun 3 pha c a vo cun s cp ca my bin p 3 pha, sau khi nng p ta c ngun 3 pha tha mn iu kin in p v dng in.Ngun 3 pha ny c a vo b chnh lu to ra dng in 1 chiu.


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(vi)

B chnh lu cu 3 pha hot ng theo nguyn l khi kch m 1 van phi pht xung kch m km vi 1 van trc v cc van c m theo th t cch nhau 600 in.Ban u kch m T1 vi gc m , dng chy t pha a vo ti.Khi van T1 c kch m th ng thi van T2 c kch m theo van T1 dn vi van T2.Sau khong thi gian 300 + .Van T3 c m ra.Do lc ny th pha b ln hn th pha a nn van T1 b t mt in p ngc lm cho van T1 t ng kha li.Lc nay van T3 s dn vi van T2.Sau khong thi gian 3.300 + th T4 c kch m lm cho van T3 dn vi van T4.Do th pha c m hn so vi th pha b nn van T2 b t mt in p ngc lm cho van T2 kha li.C nh vy, qu trnh ny tip tc xy ra v mi van c kch m sau 1/3 chu k.

(vii) Nhn xt: u im : gc m ca cc thysistor c iu chnh d dng v cng sut ca s kh ln,t trng p mch ln. Hn ch : cht lng in p u ra thp,c th c hin tng trng dn.c bit l cc thysistor chnh lu phi chu mt in p rt ln do yu cu cng ngh t ra nn vic iu chnh rt kh khn gy tn hao ln.V vy,thc t ta khng s dng mch ny. (viii)


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b, Phng n th 2:Dng mt b chnh lu cu 3 pha khng iu chnh c l b chnh lu dng cc it sau my bin p v mt b iu p xoay chiu trc my bin p

S mch chnh lu cu 3 pha khng iu chnh c:

in p v dng ri trn ti l S mch iu p 3 pha


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a) S

Cc van T1, T4 ln lt dn dng theo 1 chiu xc nh nn dng qua cp thysistor u song song ngc ny l dng xoay chiu. Cc van thysistor c pht xung iu khin lch nhau gc 180 in m bo dng qua cp van l hon ton i xng. Ta c th dng in p ra 1 pha ca mch iu p xoay chiu :

Cc mch iu p xoay chiu c nhc im c bn l trong qu trnh iu chnh, mch lun lm vic ch dng in gin on, c dng dng in v in p ra ti u khng sin. Dng in s lin tc v ng thi tr thnh hnh sin hon chnh ch khi in p ra ti ly bng in p ngun. Nh vy, khi iu chnh trn ti nhn c mt di n sng hi hnh sin. b) Quan h gia gc m v cng sut ri trn ti:Sinh vin: Phm c Thnh Lp T 50 H2 10

Khi phn tch hot ng ca s ta cn xc nh r cc giai on s c bao nhiu van dn v nh cc quy lut di y ta c th c c biu thc in p ca tng giai on,t mi tin hnh tnh ton. Di y l cc quy lut dn dng ca van trong mch iu p xoay chiu ba pha: - Nu mi pha c mt van dn th ton b in p ba pha ngun u ni ra ti. - Nu ch hai pha c van dn th mt pha ngun b ngt ra khi ti, do in p a ra ti l in p dy no c van ang dn. - Khng th c trng hp ch c mt pha dn dng. Da vo quy lut dn dng ca van trong tng giai on m ta c th xy dng c th in p ra ca mch iu p xoay chiu ba pha. Tip theo,t nhng biu thc in p ca tng giai on ,ta c th tnh ton c cc i lng nh in p, dng in, cng sut... Xt hot ng ca mch iu p xoay chiu ba pha dng 6 thysistor u song song ngc,ti thun tr u hnh sao trn v dng th quan h gia cng sut ti v gc : Cng sut ti l : P = 3.I 2 .R trong I l tr s hiu dng ca dng in ti. Dng in ny bin thin theo hai trong ba quy lut dn dng ca van nh sau : Nu mi pha c mt van dn ( hay ton mch c ba van dn) : i= U dm sin( + ) 3R

Nu ch c hai van dn (hay ton mch c hai van dn ) : i= Trong : U dm sin( + ) 2R

U dm l bin in p dy. l gc lch pha gia in p v dng in giai on ang xt.

Tu thuc vo gc iu khin m cc giai on c 3 hoc 2 van dn cng thay i theo.Sinh vin: Phm c Thnh Lp T 50 H2 11

Ta thy c 3 khong iu khin chnh :1) Khong dn ca van ng vi = 0

60 o :

Trong phm vi ny s c cc giai on ba van v hai van dn xen k nhau nh th di y :

Da vo th ta c th xc nh c biu thc lin quan h gia cng sut ra ti P v gc iu khin : P = 3.I 2 .R

3U sin ( d + = R 32 dm /3 2

+ 3

3

sin d + 42

sin d + 3 +2 3

2 3

2 + 3 2 3

sin2 d + 4

sin2 3 d ) 2 + 3

2 3U dm sin 2 = [ + ] (1) R 6 4 8

2) Khong van dn ng vi = 60

90 o :

Trong phm vi ny lun ch c cc giai on hai van dn. Ta c th in p ra di :Sinh vin: Phm c Thnh Lp T 50 H2 12

Da vo th c th xc nh c biu thc lin gia cng sut ra ti P v gc iu khin : P = 3.I 2 .R 3U sin [ d + = R + 42 dm 2 2 3 5 + 6 3 5 + 6 3

sin 2 4 d] +2 3

=

2 3U dm 3 3 [ + sin 2 + cos 2] (2) R 12 16 16

3) Khong van dn ng vi = 90

150 o :

Trong phm vi ny ch c cc giai on hai van dn hoc khng c van no dn xen k nhau. Ta c th in p ra nh di :


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Da vo th ta c th xc nh c biu thc lin h gia cng sut ra ti P v gc iu khin : P = 3.I 2 .R2 3U dm sin 2 sin 2 [ d + 4 3 d] = R + + 2 3 2 3 2 3U dm 5 3 1 = [ + cos 2 + sin 2] (3) R 24 4 16 16

Theo ba biu thc (1), (2), (3) v cho cc gi tr khc nhau v ly P = 0 l 100% ta c bng cc gi tr v th biu din quan h gia cng sut ra ti P v gc iu khin :


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0 20 30 40 50 60 70 80

P% 100 98,6 95,6 90,2 81,8 70,6 57,16 42,8

90 100 110 120 130 140 150

P% 29,3 18,1 9,7 4,3 1,3 0,1 0


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Nhn xt : Cng sut a ra ti l ln nht khi gc iu khin = 0 nhng vi =30 th cng sut ra ti cng xp x khi = 0. Trong mch iu p xoay chiu 3 pha su thysistor u song song ngc, ti thun tr u tam gic, dng in p tng pha cng nh vy.Tuy nhin do ti u tam gic nn khi mch c 3 van dn th in p ri trn in tr ti l in p dy, khi mch c 2 van dn th in p ri trn in tr ti gia hai dy l in p dy cn in p ri trn hai in tr cn li bng mt na in p dy.

Kt lun:Nh vy theo yu cu thit k ca n th in p cao p lc ln n 50kV nh vy theo cng thc tnh in p ngc ri trn cc van chnh lu th n vo khong 50kV tr ln nh vy khng c van bn dn no c th chu c mc in p nh vy c. Nh vy ta phi la chn phng n 2, c ngha l ta s dng mt mch iu p 3 pha trc my bin p ri n s mch chnh lu cu diode .


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2.2 Thit k mch lc :2.2.1 My bin p lc: a, Nhim v ca my bin p lc: Nng in p sau khi iu p ln in p 20-60kV p ng yu cu in p cao ca cng ngh ca lc bi tnh in . b, Hot ng: in p u vo sau khi iu p s c a vo cun s cp ca my in 3 pha.Sau khi c nng p ti in p U2 = m.U1 ( vi m l t s bin i ca my bin p).in p c nng p s c a ra cc cun dy th cp a vo b chnh lu. S nguyn l ca b lc bi tnh in:

2.3 Tnh chn cc thng s cho mch lcSinh vin: Phm c Thnh Lp T 50 H2 17

2.3.1 Tnh chn cho my bin p lc: Ta chn my bin p 3 pha c 3 tr u Y-Y, lm mt bng khng kh t nhin.Do thc cht ca my bin p 3 pha gm 3 my bin p 1 pha nn ta tnh chung cho 1 pha. - Yu cu ca bi : in p u ra 20-60kV : dng in 2,0A - Do h thng khi lm vic dng in i qua cc thit b bin i nn s gp nhng tn hao nht nh.V vy ta thit k h thng vi lng d tr l 10% v cng sut.Ta chn in p ti a ra ti l 66kV Dng in s l 2,2A Cng sut cc i l Pmax= 145,2 kW - in p ln nht sau khi chnh lu: Ud0 Ta c Ud0= Ud + UV + Uba + Udn Trong : Ud l in p chnh lu theo l thuyt UV l in p st trn cc van ( chn vo khong 120V) Uba l in p st p bn trong my bin p ( do ty thuc vo vt liu cu thnh v cch qun dy trong my bin p nn ta chn vo khong 5-10%) Udn l in p st trn dy qun v c tnh theo cng thc Udn= Id . Rdn= Id . .l S

coi nh rt nh

Vy ta c in p cn chnh lu theo tnh ton l: Ud0 = 66.103 + 120 + 66.103 . 5% = 69420 V in p ra thc t trn cun th cp ca my bin p l: U2 =Udo.10 = 69420 .10/9 = 77,133.103(V) = 77,133KV 9

T s bin i ca my bin p l: m = U = 77,133 : 380 = 203 P Dng in chy trong cun th cp l : I2 = Id = 2,2A Dng in chy trong cun s cp l : I1 = m.I2 = 446.6A Cng sut pha th cp l : PTC = U2 . I2 = 169,69 (KW)Sinh vin: Phm c Thnh Lp T 50 H2 18 U2

Ta chn my bin p 3 pha da trn cc thng s ca 1 pha trn. 2.3.2 Tnh chn diode +) in p u ra ca chnh lu c tnh : Ud =

3 6 U 2 .Cos U2 =

.U dmax = 0,43 . 77133 = 33167,19 V 3 6

in p ngc cc i t ln diode: Ungmax = 6 U . 2 = 6 .33167,19 = 81242,69 V Chn h s d tr in p: Ku = 1,7 Van phi chu c: Ungmax thc = 1,7 . 81242,69 = 138112,57 (V) +) Dng in trung bnh qua van:

ItbVan =

Id 3

= 2,2 / 3 = 0,73(A)

Chn van c h s d tr dng in: k = 1,2. Dng in nh mc ca van cn chn :

I dm = k .I tbth = 1,2 . 0,73 = 0,876 (A)Ta chn cng sut lm vic ca cc van l 70% trong iu kin lm mt bng cnh tn nhit v qut thng gi. Dng lm vic ca van: Itbth max = 0,876 70% = 1,25A * Vy cc thng s chn diode : Ungmax = 138112,57 (V) IVan = 1,25 (A)Sinh vin: Phm c Thnh Lp T 50 H2 19

Do khng c diode c in p cao nh vy, chng ta phi tin hnh mc ni tip cc van. Van cng sut c chn l: T thng s trn ta chn iode loi RA205420XX. Un=5400 V, U=1,45 V S lng it cn dng mc ni tip l: n=26 chic. 226,2V St p trn 3 dy it l: Ud =6 . 1,45 . 26 =

2.3.3 Tnh chn thysistor: in p ngc ln nht m thysistor phi chu l: Un max = Up = 2 . 380 = 537,4 (V) Chn h s d tr in p: Ku = 1,7 Van phi chu c: Ungmax thc = 1,7. 537,4 = 913,58 (V) Nhn xt: khi gc iu khin = 0 in p ra ti l hnh sin v nh vy,dng in qua van lc ny cng l ln nht .Vy dng in trung bnh trn cc pha l: If = Id = 169690/3. 380 = 148,8 (A) Chn van c h s d tr dng in: k = 1,2. Dng in nh mc ca van cn chn :

I dm = k .I tbth = 1,2 . 148.8= 178,62

(A)

Ta chn cng sut lm vic ca cc van l 70% trong iu kin lm mt bng cnh tn nhit v qut thng gi. Dng lm vic ca van: Itbth max = 178.62 70%= 255 (A) * Vy cc thng s chn van : Ungmax = 913,58 (V) IVan = 255 (A) T cc iu kin trn ta chn van c cc c tnh sau Unm= 1000(V) Immax=5000(A Ipik=37000(A) Ig=0,3(A)Sinh vin: Phm c Thnh Lp T 50 H2 20

Ugm=3(V) Ihmax=1(A) dng gi cho van cn dn Irmax=0,18(A) dng r Umax=1,28(V) Idmax = 1000(A) dU/dt= 750 (v/s) 2.3.4 Bo v cho van: C 2 nguyn nhn chnh phi quan tm khi tnh ton ti vic bo v: 1 - Nguyn nhn bn trong: do s tch t in tch trong cc lp bn dn.Khi kho van thyristor bng in p ngc,cc in tch ni trn i ngc hnh trnh, n nhanh chng ca dng in ngc gy nn sc in ng cm ng rt ln trong cc in cm,vn lun lun c ca ng dy ngun dn n cc thyristor.V vy,gia ant v catt ca thyristor xut hin qu in p. Ta c th th hin qu trnh bin thin ca in p v dng in trn van:

2 - Nguyn nhn bn ngoi: nhng nguyn nhn ny thng xy ra ngu nhin nh khi ng ct khng ti mt my bin p trn ng dy, khi mt cu ch bo v nhy, khi c sm st ... bo v qu in p do tch t in tch khi chuyn mch gy nn ngi ta dng mch RC u song song vi thyristor nh hnh di:Thyristor

R

C

Thng ta chn C = 0,5 F , R = 80


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Do khng c van c in p cao hn, chng ta phi tin hnh mc ni tip cc van. Khi mc ni tip cc van yu cu cn thit phi chn cc van c c tnh ging nhau, nhm m bo cho s phn b in p nh nhau trn cc van. Tuy vy, s phn b in p trn cc van khng bng nhau l thng gp. Do , cn c cc bin php phn b li in p khi cc c tnh ca van khng ging nhau. Bin php y m t hnh di y Thng chn: C = 4 F R = 30


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- Ngoi ra ngi ta cn bo v cho cc van bng cc cu ch trnh qu dng.

Vi mch iu p c cc van chu dng in nh mc ln ta chn cc cu ch c dng in nh mc: Icc1 = 1,1 . 255 = 280,5 (A) Cn vi mch chnh lu cc van chu dng in nh ta chn cc cu ch c dng in nh mc : Icc2 = 1,1 . 1,25 = 1,375 (A) - Dng Aptomat bo v: Aptomat dng ng ct mch ng lc, t ng ngt mch khi qu ti v ngn mch Tiristo, ngn mch u vo b bin i (ngn mch s cp my bin p). - Ngoi ra ngi ta cn c th lm mt bng nc vi lu lng ty theo nhit ca van cng sut.


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Chng 3 : Thit k mch iu khin3.1 Nguyn l mch to tn hiu iu khin :Chng ta to ra 2 loi in p : UAK : in p iu khin ( in p mt chiu ). Ut : in p ta ( ng b vi in p A-K ca thyristor ). Hiu in p | Uk Ut | a vo khu so snh. Uk v Ut a n u vo ca mt khu so snh bng cnh lm bin i Uk ta c th iu chnh c thi im xut gin xung tc l iu chnh c gc . Khi Uk = 0 ta c = 0 Khi Uk < 0 ta c > 0 Quan h gia v Uk nh sau : =U dk U t max

Ta ly Ukmax = Utmax. Cu trc mch iu khin s gm : Khu ng pha : c nhin v to ra in p ta ng b vi in p li, cho php xc nh c gc iu khin Khu to in p ta: To ra in p ta so snh vi in p iu khin to ra tn hiu iu khin ph hp Khu so snh : So snh in p iu khin vi in p ta xc nh gc iu khin . Ti thi im Udk=Urc th pht xung iu khin m cc van tng ng. Khu phn hi : o dng in ra trn ti v chuyn thnh tn hiu p qua v kt hp vi Ucng ngh to nn in p Uk . Nh c khu phn hi m mch lm vic chnh xc hn, thon mn dng p do cng ngh t ra. Khu to xung v khuych i xung : mch to xung c nhim v to ra cc xung c rng nht nh v phn phi xung n ng cc van. Mch khuych i xung c nhim v khuych i m bo v bin , cng sut xung iu khin ng thi cch ly mch lc v mch iu khin.Sinh vin: Phm c Thnh Lp T 50 H2 24

3.2 Tnh ton cc khu trong mch iu khin:3.2.1 Nguyn l hot ng ca s : Cp ngun 380V cho cun s cp my bin p ng pha. in p phn th cp l in p ng pha. Ta ch ly in p U21, U22 dn gii : U21 c so snh mc in p 0V, qua D1 ch ly xung vung U1 . U22 c so snh mc in p 0V, qua D1 ch ly xung vung U2. U1 v U2 c a qua b to in p rng ca ( gm R 3, R4, R5 , R6, D1, D2, D3, A3, T1,T2 ), u ra s nhn c in p rng ca U3. Uk c so snh vi U3 qua A4, cho xung ch nht vung. Xung ch nht vung ny qua D4 ch ly xung ch nht dng U4. Uk c th thay i, khi nhit tng | Uk | gim, gc pht xung ln ln lm in p u ra ca thyristor gim v ngc li, khi nhit gim | Uk | gc pht xung gim i lm in p u ra ca thyristor tng. ( Khu phn hi in p l phn hi m n nh mt in p nht nh, tng ng vi mt nhit nht nh ). Xung U4 v xung chm Ux c a qua con AND. u ra AND l xung chm. Xung chm ny c qua T3, T4 mc alinhtn khuych i cng sut ( khuych i dng ). a xung qua bin p xung BAX, cho rng xung v bin xung nh mong mun. B pht xung chm l a hi t dao ng gm R 11, R12, R13, C2 v IC A4, qua D6 ly xung chm dng.1- Khu ng pha:


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U21 = 10 2 sin t (V)

U22 = -10 2 sin t (V)

Cho qua b so snh so vi in p 0V. Tn hiu ra nh hnh v ( b so snh o ).


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Ch : Ilv < 1 mA do : U21/R1 < 1 mA R1 > 10/1 = 10 K Ta chn R1 = R2 = R1 = R2 = 15 K

2-

Khu to in p rng ca ( Uta ):

- Nguyn l hot ng : + Trong na chu k u : U1 < 0 , U2 > 0.Sinh vin: Phm c Thnh Lp T 50 H2 27

U1 < 0 nn T1 kho. Khi t C1 c np in p UC1. U2 > 0 qua it D1 v n nh in p UD3. UC1 v UD3 c a qua b tr c in p ra l U3. + Trong na chu k sau : U1 > 0 , U2 < 0. U1 > 0 nn T1 dn. UC1 phng in qua T1. in p trn t C1 nhanh chng tr v 0. U2 < 0 nn UD3 = 0 V. Do vy U3 = 0 V.

Ta c th ca khu to in p rng ca nh hnh v :

- Tnh ton mch : + Trong na chu k u : IET2 = ICT2 = (E UD2 UBE) / R5. UC1 = C I CT 2 dt = (ICT2.t) / C1 = [(E UD2 UBE).t] / (R5.C1). 1Sinh vin: Phm c Thnh Lp T 50 H2 28 1

* Khi t = T/2 th UC1 = 9 V do : [(E UD2 UBE).T/2] / (R5.C1) = 9 V Vi E = 15 V, UD2 = 9 V, UBE = 0,6 V R5.C1 = 0,006 Ta chn C1 = 0,47F R5 = 12,766 K. Chn R5=13 K V UD3 = 9 V nn U3 = UD3 UC1 = (ICT2.t) / C1. Chn R6 = 3,3 K, IR6 = IR7 = UD3 / 2.R6 = 9 / 6600 = 1,36 mA. UR7 = U2 - UD1 UD3 = 4,3 V. ( UD1 = 0,7 V ) R7 = UR7 / IR7 = 3,16 K. Vy ta chn R7 = 3,3 K Chn R3 = 10 K. IET2 = ICT2 = (E UD2 UBE) / R5 = 0,09 mA IBT2 = ICT2 / . Chn T2 l 2N2904. Chn R4 = 5 K. - Khu so snh:


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in p rng ca U3 v Uk a vo cng m A4. Khi | U3 Uk | = 0 th trig lt trng thi v c u ra U4 l chui xung ch nht. Chn R8 = R9 = 10 K Chn R10 = R8//R9 = 5K

Tn hiu in p nh hnh v :

4

Khu pht xung chm:


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- Nguyn tc hot ng : Khi UC2 t ngng lt, s chuyn trng thi. p c gi tr ngc li vi gi tr c. Sau in p trn UC2 thay i theo hng ngc li v tip tc cho ti khi cha t ngng lt khc.Sinh vin: Phm c Thnh Lp T 50 H2 31

UN = |Ung| = Umax = 13V Ung = -Umax Ungt = Umax = R11/( R11 + R12 )

dU N U UN = max dt R13C 2UN(t) = Umax = (1 (1 + e-t/R13C2)) |UN(t)| = |Ung| = |Ungt| Khi T = 2R13C2ln(1+2R11/R12). Chn tn s khu pht xung chm : f = 10 KHz. R11 = R12 T = 2,2R13C2 = 1/(10.103) R13C2 = 45,5.10-6 Chn C2 = 0,02F R13 = 2,27K. Ta chn R13 = 2,2 K Chn R11 = R12 = 10K5

Chn cng AND:

Chn IC CMOS l IC4081 c 4 cng AND c cc thng s sau Vcc = 3 - 15 V. Ta chn Vcc = E = 15 V.Sinh vin: Phm c Thnh Lp T 50 H2 32

Cng sut tiu th : 2,5 nW / 1 cng. Ilv < 1mA in p ng vi mc lgic 1 l 2 - 4,5 V.

Tn hiu in p ra nh hnh v :


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6

Khu khuch i xung v bin p xung:

Utc = Ugk = 2,5 V Itc = Ig = 0,25 A Thng th t s ca my bin p xung l k = 2 3. Chn k = 2 T ta c Isc = Itc / k = 0,125 A Usc = Utc.k = 5 V Isc = ICT4 Chn T3 l C828 c h s 3 = 10 30. Chn T4 l 2N1613 c h s 4 = 80Sinh vin: Phm c Thnh Lp T 50 H2 34

Ly 3 = 10 IET3 = ICT4 / 3 = 0,0125 A Chn UBT4 = 0,7 V R16 = UBT4 / IBT4 = 56 IBT3 = ICT4 / 3. 4 = 15,625 A. Ta c UBT3 = UBET3 + UBT4 = 0,6 + 0,7 = 1,3 (V). V Us = 2 4,5 V nn ta chn in p st trn UR15 = 1 V R15 = UR15 / IBT3 = 64 K R14 (E USC) / ISC= 80 . Ta chn R14 = 57 7 - Khu chng ngn mch lm vic: a) Nhim v: Khi xy ra hin tng phng in th khu to ra tn hiu logic a vo chn 6 ca TCA785 tt tn hiu ra Q14, Q15, ng thi a v khu to tn hiu iu khin a in p Uk v 0 trong khong thi gian tr ttr no . Sau khong thi gian tr ny mch li t ng phc hi in p pha cao p. a) S nguyn l:

Trong khu chng ngn mch lm vic c s dng 2 vi mch chuyn dng l Optocoupler PC81711NSZ v vi mch MM74HC4538.Sinh vin: Phm c Thnh Lp T 50 H2 35

- Vi mch MM74HC4538: S chn:

Section I.2 Section I.3 Section I.4

Section I.5 Section I.6 Bng chn l Xo L X X H H u vo A X H X L u ra B X X L H Q L L L Q H H H

K hiu : H - mc cao L - mc thp - chuyn t mc thp ln mc cao - chuyn t mc cao xung mc thp - mt xung mc cao - mt xung mc thpSinh vin: Phm c Thnh Lp T 50 H2 36

- Optocoupler PC81711NSZ: S chn:

Thng s k thut: + Cc gi tr cc i: Thng s Dng vo Dng vo cc i in p ngc Tn tht in p Colect Emit in p Emit Colect Dng Colect Tn tht trn colect Tng nng lng tn tht Nhit lm vic K hiu IF IFM VR P VCEO VECO IC PC Ptot Topr Gi tr 10 200 6 15 70 6 50 150 170 -30 -> +100 n v mA mA V mW V V mA mWoC oC

u vo

u ra

8 - Khi ngun :


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IC n p loi UA7815 c cc thng s l : U ngng = 35 V U ra = 0 1,5 A E = 15 V IC n p loi UA7915 c cc thng s l : Ungng = 40 V Ura = 0 1,5 A -E = -15 V UMN min = 18 V, ta thng chn UMN = 21 V Ta c UMN = Ua.2,34 = 10.2,34 = 23,4 V tho mn iu kin chn C4, C5 l t lm phng 330 F


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