1 Cu 1.nh ngha v mch hi tip. Phn loi hi tip trong mch khuch i: B K l 1 M4C tch cc c hm truyn t >1. Hi tip l ghp 1 phn tn hiu t u ra ca M4C tch cc (b K) v u vo ca n thng qua 1 M4C khc gi l mch hi tip. KKht+XraXvXkXht Phn loi hi tip: a.Phn loi theo pha ca Xht vi Xv: -Hi tip dng: Xht cng pha Xv -Hi tip m: Xht ngc pha Xv Hi tip dng lm tng h s K ca mch, hi tip m lm gim h s K. b.Phn loi theo loi tn hiu hi tip Xht: -Hi tip 1 chiu: Xht l tn hiu 1 chiu -Hi tip xc: Xht l tn hiu xc c.Phn loi theo cch mc mch gia mch K v mch ht: -Theo cch mc mch u vo ca mch K: +Hi tip ni tip: u vo mch K mc ntip vi u ra mch ht. +Hi tip //: ngc li -Theo cch mc mch u ra ca mch K: +Hi tip in p: t/hiu ht a v u vo t l vi in p ra. +Hi tip dng in: t/hiu ht a v u vo t l vi dng in ra. d.Kt hp c cch mc mch u vo v u ra ca MK ta phn thnh cc trng hp: *Hi tip ni tip in p *Hi tip ni tip dng in *Hi tip // - in p *Hi tip // - dng in Cu 2.Cc phng trnh c bn ca M4C c hi tip v ngha ca n trong trng hp c hi tip m su KKht+XraXvXkXhtKnXnXv 2 '1 .htKKK K=+l h s K ca M4C c hi tip 1 .ntphtK KKK K=+l h s K ton phn (tnh c khu ghp) ngha ca trng hp hi tip m su: Khi c ht m su th hs K ca MK c ht ch f thuc vo hm truyn t ca mch ht (Kht) m ko ph thuc vo bn than hs K (K) ca MK. iu ny c 2 ngha quan trng sau: -Mun cho MK c ht m su lm vic n nh ta ch cn chn cc linh kin ca mch ht l cc lk c cht lng cao -Ch cn thay i cc lk mc trong mch ht (thay i Kht) th ta s nhn c cc mch in c kh nng gia cng tn hiu theo cc thut ton khc nhau. Cu 4.nh hng ca mch ht m n cc tnh cht b K: -Hi tip m lm tng n nh ca hs K v: Ta c hs K ton phn ca b KD khi c htip m l: 1 .ntphtK KKK K=+ Thc hin vi phn ton phn biu thc trn theo cc tham s K, Kn, Kht Sau thay, ,n n ht htK dK K dK dK K A = = A = A v chia c 2 v cho Kn ly sai s tng i. Cui cng ta c: 1 1. .1 .tp htK K KK K K K g KA A A= =+ Ta thy sai s tng i ca hs K khi c ht m l tpKKA gim g ln so vi sai s tng i ca hs K khi ko c ht m KKA -Hi tip m ni tip lm tng tr khng vo ln g ln:Z 'v vZ g =Hi tip m // lm gim tr khng vo g ln:'vvZZg=-Hi tip m dng in lm tng tr khng ra g ln:' Zr rZ g =Hi tip m in p lm tr khng ra gim g ln:'rrZZg=-nh hng ca ht m n di ng v mo phi tuyn: +Khi c ht m th t/h t trc tip vo u vo ca b KD gim i g ln so vi khi ko c ht m. iu ny c ngha l di ng ca b K cng c tng ln g ln. (Di ng l di bin i ca t/h vo m b KD vn lm vic bt vi sai s cho php). +Hi tip m lm gim t/h t vo u vo ca b K tc l lm cho mch lm vic vng tuyn tnh ca c tuyn truyn t. Do ht m lm gim mo phi tuyn. (Mo phi tuyn l do s xhin ca nhng thnh phn hi bc cao m nguyn nhn l do s phi tuyn ca ng c tuyn truyn dt ca pt tch cc ca mch) -nh hng ca ht m n c tnh tn s v c tnh ng ca b K: +Ht m lm gim tn s ghn di, lm tng tn s ghn trn => m rng di thng cho b K. +Do t/gian xc lp tx t l nghch vi ft v st nh xungA At l thun vi fd nn ht m cng lm gim t/g xc lp v lm gim st nh xung. +Ht m lm gim hs K ca mch 3 Cu 5.Mch cung cp cho collector v cho base ca tng K dng transistor lng cc: a. Cung cp cho Collector: L t vo C 1 in p 1 chiu dng so vi E nu l T npn, v m so vi E nu l pnp Cung cp cho C c thc hin thng qua 1 st p trn in tr Rc nh hnh v: 00CC CECCU URI=Trong cc mch in c nhiu tng s dng T th thng cng ch c 1 ngun 1 chiu cp ngun cho tt c cc tng. Khi cc tng c mc // vi ngun 1 chiu. b. Cung cp cho Base: Cung cp cho B cng thng c ly t ngun cung cho C. Cung cp cho B c th thc hin theo 2 pp l pp nh dng B v nh p B. -Phng php nh dng B: Ta c 1cc BEoBoU UIR=Do BEoU 1ccBoUI constR~ =Phng php nh dng B lm dng 1 chiu BoI ko i. -Phng php nh p B: T1 !PNPT2 !NPNRcRc-Ucc+UccT1 !NPNRC2C1R1UvUr+Ucc4 1 1 BEo ccU U R I = Thng thng: 1 1 2221,BoccUI IRI I IR ~ =+Phng php ny lm cho BEoU n nh. Trong cc s trn, cc t C1, C2 c t/dng thng xc (tn hiu cn K) v chn 1c ko lm ch tnh ca tng ny nh hng ti tng pha trc v sau n. Cu 6.Hin tng tri im lm vic: T khi lm vic trong min tch cc th dng Ic ph thuc cc tham s:, ,CBo BEI U |( ) , ,c CBo BEI f I U | = . CBoI l dng in ngc chy qua mt ghp BC M, ,CBo BEI U |ph thuc nhit , v vy Ic cng b thay i theo nhit dn n im lm vic b thay i khi nhit thay i. Hin tng ny gi l hin tng tri im lm vic or tri nhit.Khi im lm vic b thay i th cc ch tiu tham s kthut ca b k b gim st. V vy cn phi n nh im lm vic cho b k, tc l n nh dng 1 chiu cho Ico. n nh im lm vic ta phi p dng cc bin php n nh dng Ico. Trong thc t c 2 phng php n nh dng Ic l pp n nh tuyn tnh v phi tuyn. T1 !NPNRcC1R2R1C2+UccUvUraIBoI1I25 Cu 7.Cc s n nh tuyn tnh, phi tuyn cho tng K: *Trong cc s tuyn tnh th im lm vic c n nh bng hi tip m 1 chiu: a.Cc s n nh bng hi tip m in p 1 chiu: S E chung S B chung: R1, R2: phn t hi tip (m, in p) C1,C2 l cc t ghp tng (thng t/h c ch xc, chn 1c) Cb: thng xc, chn 1c -> ni t B v mt xc. Lch: cun chn: thng 1c, chn xc tn s cao, chn t/h xc c ch. Quy lut n nh dng Ic ca c 2 s : | | | | | | | Transistor ong bt Ioc Rc C B BE Ct I U U U U T1 !NPNRcC2C1R1R2+UccUvUraT1 !NPNR2C1CbC2R1RcLch+UccUv Ura6 b.Cc s n nh bng hi tip m dng in 1 chiu S e chung S C chung S B chung: T1 !NPNC2CeRe R2R1RcC1+UccUvUraT1 !NPNReR2R1CC1+UccUvUra7 Trong 3 s , Re l phn t hi tip. Quy lut n nh dng Ic ca c 3 s :| || | | | Transistor ong bt Ioc e e BE Ct I I U U *Trong cc s phi tuyn, ngi ta dng cc phn t phi tuyn c tham s thay i theo nhit b li nh hng ca nhit n cc tham s ca Transistor (cc phn t phi tuyn: diode, in tr nhit) a.S n nh bng diode: S a: S b: T1 !NPNC1 C2R2CbR1RcRe+UccUvUraT1 !NPNReR3RcC2R1R2C1D+UccUvUra-Ucc8 Xt s a: Chn T v D cng c sxut t 1 loi bn dn(mt ghp BE ca T ging ht PN ca diode) => S thay i ca Ube theo nhit ging ht s thay i ca in p thng Diode l UD theo nhit Do Ube ngc chiu vi Ud nn s thay i ca Ube theo nhit s c b hon ton bi s thay i ca Ud => Ube c n nh => dng Ic n nh => im lm vic n nh.Ngun 1 chiu Ucc v R3 c nhim v phn cc cho diode thng (pcc thun) Xt s b: c tc dng n nh Ube ging s A, ngoi ra diode D cn c tc dng n nh Ube khi ngun cung cp Ucc b thay i. Ud thng ~ 0.6V (silic) = const Nu chn R2 nh -> c th coi Ube~ Ud thng = const -> ko ph thuc Ucc. Thc t c th thay R2 bng 1 cun chn Lch b. S n nh bng in tr nhit T1 !NPNR2RcC2R1ReC1D+UccUvUraT1 !NPNR2RcC2R1ReC1NTC+UccUvUra RT9 RT l in tr nhit m (khi nhit tng th RT gim) Quy lut n nh dng Ic nh sau: ( ) ( ) || | | | | | Transistor ong bt IoC T RT e BE CI t R U U U Cu 8.V s nguyn l v gii thch c im (u, khuyt im) ca cc s c bn ca tng K tn hiu nh dng Transistor (photo v) Cu 9.V s nguyn l v gii thch c im ca s Darlington v Kaskode: I.S Darlington: S Darlington chun: S Darlington b T1T2BECIb=Ib1Ic1IcIc2Ib2=Ie1Ie=Ie210 Trong cc s Dattlington, thc hin ghp 2 T thnh 1 cp t c h s K dng in ln | | | | = ~ + +1 2 1 2ciBIKI c im ca s : -u im: Ki rt ln so vi s dng 1 T -Nhc im: +c dng in d ln hn s dng 1 T v dng in d ca T1 c T2 khuch i +do 2 mt ghp B-E ca 2 T mc ni tip nn= +1 2 BE BE BEU U U , v vy mc tri in p BEUcng ln gp i so vi s dng 1 T. II.S Kaskode: T1T2BECT2T1BEC11 Gi s T1 v T2 ging nhau: h dn= =1 2s s sTa c hs K in p tng 1 mc E chung:( )= = =21 1'. .rau C v TvUK s R s ZU Do T2 l tng mc B chung nn ( )2v TZ = = =21 1drs s V vy=11uKH s K in p tng 2 mc B chung: = =2.'rau craUK s RU ->h s K in p ca mch Kaskode:= = =1 2 2. .u u u u cK K K K s RH s K dng in ca mch Kaskode:| = = =1 2 1.i i i iK K K K (do 2 iK =1) Tr khng vo: ( )= =1v be v TZ Z rTr khng ra bng tr khng ra ca tng T2 mc B chung: ( )= =2/ /ra c ce ra TZ Z R rNhn xt:Do tng T1 c=11uK(nh) -> in dung vo tng 1:= +1 1 1 1.v be u bcC C K Cnh -> tn s gii hn ca khu lc thng thp u vo tng T1 ln -> lm vic c tn s cao. M tng T2 l s B chung cng lm vic c tn s cao, do mch Kaskode lm vic c tn s cao. Mch Kaskode thc hin K in p tng mc B chung v K dng in tng mc E chung Cu 10. V s nguyn l v gii thch nguyn l lm vic ca b K vi sai. Ti sao b K vi sai li khc phc c hin tng tri (photo v) T1ReCeT2RcR2R1CbUv+UccUraU'ra12 Cu 11.Trnh by cc loi mch ghp gia cc tng: Mch ghp c nhim v truyn t nng lng ca t/h t tng pha trc -> tng pha sau sao cho hiu sut t cao nht do mc tn hiu u vo ca tng trc c th ko ph hp vi mc t/hiu vo tng sau nn mch ghp cn c nhim v dch mc tn hiu phi hp gia mc t/h ra ca tng trc vi mc t/h vo ca tng sau. I.Ghp trc tip: L ni trc tip u ra ca tng trc vi u vo tng sau: u im: n gin hiu sut cao Nhc im: in p tri nhit ca tng trc c a ti tng sau v c tng sau K ln. V vy mc tri ln Ch tnh ca cc tng lien quan n nhau. V vy vic tnh ton ch tnh tha mn cho tt c cc tng l rt phc tp. ng dng: trong cc mch t hp IC II.Ghp gin tip: a.Ghp in tr: T1 !NPNT2 !NPNR1 1kR2 1kUvUra+UccT1Rc1Rc2R1R2T2UvUra+Ucc13 R1,R2 l mch ghp -u im: To ra c 1 mc dch in p phi hp gia mc tn hiu u ra ca tng trc vi mc tn hiu vo ca tng sau. -Nhc im: +Mc tri ln +Ch tnh ca cc tng lien quan -> tnh ton phc tp +in dung vo ca tng sau kt hp vi cc in tr ghp to thnh mch lc thng thp RC lm gii hn tn s lm vic. +C tn hao nhit trn cc in tr R1, R2 -ng dng: c ng dng trong cc mch t hp IC v trong cc mch in lm vic tn s thp hoc vi tn hiu 1 chiu. b.Ghp in dung: *u im: -in p tri tng trc ko c a n tng sau -Ch tnh ca cc tng c lp nn d tnh ton. *Nhc im: -Khi khuch i tn hiu tn s thp th t in c kch thc ln, cng knh. *ng dng: -c ng dng trong cc mch ri lm vic di tn s cao. c.Ghp in cm (ghp bin p) T1Rc1Rc2T2CUvUra+Ucc14 *u im: -in p tri tng trc ko c a n tng sau -Ch tnh ca cc tng c lp -D phi hp tr khng *Nhc im: -Bin p c kch thc ln v di tn lm vic hp *ng dng: -Dng ghp trong cc mch ri lm vic vi tn hiu xc vi di tn lm vic ko qu rng. Cu 12.Trnh by khi nim v b khuch i thut ton. Cc tham s c bn ca b KTT. -B K thut ton cng ging nh cc b K thng, khc ch n cng thc hin K tn hiu (in p, dng in, cng sut) -Tuy nhin, nu nh cc b K thng thng c cc t/c v tham s ph thuc vo cu trc ca mch in th cc mch in s dng K thut ton, t/c ca mch in li ko ph thuc vo bn thn cu trc bn trong ca b K thut ton. M chph thuc vo cc linh kin mc mch ni tip bn ngoi ca b KTT.V vy ch cn thay i cc linh kin mc bn ngoi ta c th nhn c cc mch in c chc nng x l tn hiu theo cc thut ton khc nhau. B KTT c 2 u vo v 1 u ra, ngoi ra n cn c cc chn cp ngun (dng, m, ni t) T1T2L LRc2 1kUvUra+ + +-+ +PN+Ucc-UccIraNIPI|dU|NU|PU|raU15 b tn s v chnh lch 0 th b KTT cn c cc chn mc cho mch b ny. B KTT K hiu qu 2 in p t vo 2 u vo vi h s K l Ko>0 ( ) = = .ra o d o p NU KU K U U+Nu= 0NU =ra o PU K U =>in p ra ng pha vi in p vo ca P nn ca vo P c gi l ca vo thun. K hiu bi du (+) +Nu= 0PU = ra o NU K U in p ra ngc pha vi in p vo ca N nn ca vo N c gi l ca vo o. K hiu bi du (-) -B KTT l tng c cc tnh cht sau: +H s K rt ln: =oK+Tr khng vo rt ln:= vZ+Tr khng ra rt nh:= 0rZ*Cc tham s c bn ca KTT: -H s K hiu: =rhUKU = h P NU U U-c tnh tn s:=+ + +1 2 31 1 1. . ...1 1 1oK Kf f fj j jf f f -H s nn ng pha: = = = = / / 0 0P N Cm raU U U UA= A0cmrcm hU KG U constK Uor < 0 ngha ca G: cho chng ta bit cn phi t vo u vo mt in p bng bao nhiu kh c hin tng khuch i ng pha. -in tr vo hiu, in tr vo ng pha, in tr ra: ==A A= A A00||NPPUPVhNUNUIRUI 16 A=AcmVcmcmURI (cm l ng pha, vcm l vo ng pha) A=ArrarURI -Dng vo tnh, dng vo lch 0 v in p lch 0 +Dng vo tnh: +=2P NtI II+Dng vo lch 0:= o P NI I I+in p vo lch 0:= o P NU U Ukhi= 0raU Cu 15.Mch cng, tr dng khuch i thut ton: 1.Mch cng o: Ta c~ = = 0 0d N PU U UVit phng trnh dng in cho nt N: + + + + =1 21 2... 0V V Vn ran NU U U UR R R R | | = + + + | |\ .1 21 2...V V Vnra NnU U UU RR R R Nu chn= = = =1 2...nR R R R( ) = + + +1 2...Nra V V VnRU U U UR NX: mch cng o c hs K ln th ta phi chn cc in tr 1 2, ,...,nRR R nh. Nhng khi tr khng vo=1 2/ / / /... / / / /V n NZ R R R R s nh (nhc im) 2.Mch tr:-+ +R2R1RnRNUv1Uv2UvnNPUra17 Vit phng trnh dng in qua nt N: o + =110/v N ra NN NU U U UR R( )oo+ =+111NN V raNU UUVit phng trnh dng in cho nt P: ( )oo o = =+2 20.12/V p ppp vp p p pU U UUUR R M=N PU U , t (1) v (2) suy ra:( )oo oo=++2 111. . . *NP v ra v NPU U UVi mch in ny phi tnh n nh hng ca hin tng K in p ng pha vi in p vo ng pha: = =1 2 cm v vU U UT (*) suy ra h s K ng pha l: oo oo += +11Ncm P NPKt o o o o o = = +A ,N Pta c: ( )o oo o oo o o+ A= + A ~+ A + +11 1cmK (vo Anh) Ta lun mong mun Kcm cng nh cng tt. Kcm = 0 tho A =0 hay o o o = =N P Khi t (*) suy ra ( )o = 2 1 ra V VU U U Cu 16: Trnh by mch tch phn vi phn dng KTT: 1.Mch tch phn o: -+ +RN R1RpR21 VU2 VUraUo=ppR18 Nt N:+ = 0v cI I( )+ = . 0ra Nv NdU UU UCR dt ( ) + = 0ravdUUCR dt(v= = 0N PU U ) = = 1raUvRCU dt2.Mch vi phn: V= = 0N PU U -> ti nt N ta c: + = 0v radU UCdt R = vradUU RCdt Cu 17: Trnh by mch K loga v i loga Cu 18: Trnh by mch nhn tng t v mch ly tha bc 2 Cu 19: Trnh by mch chia v mch khai cn -+ +RCUvUraNPIvIc-+ +RCUvUraNP19 Cu 20: Nu khi nim phn loi ca b K chn lc. Cc ch tiu cht lng ca b K chn lc: Khi nim: -B K chn lc l b K ch thc hin K vi tn hiu c tn s nm trong 1 di tn s no v nn (loi b) nhng tn hiu c tn s nm ngoi di . - thc hin c nhim v trn th ti u ra ca b K chn lc phi c tnh cht chn lc tn s. Ti c th l: khung cng hng LC c iu chnh cng hng tn s cng tc, hoc l cc mch lc. Phn loi:-Theo tn hiu c K +K chn lc tn hiu nh: lm vic ch A, gc ct u =180o +K chn lc tn hiu ln: lm vic cc ch phi tuyn nh AB, B v C Theo cch chn lc tn s ca b K: +B K chn lc n iu chun +B K chn lc mc sole +B K chn lc i iu chun Cc ch tiu cht lng ca b K chn lc: 1.Tn s lm vic danh nh fo: L tn s trung tm ca di thng b K. ofc th c nh hoc thay i c trong 1 di thngmax minf f2.H s K: o= =| | .j ravUK K eU Trong :| | K : modul ca hs K, o gc lch pha gia Ura v Uv Ti tn s=of f th= =ax| | | |m oK K KoK c gi l hs K cng hng. 3.Di thng D: L di tn s truyn qua ca b K -Di thng mc na cng sut=0,7D Dl di tn s m nu tn hiu c tn s nm ngoi di ny th khi truyn qua b K s tng ng vi hs K nh hn2ln so vi K -Ngoi ra cn dng cc khi nim di thng cc mc khc nhau nh 0,1 0,01, D D4.Di tn s lm vic:min ax mf f nh gi rng ca di tn s lm vic ngi ta s dng h s bao tn: =axminmbfKf 5. chn lc: L tham s nh gi kh nng loi b cc tn hiu ko mong mun khi tn hiu c ch chn lc c nh gi thng qua tham s h s ch nht: = >0,10,71cnDKD. cnKcng nh (cng gn 1) th chn lc cng tt. 6.Hiu qu khuch i: = .oH KD 7.Cc tham s khc nh: mo tn hiu, kh nng lm vic n nh, h s tp m 20 Cu 21: Vn n nh ca b K Cu 22: S nguyn l ca b K chn lc ti cng hng. S khi ca cc tng K chn lc n iu chun, mc so le v i iu chun. S nguyn l: R1,R2: phn p cho bazo R3: to hi tip m, dng in, 1 chiu n nh im lm vic cho mch C3: kh hi tip m xc ko lm gim hs K ca mch C1,C2: cc t ghp tng (thng t/h cn K, chn 1c) C4: t lc ngun (C4>C1,C2,C3), thng xc xung t ko lm nh hng n ngun cung cp 1c (Ucc) Co,L: khung cng hng LC c iu chnh cng hng ti tn s cng tc fo (Co l t bin dung) S khi ca tng K chn lc n iu chun: CoL1CoL1CoL11K2KnKofofof Mc so le: CoL1CoL1CoL11K1KnK 02 0f f 02 0f f i iu chun: T1R3C3C1C2L1CoC4R1R2UvUra+Ucc21 1K1KnK0fC5L2C6L3C5L2C6L3C5L2C6L30f0f0f0f0f Cu 23: So snh v h s khuch i, di thng, chn lc v hiu qu khuch i ca cc loi khuch i chn lc n iu chun, mc sole v i iu chun. n iu chunSolei iu chun H s K = =1 2 1...nchuoi nK K K K Ko| |= | |+\ .0121nK = =0 chuoi 01 02 0 01...nnK KK K K ( )oo= =++42 21 2 1 24/| |42 1 2cs R R s R RK =21 202Cs R RK= = ,m mn c on ocK K K K( )o o o= =| | ++ |\ .12 42 2| |2 42 12R R s sKn =raoPP= == =1 2 101 02 0 01.... ...nnno nK K K K KK K K K K Di thng = = A = 1/0,72 2 1n ofD D fQ =