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7/31/2019 bo co mach on ap

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Ni dung bo co gm:Phn A: Bo co mch n p.

1. Gii thiu:2. S nguyn l:3. S lp rp:4. Gi tr cc linh kin5. Tc dng cc linh kin:6. Nguyn l lm vic.7. Cc gi tr ca U, R trong qu trnh iu chnh.

Phn B: Tr li cu hi

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PHN A:BO CO MCH N P

I. Gii thiu:Thit k mch n in p mt chiu s dng cc linh kin c b

transistor, in tr, diode, t in vv.n p lm vic ch tuyn tnh.

S nguyn l:

T2

T1

R2100k

R6560

3 2

6

7 4

R31

R4100k

T3

R51M

R7100k

RT56

R1560

DZ

VI1 VO 3

G N D

2

T4 7812

C1

Ec 14-31V(8)

(8)

(7)

(2)(1) (3)

(4)

(5)(6)

I. S lp rp:Yu cu thit k mch:a. R1,R2,,R5: 2 khuyt nm dc thng hng nhau.

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b. T1,T2,T3 : 3 khuyt nm dc thng hng ngang.c. R6 : 2 khuyt nm dc thng hng dc R3.d. T4 : 3 khuyt nm ngang.e. Dz,C1,R7,C2,Rt : 2 khuyt nm dc thng hng ngang.

T yu cu ta a n s nh sau: S lp mch:

S i dy mt di:

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IV. Gi tr ca tng linh kin:IC: HA17741.T1: n khuch i cng sut H1061.T2, T3: D468.T4: IC n p 7812.C1 = C2 = 100 F/35VR1 = 560 R2 = R4 = 100 kR3 = 1/1WR5 1 MR6 560 R7 100 kRt = 100 /5W

V. Tc dng ca tng linh kin:a. T C1 lc in p u vo. b. T C2 lc in p u ra.c. T4 l IC n p 7812 c nhim v cung cp in p n nh +12

chn s 7 ca IC HA17741.d. Dz c nhim v to nn U chun a vo chn 3 ca IC.e. R1 c nhim v phn p.f. R2 c nhim v hn ch dng vo T2.g. R3 c nhim v bo v qu ti.h. R6 hn ch dng.i. R5 R7 phn p to so snh vo chn 2 ca IC.

Cc chn ca IC: j. Chn 2 ca IC l u vo o.k. Chn 3 ca IC l u vo thun to in p so snh chun.l. Chn 4 l m ngun.m. Chn 6 l tn hiu ra ca IC.n. Ngun nui IC (+12V)

VI. Nguyn l lm vic:Mch n p c hi tip hot ng theo 1 nguyn tc chung c th biu ds khi sau:

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Khuch icng sut

Tn hiu iu khin

Hn chdng ti

Khuch i

in p chun So snh

Ly mu

in p a vochiu cha n nh Ura

Ngun 1 chiu a vo l ngun mt chiu bin i t 14V n 3HA17741 l mt b so snh. Chn 3 c a vo in p chun c ly tn p Dz. Mt phn ca in p ra c a v chn s 2 ca IC so snchun. chnh l in p trn in tr R7. R6 c dng hn ch dng

base ca T2.Gi s in p ra gim xung dn n in p trn R7 gim xung choUD = U3 U2 > 0 th in p trn chn s 6 ca IC tng, khin cho in p trn base ca T2 tng ln, dn n in p Ura tng ln.

Mch hot ng tng t vi trng hp Ura tng ln.

VII. Cc gi tr ca U, R trong qu trnh iu chnh.iu chnh mc in p ra 9V:Khi ta chn:R6 = 560R5 = 1MR7 = 100k

T1 T2 T3 Dz Uvo UraUCE 4.5V 0 2.14V 9.3V 14V 9.5VUBE 0.64V 0.71V 0.5VUCE 21.4 0 1.9V 9.5V 31V 9.5VUBE 0.5 0.6V 0.5V

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1. Khi Uvo = 14Vin th ti cc im:

(1) 12V(2) 9.2V(3) 8.8V

(4) 11V(5) 11V(6) 10.2V(7) 10V(8) 9.8VUt = 8.5V

2. Khi Uvo = 32Vin th cc im:

(1) 12V(2) 9.2V(3) 9V(4) 11.2V(5) 11V(6) 10V(7) 10V(8) 9.5VUt = 9.2V

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PHN B:TR LI CU HI

Cu 1: V v trnh by nguyn l lm vic, tc dng linh kin ca mch c ngun (dng bin p h p). Ngun c Ura = 2UV. Ngun phn p. Thit k bngun n gin nht c c 2 mc in p ra.

Tr li:

2 2 0 V

Ur

+

Bin p: Ura = 2UVN2 = 2N1

Chnh lu mt na chu k. Bin in p xoay chiu thnh in p mt chiu. +

T lc: in p ra c lc phng.

78xx

79xx

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78xx: bin p ra +xV79xx: bin p ra xV

Cu 2: Trong mch ngun dng bin p h p v ngun xung, ngi ta s dn

my loi t lc? Cho thng s k thut v cch mc?Tr li:- Ngi ta dng chung 1 loi t lc l t ha. Ty mc n nh ca im ta c th s dng cc t c in dung khc nhau vi cc cp in p ph- Cch mc:

Vi mch ngun dng bin p h p:C1: T lc sau khi chnh lu 1000 - 2000F 25V mc sau chnh lu ni C2: T lc sau n p: 470 F 16V mc sau cng ni t.Vi mch ngun xung:C5: t lc th ngay sau chnh lu. Mc sau chnh lu ni t.C6: lc in p ti, mc sau ti ni t.

Cu 3:Trong mch n p, mc darlington l mc kiu g? Tc dng ca mch?Mch darlington c my cch mc? V tng cch.

Tr li:Trong mch n p, darlington l cch mc ghp ni cc transistor. Tc dn

gia tng h s khuch i.C 3 cch mc darlington:Cch 1:

T1

T2

RINPUT

OUTPUTEc

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Cch 2:

T2

INPUT

OUTPUT

Ec

T1

RB

RC

Cch 3:

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T2

INPUT

OUTPUT

Ec

RB

RC

T1

Cu 4:Trong mch n p, ngoi n p, ngi ta cn c th s dng linh kin n n p trc tip cho tng mch c th?

Tr li:Dng IC n p.

Cu 5:Trong mch ngun xung, c bao nhiu cch mc mch lc? V mch c

th? Tr liC 2 cch mc l:

UrUcl

RtC

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+ E0

C1

Rt1

Cu 6: Nu u im v nhc im ca ngun xung, ngun dng bin p h Tr li:

Ngun xungo u im:

+ mc in p vo di rng t 80V n 240V, c a thng vo b chcu sau qua mch lc

+ Khng cn chuyn mch+ Kch thc nh gn+ Cng sut t vi ot n vi trm ot

o Nhc im:+ Gi thnh t+ Kh sa cha

Ngun dng in p h po Nhc im:

- Hn ch u vo ch c s dng 2 mc in p l 110 v 220 V- Cng knh, tn hao in nng lnloi ngun ny ngy nay ch yu c s dng trong lnh vc cng nghip

o u im:+ Gi thnh r

Cu 7: C my cch chnh lu? Mch lc ngun ? i t chnh lu dung loi gti sao?Tr li:a. Cc cch chnh lu:

+ theo pha:- chnh lu 1 pha- chnh lu 3 pha

+ Theo chu k- chnh lu c chu k: chnh lu cu, chnh lu hnh tia c chu k- chnh lu na chu k: chnh lu hnh tia c iu kin v chnh lu khn

c iu kin

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b. Mch lc ngunC 2 loi mch lc ngun

+ Mch lc 1 chiu:- S dng t ho. Tu mc n nh ca in p ra m ta c th s d

cc t c in dung khc nhau vi cc in p ph hp

+ Mch lc nhiu xoay chiu : s dng cc cun cmc. it chnh lu

C 2 loi it tip mt v tip im:+ Mch chnh lu lm vic tn s thp v dng chy qua tip gip ln, ps dng loi iot tip mt+ it tip im c din tch min tip gip nh nn khng cho c dln chy qua

Cu 8 : Trong mch hc ta o c Uce(T1) que t vo phn to nhicn que en t vo chn E vn o c ? Ti sao?Khi thay i Uvao th Uce(T1) c thay i khng? Ti sao?Tr li:- Do cu to ca Tranzitor T1 (H1061) th phn to nhit c ni vi cC ( chn gia) cho nn ta c th o c Uce ca iot ny bng cch cho que vo phn to nhit, que en vo chn E.- Khi thay i Uvao th Uce (T1) thay i v Ura khng i, V4 =V5 = 12Vnn Ve(T1) =V5 Vbe(T1) Vbe( T2) gn nh khng i. M Uce(T1)= Uvo Ve(T1) nn Uce(T1) s thay i khi Uvo thay i.

Cu 9: khi chp min tip gip BC(T1), BC(T3) th Ura= ? Ti sao?

Tr li:Khi chp lp tip gip BC th T1 ch cn l 1 iot nn Ve(T1) = Vngun

Do R5>>R3 nn Ura = VngunKhi chp lp tip gip BC(T3) th lc T3 ch cn li l mt iot nn T

khng cn tc dng n nh dong na. Khi Uvo khng i th Ura khng nhng khi Uvo tng th Ura cng tng theo ch khng n nh.

Cu 10: Trong s ngun xung, nhim v ca C5 v C6 l g? Hai t ny cth i ch cho nhau c khng? Ti sao?Mt trong hai t trn b kh th hin tng g sy ra?

Tr li:-Trong s ngun xung th nhim v ca hai t C5 v C6 lngun mt chiu

+ T C5 lc th trc in tr nhit ngay sau chnh lu.+ T C6 lc in p ti.

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- Hai t ny khng th i ch cho nhau c v:+ T C5 lc ngay sau khi chnh lu nn phi chu in p ln. in

mc ca t ny ln.+ T C6 lc in p ti thp. Nn in p nh mc ca t ny nh h

t C5.

- Khi mt trong hai t ny b kh th tr s ca t gim xung, do s nng lc ca t gim xung, in p khng n nh, v lm cc thong s u ri.Cu 11: Trnh by cch kim tra diode hai ch l ng v tnh?Tr li:Cch kim tra diode:

- Ch tnh: Do it zen c tc dng n p nn mc d dng in chn thay i nhng in p gia hai u n vn khng i . V vy ch cng h o thang o mt chiu10VDC o in p ny. Nu inn nh th it lm vic tt

- Ch ng: D ng h thang o ohm (10 ohm), nu o in tdiode phn cc thun m ng h ch mc 0 ohm hoc v cng th di b hng, cn nu khng th diode vn cn tt

Cu 12: Diode chnh lu l diode tip im hay tip mt?Tr li:Trong mch chnh lu thi diode c dng l diode tip mt, v:- Trong mch chnh lu th diode hot ng tn s thp, dng qua lp tgip ln

- Diode tip mt c din tch tip xc ln nn cho php dng in ln i q Cu 13: Trong b ngun xung, nhng linh kin g hng m mt thng nh bit c? Nu cch kim tra v thay th?Tr li:trong b ngun xung th linh kin khi hng m ta d nhn bit bng mthng c l cu ch v triac- Vi cu ch thy tinh khi hng thi qua lp thy tinh trong sut ta c thquan st c c b t hay khng? Nu b t ta thay th bng ci khc hoc trc tip bng 1 si dy( khi ny khng cn tc dng ca cu ch na)- Vi triac khi hng s b tch i ra nn ta hon ton quan st c v thath bng 1 triac mi

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