Câu hỏi ôn tập điện tử số

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  1. 1. 1 Phn ny do t copy c, cho mi ngi tham kho lun. Ai cn th nhn tin cho t (Nguyn Huy Linh nguyenhuylinha1@gmail.com) Cu 1 : Trnh by s khi v phn tch chc nng cc khi ca h thng thng tin qung b ...............................................................................................................................3 Cu 2 : cc dng tn hiu iu ch. Tn hiu iu bin v ph ca n?.............................4 Cu 3 : trnh by cu to v c ch hnh thnh ht dn trong cc loi bn dn loi n, loi p ?...............................................................................................................................6 Cu 4 : s hnh thnh mt ghp bn dn P-N. hot ng ca mt ghp di tc ng ca in trng ngoi..........................................................................................................8 Cu 5 : it bn dn ( cu to, c tnh vn-ampe, cc tham s,phn loi,ng dng ca n) .........................................................................................................................................9 Cu 6: Cu to v nguyn l hot ng ca tranzito loi PNP .......................................11 Cu 7: Cu to v nguyn l hot ng ca tranzito NPN ..............................................12 Cu 8: Cc h c tuyn tnh ca tranzisto ?...................................................................13 Cu 9: L thuyt v khuch i?(nh ngha, phn loi, cc c tuyn v tham s) ....16 Cu 10: Phn hi trong b khuch i, c im ca b khuch i khi c phn hi, cc cch mc phn hi..............................................................................................................17 Cu 11: Cc ch lm vic, cc cch cp ngun v n nh im cng tc ca tranzisto trong ch khuch i.....................................................................................19 Cu 12: Mch khuch i in tr Emito chung(EC)? (s nguyn l, tc dng ca cc linh kin, phn tch hot ng, v nh tnh c tnh bin tn s v gii thch dng ca n) .......................................................................................................................21 Cu 13: Mch khuch i in tr Colecto chung (CC), so snh mch K in tr EC v CC..................................................................................................................................22 Cu 14: c im mch khuch i cng sut; mch khuch i cng sut n c bin p ra?..................................................................................................................................24 Cu 15: Mch Khuch i cng sut y ko song song c bin p ra. ..........................25 Cu 16: Trnh by s nguyn l, nguyn l lm vic v ng dng ca mch khuch i vi sai..............................................................................................................................26 Cu 17: Mch khuch i thut ton (nh ngha, cc thng s k thut, cu trc bn trong theo s khi). .......................................................................................................27 Cu 18: Trnh by cc mch cng, tr trn khuch i thut ton. ...............................28 Cu 19: Trnh by cc mch vi phn, tch phn trn khuch i thut ton. ................29 Cu 20: Trnh by nguyn l to dao ng hnh sin dng mch khuch i c phn hi dng; phn tch iu kin cn bng bin , cn bng pha:.........................................32 Cu 21: Mch to dao ng hnh sin ghp h cm? ........................................................34 Cu 22: Mch to dao ng hnh sin kiu ba im? .......................................................34 Cu 23: To dao ng hnh sin RC trn khuch i thut ton(KTT)? ......................36 Cu 24: S khi mch ngun 1 chiu, phn tch chc nng ca cc khi? ...............38 Cu 25: Chnh lu 1 pha: Mt bn chu k, c chu k, kiu cu, so snh u nhc ca chng ..................................................................................................................................39 Cu 26: Cc mch n p mt chiu tham s.....................................................................42
  2. 2. 2 Cu 27: Mch n p mt chiu kiu b tuyn tnh, phn tch hot ng trn s khi v s nguyn l..............................................................................................................43 Cu 28 : Nguyn l n p xung .........................................................................................45
  3. 3. 3 Cu 1 : Trnh by s khi v phn tch chc nng cc khi ca h thng thng tin qung b * s khi: * chc nng cc khi : + My pht + tin tc(ngun tin tc) : mnh lnh, bi ca, hnh nh. + gia cng tin : ngun tin tc qu thit b bin i(gia cng tin) c bin i thnh tn hiu in c tn s thp(tn hiu s cp) + to sng mng : mun truyn c tn hiu s cp i cn phi c i tng truyn l 1 dao dng iu ha c tn s cao lm nhim v ti tin hoc sng mang . + iu ch : mun sng mang ti c tn hiu s cp i cn phi trn tn hiu s cp vo ti tin. Qu trnh trn, tc l qu trnh cho tn hiu s cp tc ng vo 1 tham s no ca ti tin,bt tham s phi bin thin theo quy lut ca tn hiu s cp gi l qu trnh iu ch(modulation). Sn phm ca qu trnh ny l dao ng cao tn bin iu theo dng tn hiu s cp gi l tn hiu c iu ch hoc tn hiu v tuyn in. + khuch i pht : tn hiu v tuyn in c khuch i cho ln pht vo mi trng truyn tin. Mi trng truyn tin l khng gian th thng tin l v tuyn in, mi trng l ng dy thng tin hu tuyn in (anten pht) + anten pht : ngoi tn hiu cn c cc dao ng in t khc gi l nhiu - My thu: + mch vo : tn hiu cn thu c tn s tn hiu hu ch t anten thu hoc ng dy a n mch vo loi bt nhiu.
  4. 4. 4 + khuch i cao tn : tn hiu sau khi c loi bt nhiu th vo khuch i cao tn ch khuch i khong chc ln ri a vo b trn tn. + trn tn: trn vi do ng ni b tn s (dao ng ngoi sai), ly tn s trung gian (trung tn thng = - ). + tn s trung tn l tn s n nh nn khi tn s cn thu thay i th tn s ngoi sai cng phi thay i theo. + b trn v dao ng ngoi sai lp thnh b bin tn hay i tn. v di tn s trung tn c nh nn khuch i trung tn d dng thc hin vi h s K ln v chn lc (lc nhiu) cao. Qu trnh hiu chnh tn s vo mch, mch K cao tn v mch dao ng ngoi sai din ra ng thi gi l ng chnh. Sau K trung tn tn hiu c tch sng (gii iu ch), tc l qu trnh ngc li vi qu trnh diu ch nhn c tn hiu s cp. tn hiu ny c K a n b nhn tin. Ton b cc thit b nm trn ng truyn t ngun tin n ni nhn tin lp thnh 1 knh thng tin. Cu 2 : cc dng tn hiu iu ch. Tn hiu iu bin v ph ca n? + cc dng tn hiu iu ch : - khi mun truyn tin tc i xa,do tin tc di tn s thp khng th trc tip bc x v do vic phn knh nn ngi ta phi gi tin tc trn sng mang. Vic a tin tc tc ng vo sng mang lm cho I trong tham s ca sng mang bin i theo quy lut ca hm tin tc gi l iu ch. Sng mang c dng l : = (t) l dao ng iu ha tn s cao. Tin tc S(t) = (t) = Khi iu ch : (t)= = (t) (t) + nu bin thin theo quy lut hm tin tc S(t) cn = (t) ta c tn hiu iu bin.(tn hiu b iu ch bin ) AM. + nu = cn (t) bin thin theo quy lut no ca hm tin tc S(t) ta gi y l tn hiu iu gc(tin hiu b iu ch gc) gm : Tn hiu iu tn (FM) nu tn s gc b bin thin theo quy lut hm tin tc S(t) Tn hiu iu pha (PM) nu gc pha u bin thin theo quy lut hm tin tc S(t) * tn hiu iu bin v ph ca n - tn hiu iu bin n m : Gi s = : tn tc : sng mang Ta c : (t) = [ : tn hiu iu bin = [ 1 + h. ] = [ 1 + m. ] Trong : h: hng s biu th mc thm nhp ca tin tc vo sng mang. m: l tham s. m = h. : gi l su iu ch hay l ch s iu ch tn hiu iu ch khng b mo dn : 0 m 1 th ca tn hiu iu ch:
  5. 5. 5 * Ph ca n l : (t) = [ 1 + m. ] = + m. . ] = + m . ) ] + m . ) ] Ph c dng * Tn hiu iu bin gm 3 thnh phn : thnh phn sng mang khng cha thng tin v tin tc c tn s gc , bin , 2 thnh phn bin dao ng vi tn s ( ) v i xng nhau qua l 2 thnh phn c cha thng tin ca tin tc. * Thnh phn sng mang chim nng lng ln nht nhng li khng cha thng tin cn cc phn bin th chim nng lng nh hn c 1/4 nng lng sng mang li mang theo thng tin ca tin tc do ngi ta loi b sng mang v pht i 2 bin gi l my pht iu bin
  6. 6. 6 cn bng hoc ch pht i 1 bin gi l my pht iu bin n m. Vi cc loi my pht ny va tng c li thng tin, va m bo bo mt thng tin. * Tn hiu iu bin a m: - Trng hp tn hiu s cp khng phi ch l mt tn s = 2 f , m l 1 di tn s t n (hay ) tc l (t) = th thc hin cc bin i ton hc tng t ta c biu thc ca tn hiu iu bin: = + + ) ] : ch s iu bin thnh phn = h ch s iu bin ton phn: m = vi 0 m 1 * Ph ca n s gm sng mang, mt gii bin trn l [ ] v gii bin di [ ] c dng B rng ca ph tn hiu iu bin : = ) - ) = 2 hay = 2 Nu loi b i thnh phn sng mang th s c tn hiu iu bin cn bng, cn nu loi b thm 1 bin ch cn li 1 bin th c tn hiu n bin. Cu 3 : trnh by cu to v c ch hnh thnh ht dn trong cc loi bn dn loi n, loi p ? * bn dn loi n : - nu ta trn tp cht thc nhm 5 ca bng h thng tun hon Mendelep vo mng tinh th ca bn dn thun th 1 nguyn t tp cht vi 5 nguyn t lp ngoi cng s c 4 in t tham gia lin kt vi 4 nguyn t cht bn dn thun,cn li 1 in t t do. Ngi ta trn nguyn t AS thuc nhm 5 vo trong mng tinh th ca bn dn thun ca nhm 6 vi nng c nguyn t/ . Khi to thnh nguyn t tp cht do AS c 5 in t lp ha tr ngoi cng m mi lin kt mng ch cn 4 nn 1 in t AS s tha ra, n lin kt rt lng lo vi ht nhn v d dng b ion ha ngay thng to thnh in t t do. ng thi vi n s ion ha nh ca bn dn thun cng xy ra v to thnh cc cp e v l trng => kt qu c 1 bn dn tp cht trong cha rt nhiu in t t do cn s l trng th t hn, bn dn ny c gi l bn dn loi n hay l bn dn cho in t.
  7. 7. 7 * bn dn loi p : ta trn vo mng tinh th bn dn thun nhng nguyn t nhm 3 ca h thng bng tun hon Mendeleep vi nng t nguyn t/ th trong nguyn t ca mng tinh th s thiu 1 in t lin kt , in t ny s c ly t 1 nguyn t bn dn no v li 1 l trng c to thnh 1 nguyn t bn dn th to ra 1 l trng(do nhng nguyn t nhm 3 ch c 3 in t lp ha tr ngoi cng m mi lin kt mng li cn n 4 ) => s l trng rt ln. ng thi vic ion ha cc bn dn thun cng xy ra to thnh tng cp in t v l trng => kt qu: nhiu l trng, t in t. Cht bn dn ny lun c xu hng nhn in t gi l bn dn loi p hay l bn dn nhn in t. + gin nng lng: * bn dn thun : l nhng cht thuc nhm 4 trong h thng bng tun hon Menddeleeep nh Geemani (Ge), silic (Si). Trong cu to nguyn t ca chng c 4 in t lp ngoi cng. Trong iu kin thng chng lin kt cht vi ht nhn nn nhit thp khng c in t t do nn chng cch in. khi cao hoc do 1 tc ng no s lm cho cc in t tch khi cu trc mng tinh th tr thnh in t t do v tham gia dn in lc chng tr thnh cht bn dn. Mi khi to ra 1 in t t do th ch in t ra i to thnh 1