ky thuật mạch điện tử P I-Trương Văn Cập -Điện tử tương tự

8/7/2019 ky thut mch in t P I-Trng Vn Cp -in t tng t

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HC VIN K THuT QUN S

B MN C S K THUT V TUYN - KHOA V TUYn IN T

Pgs . ts trng vn cp (ch bin)

Ths. Nguyn Duy chuyn - ts. trn hu v

Ths. Luyn quang minh - ts. nguyn huy hong - ts. T ch hiu

K thut mch in tPhn mt

H ni - 2008


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Mc lc

TrangTULi ni uUT.......................................................................................................9TUChng 1 Nhng khi nim chung v cc vn c s ca

mch in tUT ................................................................................................11TU1.1 Khi nim v mch in t v nhim v ca n UT ...............11TU1.2. Hi tip trong b khuch i UT ...................................................12

TU1.2.1. nh ngha v phn loiUT ......................................................................12TU1.2.2. Phng trnh c bn ca mng bn cc c hi tip.UT ...........................13TU1.2.3. Phng php phn tch b khuch i c hi tip UT ..............................14

TU1.3. nh hng ca hi tip m n cc tnh cht ca bkhuch i.UT ................................................................................................18

TU1.3.1. nh hng ca hi tip m n n nh ca h s khuch iUT .....18TU1.3.2. nh hng ca hi tip m n tr khng vo.UT ..................................19TU1.3.3. nh hng ca hi tip m n tr khng raUT .....................................20TU1.3.4. nh hng ca hi tip m n di ng v mo phi tuyn ca bkhuch i.UT....................................................................................................22

TU1.3.5. nh hng ca hi tip m n c tnh tn s v c tnh ng cab khuch i.UT...............................................................................................23

TU1.4. n nh ch cng tc cho cc tng dngtranzistor lng cc.UT.........................................................................29

TU1.4.1. t vn UT ...........................................................................................29TU1.4.2. Hin tng tri im lm vic UT ............................................................30TU1.4.3. Cc s n nh tuyn tnh. UT .............................................................32TU1.4.4. Cc s n nh phi tuynUT ...............................................................36

TU1.4.5. n nh im lm vic trong cc mch t hp tng t.UT....................37TU1.5. n nh ch cng tc cho cc tng dngtranzistor hiu ng trng.UT ...........................................................39

TU1.5.1. Ch tnh . UT........................................................................................39TU1.5.2. Cc s n nh im lm vic. UT.......................................................39

TUChng 2 cc s c bn ca tng khuch i tn hiunh dng tranzistorUT.............................................................................42

TU2.1. cc tham s c bn ca b khuch i.UT .............................43


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TU2.2. cc s c bn dng mt tranzistor lng ccUT ......45TU2.2.1. Mch in cc b khuch i UT .............................................................45TU2.2.2. Cc tham s c bn ca tng s : UT ...................................................47TU2.2.3. Cc cng thc n gin thng dng UT ...................................................52TU2.2.4. Nhn xt chung: UT..................................................................................53

TU2.3. Cc s b khuch i dng tranzistor hiu ngtrng (FET)UT ...............................................................................................53TU2.4. cc b khuch i dng nhiu tranzistor. UT......................56

TU2.4.1. S Dalington:UT .................................................................................56TU2.4.2. S tranzistor bUT ..............................................................................59TU2.4.3. S Kaskode UT ....................................................................................60TU2.4.4. Mch kt hp FET - tranzistor lng cc.UT ..........................................61

TU2.5. b khuch i vi sai. UT ......................................................................62TU2.5.1. S v nguyn l lm vic:UT ..............................................................63TU2.5.2 Cc tham s c bn ca b khuch i vi sai: UT .....................................65TU2.5.3. Hin tng tri.UT...................................................................................69TU2.5.4. Mt s s b khuch i vi sai.UT .....................................................70

TUChng 3 B khuch i thut ton v ng dngUT ..................75TU3.1 Nhng vn chung v b khuch i thut ton UT ......75

TU3.1.1 Cc tnh cht v tham s c bn ca b khuch i thut ton.UT ..........75TU3.1.2. Cc s c bn ca b khuch i thut ton UT ..................................81

TU

3.1.3. nh hng ca dng in tnh, ca in p lch khng, ca hintng tri im cng tc ca b KTT v cc bin php b. UT ......................86TU3.1.4. n nh cng tc ca b KTT v cc bin php b tn s.UT .............89TU3.1.5. Cu trc bn trong ca b khuch i thut ton. UT ..............................97

TU3.2 Cc mch tnh ton v iu khin tuyn tnh dngkhuch i thut tonUT .....................................................................103

TU3.2.1 Khi nimUT ..........................................................................................103TU3.2.2 Mch cng v mch trUT .....................................................................104TU

3.2.3 Mch bin i tr khngUT

....................................................................109TU3.2.4 Mch tch phnUT ..................................................................................123TU3.2.5 Mch PI (Propotional - Integrated)UT........................................................2TU3.2.6 Mch vi phn UT .....................................................................................129TU3.2.7 Mch PID (Propotional Integrated - Differential)UT ..........................130


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TU5.3. TNG KCS TN S THP DI RNGUT ...........................................212TU5.3.1. Tng KCS n UT................................................................................212TU5.3.2. Tng KCS mc y ko UT..................................................................214

TU5.4. TNG KCS CAO TN DI RNGUT ...................................................218TU5.4.1. S khi tng qut ca tng KCS cao tn di rng UT.....................218TU5.4.2. Mt s c im ca tng KCS cao tn di rngUT ...........................219

TU5.5. TNG KCS TI CNG HNG. UT ......................................................221TU5.5.1. S v nguyn l lm vic. UT ............................................................221TU5.5.2. Quan h nng lng trong tng KCS UT .............................................222

TUChng 6 Cc b to dao ng UT ..........................................................226TU6.1 cc vn chung v to dao ng UT ....................................226

TU6.1.1 Nhim v, yu cu v phn loi.UT........................................................226TU6.1.2 iu kin to dao ng v c im ca mch to dao ng. UT...........227TU6.1.3 n nh bin v tn s dao ng.UT..................................................231

TU6.2 B to dao ng RC. UT......................................................................232TU6.2.1 c im chung ca cc b to dao ng RCUT ...................................232TU6.2.2 B to dao ng dng mch di pha trong khu hi tip. UT ...................232TU6.2.3 B to dao ng dng mch lc T v T- kp trong khu hi tip. UT.....234TU6.2.4 B to dao ng dng mch cu Vin trong khu hi tip. UT...............236

TU6.3 B to dao ng LC UT .......................................................................238TU6.3.1 Cc loi mch to dao ng LCUT.........................................................238TU6.3.2 Vn n nh bin trong cc b to dao ng LC. UT ....................241TU6.3.3 Vn n nh tn s trong cc b dao ng LC: UT ............................242

TU6.4. B to dao ng thch anhUT...................................................246TU6.4.1 Tnh cht vt l ca v s tng ng ca thch anh.UT ...............246TU6.4.2 Nhng tc ng nh hng ti n nh tn s ca b dao ngthch anh v bin php khc phc. UT .............................................................250TU6.4.3 Mch in b to dao ng thch anh UT ...............................................251

TU6.5 To dao ng bng phng php t hp tn s. UT ...........255TU6.5.1 t vn . UT .........................................................................................255TU6.5.2 T hp tn s bng phng php kt hp.UT .........................................255TU6.5.3 T hp tn s theo phng php t ng iu chnh tn s.UT .............260TU6.5.4 T ng iu chnh theo tn s (TT). UT .............................................261TU6.5.5. T ng iu chnh tn s theo pha (TF) UT.......................................266TU6.5.6 To mng tn s n nh bng vng kho pha PLL. UT .........................277

TUChng 7 iu ch dao ng cao tn UT ............................................278


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TU7.1. Khi nim chung:UT............................................................................278TU7.2. iu ch bin UT .............................................................................278

TU7.2.1. Khi nim v iu ch bin UT .........................................................278TU7.2.2. iu ch li mt v iu ch baz (cc gc ) UT...............................283TU7.2.3. iu ch Ant v iu ch colect (cc gp)UT ..................................285TU7.2.4. iu ch bin s dng mch nhn tng tUT.................................288TU7.2.5. iu bin cn bng.UT ..........................................................................289

TU7.3. iu ch tn s UT .............................................................................291TU7.3.1. Khi nim c bn.UT.............................................................................291TU

7.3.2. Bin php k thut thc hin iu ch tn s.UT

..................................293

TU7.4. iu ch n binUT...........................................................................298TU7.4.1. Bn cht ca iu ch n bin UT ........................................................299TU7.4.2. Nhng c im ch yu ca thng tin n bin UT .............................301TU

7.4.3. Cc phng php truyn tin tc khi iu ch n bin.UT

...................302TU

7.4.4. Cc phng php to tn hiu n bin. UT ..........................................303

TUChng 8 tch sng tn hiu v tuyn UT .........................................310TU8.1. Khi nim chungUT.............................................................................310TU8.2. Tch sng bin (TSB) UT ...........................................................310

TU8.2.1. Cc ch tiu cht lng ca TSB:UT ...................................................311TU8.2.2. Phn tch ch ca b tch sng dng it. UT ..................................312TU8.2.3. Tch sng ng bUT............................................................................321TU8.2.4. Hin tng phch v hin tng chn p tn hiu trong b tch sngiu bin.UT ....................................................................................................322

TU8.3. Tch sng tn hiu xung.UT ..........................................................324TU8.3.1. Tch sng xung v tuyn. UT.................................................................324TU8.3.2. Tch sng xung th tn (Tch sng nh).UT ........................................326

TU8.4. tch sng phaUT .................................................................................327TU8.4.1. Cng dng, nguyn l tch sng pha. UT...............................................327TU8.4.2. Cc dng s tch sng pha. UT..........................................................328

TU8.5. B tch sng tn s UT ....................................................................330TU8.5.1. Cng dng, nguyn l tch sng tn sUT ............................................330TU8.5.2. Cc dng s b tch sng tn s.UT .................................................330

TUChng 9 Trn tnUT......................................................................................338TU9.1. L thuyt chung v trn tn UT ................................................338

TU9.1.1. nh ngha: UT .......................................................................................338


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TU9.1.2. Nguyn l trn tn UT ............................................................................338TU9.1.3. Phn loi UT ...........................................................................................339

TU9.1.4. ng dngUT...........................................................................................340TU9.2. H phng trnh c trng v cc tham s c bn. UT .340TU9.3. Mch trn tn. UT ...............................................................................344

TU9.3.1. Mch trn tn dng it.UT ..................................................................344TU9.3.2. Mch trn tn dng phn t khuch i UT ...........................................347

TU9.4. Nhiu trong mch trn tn. UT...................................................355TU9.5. Trn tn siu cao tn UT.............................................................357

TU9.5.1. Trn tn trn n thp 3 cc.UT ............................................................357TU9.5.2. B trn tn dng n thp 2 cc. UT ......................................................358TU9.5.3. B trn tn dng it bn dn. UT..........................................................361TU9.5.4. Chn ch cng tc cho b bin tn dng it bn dn. UT ...............367TU

9.5.5. S cu trc ca b trn tn dng it bn dn.UT

............................369TU9.5.6. B trn tn cn bng dng it bn dnUT ...........................................374

TUChng 10 Mch cung cp ngun UT......................................................379TU10.1. Khi nimUT ..........................................................................................379TU10.2. Bin p ngun v chnh luUT...................................................380

TU10.2.1. Mch chnh lu na sng.UT...............................................................382TU10.2.2. Mch chnh lu ton sng.UT..............................................................385TU10.2.3. Ti ca b chnh lu.UT ......................................................................388TU

10.2.4. Mch bi p.UT

...................................................................................390TU10.2.5. Khu lc trong cc b chnh lu.UT....................................................390

TU10.3. n pUT ...................................................................................................393TU10.3.1. Mch n p dng it Zener. UT..........................................................393TU10.3.2. Mch n p dng it Zener vi mch lp emito u ra. UT ............395TU10.3.3. Mch n p c hi tip. UT ..................................................................397

TU10.3.4. n p xung.UT ....................................................................................404TU10.4 Chnh lu o.UT ................................................................................405TU10.5. Bin i in p mt chiu v b ngun khng dngbin p ngun. UT ........................................................................................407

TU10.5.1. Bin i in p mt chiu UT .............................................................407TU10.5.2. B ngun khng dng bin p ngun. UT ............................................408

TUTi liu tham khoUT ..................................................................................410


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Li ni u

B sch K THUT MCH IN T bao gm hai phn c dng lm

ti liu ging dy mn hc cng tn s dng Hc Vin K thut Qun S. Phn

I bao gm cc ni dung m rng dng chung cho nhiu chuyn ngnh c linquan. Phn II gm cc ni dung chuyn su, s dng ring cho hai lnh vc vi

c th khc nhau ca cc chuyn ngnh in t vin thng v Raa, Tn la,

Tc chin in t.

Ngoi vic dng lm ti liu hc tp v tham kho cho cc hc vin chuyn

ngnh V tuyn in t, b sch cng c th dng lm ti liu tham kho b ch

cho cc k s, cn b k thut ca cc ngnh c lin quan n k thut in v

in t.Tham gia bin son b sch c cc ging vin thuc B mn c s k thut

v tuyn - Khoa v tuyn in t .

Trong qu trnh bin son chng ti nhn c nhiu kin ng gp rt

b ch. Chng ti xin by t lng bit n chn thnh v s gip qu bu .

Mc d c nhiu c gng, nhng v xut bn ln u nn chc chn s

cn nhiu hn ch v thiu st. Chng ti rt mong nhn c cc kin ng

gp qu bu gi v Khoa VTT Hc Vin KTQS.


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Chng 1Nhng khi nim chung v cc vn c s

ca mch in t

1.1 Khi nim v mch in t v nhim v ca nCc mch in t c nhim v gia cng tn hiu theo cc thut ton khc

nhau. V d: tn hiu c th c khuch i, iu ch, tch sng, chnh lu, nh,

o, truyn t, iu khin, bin dng, tnh ton

Trn quan im k thut, ngi ta phn bit hai loi tn hiu: tn hiu tng

t v tn hiu s. gia cng chng, c hai loi mch in t c bn tng ng:

Mch tng t v mch s. Trong thi gian gn y, k thut x l tn hiu s

pht trin mnh m v ng vai tr rt quan trng trong vic gia cng tn hiu.Tuy vy mch s cng khng th thay th hon ton mch tng t v trong thc

t c nhiu thut ton khng th thc hin c bng cc mch s hoc nu thc

hin bng mch tng t th tin li hn. Ngay c trong h thng s, cng c

nhiu phn t chc nng tng t mt s khu no .

Mch s c nghin cu trong mn hc k thut s, cn i tng ca

mn hc k thut mch in t ch yu ch nghin cu cc mch tng t. Trong

thi gian gn y, cc mch t hp tng t ng vai tr rt quan trng trong

k thut mch in t. Chng khng nhng m bo tt vic tho mn cc chtiu k thut m cn c tin cy cao v gi thnh h. Trong , s ra i ca b

khuch i thut ton nh du mt bc ngot quan trng trong qu trnh pht

trin ca k thut mch in t. Trc y, gia cng tn hiu theo nhiu thut

ton, c v s cc mch chc nng khc nhau. Ngy nay, c th dng mt s

lng hn ch cc mch chc nng dng khuch i thut ton thc hin nhiu

thut ton khc nhau nh mc mch hi tip ngoi thch hp.

Xu hng pht trin ca k thut mch tng t l nng cao tch hp ca

mch (gim s chng loi nhng li tng kh nng s dng ca tng chng loi).Khi tch hp tng th c th ch to nhng h thng c chc nng ngy cng

hon thin hn trn mt chp. Tm li, c hai hng pht trin ca k thut mch

tng t, l: Gim nh kch thc bn trong ca mch trong ch to v tng

tnh ph bin ca mch trong ng dng.


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1.2. Hi tip trong b khuch i1.2.1. nh ngha v phn loi:

Hi tip l qu trnh ghp mt phn tn hiu ra ca mng 4 cc tch cc vu vo thng qua mt mng 4 cc khc gi l mng hi tip (hnh 1.1). Hi tipng vai tr rt quan trng trong k thut mch tng t. N c kh nng lm

thay i thut ton khi gia cng tn hiu ng thi cho php ci thin cc tnhcht ca b khuch i.

htX

hvXrXK

htK

Hnh 1.1:S khi b khuch i c hi tip.K: Khu khuch i; KBhtB: khu hi tip; XBvB: tn hiu vo; XBrB tn hiu ra; XBhB : tn

hiu hiu (hoc tng); XBhtB: tn hiu hi tip.

C th phn loi hi tip theo mt s cn c nh sau:

a) Theo pha gia htX v vX ta c hai loi:

-Hi tip m:htX ngc pha vi vX . Khi hX l tn hiu hiu, nh hn

vX v h s khuch i ca mch gim.

Hi tip m c th ci thin c cc tnh cht ca b khuch i.

-Hi tip dng:htX

B

B ng pha vi vX lm cho h s khuch i ca mch

tng. Hi tip dng c s dng to dao ng v s c xt n trong

chng 6 ca ti liu ny.

b) Cn c vo tn hiu hi tip ta c hai loi hi tip mt chiu v hi tip

xoay chiu.

-Hi tip m mt chiu c dng n nh ch cng tc.

-Hi tip m xoay chiu c dng n nh cc tham s ca b khuchi.

c) Mch in ca b khuch i c hi tip c phn thnh 4 loi:

-Hi tip ni tip - in p: tn hiu hi tip a v u vo ni tip vi

ngun tn hiu ban u v t l vi in p ra .


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-Hi tip song song - in p: Tn hiu hi tip a v u vo song song

vi ngun tn hiu ban u v t l vi in p ra.

-Hi tip ni tip - dng in: tn hiu hi tip a v u vo ni tip vi

ngun tn hiu ban u v t l vi dng in ra.

-Hi tip song song - dng in: tn hiu hi tip a v u vo song songvi ngun tn hiu ban u v t l vi dng in ra.

1.2.2. Phng trnh c bn ca mng bn cc c hi tip.

S khi ton phn ca b khuch i c hi tip ch ra trn hnh 1.2.

htX

hXvXr

XK

htK

nKnX

+

Hnh 1.2. S khi ton phn ca b khuch i c hi tipKBnB: Khu ghp vi ngun tn hiuXBnB: tn hiu ngun a n b khuch i

Gi thit cc khi u l cc h tuyn tnh v tn hiu ch chy theo chiu

mi tn. T s hnh 1.2 ta rt ra cc quan h sau:

hththnnv KXXKXXKXX .;.;. === v htvh XXX = ( chng ny ch

yu xt hi tip m). T cc quan h ny, ta c th rt ra phng trnh c bn ca

mng bn cc c hi tip:

htV

r

KK

K

X

XK

+==

1(1.1)

1

nrtp n

n ht

K KX K K K

X KK = = =

+(1.2)

Trong : K - Hm truyn t ca mng bn cc tch cc c hi tip, v:

tpK - Hm truyn t ton phn

Gi:htV KKK .= l hm s truyn t vng, v:

Vht KKKg +=+= 1.1 l su hi tip.


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Nhvy c th xy ra hai trng hp:

* 11 , ta c hi tip dng v h s

khuch i ca mch tng so vi khi khng c hi tip.

* KKKKg ht += ,11 , tng ng ta c hi tip m.Trng hp c bit, khi hm s truyn t vng 1>>VK tc l

VV KKg += 1 , t (1.1) v (1.2) suy ra:

htV

r

KX

XK

1= v:

ht

n

n

rtp

K

K

X

XK = (1.3)

Trng hp c bit ny c gi l hi tip m su.

Nh vy, trong mt h thng khp kn vi hi tip m su, khi hm s

truyn t vng c gi tr rt ln, th hm truyn t ca n hu nh khng ph

thuc vo cc tnh cht ca b khuch i m ch ph thuc vo tnh cht ca

khu hi tip. S thay i cc tham s ca phn t tch cc v tp tn ca n

khng nh hng ti cc tnh cht ca b khuch i c hi tip. V vy c th

la chn khu hi tip mt cch thch hp thay i thut ton khi gia cng tn

hiu. y chnh l mt ng dng rt quan trng ca hi tip m.

1.2.3. Phng php phn tch b khuch ic hi tip:

nhn bit c nguyn tc lm vic ca cc mch in t mt cch

thun tin v nhanh chng, cng nh c th d dng chuyn tt c cc mch c

hi tip v dng cu trc chun (th hin trn hnh 1.2), trong mc ny s trnh

by phng php phn tch khi (thng dng trong k thut iu khin) phntch b khuch i c hi tip. Lu thut ton thc hin qu trnh phn tch

bao gm 6 bc nh ch ra bng 1.1.

* Bc 1: Xc nhX .


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- Chn i lng ra l in p nu mch c hi tip in p hoc l dng

in nu mch c hi tip dng in.

* Bc 2: Chn hVn XXX ,, .

- Vi hi tip ni tip: chn nX l in p khng ti ca ngun tn hiu.

Biu din ngun tn hiu bng s tng ng in p v chnVh XX , l in

p.

- Vi hi tip song song: chnnX l dng in ngn mch u ra ca

ngun tn hiu. Biu din ngun tn hiu bng s tng ng dng in..

ChnVh XX , l dng in.

Thng chnhX cng th nguyn vi nX (khng bt buc).

* Bc 3: Xy dng h hai phng trnh m t qu trnh x l tn hiu trong

cc khu ca b khuch i c hi tip.

)(1 hXfX = v ),(2 XXfX nh =

Ring phng trnh 2f nu p dng nguyn l xp chng xy dng th

c th thc hin n gin hn.

* Bc 4: V s khi theo cu trc chun (hnh 1.2).

* Bc 5: Xc nh cc h s truyn t nht KKK ,, v su hi tip

htKKg += 1 .

* Bc 6: Xc nh cc thng s cn thit khc.


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Bng 1.1: Lu thut ton phn tch b khuch i c hi tip.

Bt u

Xc nh Xr

Chn Xn, Xv, Xh

Xy dng h phng trnhXr= f1(Xh)

Xh = f2(Xn, Xr)

V s cu trc chun(Hnh 1.2)

Xc nh cc h s K, Kht, Kn, g

Xc nh cc thng s cn thit khc

Kt thc

Xt mt v d n gin minh ho phng php phn tch xt. Trn

hnh 1.3a trnh by mt b khuch i dng tranzistor lng cc mc E chung, c

in tr ER ng vai tr hi tip m dng in; hnh 1.3b l s tng ng

ca n.


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1

+ +n be E R r R K =

htI

BI CIVInU

)C

)a )b

nU

rU

nU

CI

BI

EI EInR

ER CR

CI

BI

berB

ICR

C

E

CCU+ B

nR

ER

+ +

E

n be E

R

R r R

Hnh 1.3: a) B K mc E chung c hi tip m dng in.

b) S tng ng.c) Cu trc chun m t theo lu thut ton.

Bc 1: ChnCr IX = v mch c hi tip m dng in.

Bc 2: Chnnn UX = v mch c hi tip ni tip.

ChnBh IX = s thun li hn cho vic phn tch .

Bc 3:BC II = v

beEn

ECnB

rRR

RIUI

++

= (Dng nguyn l xp chng)

Bc 4: V hnh 1.3c.

Bc 5:1

; ; En ht E be n E be n

R K K K

R r R R r R= = =

+ + + +

Bc 6: Nu chn ER ln th: 1. >>++=

Eben

Eht

RrR

RKK

do 1>> .

Nhvy: 1>>g v = =n n nr Cht E

K X U X I

K R

V h s khuch i in p ca mch:

E

C

n

CC

n

ru

R

R

U

RI

U

UK == (1.4)


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18

1.3. nh hng ca hi tip m n cc tnh cht ca

b khuch i.

1.3.1. nh hng ca hi tip m n n nh ca h s khuch i:

Trong thc t, h s khuch i l mt i l

ng ph thuc vo nhit ,vo s thay i ca in p ngun cung cp, vo thi gian s dng cng nh vo

tp tn ca tranzistor. Bng tnh ton sau y ta c th chng minh rng: S

dng hi tip m c th khc phc c cc nh hng trn v m bo h s

khuch i n nh hn.

Trc ht, ly vi phn ton phn hai v biu thc (1.2) v b qua cc thnh

phn v cng b bc cao, ta nhn c biu thc:

dKKK

KdK

KK

KKKdK

KK

KdK

ht

nht

ht

nn

ht

tp 22 )1()1(

)1(

1 ++

+

++

Vi sai s nh ta c biu thc gn ng i vi khng n nh tng i

ca hm s truyn t:

K

K

KKK

K

KK

KK

K

K

K

K

htht

ht

ht

ht

n

n

tp

tp

+

+

+

1

1

1

(1.5)

T (1.5) ta rt ra kt lun: Sai s tng i ca h s khuch i khi c hi

tip m nh hn g ln ).1( htKKg += so vi sai s tng i ca n khi khng c

hi tip. Tuy vy, kt lun ny ch c ngha khi cc h snK v htK n nh. Do

vy, khi cn h s khuch i n nh, phi la chn cc phn t ca cc mch

hi tip v mch ghp vi ngun tn hiu c chnh xc cao.

T , kt hp vi biu thc (1.3) suy ra: Hi tip m gi cho quan h

ht

v

K

XX n nh (tc l

vX

X hng s). i lngvXX ; c th l in p


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19

hoc dng in. Nh vy, tu thuc vo loi mch hi tip, ta c cc ng dng

khc nhau nh ch ra bng 1.2.

Bng 1.2: ng dng hi tip m n nh cc tham s ca b khuch i

Loi mch hi tip i lng c nnh

Loi mch ngdng

Hi tip ni tip - in p H s khuch i in

pVr UU

Khuch i in p

Hi tip ni tip - dng in in dn truyn t

Vr UI

Bin i IU

Hi tip song song - in p in tr truyn t

Vr IU

Bin i t UI

Hi tip song song - dng

in

H s khuch i dng

in Vr II

Khuch i dng

in

1.3.2. nh hng ca hi tip m n tr khng vo.Hi tip m lm thay i tr khng vo ton phn ca b khuch i. S

thay i ny ch ph thuc vo phng php mc mch hi tip v u vo (ni

tip hoc song song). V vy ta ch cn phn bit hai trng hp nh ch ra trn s

tng ng u vo. (Hnh 1.4.)

a) Tr khng vo ca b khuch i c hi tip m ni tip (hnh 1.4a) .

Coi ni tr ca s tng ng in p ca khu hi tip c gi tr kh

nh (vi sai s c th chp nhn c): hrht rr


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20

hr

hU

ht rK X

r h tr

1

1

vI

ht rK X

rhtr

K

htK

vI

hI

hr hU hr

K

htK

vU

1

1

vU

a) b)

Hnh 1.4. S tng ng u vo b khuch i c hi tip

a) Hi tip ni tip; b) Hi tip song song.

b)Tr khng vo ca b khuch i c hi tip m song song ( hnh 1.4b.)

Coi ni tr ca s tng ng dng in ca khu hi tip c gi tr kh

ln:hrht rr >> , ta c th tnh c dn np vo ton phn khi c hi tip m:

(1 )V h ht r h ht V V

V V v

I I K X I K K Y gY

U U U

+ + = = =

1 vV

v

ZZY g

= (1.7)

T (1.6) v (1.7) ta nhn thy: Hi tip m ni tip lm tng tr khng vo

ca phn mch nm trong vng hi tip g ln v hi tip m song song lm gim

tr khng vo cng by nhiu ln.

1.3.3. nh hng ca hi tip m n tr khng ra:

Hi tip m cng lm thay i tr khng ra. Ngc li vi tr khng vo, s

thay i ny ch ph thuc vo phng php mc u ra ca khu khuch i vi

u vo ca khu hi tip (hi tip in p hay hi tip dng in). Nh vy, ta

cng ch cn phn bit hai trng hp nh ch ra trn s tng ng u ra.

(Hnh 1.5.)


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21

vhtr

2

K

rr

2

2

2

htK

vhtr

hKX t

R

K

tr

U

hKX

htK

rr

Rtr

U

a) b)

Hnh 1.5. S tngngu ra ca b khuch i c hi tip

a)Hi tip in p; b) Hi tip dngin. tnh trkhng ra, ta cn xc nh cc i lng: in p ra hmch

rhU

v dng in ra ngn mch rngI :

rng

rh

rI

UZ = (1.8)

a) Trkhng ra ca b khuch ic hi tip m in p (hnh 1.5a).

Ta c nhn xt: Khi ngn mch u ra 22 s hnh 1.5a, s khng c

in p a v u vo khu hi ti p - mch hi ti p khng c tc dng v

vh XX = , nh vy:

r

V

r

hrng

r

KX

r

KXI == v

ht

v

vhtr

Vhthrh

KK

KX

rr

rKXU

+

+=

1

Vi gi thit ni trca s tng ng in p ca khu khuch i c

gi tr nh )( htrr vr


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22

b)Tr khng ra ca b khuch i c hi tip m dng in (hnh 1.5b).

Ta cng nhn xt: Khi h mch )(22 = tR s hnh 1.5b s khng c

dng in a v u vo khu hi tip - tc l mch hi tip khng c tc dng

v ;vh XX = Nhvy:

v:

ht

v

vhtr

rhrng

rvrhrh

KK

KX

rr

rKXI

rKXrKXU

+

+=

==

1

.

Vi gi thit ni tr ca s tng ng dng in ca khu khuch i

c gi tr ln )( vhtr rr >> . Nhvy:

(1 )rhr ht r r

r ng

U Z KK r gZ

I = + (1.10)

Trong :rvhtrr rrrZ +=

Nh vy, hi tip m in p lm gim tr khng ra ca b khuch i g

ln, trong khi hi tip m dng in lm tng tr khng ra ca b khuch i

cng by nhiu ln.

1.3.4. nh hng ca hi tip m n di ng v mo phi tuyn ca b

khuch i.Khi khng c hi tip m, ton b tn hiu vX c a n u vo b

khuch ivh XX = ; ngc li, khi c hi tip m, ch c mt phn tn hiu vX

c t vo u vo b khuch i:

hhtvrhtvh XKKXXKXX ==

Nhvy:g

X

KK

XX v

ht

vh =+

=1

(1.11)

iu chng t: Nh hi tip m, di ng ca b khuch i c m

rng ra. Cng t (1.11), mo phi tuyn do cong ng c tuyn truyn t

ca b khuch i gy ra, tng ng cng gim i. l mt u im ln ca hi

tip m nhm nng cao cht lng cho b khuch i.


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23

1.3.5. nh hng ca hi tip m n c tnh tn s v c tnh ng

ca b khuch i.

1.3.5.1. c tnh tn s v c tnh ng ca b khuch i:

a)c tnh tn s:

c tnh tn s l c tnh biu din mi quan h gia mun v gc phaca hm s truyn t phc ca b khuch i theo tn s.

minh ho, ta xt mt v d n gin khi kho st b khuch i di rng

trnh by trn hnh 1.6, vi gi thit h s khuch i l tng ca n khng ph

thuc vo tn s v l hng s:

2( ) 0

1u f u

U K K const

U= = =

uoK

2R

1R

1C

2C

2C2

2

vU

1

1C

1R

1

1

1

1

1UrU

rU

2R

1uoK U

)a

)b

vU

1U 2U

1 22

2

2

Hnh 1.6.B khuch i di rnga) S khi; b) S tng ng 1 1 2 2// ; //v rR R R R R R = =

Trong thc t, b khuch i lm vic, cn c mch ghp vi ngun tn

hiu u vo v u ra mc vi ti. Hm s truyn t ca cc mch ny ph

thuc vo tn s do s c mt ca cc phn t in khng ( hnh 1.6 l cc t

in 21,CC ). Do vy cn c vo s tng ng hnh 1.6b, ta xc nh c

hm s truyn t phc:

1

( ) ( )1 (1 ) (1 )

uo dr

j Pv v t d

K PT U UUr

K K U U U PT PT = = = = + + (1.12)Trong :


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24

1 1

2 2

1 1

2

1 1

2

d

d d

t

t t

T R Cf

T R Cf

= = =

= = =

T (1.12) ta xc nh c mun ca hm s truyn t:

( ) 2 2 2 2

/

1 ( ) 1 ( ) 1 ( / ) 1 ( / )

uo d uou j

t d

K T K f fdK K

T T f ft f fd

= =

+ + + +(1.13)

V gc pha ca hm truyn t:

dt TarctgTarctg

=2

(1.14)

Theo (1.13) v (1.14) ta xy dng cc c tuyn bin tn s v c tuyn

pha tn s ca b khuch i nhch ra trn hnh 1.7.

Trn hnh 1.7, ta gi:

222

1

2 CRf tt

== (1.15)

l tn s gii hn trn, v:112

1

2 CRf dd

== (1.16)

l tn s gii hn di.Thng c: 2211 CRCR >> nn td ff


24/408

25

uK

uoK

090

00

090

f

df

uoK7,0

tf

d

ft

f

Hnh 1.7. c tuyn bin (a) v c tuyn pha (b) ca b khuch i

b) c tnh ng ca b khuch i.

Cho tc dng vo u vo b khuch i mt xung ch nht l tng, th

u ra ta nhn c dng xung nh ch ra hnh 1.8.

t

t

vU

)a

)b

A

raU

uo vK U=

1trt

Xt2tr

t

Hnh 1.8 Dng xung vo (a) v ra (b) ca b khuch idi rng.c tnh ng ca b khuch i c nh gi qua cc tham s sau ca

xung ra:- Thi gian xc lp xt - xt nh xung A - Thi gian tr (trc v sau)

21; trtr tt


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26

c tnh tn s v c tnh ng c quan h cht ch vi nhau. Tht vy,gia cc tham s ca c tnh ng nu v cc tham s ca c tnh tn s( dt ff ; ) c mi lin quan rng buc.

Gi thit a vo u vo b khuch i hnh 1.6 hm n v 1(t).T biu thc (1.12), ta xc nh c hm qu

)(Ph

)1)(1(1.)()(

dt

duoPP

PTPTTK

PKh

++==

Bin i ngc Laplas ta c://

( )

( )

(1 / )

tt Tt Td

uot

t d

K e eh

T T

=

(1.17)

Vi b khuch i di rng .td TT >> * Khi xt qu trnh qu xc nh xt , c th coi dTt>= Nn: 35,0. Btx (1.19)

Nh vy, thi gian Xt ca qu trnh qu ca mt b khuch i di rng

t l nghch vi tn s gii hn trn ca b khuch i .

* quan st xt nh xung, ta xt c tnh xung trong khong thi gian

tng i di, v qu trnh xt nh din bin chm. Nh vy biu thc (1.17) vi

tTt>> c th vit gn ng:

dTt

uot eKh/

)(

(1.20)

Vi rng xung khng qu ln dT


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27

d

d

fT

2==

Nh vy, xt nh xung t l thun vi tn s gii hn di ca b

khuch i.

1.3.5.2. nh hng ca hi tip m n cc c tnh tn s v c tnhng.

a) Trng hp vi b khuch i di rng.

Trong trng hp ny, hm s truyn t (1.12) c cha 2 im cc: mt

im cc tn s thp1 1/ dP T= v mt im cc tn s cao tTP /12 = , vi

cc tn s ny cch kh xa nhau )( dt ff >> . th Bode ca b khuch i

c biu din bi ng lin nt trn hnh 1.9 (a: c tuyn bin ; b: c

tuyn pha).

Thay biu thc (1.12) vo (1.1) ta nhn c hm s truyn t ca bkhuch i c hi tip:

(1 )(1 )

uo dp

d t uo ht d

K PTK

PT PT K K PT =

+ + +(1.21)

Ln lt xt hai trng hp:

* tn s tng i thp )( tff


27/408

28

Trong : ; ; 1uouo d d uo ht

K K T gT g K K

g = = = +

Nhvy: gfddf /= (1.23)

Tc l hi tip m lm gim tn s gii hn di gln.

* tn s t

ng i cao fdf >>( hay )1>>dT Nhvy:

(1 ) 1

uo uop

d t uo ht t

K PTd K K

PT PT K K PTd PT

=

+ + +(1.24)

Vi gTT tt /= hay t t f gf = (1.25)

Tc l, hi tip m lm tng tn s gii hn trn g ln.

T (1.23) v (1.25) ta kt lun: hi tip m c tc dng m rng di thng

ca b khuch i. Vi khuch i di rng tdt fffB = th di thng tng ln

g ln. Tng ng mo tn s v mo pha cng gim. Tuy nhin khi c hi tip

m, phi chp nhn h s khuch i trong di thng gim.

g

K

KK

KK uo

htuo

uouo =+

=1

(1.26)

Ni cch khc, t (1.25) v (1.26) ta c:

1.. fBKBK uouo == (1.27)

y 1f l tn s ti 1(0 )u K db= , gi l tn s n v.

Nhn xt: Tch BKuo ca mt b khuch i khng ph thuc vo su

hi tip, n l mt hng s v bng tn s n v ca b khuch i .Vi quan h gia c tnh tn s v c tnh ng nh ch ra trn, ta nhn

thy khi c hi tip m, thi gian xc lpxt v st nh xung s cng gim g

ln so vi khi khng c hi tip.

b) Trng hp vi b khuch i tng qut.

Ta xt trng hp khi hm s truyn t ca b khuch i vit di dng:

)1(...)1)(1(21

0

n

PTPTPT

KK

+++= (1.28)

Tc l c th c n im cc 1i

i

PT

= vi )...,3,2,1 ni = .

Trng hp ny c ngha thc t ln, v hu nh tt c cc b khuch i

u c hm s truyn t gn vi dng (1.28). Tin hnh nh gi nh hng ca

hi tip m n cc tnh cht ca b khuch i cng c thc hin theo phng


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29

php nu trn (xem [1]). Trong mc ny ta khng nghin cu chi tit, ch nu

mt vi nhn xt sau:

* Trng hp hm s(1.28) c hai im cc:

Khi tng dn su hi tip g , h s phm cht Q ca b khuch i tng,

tng ng h s suy gim QD 2/1= s gim, cc im cc 21 ;PP c th chuyn t

gi tr thc thnh gi tr phc, c tnh xung c thi gian xc lp ngn, nhng thay

vo c hin tng dao ng (hnh 1.10 v 1.11).

Hnh1.10 th Bode ca Hnh1.11. c tuyn qu

h thng tuyn tnh bc 2. ca h thng tuyn tnh bc 2.

* Trng hp nhiu im cc:

Vi h thng tuyn tnh bc 2, cc nghim ca phng trnh c trng u

nm bn tri mt phng phc vi mi gi tr ca D nn h thng lun n nh. Tuy

nhin khi D nh, mc an ton ca h thng khng cao.

Ngc li cc h thng tuyn tnh bc cao )2( >n , qu tch nghim ca ph-

ng trnh c trng c th ct trc tung trong mt phng phc lm cho h thng

mt n nh v cn phi s dng cc bin php m bo tnh n nh ca h thng.

1.4. n nh ch cng tc cho cc tng dngtranzistor lng cc.

1.4.1. t vn :

Trong cc tng khuch i tn hiu nh, im lm vic nm trong min tch ccca Tranzistor lng cc (vi cc tng khuch i tn hiu ln ta xt ring trongchng khuch i cng sut). ch tnh , ngha l khi cha c tn hiu


29/408

30

vo, trn cc cc ca Tranzistor c cc dng in tnh :000 ;; EBC III cng nhin

p mt chiu gia cc cc:00 ; CEBE UU . im lm vic ng vi ch ny gi l

im lm vic tnh .

Khi c tn hiu vo th cc in p v dng in thay i xung quanh gi tr

tnh . m bo cho cc tng lm vic bnh thng trong cc iu kin khcnhau, ngoi vic cung cp in p thch hp cho cc cc, cn cn phi n nh

im lm vic tnh chn. Trong trng hp ngc li, khi im lm vic tnh

khng n nh, ch tiu cht lng ( mo, h s khuch i in p v dng

in, cng sut ra...) ca tng s b gim.

Thc cht ca vn n nh l lm cho im cng tc tnh t ph thuc

vo tp tn tham s ca tranzistor, vo nhit , vo in p ngun cung cp.

C th l p dng cc phng php gi cho dng colecto 0CI khng i.

1.4.2. Hin tng tri im lm vic:

T s tng ng ca tranzistor lm vic trong min tch cc, ta nhn

thy c tnh ca tranzistor c c trng ch yu bi cc tham s: H s truyn

t dng in hoc ; Dng in ngc (0CBI hoc )00 CBCE II = v in p

BEU . S thay i cc tham s ny l nguyn nhn dn n s thay i im lm

vic tnh , tc l khi c s thay i ca nhit , in p ngun, tp tn catranzistor v ca cc linh kin khc th s dn ti hin tng tri im lm vic

tnh .

nh gi nh hng ca cc tham s c trng nu trn ti s thay

i ca dng in colectoCI , ta xc nh lng bin i Ic qua biu thc vi

phn ton phn:

+

+

ccU

U

Icc CB

CB

BE

BE

0

0

(1.29)

Trong (1.29), cc lng bin i ,, 0CBBEU gi l cc i lng tri,

cn cc o hm ring0

, ,

BE CB

Ic Ic Ic

U I l cc h s, c trng cho kh nng n

nh ca mch.


30/408

31

Hin tng tri c tc hi nht l tri nhit.

Bng (1.3) cho v d minh ho hin tng ny.

Bng (1.3): S ph thuc cc tham s ca tranzistor vo nhit

Nhit (P0 Pc)

Thams

LoiTranzistor -65 +25 +70 +150

Silic


31/408

32

cIBEU

2t 1t

cI

BEU

BEOU

E

C

BEU

)a)b

0

Hnh 1.12. a) S tng ng ca tranzistor do tri nhit BEU ;

b) S thay i im lm vic do tri nhit

nh hng ca nhit n ch cng tc ca tranzistor qua hnh 1.12b,

c th c gii thch nh sau: mun cho dng cI n nh, th khi nhit thay

i mt lng 0t , BEU cng phi thay i tng ng theo biu thc:

0

02,5BE mvU t

C =

Ngc li, nu gi BEU c nh th dng colecto s thay i mt lng:

0exp 1BEC E

T

U

U

(1.31)

Trong ,EoI l dng emito khi 0= BEU .

nh gi mc nh hng ca in p tri n in p ra, ta dng h

s khuch i in p tri trK , xc nh theo biu thc:

BE

COtr

U

UK

= (1.32)

ViCOU l lng bin i in p mt chiu u ra.

1.4.3. Cc s n nh tuyn tnh.

Cc s n nh tuyn tnh c dng ph bin nht hin nay l s hi

tip m mt chiu nhm bin i thin p ca tranzistor sao cho c th hn ch s

di chuyn im lm vic tnh trn c tuyn ra, gy nn bi cc yu t mt n

nh.

a) n nh im lm vic bng hi tip m in p mt chiu (hnh 1.13).


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Hnh 1.13.

Cung cp v n nh im lm vic bng hi tip m in p mt chiu

a) Mch E chung; b) Mch b chung; c) S tng ng.

Trn cc s 1.13a v b, vic cung cp thin p cho baz cng c ly t

ngun chung vi mch cung cp in p mt chiu cho colecto (CC

U ), tuy nhin

in tr 1R lm nhim v a thin p v baz c ni vi colecto ch khng phi

vi ngun cung cp, dn in p hi tip v u vo.

Nguyn tc n nh c th c gii thch nh sau: Khi dng in mt chiu

COI tng (hoc gim) do cc nhn t khng n nh gy ra, th in p

CEOU gim

(hoc tng), dn ti thin pBEO

U gim (hoc tng) theo v iu ngn cn s

bin i caCOI . Nhvy im lm vic tnh n nh hn.

S dng ngun in p tng ng thay cho mch cung cp thin p mt

chiu, ta c s tng ng hnh 1.13c. Khi b qua dng in ngcCBO

I c

gi tr kh nh, ta tnh c:

)(1

21

2

21

12

21

2RIRIU

RR

R

RR

RRI

RR

RUU

BoCCoCCBoCEoBEo

+

=

+

+

=

Thay CoBoII = , ta xc nh c dng colecto:

[ ]

1

21)1(

RR

RRUUI

C

BEoCC

Co

+

+

=

(1.33)

Theo (1.33), mch lm vic n nh khi s dng hi tip m in p mt

chiu, cn phi m bo cc iu kin sau:


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34

- Chn tranzistor c ln, sao cho:

1cR R >> v coI t ph thuc vo .

- Chn 211 RR+ c gi tr nh ( 2R chn ln).

Vi cc iu kin trn, h s khuch i in p tri c xc nh gn

ng theo biu thc:

2

11R

R

U

UK

BE

Ctr +

= (1.34)

S ny c u im: 1R v 2R ln nn cng sut tn hao nh, Khi =2R

th:

1.RIUU BoCEoBEo = (1.35)

b) n nh lm vic nh hi tip m dng in mt chiu. (Hnh 1.14).

in trE

R mc trn emitong vai tr phn t hi tip m dng in.

Nguyn tc n nh c gii thch nh sau:

KhiCoI tng (hoc gim) do hin tng tri, th in p h trn in tr ER

).( EEoEo RIU = tng (hoc gim), dn ti thin p gia baz v emito

)( EoBoBEo UUU = gim (hoc tng). iu ngn cn s thay i ca dng

colecto v nhvy im lm vic tnh n nh hn.

Theo s tng ng hnh 1.14d, ta xc nh c quan h:

BEoCCnBoEoE URR

R

URIIR +=+ 212

Vi 2121

RR

RR

Rn +=

Dng in baz ch tnh c xc nh t quan h ca cc thnh phn

dng in trong tranzistor:

CBoEoBo III = )1( (1.36)

Nhvy:

2 2 1 2

1 2 1 2 1 2

1 2

1 2

(1 )

( )(1 )

CC BEo CBo n CC BEo CBo

Co Eo

E nE

R R R RU U I R U U I

R R R R R RI I

R RR R

R R R

+ ++ + +

= =+

+ + +

Vi:

+

=1

11 . (1.37)


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35

Hnh 1.14. Cung cp v n nh im lm vic bng hi tip m dngin mt

chiu.a) Mch E chung; b) Mch B chung; c) Mch C chung;

d) S tngng mt chiu ca mch E chung.

Tng t, theo (1.37), mch lm vic n nh khi s dng hi ti p m

dng in mt chiu, cn phi m bo cc iu kin sau:

- Chn in trhi tip ER c gi tr ln, khi )1( +>> nE RR . v CI t

ph thuc vo .

- Coi dng in dCBoI nh th ch c s thay i ca BEoU nh hng ti

dngCoI . nh hng ny cng nh nu BEU nh.

H s khuch i in p tri c xc nh theo biu thc:

E

C

BE

CCo

BE

Cotr

R

R

U

RI

U

UK

=

=

.(1.38)

Nh vy h s khuch i in p tri bng h s khuch i in p tn hiu.


35/408

36

Hin nay, cc s n nh im lm vic dng hi tip m dng in mt

chiu c ng dng rng ri. Chng cho php cng tc n nh khi h s khuch

i dng in thay i kh nhiu v nhit bin thin trong phm vi t

cc 00 10030 . Tuy nhin, nhc im ca cc s ny l h s khuch i gim.

trnh hi tip m xoay chiu, thng mc song song vi ER mt t in EC c

tr s in dung ln.

1.4.4. Cc s n nh phi tuyn:

Trong cc s ny, ngi ta dng phng php b nhit nh cc phn t

c tham s ph thuc vo nhit chng li ngay hin tng tri nhit.

Phng php ny thng dng trong cc mch t hp v trong cc tng khuch

i cng sut. Cc phn t c tham s ph thuc vo nhit nh it, tranzistor,

in tr nhit... dng n nh im lm vic, c trnh by trn hnh 1.15.

Trn hnh 1.15a, nu it D v tranzistor T u c ch to t cng mt

loi bn dn nh nhau v nu nhit mt ghp ca chng ging nhau, th c

tnh nhit ca BEU v DU c th b tr ln nhau do chng mc ngc chiu nhau.

it c phn cc thun ging nh mt ghp EB ca tranzistor, ta dng

thm mt ngun ph E . S hnh 1.15b cng lm vic theo nguyn tc tng

t. Tuy nhin s ny cn n nh c thin p cho transistor, do mch phn

p gm 21;RR v mt s it mc ni tip c phn cc thun. Khi 12 RR


36/408

37

Hnh 1.15. Cc s n nh phi tuyn

a) S bBE

U

b) S b BEU

v b ngun cung cp; c) S b dng in tr nhit.1.4.5. n nh im lm vic trong cc mch t hp tng t.

Trong cc mch t hp tng t, n nh im lm vic, thng dng cc

ngun dng in, v phng php ny d thc hin di dng mch t hp.

Ngun dng in dng n nh im lm vic trong cc mch t hp

tng t c trnh by trn hnh 1.16.

Gi thit cc tranzistor21

;TT ca s hnh 1.16a c tham s hon ton

ging nhau v cng nhit , khi v21 BEBE

UU = nn21 BB

II = v21 CC

II = ; nh

vy:

2

2222111

22

C

CBCBBC

IIIIIIII +=+=++=

T suy ra (vi 1>> )

1

1

221

II

IC

+

=

(1.39)


37/408

38

ER

2CI2CI 1I1I R

P

1CI

1BI 2BI

ccU+

1CI

1T1T 2T 2T

1BEU2BEU

)a )b Hnh 1.16Mch in dng ngun dng n nh

2CI

a) Mch n gin; b). Mch dng khi yu cu dng2C

I nh.

Nh vy, c th dng 1I iu khin gi tr ca dng in 2CI . c c

1I n nh, cch n gin nht l ni imP ca mch t hp tng t vi ngun

in p mt chiu n nh )( CCU+ qua in tr RUIR CC1( do BEU kh nh).

Biu thc (1.39) c p dng khng nhng cho thnh phn mt chiu mcn cho c thnh phn tn hiu. Khi yu cu cn c ngun dng b th yu cu R cn c gi tr ln (khi kh ch to v n chim th tch ln), ta chuyn sang sdng s hnh 1.16b. Lc ny, theo biu thc gn ng i vi c tuyn vo

ca transistor ta c:

=

=T

CEBEEbh

T

BEEbhC

U

IRUI

U

UII 22122 expexp

2 2 11

2 2

exp / exp

exp

E C C BEEbh

E CT T

T

R I I UI

R IU U

U

= = (1.40)

NhvyT

CE

CC

C

U

IR

I

I

I

I 22

2

1

2

1 exp= vi mvUT 26 l in p nhit.

Nh vy, Nh mc thm in tr hi tip 2ER nn 12 IIC


38/408

39

1.5. n nh ch cng tc cho cc tng dng

tranzistor hiu ng trng.1.5.1. Ch tnh .

V nguyn tc, vic cung cp v n nh ch cng tc i vi cc tng

dng tranzistor hiu ng trng cng tng t nh i vi tranzistor lng cc.Trong trng hp khuch i tn hiu, FET lm vic trong min tht. ch tnh , trn cc cc ca FET c cc dng tnh SoGoDo III ,, v gia cc cc c cc

in p mt chiuGSoU v DSoU . Thc cht ca vn n nh cng l p dng

cc bin php gi cho dng in mngDoI t bin i theo nhit , theo thi

gian, t ph thuc vo tp tn tham s ca FET v s bin i ca ngun cung

cp.

So vi tranzistor l

ng cc,FET

c

u im c bit l khng yu cu dngvo ln (tr khng vo cc ln), do vy ngun tn hiu ch cn cng sut nh, tc

l h s khuch i cng sut ca tng dng FET ln. Ngoi ra, in dung ghp

hi tip nh nn lm vic n nh.

Nhc im ca FET l h dn nh v cc tham s: PU (in p tht) v

DSSI (dng mng bo ho) c tp tn ln.

1.5.2. Cc s n nh im lm vic.

Cc s n nh im lm vic khi dng FET cng tng t nh khi dng

tranzistor lng cc. Ta ch trnh by ngn gn hai loi s dng hi tip m.a) n nh im lm vic dng hi tip m dng in. (Hnh 1.17)

)a )c

DDU+

sR

DR

GR

DDU+

SC SR

1RDR

2RSC

)b

DDU+

SR

1RDR

2RSC

3R

Hnh 1.17. Cung cp v n nh cho cc tng dngFET vi hi tip m dng

in. a) Mch dngJFET; b) Mch dng FETMIS knh c sn; c) Mch dng

FETMIS knh cha c sn.


39/408

40

in tr mc cc ngun (Source) SR ng vai tr phn t hi tip m

dng in. trnh hi tip m xoay chiu, c th mc song song viSR mt t

in SC c in dung ln. i vi FET t ngt (knh cha c sn) (hnh 1.17c),

thin p c dn vo cc ca (Gate) nh mch phn p 21 ;RR ly t ngun

cung cp in p mt chiu cho cc mng (Drain). i vi FETMIS t dn

(Knh c sn)( hnh 1.17b), hoc JFET (hnh 1.17a), c th b 1R ( =1R ), lc

in p cc ca so vi cc ngun chnh l mch thin p t cp, vi gi tr:

SDGS RIU =

tr khng vo ca tng khng b gim ng k cn chnGR hoc 2R c

gi tr ln (c ), hoc mc thm 3R nh s (hnh 1.17c).

Cc s hnh 1.17 u c th biu din bi s tng ng hnh 1.18a.

)a

sR

DR

GR SC

GU

DI

DSSI

FET MOS FET

GSU

GUPU

)b

Hnh 1.18. a) S tng ng ca mch trn hnh 1.17

b) c tuyn truyn t ( )GSUD fI = vi21

2

RR

RUU DDG +

= .

Nu coi dng cc ca 0GI , ta c:

SDGGS RIUU .' = (1.41)Biu thc (1.41) cho bit dc ca ng in tr SR trn c tuyn

truyn t (hnh 1.18b)

SGS

D

RdU

dItg

1== (1.42)


40/408

41

Nh vy, thin p v in trS

R phi c chn sao cho dng mngDI t

thay i khi thay FET , v im lm vic nm trong on thng ca c tuyn

truyn t.

b) n nh im lm vic dng hi tip m in p (hnh 1.19).

Hnh 1.19. Cung cp v n nh cho cc tng dng MOS-FET vi hi tip m

in p.

Thin p ly t cc mng qua b phn p in tr, trnh hi tip m xoay

chiu, mc thm t in C c gi tr in dung ln.


41/408

42

Chng 2cc s c bn ca tng khuch i tn

hiu nh dng tranzistor

Khuch i tn hiu l mt trong cc chc nng quan trng nht ca mchin t. Cc b khuch i gm phn t khuch i (phn t tch cc) mc vi

mt s phn t th ng khc. Vic phn loi cc b khuch i c th da trn

cc yu t sau:

- Theo tham s ca tn hiu cn khuch i: ta c khuch i tn hiu nh,

khuch i tn hiu ln, khuch i tn s thp, tn s cao; Khuch i tn hiu

lin tc, khuch i tn hiu xung.

- Theo dng phn t dng khuch i: theo cch phn loi ny ta c cc

b khuch i dng n in t chn khng v dng dng c bn dn, vi mch...- Theo phng php khuch i dng iu khin qu trnh bin i nng

lng thnh tn hiu c ch u ra.

Trong chng ny chng ta ch xt cc b khuch i tn hiu lin tc

(Analog) c bin nh dng tranzistor lng cc hoc FET.

Cc b khuch i tn hiu nh dng tranzistor l phn t tch cc (phn t

khuch i) c mc vi mt s phn t th ng khc. Khi im cng tc c

la chn ph hp, cc phn t ny c biu din bng mt mng bn cc tuyn

tnh. Trong thc t ngi ta hay dng s tng ng hnh v cc tham shn hp H , dn np Y phn tch tnh ton cc mch khuch i dngtranzistor. Vic s dng h thng tham s no tu thuc vo nhiu yu t khc

nhau.

u vo v u ra ca cc mng bn cc b khuch i, u phi ghp vi

cc mch pha trc v pha sau. u vo ghp vi ngun tn hiu (ngun dng

nI hay ngun in p nU ) c tr khng nZ (hoc dn np nY ). Cc mch pha sau

c m t bng mt ph ti c tr khng )( tt YZ . Tu thuc vo tng quan

gia tr khng ca ngun tn hiunZ vi tr khng vo ca mng bn cc vZ v

tr khng ph titZ vi tr khng ra ca mng bn cc raZ , ngi ta c th biu

din phn t tch cc (khuch i) bng s ngun dng hoc ngun p. Trong


42/408

43

cc tnh ton sau ny ta coi tr khngn

Z vt

Z ch l thun tr ngha l ch cn

R

vt

R .

2.1. cc tham s c bn ca b khuch i.Mt b khuch i - mt mng bn cc - c th c c trng bng phng

trnh tuyn tnh ch mi quan h gia cc i lng1 1, 2, 2

,U I U I hnh 2.1a. Khi s

dng tham s h , ta c h phng trnh :

1 11 1 12 2

2 21 1 22 2

U h I h U

I h I h U

= +

= +

(2.1)

S tng ng ca tng khuch i tn hiu nh khi c dng hnh2.1b.

Hnh 2.1. Mng 4 cc khuch i (a) s tng ng (b)

xc nh cc tham s theo nh lut Kic Khp:

1 11 1 12 2

2

21 1 22 20

t

U h I h U

Uh I h U

R

= +

= + +(2.2)

T (2.2) ta xc nh tr khng u vo:

111 12 21

11

1 2222

1 1

t

vao

t

t

hR hU h hZ h

I R hh

R

+= = =

++

(2.3)

Vi11 12 12 21

h h h h h =

Tr khng u ra c xc nh vi iu kin cho 0=n

U .


43/408

44

Khi ta c h phng trnh :

11 1 12 2 n 1

2 21 1 22 2

0= h I +h U +R I

I = h I +h U(2.4)

Suy ra:

1120

2 22.n

nra U

n

h RUZ I h h R

= += + (2.5)

T (2.2), thaytRIU 22 = , ta c h s khuch i dng in:

2 21

1 221

ra

t

I hZ

I h R= =

+ +(2.6)

H s khuch i in p:

2 11 21 1

111 11.

.

tu i

t v

t

RU h h RK K

hU h h R Z h

R

= = = =+

(2.7)

V h s khuch i cng sut: 2212 2

1 1 22 11

2

21

2

22

.

. (1 ).( . )

(1 ) .

ra t p u i

vao t t

t

t v

P h RU I K K K

P U I h R h h R

h R

h R Z

= = =+ +

=+

(2.8)

y l nhng tham s c trng c bn nht ca mng bn cc, trong thc

t, nht l khi lm vic siu cao tn, ngi ta phi thc hin phi hp tr khng

u vo v u ra. Tc l thc hin iu kin:

n v vao

t ra ra

R R Z

R R Z

= =

= =

Khi phi hp tr khng:

11

22

nopt opt

h hR R

h

= = (2.9)

11

22.

topt opt

hR R

h h= =

(2.10)

V: ( )2

2211

2

21

maxhhh

h

KP += (2.11)

Theo c im ngi ta c th s dng ba cch mc: Emito chung, baz

chung v colecto chung. Sau y ta s a ra cc tham s ca cc b khuch i

mc theo cc s ny v nhn xt chng.


44/408

45

2.2. cc s c bn dng mt tranzistor lng cc2.2.1. Mch in cc b khuch i

chng trn, khi phn tch cc mch cung cp cho tranzistor, ta a racc cch mc c bn. Trong thc t, cc mch khuch i dng tranzistor c ti

in tr c v theo hnh 2.2. Cc s ny, khng ph thuc nhiu lm voc tnh ph ti l ti in tr (phi chu k) hay ti cng hng.

3R

1R

2R

4R

1C1C

1C

2C

2C

2C

3C

3C 1R2R

2R

1R

3R

3R4R

)a

)b

)c

ccU+ ccU+

ccU+

1U

1U

1U2U

2U

2U

Hnh 2.2. Ba s c bn dng tranzitor

Trong cc s trn, vic tnh ton cc gi tr in tr, t in m bo

im lm vic ca tranzistor v t h s khuch i cao nht thng da theo

cc ch c gii thiu trong s tay linh kin v theo kinh nghim ca

ngi thit k.

Trong hnh 2.2a, khi s dng t in 2C ngn mch cc thnh phn xoaychiu th khu

23CR c tc dng n nhit cho b khuch i. Trong mt s trng

hp, s dng tnh cht ca hi tip m (c bit l tnh n nh) ngi ta

khng mc t 2C . Khi mch c hi tip m dng in qua 3R . Cc tham s ca

mch hi tip m c nghin cu chng trc.

Vic la chn chnh xc im lm vic m bo cho b khuch i c tham

s c tr

ng tt nht v vic gi nh coi phn t khuch i l phn t tuyn tnhc chnh xc hn. Tuy nhin trong thc t do c tuyn von-ampe ca

tranzistor phi tuyn nn vn tn ti mo khng ng thng trong qu trnh

khuch i. Thng thng c th coi quan h gia dng u vo )(1 bii v dng ra


45/408

46

)(2 cii l tuyn tnh nn nguyn nhn gy mo tn hiu ch yu do quan h phi

tuyn ca c tuyn vo: )( beb ufi = .

Gi s a ti u vo in p iu ho, khi in p trn cc baz -

emit:

tUUu BEBEobe sin+= (2.12)

y: BEoU - in p mt chiu trn cc baz, nhm xc nh im cng tc

ca tranzistor.

BEU - Bin in p xoay chiu t vo.

Phng trnh ca c tuyn vo:

T

beob

UuIi exp (2.13)

Thaybeu vo ta c:

exp exp sinBEo BEb oT T

U Ui I t

U U

=

(2.14)

tT

BEooBo

U

UII exp=

Khai trin (2.14) theo hm Taylor v ch xt mo phi tuyn do hi bc hai

gy ra, ta c:

( )

++= t

U

Ut

U

UIi

T

BE

T

BEBob 2cos1

4sin1

2

2

(2.15a)

Vy ngoi thnh phn tn s c bn , u ra cn c thnh phn dng

in c tn s l hi bc 2, chnh thnh phn ny do c tuyn u vo phi tuyn

gy ra v xc nh h s mo phi tuyn theo biu thc:

%1004 T

BEm

U

Uk = (2.15b)

Vy h s mo phi tuyn ph thuc vo bin in p u vo BEU ,

%1


46/408

47

Nh vy gii hn ca chng ny chng ta ch xt cc b khuch i tn

hiu nh, khi ta c th coi cc b khuch i lm vic trong ch tuyn tnh.

(Cc tham s ca chng khng i trong phm vi lm vic). Cn khi in p vo

ln hn mt cht ta phi s dng cc b khuch i c hi tip m hay l cc b

khuch i vi sai.

2.2.2. Cc tham s c bn ca tng s :

Cc tham s c bn ca cc b khuch i c tnh ton theo cc bng

(2.1); (2.2); (2.3); (2.4). Cc bng ny cho php tnh ton mt cch y nht

cc tham s c trng ca cc b khuch i.

Trong cc cng thc, thng nht cc k hiu, ta cn lu :

Eo

Te

IUr = : in tr khuch tn emit (it emito- baz)

EoI : Dng in emit ti im cng tc

1

1)1(

12 =

++

==

eb

ebebe

C

CCr

f (2.16)

ER : in tr mc cc emit thc hin hi tip m dng in.

i vi mch baz chung, trong mt s trng hp ci thin tham s ca

s , ngi ta mc in tr BR (c tc dng vi c thnh phn xoay chiu) nh

l mt mch hi tip trnh by trn hnh 2.3.

raUvaoU

BR

n

U

nR

tR

Hnh 2.3. S tng ng xoay chiu

ca mch baz chung c hi tip


47/408

48

Bng 2.1 Cc tham s ca s emit chung.

MchEC

Tham sh Tham sy

uK

11

21

11

21

~

h

Rh

hRh

Rh

t

t

t

+

t

t

t

Ry

RyRy

21

22

21 ~1

+

iK 2122

21 ~1

hRh

h

t+

11

21

11

21 ~y

y

yRy

y

t+

VZ 1122

11 ~1

hRh

hRh

t

t

++

1111

22 1~1

yyRy

Ry

t

t

++

raZ n

n

Rhh

Rh

22

11

++

n

n

Ryy

Ry

+

+

22

111

Tham s vt l

( )0

1;

1 11

tu

e bb b c t

bb e b e e

RK

r r C RS

r r C r

+ + + +

0

1iK

S

+

( )( )0

V 0

1 11

1

1 1

bb b c t

bb e b e e

bb e

b c t

b e e

r C RS

r r C r Z r r

C RS

C r

+ + + + + +

+ +

( )

( )

( )

( )

0 0

R

0

0

11 1

1 21

1 2

bb n

bb n e bb n ee

e bb n b ce n

e n b e

r RS

r R r r R r rZ

r r R C r RS

r R C

+++ + + + + +

++ + +

+ +


48/408

49

Bng 2.2 Tham s b khuch i mcemit chung c hi tip m dng in.

MchEC c

hitip

Tham sh Tham sy

uK

E

t

EtE

t

Rhh

Rh

RhhRRhh

Rh

)1(~

)1()(

2111

21

211122

21

++

++++

E

t

tE

E

Ry

Ry

RyRy

Ry

21

21

2221

21

1~

1

+

++

iK 2122

21 ~)(1

hRRh

h

Et ++

11

21

11

21

~

)(

y

y

RRyy

y

Et ++

VZ

Ei

t

EtE

RKh

Rh

RhhRRhh

)1(~

1

)1()(

11

22

211122

++

+++++

11

21

11

2221

1~

1

y

Ry

yRy

RyRy

E

t

tE

+

+++

Tham s vt l

( )( )0

1;

1 11

tu

e E bb E t b c

bb e E e b e

RK

r R r R R C S

r r R r C

+ ++ + + + +

0 ;1

iK

S

+

( )( )( )( )0

V 0

1 11

1 ;

1 1

bb e t b c

e E e b e

e E

t b c

e b e

r r R C S

r R r C Z r R

R CS

r C

++ + + + + +

+ +


49/408

50

Bng 2.3. Cc tham s ca s b khuch i mc colecto chung.

MchCC

Tham sh Tham sy

uK

t

t

t

t

Rhh

Rh

Rhhhh

Rh

)1(

)1(~

~)1((

)1(

2111

21

122111

21

+++

++++

E

t

t

t

Ry

Ry

Ryy

Ryy

21

21

2111

2111

1~

(1

)(

+

+++

iK )1(~1

121

22

21 hRh

h

t

++

+ )1(~

11

21

11

2111

y

y

yRy

yy

t

++

+

VZ

t

t

t

Rhh

Rh

Rhhhh

)1(~

1

)1(

2111

22

122111

++

+++

t

t

t

R

y

y

y

yRy

Ryy

11

21

11

11

2111

11

~

~)(1

++

+++

raZ

n

n

n

n

Rhh

Rh

Rhhhh

Rh

2221

11

221221

11

)1(~

1

+++

+++++

2111

11

2111

11 1~1

yy

Ry

yRyy

Ry n

n

n

++

+++

Cc tham s vt l:

( )( )tebbtbbte

tu

RrrRrS

S

Rr

RK

+++ ++

++

+

0

0

11

11

( )

++

++

++

+

eeb

tcb

i

rC

RCS

S

K

11

11

1 00

( )( )( )( )

++

++++

+++

e

t

eb

cb

tebb

tbb

etV

r

R

C

CS

Rrr

RrS

rRZ

11

11

1 00

( )

+

++

+++

+

++

e

n

eb

cb

enbb

nbb

ner

r

R

C

CS

rRr

RrS

RrZ

11

1

11

1

0

0

0


50/408

51

Bng 2.4 Cc tham s ca s b khuch i mc baz chung.

MchBC

Tham sh Tham sy

uK ttBB

tB

Rh

h

RRhhRh

RRhh

11

21

2211

2221

~)((

)(

+++

ttBt

RyRRyy

Ry21

22

21

~)(1 +++

iK

21

21

2221

21

1~

~)(1

h

h

RRhhh

hh

Bt

+

+++

)~

)(

2111

21

2111

2221

yy

y

RRyyy

yy

Bt

+

+++

VZ

21

11

122221

221122

1~

)1)]((1[

)1()((

hRh

hRRhhh

hRhRhh

B

tB

Bt

++

+++++++

2111

11

1211222111

121122

1~

~))](([

)(1)(

yyRy

yyRRyyyy

yyRRyRy

B

tB

BtB

++

+++++++

raZ nB

nB

RRhh

hRRh

+++++

(

)1(

22

2111 )(

)(1

22

211111

nB

nB

RRyy

yyRRy

+++++

Tham s vt l:

( )

( )

( )

( ) ( )( )

0

0

0

0

0

V

0

0

0

R

0

1;

11

1

11 1

1

11

11 1

1

1

1 2

tu

bb B b c t e

bb e b e e

i

b c t

b e e

bb b c t

bb e b e eBe

b c t

b e bb e

bb B e E e

e n

RK

r R C RrS

r r C r

KC R

SC r

r C RS

r r C r RZ r

C RS

C r r

r R r RrZ

r R

+

++ +

+

+ + +

++ +

+ + + + + +

+ + + +

+ + ( )( ) ( )0

11

1 12

B n

bb B e E

B b c

b e

R RSr R r R

R CS

C

+++ + + +

+ + +


51/408

52

2.2.3. Cc cng thc n gin thng dng

Trong thc t ngay c khi thit k cc thit b in t v nht l nh gi

nh tnh cc s , ngi ta s dng cc cng thc sau:

a) Vi mch emit chung

H s khuch i in p:

ttu SRRyK = 21 (2.17)

H s khuch i dng in:

oiy

yK =

11

21 (2.18)

in tr vo:

beV ry

Z 22

1 (2.19)

in tr ra:

cera ry

Z =22

1 (2.20)

b) Vi mch baz chung:

H s khuch i in p:

ttu SRRyK = 21 (2.21)

H s khuch i dng in:

11121

21 +

yy

yK

i

(2.22)

in tr vo:

Syyy

RyRV

111

212111

11 +

+= (2.23)

in tr ra:

22

1

yRra =

c)Vi mch emit chung:

H s khuch i in p:

1)(1

)(

2111

2111 =++

+=

t

tu

Ryy

RyyK (2.24)

H s khuch i dng in:


52/408

53

)1()1(11

21 +=+= oiy

yK (2.25)

in tr vo ca mch cng tng t nh in tr vo ca mch emito

chung c hi tip m su.

in tr ra ca mch rt nh v ph thuc vo in tr ngun tn hiu, cctham s ca tranzistor.

2.2.4. Nhn xt chung:

Trong cc s b khuch i dng mt tranzistor lng cc, s mc

emito chung hay c s dng nht. Nh c h s khuch i in p, dng in

ln, in tr ra v in tr vo t kh cao. Tuy nhin do hi tip qua 12y nn

n nh ca s khng cao, nht l siu cao tn.

S baz chung c h s khuch i in p, tr khng ra gn ging s

emito chung nhng h s khuch i dng (xp x bng 1) v tr khng vo qu

nh (dng vo chnh l dng emito). Tuy vy n c n nh cao.

S colecto chung c h s khuch i in p nh (coi nh hi tip m

ton phn). in tr ra nh nht trong ba loi s . N c s dng phi hp

vi cc ph ti c tr khng nh.

Tu thuc vo yu cu ca tng mch m ta s dng loi mc v loi

tranzistor no. Tuy vy, khi s dng s dng mt tranzistor khng bao gi

trnh c cc nhc im. Ngi ta s dng s b khuch i dng nhiu

tranzistor khc phc cc nhc im ny.

2.3. Cc s b khuch i dng tranzistor hiu

ng trng (FET)

Hin nay cc bn dn trng (FET) c s dng rt rng ri. Ngi ta s

dng chng trong cc mch c lp, hoc l cc tng u tin ca cc IC . S mc thng dng nht i vi cc bn dn trng l s cc ngun (S) chung v

cc mng (D) chung. (Hnh 2.4).


53/408

54

)a )b

VU VU

1C 1

C

2C

2C

1R

1R 2R

2R

tR

3C

rU

rU

Hnh 2.4. B khuch i dng FET;a) S chung; b) D chung.

Ch lm vic ca bn dn trng c xc nh t c im ca bn dn

trng c dng u vo rt nh (coi bng 0). Thng thng, in p ban u (im

lm vic)gsU do in tr mc ti cc ngun xc nh. Trong mt s trng hp

ngi ta c th s dng mch phn p u vo. Vic tnh ton cc tham s ca s

trn c s phn tch cc s tng ng ca chng.Hnh 2.5a l s thay th tng ng ca b khuch i dng FET mc

theo s cc ngun chung (SC). Trong s ny chng ta phi lu nh hng

in dung ca tr khng ph ti. V h dn ca tranzistor trng rt nh hn so vi

ca bn dn lng cc nn in tr ph tit

R ca chng ln hn nhiu v dn ti

nh hng ca in dung ph ti cng tng cao.

Hnh 2.5. S tng ng ca b khuch i dng FET;

a) S chung

b) D chung


54/408

55

H s khuch i in p:

t

uo

L

uou

j

K

RCj

K

U

UK

/111

2

+=

+=

&

&& (2.26)

Gi tr h s khuch i in p vng tn s thp:

mtuo gRK = (2.27)

Vi:dstt rRR //=

Tn s gii hn xc nh khi h s khuch i in p gim 3 dB:

tt

tCR

=1

(2.28)

Vi:gddstt CCCC ++=

(Gi thit rng 1>>uoK ).

Vi s b khuch i mc cc mng chung (ti cc ngun) ta c s

tng ng trn hnh 2.5b.

H s khuch i in p:

2

1

1

2

1

1

.

j

j

KU

UK uou

+

+==

&& (2.29)

Vi gi thitdst rR


55/408

56

vng tn s thp (m

ragsmg

RCg1

), =>> .

Hnh 2.6 gii thiu mt s b khuch i s dng FET mc theo s

cc mng chung. Mch ny c s dng u vo cc xil.

in tr 2R , t in 1C v cc it 21 , DD chng qu ti; 54 ; RR xc

nh im lm vic. in tr 1R c nh hng n in tr vo ca mch.

Mch ny cho dng in u vo c A1410 v PFCvao 4,0= .

1C

2C

1D

2D

Vao

ra

1R

2R 3R

5R

4R

ccU+

ccU Hnh 2.6. B khuch i s dng u vo cc xil.

2.4. cc b khuch i dng nhiu tranzistor.Trong nhiu trng hp, gii quyt mt yu cu c th no , v d tng

h s khuch i dng in, tng n nh s hoc tng tr khng vo, ngi

ta c th s dng nhiu tranzistor trong mt tng khuch i hoc s dng kt hp

nhiu cch ni tranzistor vi nhau. Di y ta s xt mt s s c th.

2.4.1. S Dalington:

t c h s khuch i dng ln v in tr u vo cao, ngi ta s

dng hai tranzistor ni vi nhau nh hnh 2.7. S ny c gi l s

Dalington. n gin ta gi thit h s khuch i dng mt chiu ca cc

tranzistor ging nhau, cng tng t nh h s khuch i dng xoay chiu tn

hiu nh bng nhau. Trn quan im mch in, cc tranzistor ny tng ng

nh mt tranzistor.


56/408

57

T1T

2T

I1BI

( )[ ] BC II 211 1 ++=( ) BI211 +

( )BI11 + ( )( ) BE II 21 11 +=

E

B

C

B

E

C

E

1BBr 1B EB

Z

1EBC 1

CBC

( )21

1

1

+

E

EB

r

U

1E 12 BBr 2B( )( )21 11 ++Er

( )11 +Er

22 CBC

22 EBC

E

EB

r

U221

2

Er

( )22

1/ +E

r( )1

21

+Er

( )( )211

11

++Er

)a

)b

+

Hnh 2.7.S Dalington v mch tng ng ca chng.

Dng emit ca tranzistor t

ng

ng

c k hiu EI , cn dng emitca 1T:

2

11 +

= EEI

I (2.33)

Dng Emito ca 2T:

EE II =2 .

H s khuch i dng ngn mch ban u (tnh ):

2121121 )1( ++==h (2.34)

Gn bng tch hai h s khuch i; Trong khi in tr vo tnh :

)1(2)1(2)1(2)1(

)1()1()1(

221

2

221

1

212

1

//

+=+++++

++++++=

E

E

T

E

T

BBBB

E

T

BB

E

T

BB

rI

U

I

Urr

T

Ur

I

Urh

(2.35)

Nhvy in tr vo tnh tng ln hai ln.H dn ca tranzistor tng ng:

ET

E

rU

I

h

hy

22.

111

21

21

=

+==

Ch bng mt na h dn ca tranzistor c dng emito l EI .


57/408

58

vng tn s cao (hnh 2.7b), cc thnh phn in dung trong cc mch

1EBZ v 2EBZ to ra s ph thuc tn s ca h s khuch i. Nu coi tn s gii

hn ca cc tranzistor ging nhau, th tn s gii hn ca tranzistor tng ng

ph thuc vo tn s gii hn ca cc tranzistor ring bit. Khi ta xt vng tn

s thp v tn hiu nh ta bit EBC ch yu do in dung mch ghp emito -baz, in dung ny khng ph thuc vo dng emit. Do vy tn s gii hn ca

tranzistor11 ( )B

T f nh hn )1( 2+ ln tn s gii hn ca 22 ( )BT f . Tn s gii hn

ca mch Dalington do tn s gii hn 1Bf quyt nh.

Tuy vy mch Darlington cng c mt s nhc im sau:

-S Darlington c dng in d ln hn s dng mt tranzistor v

dng in dca 1T c 2T khuch i ln.

-Cc in p baz- emit ca hai tranzistor trong mch Darlington mc lintip nhau. V vy dng emito ca 1T ph thuc vo h s khuch i dng ca 2T,

in th mt chiu baz - emito cng nh mc tri ca in p ny ln gp i so

vi trng hp dng mt tranzistor.

-H s tp m ca b khuch i Darlington trong trng hp in tr

ngun c chn m bo h s tp m tt nht cng ln hn h s tp ca

mt tranzistor.

H s tp ca tng th hai kh ln, trong khi h s khuch i cng sut

ca tng u li nh. V vy h s tp ca n kh ln. khc phc mt phn cc nhc im ny, ngi ta dng cc s

Darlington theo s (2.8a;b). Trong s (2.8a), 1R cn la chn in th

trn baz - emito khong V)7,05,0( v dng chy qua n khong 2 ln

nh hn dng baz ca 2T. Khi dng emit ca 1T in p baz - emito v s

ph thuc nhit ca n gn nh khng ph thuc vo dng emit ca 2T. Ta

cng c th chn in tr ngun cho 2T vi mc ch gim nh tp m.

Nu s dng ngun dng (kt hp 1R ) th ta s c hiu qu ln hnnhiu.(Hnh 2.8b).


58/408

59

1R1R

B

C

E

E

C

B1T

2T 2T

1T

3T

)a )b Hnh 2.8. Cc s Dalington nhm khc phc hin tng tri v h s

tp m ln.2.4.2. S tranzistor b

Trong mt s mch in cn c h s khuch i dng v dng in ln,

ng thi s dng cch mc ni cc tranzistor tng ng khc loi n-p-n vp-

n-p. V d nh cc mch khuch i cng sut, ngi ta s dng cc mch

tranzistor b. (Hnh 2.9 a;b).Nu coi 1T c cc dng in: ( )1 11 ,+ B B B I , I I th 2T c cc dng tng

ng:

BE

BC

BB

II

II

II

122

212

12

)1(

+=

=

=

B

C

E

B

E

C CC

B B

EE

)a )b

Hnh 2.9. S Dalington b

Cc dng in ca tranzistor tng ng:

1BB II = (2.36)

BC II 12 )1( += (2.37)[ ]1 2 2 11 (1 ) . E E C B I I I I = + = + + (2.38)

Cc tham s tng ng ca chng:

- H s khuch i dng ca tranzistor tng ng:

1221 )1( +==h (2.39)


59/408

60

- H dn ca tranzistor tng ng:

Ery

21 (2.40)

- in dn ra:

Erh

122 2

(2.41)

- in tr vo:

)1(111

++= EBB rrh (2.42)

2.4.3. S Kaskode

kt hp u im ca hai loi s mc emito chung v baz chung

ngi ta s dng s Kaskode. Tng khuch i ny dng hai tranzistor:

tranzistor u mcEC, ci sau mcBC theo s hnh 2.10.

1R

2R

CR

1T

2T

ccU+

)a )b

BC

raU

raU

vaoU

1R

2R

4R

3R

5R

6R

7R

1C 2C

raU

ccU+

vaoU

1T

2T

Hnh 2.10. S Kaskode- a)Dng tranzitorcng loi; b) Dng tranzitor b

Khi dng s Kaskode b vi tranzitor khc loi (hnh 2.10b), c th h

thp gi tr ca ngun U BccB so vi s dng tranzitor cng loi (hnh 2.10a)

V bn dn 2T mc theo s baz chung, do vy theo cng thc (2.23) ta

c:

2

2

1)(

STRV =

Ph ti ca tranzistor th nht bao gm in tr ra ca bn thn1

T v in

tr )( 2TRV v cc phn t khc mc ngoi (nu c). Gi tr )( 2TRV rt nh so vi


60/408

61

cc gi tr khc, do vy n quyt nh in tr ph ti ca 1T. H s khuch i

in p ca tng th nht:

11

.)(2

12101 ==S

STRSK V (2.43)

iu ny c ngha l tng th nht khng khuch i in p. N ch khuch

i dng in v m bo in tr vo ln nhmch emito chung thng thng.

Cn 2T lm nhim v khuch i in p:

7202 RSK (2.44)

V tt nhin n khng khuch i dng in. H s tp m ca mch tng

ng nh ca mt tranzistor.

V tnh cht n nh, ta thy mch Kaskode c in dung Miller trong mch

vo 1T nh (v 101 K ). Do vy tn s gii hn trn ca n (khi nR nh) cng bc

vi tn s gii hn trn ca mch baz chung, trong khi tr khng vo li ln

hn tr khng vo ca mch baz chung. l u im c bn ca mch

Kaskode, c ngha l n cho php lm vic tn s cao vi n nh ln.

2.4.4. Mch kt hp FET - tranzistor lng cc.

tn dng tnh cht u vit ca FET l tr khng vo rt ln (Dng in

vo rt nh) ngi ta thng s dng chng trong cc mch u vo, theo s

cc ngun (Source) chung hoc cc mng (Drain) chung.

ccU+

3C

5R

1R

2R

)a

3C

ccU+

3R

2R1R

2T

1C

2C

4C4R

6R

7R

4R

3R

1C

2C

6R

5R

)b

Hnh 2.11. S khuch i dng tranzitor kt hp.


61/408

62

b vo h s khuch i ca FET nh, ta mc tranzistor lng cc c h

dn ln u ra ca FET. Cc s (2.11) gii thiu mt s s kt hp ny.

Trn s (2.11a) nu tng u vo s dng tranzistor trng knh n th

it ca - knh nhn in th khong V1 . in th ny c th c hnh thnh

nh sau: Dng cc mng DI to ra trn in tr 4R in p V5 . Nh b chia p

21 ;RR v qua 3R a vo cng in th V4 . Do in tr u vo ca FET rt ln

nn thc t khng c dng chy qua3R .

H s khuch i in p ca tng FET:

1

1

011 Sm

tm

Zg

RgK

+ (2.45)

y:1 5 2

1 4 3

// //

//( )

=

= t ds BE

S C

R R r R

Z R jX

S (2.11b) c s dng hi tip m su qua cc mch625 ;; RRR . Nh

mch 44 CR c th tng in p h trn in tr 3R v nh vy lm cho h s

khuch i ca tranzistor trng t c ln hn. C th tnh h s khuch i

theo biu thc:

26

5

0//1 RR

R

K += (2.46)

2.5. b khuch i vi sai.

Trong cc b khuch i tn hiu xoay chiu, ngi ta khng quan tm n

hin tng tri, v qua phn t ghp in dung, tri khng c a n u ra.

Tri ch lm thay i h s khuch i ca mch. nh hng ny c th khc

phc c bng hi tip m.

Ngc li, trong cc b khuch i tn hiu mt chiu, tri cng c

khuch i v a n u ra nh tn hiu. V vy trong trng hp ny phi tm

cch gim tri. Trong thc t khng th tc ng trc tip vo tranzistor gim

tri c, nn ngi ta dng b khuch i vi sai. B khuch i vi sai khuch i


62/408

63

hiu hai in p vo ca tranzitor. Do b khuch i vi sai c mc tri rt thp.

Trng hp mch hon ton i xng th tri c kh hon ton. pht huy u

im ngi ta khng nhng dng b khuch i vi sai khuch i hiu hai

in p m cn dng khuch i mt in p. in p c a n mt u

vo, u vo th hai

c ni t.2.5.1. S v nguyn l lm vic:

S b khuch i vi sai dng tranzitor lng cc v FET c trnh by

trn hnh2.12.

Hnh 2.12. B khuch i vi sai dng tranzitor lng cc (a) v

tranzitor trng FET (b)

B khuch vi sai l mt b khuch i tn hiu mt chiu i xng, c hai

u vo v hai u ra. iu c bit ng ch l b khuch i ny, nu inp vo hiu UBdB= U Bv1B-UB v2B c khuch i ln K BudB ln th cc in p ging nhau

hai u vo, gi l in p vo ng pha U BcmB, ch c khuch i ln K BcmB ln, vi

KBcmB


63/408

64

12

22

v cm d

v cm d

U U U

U U U

= +

=

a) Xt trng hp khi ch c in p hiu hai u vo

0du v UB

cmB

= 0; khi 1 2 2v v dU U U= = .V gi thit mch hon ton i xng nn in th emit (im P hnh 2.12)

lun lun khng i. Khi c th quy v s tng ng hnh 2.13b l s

emit chung hoc ngun chung c in p vo trn mi tranzitor l / 2dU .

tnh h s khuch i K BuB,KBi Bv tr khng vo ZBv B, tr khng ra Z BrB c th p dng

cc biu thc ca s emit chung v Source chung bit (xem bng 2.5)

Ring i vi h s khuch i hiu:

2 1 rdr rud

d d

UU UK

U U

= = (2.49)

vi gi thit U BcmB =0

tn s thp cng nh tn s cao KBud B u c gi tr ging nh mch emit

chung hoc Source chung. Trng hp cn ly tn hiu trn mt u ra so vi t,

ta c h s khuch i i vi mt u ra:

11 22 2

rd ud rud ud

d d

U KUK K U U= = = = bng mt na so vi khi ly in p ra i

xng.

b) Xt trng hp ngc li: 0cm

U v 0d

U = :

Khi ch c in p vo ng pha U BcmB= U Bv1 B=U Bv2 B, th mch lm vic ch

khuch i tn hiu ng pha. Lc ny c hai tranzitor u c iu khin bi

mt in p c bin v pha nh nhau. Do mch i xng, nn dng in trn

cc cc tng ng ca tranzitor bng nhau. V c th quy v s tng nghnh 2.13a. V in th emit ca tranzitor bng nhau, nn trong s tng

ng khng v dy ni hai emit. Tch s tng ng thnh hai na i

xng, mi na tng ng vi mt mch emit chung c in tr emit l 2R BEB


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65

hoc mt mch source chung c in tr ngun l 2R BsB v c hi tip m dng

in.

ccU+ccU+

CCU

0CC EU I R +

2vU1v

U2v

U 2vU

1Ur1rU 2rU

2rU

CR CR CR CR

2 ER 2 ER

cmU

dU)a )b

Hnh 2.13. S tng ng ca b khuch i vi sai dng tranzistor lng cc.

a) i vi in p vo ng pha; b) i vi in p vo hiu

H s khuch i tn hiu ng pha K BcmB c th suy ra t biu thc c i

vi s emit chung v source chung c hi tip m dng in. in tr hi

tip 2RBEB hoc 2RBSB cng ln th KBcmB cng nh. Nu cc in tr ny rt ln (tng

ng UBEEB cng rt m) th dng emit ca hai tranzistor hu nh khng i trong

qu trnh lm vic, ngha l trn u ra in p hu nh khng i, do K BcmB tin

ti khng.

Tm li s khc nhau c bn gia ch khuch i tn hiu hiu v ch

khuch i tn hiu ng pha l ch: ch khuch i tn hiu hiu, RBEB v RBSB

khng c tc dng hi tip m; ngc li ch khuch i tn hiu ng pha

chng c tc dng hi tip m ln lm cho h s khuch i tn hiu ng pha

gim .2.5.2 Cc tham s c bn ca b khuch i vi sai:

Cc tham s c bn ca b khuch i vi sai c ch ra theo bng 2.5


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66

Bng 2.5. Cc tham s c bn ca b khuch i vi sai

V d bng sTham

s

Tranzistor lng

cc

Tranzistor trng (FET)

Lng

cc

FET

KBud B(1)( // )

(1)C ce

n be

R r

R r

+ S(RBDB// r BdsB) 180 9

KBcmB 2(1 )+ + +C

n be E

R

R r R

1 2

D

s

SR

SR

+

14

1

4

CMMR 11 2 (2)

2

E

n be

R

R r

+ +

[ ](3)1

1 22

s sSR SR+ 400 20

rBdB 2rBbeB 2rBgsB 5k

rBcmB [ ]1

2(1 )2

be Er R+ + 21

(1 ) 22 2

s ds gs s

ds D s

SR rr R

r R R

+ + + +

1

(4 )

VC ( // )2

bci C ce

CS R r ( // )

2

gd

D ds

CS R r 450pF 22,5pF

rBrdB 2rBceB 2rBdsB 100k 100k

Trong bng 2.5: cho RBcB= RBDB=5k ;

( ) ( )10 ; 0; 50 ; 100; 2,5 ; ; 40 ; 2

E s n ce ds be gs T FET mA mAR R k R r r k r k r S S

V V= = = = = = = = = =

(1) Nu 1; ; /c ce be T E R r r U I >> > ;

(3) dsr ;

(4) B qua C BbeB hoc CBgs;B

Trong b khuch i vi sai, ngi ta cn a ra khi nim v h s nn tn

hiu ng pha G (CMRR) l t s gia h s khuch i hiu v h s khuch i

ng pha:

1( ) ( )ud

cm

KG CMRR dB

K= (2.50)


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c h s nn tn hiu ng pha ln, chn R BEB hoc RBSB ln. Tuy nhin

khng th chn RBEB, R BSB qu ln sao cho I BE.B RBEB hoc IBS.B RBSB nh hn (10 15 )V

m bo iu khin v cng sut tn hao trn in tr v iu kin v ngun cung

cp - UBEEB. V vy trong thc t, thay cho R BEB v R BSB ngi ta dng mt ngun dng

c in tr trong ln v h p trn n nh (hnh 2.14). Trn s hnh 2.14 ta c

C EI I , tr s ca n thay i c nh thay i UBBB,UBCC B v ER . Bng s ny

G c th t ti gi tr ( 60 80 )dB.

Trong thc t thng hay gp trng hp in p t vo b khuch i vi

sai gm c hai thnh phn dU v cmU , lc in p ra

2 1r ud d cm cmU K U K U = + (2.51)

Trong b khuch i vi sai, ngi ta phn bit tr khng vo hiu:/d d V Z U I =

v tr khng vo ng pha:

/cm cm cm Z U I =

T s tng ng hnh

2.13 ta thy rng ch khuch

i hiu, ngun tn hiu mc ni

tip vi cc u vo ca tranzistor,

do tr khng vo hiu tng ln

hai ln so vi tr khng vo ca

mch emit chung hoc Source

chung n gin. Ngc li, tr

khng vo ng pha gim i hai ln

so vi tr khng vo ca mch

emit chung hoc source chung c

tr khng hi tip l 2RBEB hoc 2RBSB.

Thng tr khng vo ng pha 100cm dZ Z> .

ccU+

CCU

CR CR

)a )b

1E2EI

ER

CI cor

P

BU

Hnh 2.14. B khuch i vi sai dngngun dng trong mch emit:

a) s ; b) s tng ng , rco cxc nh theo hnh 2.14a


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in tr ra ca mch chnh l in tr rBCOB nm gia colect TB3 B v t, n c

tr s ng bng tr s r BCOB ca s emit chung hoc source chung c in tr

hi tip RBEB // rBd1 B, trong rBd1 B l in tr ra khuch tn emit ca TB3 B; 1 1/r T Er U I

(i vi tranzistor lng cc) v1

1/dr S (i vi fet).

nh gi di ng ca tn hiu vo b khuch i vi sai, ta xt c tuyn

truyn t tnh ( )1 2c V V I f U U = ca n.

Khi gi thit ex

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