Câu hỏi ôn tập điện tử số

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<ol><li> 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 </li><li> 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 </li><li> 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. </li><li> 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: </li><li> 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 </li><li> 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 =&gt; 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. </li><li> 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 ) =&gt; 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 =&gt; 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 l trng mang in tch dng (+) c gi tr bng in t = . Do s in t bng s l trng. s ht dn trong bn dn thun thng rt ngho nn. </li><li> 8. 8 Cu 4 : s hnh thnh mt ghp bn dn P-N. hot ng ca mt ghp di tc ng ca in trng ngoi. * s hnh thnh mt ghp bn dn P-N: - Ngi ta dng 2 thanh bn dn loi p v loi n ghp vi nhau bng mt cng ngh c bit th mi trng 2 bn mt ghp c 1 tnh cht c bit gi l mt ghp bn dn. - do tip xc nn cc ht a s ca 2 thanh bn dn s khuch tn sang nhau =&gt; p n =&gt; n p c trng cho s khuch tn ta k hiu dng khuch i c chiu t p n. Do s khuch tn s to nn cc min ion dng (+), ion m (-) 2 bn mt ghp trong min ny rt ngho nn ht dn a s v in tr ln, do cc tch t 1 in trng hng t n p gi l ETX. in trng ny c khuynh hng lm gim ng thi kch thch cc ht thiu s t n p c trng bi ( I tri ). Qu trnh din ra cho n khi = th mt ghp c hnh thnh v trn xut hin 1 in th tip xc c gi tr xc nh. * Hot ng ca mt ghp di tc ng ca in ngoi. - Khi in trng ngoi tc ng theo chiu thun ( mt ghp phn cc thun ) Dng (+) vo p v m (-) vo n + Do in trng ngoi c hng lm suy gim in trng tip xc nn hin tng khuch tn tip tc xy ra, dng khuch tn c gi l dng in s tng ln rt ln ta gi l dng in thun i qua mt ghp. ng thi in trng ngoi cng lm cho gim i nhng chng c gi tr rt nh nn din bin ny khng ng k. - Khi in trng ngoi tc ng theo chiu ngc ( mt ghp phn cc ngc) m (-) vo p v dng (+) vo n </li><li> 9. 9 - Khi phn cc ngc th I ngoi s tc ng cng chiu vi ITX lm cho gim v 0 cn tng ln l dng in ngc. Dng in ny khng th tng ln v chng c to nn bi cc ht thiu s, thng th I ngc ny rt nh. Trong nhiu trng hp ta coi n khng nh k. Cu 5 : it bn dn ( cu to, c tnh vn-ampe, cc tham s,phn loi,ng dng ca n) * Cu to - it bn dn c to bi 1 mt ghp bn dn p-n vi mc ch s dng n nh mt van in. cc in cc ni vi thanh bn dn gi l cc cc ca it + cc ni vi thanh p gi l ant + cc ni vi thanh n gi l catt * c tnh vn- ampe - c tnh V/A ca it l quan h gia dng in qua it v in p mt chiu t ln n. + c tnh vn- ampe c chia lm 3 on r rng on 1 : c tuyn thun : khi in p tc ng theo chiu thundng in tng rt nhanh. on 2: c tuyn ngc: Khi in p ngc tc ng ln it th dng in ngc tng ln nhng rt chm cho n 1 gi tr in p ngc tng ln t ngt, in p gi l in p ngc cho php. on 3: khi in p ngc ln hn in p ngc cho php th dng in ngc tng ln t ngt gi l dng nh thng it. khi in p trn it khng tng na mt s loi mt ghp khi b nh thng s b ph v hon ton, khi in p trn mt ghp gim v 0, it nh 1 dy dn. cn 1 s loi mt ghp th khi in p ngc gim i mt ghp li c phc hi v </li><li> 10. 10 ngay c khi b nh thng th cng khng b h hng in p trn n lun duy tr n nh. Ngi ta dng mt ghp loi lm it n p. K hiu it n p: Phn loi: - Theo cu to: + it tip im + it tip mt - Theo cng sut: + it cng sut nh: 1A + it cng sut ln : vi chc A + it cng sut va : 5A - Theo tn s: + it cao tn + it m tn - Theo ng dng: + it chnh lu + it tch sng + it vanicag * ng dng ca it : c ng dng trong rt nhiu lnh vc : - Chnh lu, tch sng, bin tn, chn knh (Dv), n p (Dz) * Cc tham s c bn ca it: - Dng thun cc i , l dng thun m it cn chu c khi n cha b thng (v nhit) hay cn gi l dong thun cho php - Cng sut cc i cho php trn it khi it cha b thng - in p ngc cc i cho php in p phn cc ngc cc i ca it khi cha b nh thng - Tn s hot ng (tn s gii hn ) ca it: l tn s ln nht m ti it cha mt tnh cht van (do in dung k sinh) - in p m ca it: l in p dng thun qua it t 0,1 - in dung mt ghp - in tr mt chiu c xc nh ti mt im trn c tuyn = = - in tr xoay chiu R ca it c xc nh ti 1 im trn c tuyn: R = = S = = S in dn ca it, S = tan </li><li> 11. 11 Cu 6: Cu to v nguyn l hot ng ca tranzito loi PNP * Cu to: tranzito thun - c to bi 3 thanh bn dn, ghp xen k vi nhau to thnh 2 mt ghp lin tip theo th t sp xp: p-n-p - Thanh bn dn th nht (p) c kch thc v nng hp cht ln nht c gi l min Emit hay min pht. in cc ni vi n c gi l cc Emito (E) - Thanh bn dn gia khc loi vi 2 thanh bn (n) , c kch thc v nng tp cht nh nht gi l min baz hay min gc. in cc ni vi n gi l cc baz hay cc gc (B) - Thanh bn dn th 3 (p) cng loi vi thanh bn dn th nht c kch thc v nng tp cht mc trung bnh gi l min colecto hay min gp. in cc ni vi min ny gi l in cc colecto hay cc gp (C) * Hot ng: - tranzito hot ng ta phi t cc in p 1 chiu vo cc chn ca tranzito + tranzito lm vic ch khuch i th phi tha mn phi c phn cc thun v phi c phn cc ngc. </li><li> 12. 12 - Do in trng tc ng trn l phn cc thun nn cc ng t chnh t min E l cc in t t do s tro sang min B to thnh dng l dng cc pht - Khi sang min B th mt s t ti hp vi ng t chnh ca min B to thnh dng in . do min B c kch thc nh v nng ht dn cng nh nn dng rt nh cn a s cc ht c chuyn t E qua B s n im tip gip lc n ng vai tr ht thiu s ca min B nn in trng ngc ca vn qua min C to thnh dng cc gp Nh vy ta c : = + nh gi s truyn t dng ngi ta dng h s truyn t dng = v nh gi h s khuch i = = &lt; 1 ( ) Cu 7: Cu to v nguyn l hot ng ca tranzito NPN * Cu to: tranzito ngc - c to bi 3 thanh bn dn, ghp xen k vi nhau to thnh 2 mt ghp lin tip theo th t sp xp n-p-n - Thanh bn dn th nht (n) c kch thc v nng tp cht ln nht c gi l min Emito hay min pht. in cc ni vi n gi l cc Emito (E) - Thanh bn dn gia khc loi vi 2 thanh bn (p) c kch thc v nng tp cht nh nht gi l min baz hay cc gc (B) - Thanh bn dn th 3 (n) cng loi vi thanh bn dn th nht, c kch t...</li></ol>