Chng I

Mn hc: in t cng nghip

Chng I

Cc phn t c bn trong mch in t

1.1 Khi nim v cht bn dn

1. Khi nim

Cht bn dn l cht trung gian gia cht dn in v cht cch in. Cht bn dn thng gp trong k thut l Gcmany (Ge), Silc (Si), Indi (In)

2. c im ca cht bn dn.

- in tr ca cht bn dn gim khi nhit tng v ngc li.

- in tr ca cht bn dn thay i theo tinh khit ca cht bn dn do nu pha tp cht vo trong cht bn dn th c th thay i c in tr ca cht bn dn.

- Gi tr in tr ca cht bn dn thay i theo cng ca nh sng chiu vo.

1.2. Cc phn t bn dn

1.2.1. Cc cht bn dn thng dng. Ge 1. Cht bn dn loi P (Positive)(Cn c gi l bn dn dng hay bn dn l trng) - Nu pha vo tinh th Ge tinh khit 1 lng nh In, (cht In c 3 in t ho tr lp ngoi Ge In Ge cng) th s tng mt l trng ln rt nhiu trong mng tinh th. - Trong mng tinh th Ge - In mi nguyn tca In lin kt vi 4 nguyn t Ge xung quanh Ge Inbng 3 vng lin kt, nn cn thiu 1 in t,

tc l n d ra 1 l trng. Vy l trng ny khng phi do 1 in t tch ra khi vng lin Mng tinh th Ge - Inkt m do c s pha trn ca tp cht.

- Cht bn dn c mt l trng nhiu hn hn mt in t t do nn cht bn dn ny

gi l cht bn dn l trng Ge2. Cht bn dn loi N (Negative) (Cn c gi l bn dn m hay bn dn in t) - Nu pha vo tinh th Ge tinh khit 1 lng Ge As Ge nh As, (cht As c 5 in t ho tr lp ngoi cng) th s tng mt in t ln rt nhiu trong mng tinh th.

- Trong mng tinh th Ge - As mi nguyn t Ge As ca As lin kt vi 4 nguyn t Ge xung quanh

bng 4 vng lin kt, nn cn tha 1 in t th

Mng tinh th Ge - As

5 lin kt yu t vi ht nhn v d dng tch ra khi lin kt. Vy in t ny l do c s pha trn ca tp cht.

- Cht bn dn c mt in t nhiu hn hn mt l trng nn cht bn dn ny gi l cht bn dn in t

1.2.2 Diode (it) bn dn

1. Cu to

a. K hiu v hnh dng ca it

* K hiu * Hnh dng

b. Cu to

- it c cu to gm 2 ming bn dn loi P v N ghp li vi nhau. gia 2 cht bn dn hnh thnh lp tip gip P - N. iu kin bnh thng th lp tip gip nh hng ro ngn cch khng cho in t v l trng ti hp vi nhau.- hnh v k hiu quy c: u P ca ming ghp gi l Ant (A), u N ca ming ghp gi l Katt (K).

Lp tip gip

2. c tnh Vn - Ampe.

* Khi nim: Biu th quan h bng th gia dng in chy qua it v in p t gia Ant v Katt ca it .

* th

- Phn cc thun: t dng ngun vo dng ca it (Ant), cn m ngun vo m ca it (Ktt), lc ny it m cho dng in chy qua.

- Phn cc ngc: t dng ngun vo m ca it (Ktt), cn m ngun vo dng ca it (Ant), lc ny it kho khng cho dng in chy qua. Ith

Ung Ungmax Ung cp 0 * Nhnh Thun:- Phn p thun, it m cho dng in chy qua.

- Khi t 1 in p l Um vo it th ng c tnh c dng Parabol (1).

- Khi tng in p thun th ng c tnh gn nh ng thng (2)

Lc ny in tr thun ca it rt nh.

* Nhnh ngc:

- Phn p ngc, it kho, ch c dng in r rt nh chy qua.

- Khi tng in p ngc th dng r tng chm. Khi in p ngc mc cho php th dng r gn nh khng i, ng c tnh l ng (3)

- Khi tng in p ngc n gi tr Umax th dng r tng nhanh (4) v cui cng it b nh thng (5)

* Kt lun: - Mun it m th phn cc thun

- Mun it kha th phn cc ngc bng cch t in p ngc ln D nhng khng t in p qu ln s ph hng D.

3. Tnh chnh lu ca it bn dn.

- Chuyn tip P - N l b phn quan trng nht ca tip xc gia 2 bn dn khc loi. Tu theo in p t vo theo chiu thun hay ngc m n c c tnh khc nhau.

+ Khi phn cc thun, chuyn tip P - N hp, in tr R nh, dng in I ln v tng nhanh theo in p U.

+ Khi phn cc nghch, chuyn tip P - N m rng, in tr R rt ln, dng in rt nh v t thay i theo in p U.

- Nh vy: Chuyn tip P - N dn in theo 2 chiu khng ging nhau, nu c in p xoay chiu t vo th n ch dn in ch yu theo 1 chiu.

Ta gi l tnh cht van hay c tnh chnh lu ca it.

4. c tnh k thut.

Mt it c cc thng s k thut cn bit khi s dng l:

+ Cht bn dn dng ch to.

+ Dng in thun - nghch cc i

+ Dng in bo ho

+ in p nghch cc i.

5. Cch th it

* Kim tra cht lng it: Dng ng h vn nng (thang o in tr) hoc ng h mmt kim tra

- Cch th:

+ Vn nm xoay v R x 1+ u 2 que o vi 2 cc it. Ln lt o u que o tm c 2 gi tr in tr khc nhau.

+ in tr c gi tr khong vi trm kil m gi l in tr ngc Rng

+ in tr t vi chc n vi trm m c khi ti vi K( gi l in tr thun Rth

- Kt lun:

+ Rng cng ln hn so vi Rth cng tt+ Nu Rng ( Rth th it b hng

* Xc nh cc cho it

Nu it cn tt nhng khng c k hiu phn bit cc th dng ng h vn nng (thang o in tr) hoc ng h mmt xc nh.

- Vn nm xoay v R x 1

- u 2 que o vi 2 cc phn cc cho it

- Nu gi tr in tr o c khong vi trm kil m th c ngha l lc ny it ang c phn cc ngc. Ta kt lun:

Chn u v pha cc dng pin trong ng h l cc m it (K) Chn u v pha cc m pin trong ng h l cc dng it (A)

- Ngc li: Nu gi tr in tr o c ch khong vi chc n vi trm m th c ngha l lc ny it ang c phn cc thun. Ta kt lun:

Chn u v pha cc dng pin trong ng h l cc dng it (A)

Chn u v pha cc m pin trong ng h l cc m it (K)

nh du cc cc thun li trong qu trnh lm vic sau ny.

6. Cc loi it khc

a. it Zener

- it Zener c cu to ging it thng nhng cc cht bn dn c pha tp cht vi t l cao hn it thng. it zener thng c s dng trong cc mch n p.

- K hiu:

Hnh dng

b. it pht quang (LED)- mt s cht bn dn c bit khi c dng in i qua th c hin tng pht ra nh sng (bc x quang)- Tu theo cht bn dn m nh sng pht ra c mu khc nhau. Da vo tnh cht ny ngi ta ch to ra cc n LED khc nhau.

- K hiu:

c. it bin dung- L loi linh kin bn dn 2 cc c chuyn tip P - N c ch to 1 cch c bit sao cho in dung ca n thay i nhiu theo in p ngc t vo.- ng dng: mch t ng iu chnh tn s cng hng, trong cc b khuch i tn s hoc nhn tn

- K hiu:

d. it Tunen- L loi linh kin bn dn 2 cc c chuyn tip P - N c nng tp cht cao.

- ng dng: Dng trong cc mch khuch i, to dao ng siu cao tn.- K hiu:

1.2.3. Tranzitor

1. Cu to- TZT l 1 linh kin in t c ghp li t 3 lp bn dn, sao cho 2 lp lin nhau khc loi, to thnh 2 tip gip P - N

- Tu theo cch sp xp cc vng bn dn m ta c 2 loi TZT l loi PNP v NPN.

- Min th nht ca TZT c gi l min Emittor, min ny c pha vi nng tp cht ln nht. Cc ni vi min ny c gi l cc Emittor (k hiu: E)- Min th hai ca TZT c gi l min Base, min ny c pha vi nng tp cht nh nht. Cc ni vi min ny c gi l cc Base, (k hiu: B)

- Min th nht ca TZT c gi l min Colector, min ny c pha vi nng tp cht trung bnh. Cc ni vi min ny c gi l cc Colector (k hiu: C)

- Loi TZT PNP c gi l TZT thun, gm 1 ming bn dn N gia v 2 ming bn dn P 2 bn. (hnh a). K hiu vi mi tn cc E c chiu i vo ch chiu i ca dng in trong cht bn dn. (hnh c)- Loi TZT NPN c gi l TZT ngc, gm 1 ming bn dn P gia v 2 ming bn dn N 2 bn. (hnh b). K hiu vi mi tn cc E c chiu i ra ch chiu i ca dng in trong cht bn dn. (hnh d)

P N P N P N

E C E C

B B

Hnh a Hnh b

2. Nguyn l lm vic

- Xt 1 TZT thun PNP u theo kiu mch cc gc chung.

+ Ngun E1 phn cc thun cho tip gip E - B

+ Ngun E2 phn cc nghch cho tip gip B - C.

C phn cc ng nh vy th TZT mi lm vic c.

- Tip gip EB c phn cc thun nn l trng t vng pht chy sang vng gc, in t tha t vng gc chy sang pht ti hp vi l trng.

- Ngi ta ch to sao cho s l trng vng pht rt nhiu so v s in t tha vng gc.

- Cc in t t cc m ngun E1 ti b xung.

- Dng in t t cc m ngun E1 vo vng gc sinh ra dng gc t cc gc ra m ngun 1 TZT lm vic c th ng thi phi c cc iu kin sau:

+ C ngun 1 chiu E1 phn cc thun cho lp tip gip E - B

+ C ngun 1 chiu E12 phn cc ngc cho lp tip gip B C

+ Ngun E2 > E11.2.4. Thyritor

1 TRT m cho dng in chy qua th phi c cc iu kin sau:

+ Phn cc thun cho TRT (Cc dng ngun ni vi Anot, cc m ngun ni vi Ktt)

+ C xung dng t vo cc iu khin G

1.2.5. Triac

Triac c tc dng nh 2 TRT mc song song ngc, cho nn n m cho dng in i qua theo c 2 chiu, nu c xung t vo cc iu khin G

1.2.6. in tr:

in tr thng Bin tr

1.2.7. T in

T thng T ho T xoay

Chng II mch khuch i

2.1. Mch khuch i dng TZT thng

1. Tng khuch i mc E chung

a. S mch: Gm c

- R1: in tr nh thin m bo cho tip gip pht gc c phn cc thun vi in p c 0,1 n 0,3 Vn

- R2: in tr ti (tr gnh) a in p t ngun E2 n cc gp tip gip gc gp c phn cc nghch.

- C1, C2 l t lin lc vi tng trc v tng sau.

- IB l dng in vo - IC l dng in ra

- UV l in p o, UR l in p ra

* Tn hiu vo a ti 2 cc gc pht, tn hiu ra ly 2 cc gp pht. Cc pht E tham gia c mch vo v mch ra nn c gi l mch khuch i mc EC

b. Nguyn l lm vic

* Khi a tn hiu xoay chiu UV vo u vo ca mch ta thy:

+ TZT lm vic, tip gip EB phi c phn cc thun nn c dng IE mt chiu t dng ngun E1, E2 chy t chn E qua chn B. Ti y, mt phn nh l

dng IB i qua R1 v m ngun E1, cn phn ln l dng IC chy qua R2 v m ngun E2 (IE = IB + IC)

- Trong na chu k dng ca tn hiu vo, in p dng ca tn hiu lm cho in p ti chn B l UB bt m hn (dng ln), ( in p UBE gim, ( dng IB v IC u gim. St p trn R2 gim i lm cho UC tng (theo chiu m), tc l m hn.

- Trong na chu k m ca tn hiu vo, in p m ca tn hiu kt hp vi in p m ngun E1 lm cho in p ti chn B l UB cng m hn, ( in p UBE tng, ( dng IB v IC u tng. St p trn R2 tng ln lm cho UC gim (theo chiu m), tc l dng ln.

* Nh vy in p ra ngc pha vi in p vo. Nu thay i UV (l UBE) th dng IB, IE thay i dn n IC thay i theo

- Trong mch mc EC, dng in ra ln hn dng vo t hng chc n hng trm ln.

- Mch din ny c ng dng rt rng ri

2. Tng khuch i mc CCa. S nguyn l: Gm c

- R1: in tr nh thin m bo cho tip gip pht gc c phn cc thun- R2: in tr ti (tr gnh)

- C1, C2 l t lin lc vi tng trc v tng sau.

- IB l dng in vo - IC l dng in ra

- UV l in p o, UR l in p ra

* Tn hiu vo a ti 2 cc gc gp, tn hiu ra ly 2 cc gp pht. Cc gp C tham gia c mch vo v mch ra nn c gi l mch khuch i mc CC

b. Nguyn l lm vic

* Khi a tn hiu xoay chiu UV vo u vo ca mch ta thy:

+ TZT lm vic, tip gip EB phi c phn cc thun nn c dng IE mt chiu t dng ngun E1, E2 chy t chn E qua chn B. Ti y, mt phn nh l dng IB i qua R1 v m ngun E1, cn phn ln l dng IC chy v m ngun E2 (IE = IB + IC)

- Trong na chu k dng ca tn hiu vo, in p dng ca tn hiu lm cho in p ti chn B l UB bt m hn (dng ln), ( in p UBE gim, ( IE gim. St p trn R2 gim i lm cho UE dng hn c trc khi c tn hiu vo.

- Trong na chu k m ca tn hiu vo, in p m ca tn hiu kt hp vi in p m ngun E1 lm cho in p ti chn B l UB cng m hn, ( in p UBE tng, ( IE tng. St p trn R2 tng ln lm cho UE cng m hn.

* Nh vy in p ra ng pha vi in p vo, ng thi khuch i in p

km KU < 1, v: UV = UR2 + UEB v UR = UR2 do vy

- Trong mch ny, khuch i dng in KI t hng chc n hng trm ln.

- Mch din ny c ng dng rt rng ri

3. Tng khuch i mc BC

a. S nguyn l: Gm c

- R1: in tr nh thin m bo cho tip gip pht gc c phn cc thun- R2: in tr ti (tr gnh)

- C1, C2 l t lin lc vi tng trc v tng sau.

- IE l dng in vo - IC l dng in ra

- UV l in p o, UR l in p ra

* Tn hiu vo a ti 2 cc pht gc tn hiu ra ly 2 cc gp gc. Cc gc B tham gia c mch vo v mch ra nn c gi l mch khuch i mc BC

b. Nguyn l lm vic

* Khi a tn hiu xoay chiu UV vo u vo ca mch ta thy:

+ TZT lm vic, tip gip EB phi c phn cc thun nn c dng IE mt chiu t dng ngun E1, chy t chn E qua chn B. Ti y, mt phn nh l dng IB v m ngun E1, cn phn ln l dng IC chy v m ngun E2 (IE = IB + IC)

- Trong na chu k dng ca tn hiu vo, in p dng ca tn hiu phi hp vi in p dng ngun E1 lm cho in p ti chn E l UE dng hn UB, ( in p UBE tng, ( IE cng tng. St p trn R2 tng ln lm cho in p ti chn C (UC) gim ( Theo chiu m), tc l tng ln, ( in p ra UR tng ln .

- Trong na chu k m ca tn hiu vo, in p m ca tn hiu lm cho in p ti chn E l UE bt dng hn so vi UB, ( in p UBE gim, ( IE gim, IC gim theo. St p trn R2 gim nn lm cho in p ti chn C (UC) tng ln ( Theo chiu m), tc l gim i, ( in p ra UR m i.

* Nh vy in p ra ng pha vi in p vo, ng thi khuch i dng in km KI < 1, v:

- Trong mch ny, khuch i in p KU t hng trm n hng ngn ln.

- Mch din ny c ng dng trong tng dao ng ca my thu dao ng n nh.

2.2. Ghp gia cc tng khuch i

1. Ghp trc tip

Gii thch s :

- R1: in tr nh thin v n nh nhit cho TZT T1

- R2: Va l in tr ti (tr gnh) co TZT T1, va l in tr nh thin cho TZT T2. Phi chn R2 c tr s sao cho in p UC1 khng qu thp nhng in p UB2 khng qu cao v UB2 cao s lm hng n T2. ng thi IB2 khng ln qu.

Hnh a

- Thng thng ngi ta chn R2 = 5 n 10 K( th UB2 = 0,2 n 0,3 V nhng lc ny UC1 qu thp nn s khng gy mo v khuch i thp. (hnh a)

Hnh b

- khc phc nhc im ny trong mch cc pht ca T2 nn mc thm in tr R3 nng UE2 ln c 1 n 2 V, nh c th dng R2 tng i nh UC1 = 1,2

n 2,2 V m bo UB2 = 0,1n 0,2 V. (Hnh b)

- T C1 v C2 kh cc hi tip m i vi m tn trong mch.2. Ghp bng in dung - in tr

a. Gii thch s

- Cc t C1, C2, C3 l t lin lc, dn tn hiu ra vo gia cc tng (dng t ho)

- Tng 1:

+ R1, R2 l in tr thin p. R1 iu chnh, chn ch lm vic cho n T1

+ RE1: n nh nhit cho n T1.

+ CE1: lm ngn mch tn hiu xoay chiu trnh hi tip m.

+ RC1: in tr ti ca n T1.

- Tng 2:

+ R3 l in tr thin p, c mc theo kiu n nh nhit. ng thi R3

iu chnh, chn ch lm vic cho n T2.

+ RC2: in tr ti ca n T2.

b. Nguyn l hot ng

- T C1 dn tn hiu ra xoay chiu ca tng trc vo cc gc v pht ca n T1. Qua T1 tn hiu c khuch i ln.

- Tn hiu ra ly cc gp v pht ca n T1 a qua t C2 vo cc gc v pht ca T2 . tng ny tn hiu li c khuch i ln 1 ln na.

- Tn hiu ra ly cc gp v pht ca n T2 a qua t C3 vo cc gc v pht ca tng tip theo. C nh th tn hiu s c khuch i ln nhiu ln.

* Ch : - Cc t C1, C2, C3 l t lin lc, dn tn hiu xoay chiu ra vo gia cc tng ng thi ngn tn hiu 1 chiu gia cc tng. Do vy ngi ta dng t ho.

- Trong mch dng 2 TZT thun, m TZT thun m th in th UB2 bao gi cng t m hn UC1 nn ngi ta u chn t nh sau:

+ u cc dng ca t v pha cc gc ca n tng sau:

+ u cc m ca t v pha cc gp ca n tng trc:

c. Nhn xt.

- u im:+ S n gin, tin cy ln, linh kin gn nh, d thc hin n nh im lm vic.

+ Gi thnh h

- Nhc im:

+ Khng phi hp c tr khng gia cc tng.

+ Kh khai thc khuch i cng sut ln nht.. Mi tng ch K c 20 n 100 ln.

3. Ghp bng bin p.

a. S mch in

* Tng 1:

- R1 l in tr nh thin, xc nh ch lm vic cho TZT T1

- R2, R3 l in tr phn p, n nh nhit cho TZT T1

- C1 l t lin lc vi tng trc.

- C2 v C4 l t kh hi tip m i vi xoay chiu

* Tng 2:

- R5 l in tr nh thin, xc nh ch lm vic cho TZT T2

- R4 l in tr phn p cho TZT T2.

- R6 n nh nhit cho TZT T2

- C3 l t kn mch vo tng kch ng. Nh c C3 v C4 m ton b tn hiu t th cp my bin p BA1 c a vo cc gc pht ca n T2.

b. Nguyn l lm vic.

- Tn hiu vo c t C1 dn vo cc gc pht TZT T1 v c K ln. Khi dng IC1 ca TZT T1 chy qua cun s cp ca my bin p BA1, n cm ng sang cun th cp. Tn hiu ra t cun th cp BA1 s c dn vo tng T2.

- Ti tng ny tn hiu li c K ln. Sau khi tn hiu ra t T2, qua s cp BA2, cm ng sang th cp BA2, ri c a vo tng tip theo na hoc ni vi ti.

- Nu my bin p BA2 c cun s cp gm W1 vng dy v cun th cp gm W2 vng dy th t s BA l;

KBA = hay

- V tr khng ra ca TZT tng trc ln hn tr khng vo ca TZT tng sau khong 10 n 50 ln nn s vng dy s cp BA thng gp 3 n 7 ln s vng dy ca th cp BA.

- Nh vy, nu tnh ton v iu chnh t s vng dy s - th ca cc my bin p sao cho ZV ca tng sau bng ZR ca tng trc, th s t c K cng sut ln nht.

c. Nhn xt

* u im:

- D dng phi hp c tr khng gia cc tng, chn t s BA thch hp.

- C th chn c h s K ln nht, ti 10.000 ln.

* Nhc im:

- Cng knh, gi thnh cao, lp rp phc tp.

- Gy mo tn s.

* ng dng: Thng c ng dng cho cc tng K cui v trc cui.

2.3 Khuch i cng sut

1. Khuch i cng sut n ch A

* Gii thch s

- C1 l t ni tng (t lin lc). CE l t kh hi tip m.

- RE l in tr n nh nhit cho n.

- R1 l in tr nh thin v xc nh im lm vic cho n.

- R2 l in tr phn p.

- BA ly tn hiu ra ti (loa)

* H s K ca tng: KP = (2 .

Trong + ( l h s K tnh ca TZT mc kiu EC.

+ RT l in tr ti (trn th cp my bin p) quy i v s cp.

+ RV l in tr vo ca tng.

2. Khuch i cng sut y ko lm vic ch B

* S nguyn l

- Mch dng 2 TZT thay phin nhau lm vic c 2 na chu k.

- My bin p t ngu BA1 lm nhim v to ra 2 tn hiu U1 v U2 bng nhau v ln nhng ngc pha nhau.

- E l ngun phn cc cho TZT T1, T2.

- R1, R2 to ra st p

- My bin p BA2 ly in p ra ti (Loa)

* Nguyn l lm vic: t tn hiu xoay chiu vo s cp my bin p BA1, tn hiu c cm ng sang cun th cp BA1. Ti y ta c:

- Na chu k u: UT1 dng hn UT2 :

+ Nh vy UB1 nhn c tn hiu dng s dng hn so vi UE1 do vy TZT T1 kho.

+ Ngc li UB2 nhn c tn hiu m s m hn so vi UE2 do vy TZT T2 m.

+ Dng in i t chn E2 qua tip gip E2B2, qua chn C2 , qua s cp my bin p BA2, cm ng sang th cp BA2 v ra ti.

- Na chu k sau: Ngc li, UT2 dng hn UT1 :

+ Lc ny T1 m, T2 kho.

+ Dng in i t chn E1 qua tip gip E1B1, qua chn C1 , qua s cp my bin p BA2, cm ng sang th cp BA2 v ra ti.

- Kt lun: Mch ny c dng K vi cng sut ln, hiu sut cao.Chng iii ngun mt chiu

3.1 Khi nim chung

- Trong thc t (lnh vc in t) trn cng mt mch thng c nhiu dng in khc nhau. C c dng 1 chiu, dng xoay chiu, v trong nhng dng xoay chiu cng c th c tn s khc nhau.

- Nu ni mch ny vo tng sau th tt c cc thnh phn u vo tng sau. Nhng c nhng trng hp ch cn a sang mch sau 1 hoc vi dng in nht nh cn nhng dng khc th khng cn thit a sang, hoc nu a sang th s gy tc hi n mch sau.

- V vy cn phi la lc nhng dng bng nhng b lc.

B lc l nhng mch in cn c theo yu cu ca s la lc, v tnh nng ca cc phn t c bn ca mch in sp xp b tr nhng phn t gia mch trc v mch sau nhm mc ch ch a sang mch sau nhng dng in cn thit.

3.2 Lc cc thnh phn xoay chiu ca dng in ra ti.1 Lc bng t in C

* Tnh nng c bn ca t in l:- Khng cho dng in 1 chiu qua.

- Ch cho dng in xoay chiu qua. Dng c tn s thp qua kh, c tn s cao qua d. Dng c tn s cng cao qua cng d.

*Hnh v

O t

h/v Lc bng t nn na chu k th

- Nhn vo hnh v ta thy: Tn hiu xoay chiu sau khi qua Diot s c nn thnh tn hiu 1 chiu. Tuy nhin s cn st li vi thnh phn xoay chiu. Cc phn t xoay chiu ny s chy qua t in C v ngun, cn cc tn hiu 1 chiu qua ti R.

- th tn hiu 1 chiu qua ti nh hnh v.

2 Lc bng cun dy L

* Tnh nng c bn ca cun dy l:

- Cho dng in 1 chiu qua.

- Cho dng xoay chiu qua nhng dng c tn s thp qua d, tn s cao qua kh. Dng c tn s cng cao qua cng kh, gn nh b cn li.

* Hnh v

- Nhn vo hnh v ta thy: Tn hiu xoay chiu sau khi qua Diot s c nn thnh tn hiu 1 chiu v c th s cn st li vi thnh phn xoay chiu. Nhng ch c cc tn

O t

h/v Lc bng cun dy nn na chu k thch c cc tn hiu 1 chiu qua c cun cm L v qua ti R. Cn cc tn hiu xoay chiu b cn li.

- th tn hiu 1 chiu qua ti nh hnh v.

3.3 B lc hnh L ngc v hnh (.

1 B lc hnh L ngc (Lc tn s thp)

- y l 1 dng b lc hn hp. Gm c cun cm L mc ni tip vi ti, t in C u song song vi ti.

- Theo c tnh ca t in v cun cm th:

+ Cc sng iu ho c tn s thp qua cun cm d dng, qua t in kh v th cc sng ny t ngun d dng qua cun cm L qua ti, cn phn r qua t in rt nh.

+ Ngc li, cc sng iu ho bc cao c tn s ln s b cun cm L cn li, cc sng ny ch yu i qua t in C.

- Kt qu l trn ti ch gm cc sng iu ho bc thp. c bit thnh phn 1 chiu hon khng suy gim qua cun cm L v khng r nhnh qua t C nn c bo ton chuyn n ti.

- Hnh v:

H/v: B lc B lc mc vi ti

2 B lc hnh (.

- y l 1 dng b lc hn hp. Gm c cun cm L mc ni tip vi ti, t in C1 u song song vi ngun t in C2 u song song vi ti.

- Kt qu l trn ti ch gm cc sng iu ho bc thp v tn hiu 1 chiu

H/v: B lc B lc mc vi tiChng IV

CHnh lu v nghch lu

4.1 Khi nim chung v chnh lu (CL) v nghch lu (NL)

1. Khi nim v CL

- CL l bin tn hiu xoay chiu u vo thnh tn hiu 1 chiu u ra. Mt thit b lm nhim v CL c gi l 1 b CL.

- Cc linh kin in t lm nhim v CL thng dng nht l Diot (CL khng iu khin), hoc dng Thyristor (CL c iu khin)

- Cch phn loi n gin nht c thit lp nh sau:

1 pha

Hnh tia

Khng iu khin

S CL : 3 pha

n pha

Hnh cu

C iu khin

VD:S CL 1 pha hnh tia c iu khin

S CL 3 pha hnh cu khng iu khin

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

S khi 1 b CL gm: vo CL ra

2. Khi nim v nghch lu (NL)

- Nghch lu l qu trnh bin i tn hiu 1 chiu u vo thnh tn hiu xoay chiu u ra (Ngc li vi CL)

- C 2 loi s NL chnh:

+ S NL lm vic ch ph thuc vo in p xoay chiu ca li in.

+ S NL lm vic ch c lp (ngun in c lp nh acquy, my n)

- Trong mch NL cc linh kin in t chnh lm nhim v NL thng dng l cc n c iu khin nh: Thyristor hay Tiratron

4.2 Chnh lu dng it 1. Chnh lu cu 1 pha ti thun tr

* S nguyn l (h.a)

Hnh a

Gm c

- 1 my bin p 1 pha.

- 1 b CL gm 4 Diot mc thnh 1 mch vng khp kn gi l cu CL

- Ti l R

* Nguyn l lm vic :

- a tn hiu xoay chiu vo phn s cp MBA, trn cun th cp s c cm ng 1 tn hiu xoay chiu. a tn hiu ny vo b CL. Xt qu trnh lm vic ca 1 chu k:

+ Na chu k u, gi thit dng a, m x.

Trn cu CL c cc Diot D1 v D3 c phn cc thun nn dn. Cn D2 v D4 phn cc ngc nn kho.

Dng in qua ti c chiu l:

(+) a ( D1 ( M ( R ( N ( D3 ( (-) x

+ Na chu k sau, tn hiu vo c chiu ngc li, dng x, m a

Lc ny, trn cu CL cc Diot D1 v D3 c phn cc ngc nn kho. Cn D2 v D4 phn cc thun nn dn.

Dng in qua ti c chiu l:

(+) x ( D2 ( M ( R ( N ( D4 ( (-) a

* Nh vy c 2 na chu k, tn hiu qua ti u c chiu t M n N, m bo l tn hiu 1 chiu. Cc chu k sau tng t (tn hiu vo l hnh sin, ra l nhng na hnh sin lin tip).

* th tn hiu dng in v in p vo ra (h b)

u,i

Hnh b

O T/2 3T/2

T (t)

u,iRa u i

O

T/2 T 3T/2 (t)

Nhn vo th tn hiu ra ta thy: tn hiu rt mp m khng bng phng, tn hiu trn ti bng phng ngi ta phi s dng cc b lc. (Xem chng III)

2. Chnh lu cu 1 pha ti R + L ni tip

* S nguyn l (hnh c)

Gm c

- 1 my bin p 1 pha.

- 1 b CL gm 4 Diot mc thnh 1 mch vng khp kn gi l cu CL

- Ti l cun dy mc ni tip vi in tr R (Chng hn l 1 ng c) c tng tr l ZT Hnh c

* Nguyn l lm vic :Tng t nh s hnh a

Ch : Do trn cun dy c in cm L nn khi dng in qua cun dy, n s bin thin chm hn so vi khi i qua in tr. Ngha l dng qua ph ti s tng chm v gim chm nn khng bao gi dng v 0. Tng t nh qua 1 b lc.

th tn hiu dng in v in p ( Hnh d) u,iV

O T/2 3T/2

T (t)

u,iRa u i

Hnh d O

T/2 T 3T/2 (t)

3. Chnh lu cu 3 pha ti thun tr (Cu Lavinp)

Hnh e

* S nguyn l (hnh e)

Gii thch s , gm:

- 1 my bin p 3 pha: - s cp c th ni Y hoc (, trn s ny l ni Y

- th cp ni hnh Y

- Cu chnh lu gm 6 it chia lm 2 nhm:

+ Nhm 1 gm D1, D3, D5 u chm Katt (K)+ Nhm 2 gm D2, D4, D6 u chm Ant (a)

- Ti l R1 u vo 2 u chm A v K

b. Nguyn l lm vic:

th tn hiu vo t cun th cp my bin p ti cu chnh lu nh hnh f1.

Xt tng khong thi gian nh c tn hiu xoay chiu trn mch th cp t vo chnh lu, ta c

- Khong thi gian t 0 - t1+ Tn hiu vo c pha c dng nht; pha b m nht. Trn cu chnh lu c cc it D5 v D4 dn, cc it cn li u kho.

+ Dng in qua ti c chiu l:

(+)c( 5 ( M ( R1 ( N ( 4( (-)b- Khong thi gian t :t1- t2 Tn hiu vo c pha a dng nht, pha b m nht.

+Trn cu chnh lu c cc it D1 v D4 dn; cc it cn li u kho.

+Dng in qua ti l:

(+)a( 1 ( M ( R1 ( N ( 4 ( (-) b- Cc khong thi gian tip theo: tng t.

+Tn hiu 1 chiu qua ti c biu din trn th hnh f2 u,iV ua ub uc

O T/2

T (t)

Hnh f1

u,i ra

Hnh f2

0 t1 t2 t3 t4 t5 t6 T

- Kt lun :

+ Tn hiu vo l xoay chiu nhng tn hiu ra l mt chiu v chy t M( N.

+Trong mi khong thi gian u c 2 it thuc 2 nhm dn nn thi gian chuyn mch l tc thi.

Mun tn hiu ra bng phng ngi ta phi s dng b lc( xem chng III)

c. u im v ng dng:

-Tn hiu ra bng phng hn so vi chnh lu 1 pha do vy b lc n gin hn.

- in p ngc trn mi D nh.

- Cng sut ca my bin p gim nh.

ng dng rng ri trong mch chnh lu cng sut ln.

4. Chnh lu cu 3 pha ti R + L

a S nguyn l (hnh g)

Hnh g

Gii thch s :

- Ngun v cu chnh lu nh hnh v e.

- Ti gm cun dy mc ni tip vi in tr R1 Tng tr ca ti l Zt.

b. Nguyn l lm vic:

Tng t nguyn l lm vic ca cu chnh lu v trn hnh e.

( Ch :

- Dng in qua ti trong tng thi im s chy t pha dng nht qua 1 D nhm l, qua in tr, qua cun dy, qua 1D nhm chn v pha m nht.

- Do cun dy c cm khng XL nn s chuyn mch dng in l khng tc thi m phi mt mt khong thi gian chuyn mch nht nh l 2((( (hoc T/6)

- Trong thi gian chuyn mch, dng qua ti s bin thin chm (tng chm v gim chm) nn dng bng phng hn so vi mch CL hnh e.

* th tn hiu qua ti nh hnh v h u,iV ua ub uc

O T/2

T (t)

u,i ra

Hnh v h

0 t1 t2 t3 t4 t5 t6 T

4.3 Chnh lu dng Thyristor ( CL c iu khin)

*C bao nhiu mch CL dng D th c by nhiu mch tng ng dng TRT

- Thyriston dn in phi c cc iu kin sau:

+ Phn cc thun cho Thy (A > (K

+ C dng iu khin a vo cc G

Do vy, dng TRT chnh lu c gi l chnh lu c iu khin.

Cc mch chnh lu dng TRT c ng dng rng ri trong in cng nghip nh : iu chnh v cp tc ng c; t ng ho iu chnh in p cho cc thit b c yu cu; iu khin ng m cc kho in t ch to b nghch lu, b bin tn v.v...1. CL iu khin cu 1 pha, ti R + L

a. S nguyn l (hnh a)

Hnh a

Gii thch s gm :

- 1 my bin p 1 pha, cp ngun xoay chiu vo cu chnh lu.

- Cu chnh lu gm 4 TRT mc thnh mch vng khp kn (nh cu chnh lu dng it).

- Ph ti gm cun dy L ni tip in tr R, tng tr ca ph ti l Zt. b. Nguyn l lm vic:

- t tn hiu xoay chiu trn mch th cp my bin p vo cu chnh lu.

- m TRT cn c b pht xung a xung n cc iu khin (cc ca G). Gc m ca TRT l gc ( . Khi thay i gc ( ta s thay i c khong thng ca TRT (( ( () v t dng chnh lu v in p chnh lu cng c thay i .

- Xt na chu k u, gi s tn hiu vo c chiu dng a; m x .

+ Trn cu chnh lu c TRT Ti1 v Ti3 m, cn Ti2 v Ti4 kho.

+ Dng in qua ti c chiu l:

(+)a ( Ti1 ( M ( Ti ( N ( Ti3 ( (-)x

- Xt na chu k sau; tn hiu vo i chiu

+ ... ngc li, Ti2 v 4 m; cn Ti1 v 3 kho.

+ Dng qua ti c chiu l:

(+)x( Ti2 ( M ( Ti ( N ( Ti4 ( (-)a

Nh vy , c 2 na chu k, tn hiu qua ti u c chiu t M ( N ( m bo l 1 chiu .

* Ch : Do trn cun dy c cm khng XL nn dng in qua cun dy s bin thin chm hn so vi in p (tng chm v gim chm), v vy dng tng i bng phng.

u,i vo

O T/2

T (t)

Hnh b1

xung

Hnh b2 ( 2(

( ( (

u,iRa u i

Hnh b3 O ( ( - ( ( 2( (

* Tn hiu vo, ra nh hnh v b.

b1 : tn hiu xoay chiu mch th cp my bin p.

b2 : tn hiu xung t vo cc G.

b3 : tn hiu dng v p trn ph ti .

Trn TRT cn c gi tr in p ngc ( khng xt y).2. CL iu khin cu 3 pha ti R + L

a, S nguyn l (hnh c)

Hnh c

Gii thch s gm:

- 1 My bin p 3 pha

- 1 Cu chnh lu gm 6 TRT mc theo mch cu c chia lm 2 nhm.

+ Nhm 1 (l) gm 3 TRT 1; 3; 5 u chm Katt.

+ Nhm 2 ( chn) gm 3 TRT 2; 4; 6 u chm Ant.

- Ph ti gm cun dy ni tip vi in tr c tng tr l Zt b. Nguyn l lm vic

- t tn hiu xoay chiu mch th cp my bin p vo 2 u cu chnh lu.

- a xung iu khin vo cc G ca TRT m TRT.

TRT no va c phn cc thun va c xung iu khin th n s m.

Xem th tn hiu vo cu chnh lu v tn hiu xung, nh hnh v d.

Xt tng khong thi gian lm vic ca tn hiu vo cu chnh lu v tn hiu xung.

- Xt khong thi gian t1- t2 ta c :

+ Tn hiu vo dng nht pha a; pha b m nht.

+ Ti thi im t1 cp xung ( trn cu chnh lu c TRT (1) v (4) m cn cc TRT 2, 3, 5, 6 u kho.

Dng in qua ti l:

(+) a ( TRT1 ( M ( Ti ( N ( TRT4 ( (-)b.

- Tng t, xt cc khong thi gian: t2- t3 ; t3- t4 ; t4- t5 ; t5- t6 ; t6 - T - t1, ta u thy trong tng khong thi gian ny dng in u i t : pha dng nht( TRT nhm l( M( ti ( N ( TRT nhm chn ( pha m nht.

Kt lun: Tn hiu qua ti lun i theo chiu t M( N( m bo l mt chiu.

Tn hiu dng v p trn ti nh hnh v d .Ngoi ra trn TRT cn c cc gi tr in p ngc ( khng xt y).

u,i vo ua ub uc 0 T/6 T/2 2T/3 T

T/3 2T/6 T (t)

u xung

0 t1 t2 t3 t4 t5 t6 T (t) u,i ra u ti

i ti

0 t1 t2 t3 t4 t5 t6 T (t) Hnh v d

4 . 4 Nghch lu

1. Nghch lu ph thuc cu 3 pha dng Thyristor

a. S nguyn l: Hnh a

Xt s nn in cu 3 pha cung cp cho 1 my in 1 chiu, mch lm vic ch thun nghch va chnh lu va nghch lu.

Hnh a

Gii thch s :

- 1 my bin p 3 pha:

+ S cp u Y, ly ngun t li in xoay chiu

+ Th cp u Y cp ngun cho cu TRT.

- B chnh lu v nghch lu gm 6 TRT mc theo hnh cu ( nhm 1 ....; nhm 2.....)

- Cun cm Ld lm nhim v san phng dng in a vo cu NL.

- My in M c th lm vic ch ngun hoc ti.

b. Nguyn l hot ng:

* ch chnh lu: (Tng t mch chnh lu cu 3 pha, ti R+L)

- Lc ny my in M lm vic ch ng c (ti) tiu th in 1 chiu (+) M; (-)N.

- in p chnh lu U cao hn sc in ng E ca my in.

* ch NL.

Nu o chiu sc in ng E ( bng cch i chiu dng kch t ), v tng gc m ( ca TRT sao cho ( ( (/2 cho E c gi tr cao hn U ( U c gi tr m nhng dng in qua TRT vn gi nguyn nh c .

Lc ny cng sut ca c my in 1 chiu v mch xoay chiu u i du. ng thi my in s lm vic ch ngun cng sut, cn li in xoay chiu s tr thnh ti.

Kt lun:

Nh vy mch NL ph thuc chnh l mch chnh lu trong in p 1 chiu sau chnh lu s c i du v gc m ( ca thy phi tho mn iu kin:

((( ( ( ( ( hoc ((( ( ( ( (((2. Nghch lu c lp in p cu 1 pha.

a. S nguyn l (hnh v b ).

Gii thch s :

+ E: Ngun ngoi 1 chiu cung cp cho b NL.

+ B NL gm 4 TRT mc theo mch cu kn .

+ Cc it 1 - 4 mc ngc chiu vi cc TRT thng dng cho ti trong khong thi gian dng in ngc chiu vi chiu dn ca cc TRT (ngha lTRT ng)

+ Ti R+L ti c tnh cm khng, tng tr ca ti l Z T. b. Nguyn l lm vic:

- Xt ti thi im cp xung TRT1 v TRT3 m.

+ Lc ny s c dng qua ti, chiu dng in l:

(+) E ( TRT1 ( im M ( ti ( N ( TRT3 ( (-)E

+ in p trn ti bng in p ngun E

- Khi TRT 2 v 4 cha m m TRT1 v 3 ng, do ti c tnh cm khng nn dng qua cun dy bin thin chm (gim chm, tng chm). V vy dng in ny s chy qua cc it 1,3 v ngun. Chiu dng in nh sau:

ti( M (1( (+) E( (-) E (3 (N( ti

Chnh ti thi im thng diot 1, 3 lm thay i cc tnh in p ra trn ti.

Nng lng c tch lu trong cun cm L t giai on trc s c tr li ngun cung cp . Do vy xut hin trng hp tn hiu qua ti v O.

- Thi im tip theo cp xung TRT2 v 4 m, TRT1 v 3 ng hn.

+ Qu trnh li xy ra tng t s c dng qua ti li tiu th nng lng t ngun .

+ Khi TRT1 v 3 cha m , TRT2 v 4 ang ng th cun dy li phng tr nng lng qua it 2, 4 v ngun.

Cc thi im tip theo tng t ; cc cp D v cp TRT ln lt ng - m.

c. Kt lun

Ngun cung cp E l 1 chiu nhng tn hiu trn ti qua cc thi im khc nhau u l xoay chiu.

th m t nguyn l hnh thnh dng in p trn ti nh hnh v c

u,i ti u ti i ti

0 M N M E

t1 t2 t3 t4 t5 t6 T

D 2, 4 m

Ti2, 4 m Ti1,3 m Ti1,3 m Ti2,4 m

Hnh v b

Tham kho: 3. Nghch lu c lp in p cu 3 pha

a. S nguyn l (h. v a)

b. Nguyn l lm vic

h. b (/3 2(/3 ( 4(/3 5(/3 2( 7(/3 8(/3 3( Ti 600 1200 1800 2400 3000 3600 4200 4800 5400

mng

1

2

3

4

5

6 T/2 T 3T/2 (t)

h.c UA

O T/2 T 3T/2 (t)

h.d IA

O T/2 T 3T/2 (t)

t

Hnh b: Ti mi thi im c 3 Ti m v c 600 li c chuyn mch

Hnh c, e, f: Tn hiu in p cc pha tng ng: A, B, C

Hnh d: Tn hiu dng in pha A

h.e UB

O t

h.f UC

O t

h. g S tng ng ca ph ti tng thi im khc nhau:

h. g1: Thi im 0 n 600

h. g2: Thi im 600 n 1200

h. g3: Thi im 1200 n 1800 K

A

I

I

M

L trng

in t

N

P

1

2

3

4

5

Uth

Ing

Um

P

P

N

N

PAGE 3Hong Hnh

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