tracnghiemktdt3

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By Kht Vng SngEmail: [email protected] Website:WWW.BeautifuLife.Cwahi.net

A. Cu hi phn BJT. 3.1. Gii thiu v BJT. 1. Ba in cc ca BJT l g ? a. pht [emitter], gc [base], gp [collector]. c. ngun [source], cng [gate], mng [drain].

b. T1, T2, T3. d. emitter, gate, collector.

2. Mi tn trong k hiu mch ca BJT lun lun ch vo loi vt liu no ? a. dng P; b. dng N; c. dng base; 3. Cc BJT c phn loi thnh . . . . a. cc dng c PPN v PIN. c. cc dng c NNP v PPN.

d. dng PN.

b. NPN v PNP. d. dng N v dng P.

4. K hiu mch ca transistor PNP l . .b . . 3.2. Cu to ca BJT. 5. C bao nhiu tip gip PN trong BJT? a. 0. b. 1. c. 2. 6. Loi vt liu no l vng base ca transistor PNP? a. dng P. b. dng N c. dng base.

d. 3.

e. 4.

d. dng PN.

7. So vi vng collector v emitter, vng base ca BJT l . . . . a. rt dy. b. rt mng. c. rt mm. d. rt cng. 8. Trong mt BJT, dng base l . . . . . . . . . . . . . . . . . . khi c so vi hai dng collector v emitter.. a. nh. b. ln. c. nhanh. d. chm. 9. Mt BJT c cu to vng base ca n rt mng v . . . . .. a. c pha tp m. b. c pha tp nh vng collector. c. c pha tp long. d. c pha tp nh vng emitter. 10. Dng collector ca BJT lun lun . . . . a. nh hn nhiu so vi dng emitter ca BJT. c. bng dng emitter.

b. nh hn so vi dng base. d. bng dng emitter tr dng base.

11. Trong hot ng thng thng ca transistor NPN, phn ln in t di chuyn vo cc emitter .

Email: [email protected] .... a. ra khi transistor thng qua cc collector. c. s c hp th bi transistor.


b. ra khi transistor thng qua cc base. d. khng phi cc trng hp trn.

12. Phng trnh no biu din quan h ng gia cc dng base, emitter, v collector ? a. IE = IB + . b. IC = IB + IE. c. IE = IB + IC. d. IB = IE + IC. 13. T s ca dng collector v dng base c gi l . . . . . . . . a. rho b. pi c. omega d. beta

e. alpha.



3.3. Chuyn mch bng BJT. 14. Khi mt chuyn mch bng BJT ang dn bo ho, th VCE xp x bng . . . . . . . . a. VCC; b. VB; c. 0,2V; d. 0,7V. 15. Khi mt chuyn mch bng BJT ang dn, th dng collector s c gii hn bi . . . . . . a. dng base; b. in tr ti; c. in p base; d. in tr base. 16. Khi mt chuyn mch bng BJT ngng dn, th VCE xp x bng . . . . . . a. VCC; b. VB; c. 0,2V;

d. 0,7V.

3.4. Trang s liu v cc thng s ca BJT. 17. Ba thng s quan trng ca BJT l beta, cng sut tiu tn ln nht, v . . . . . . . . a. rho nh nht; b. pi nh nht; c. dng collector nh nht; d. dng gi nh nht. 18. Thng s hfe s bng vi . . . . . . . . ca transistor. a. alpha; b. beta; c. dng collector ln nht; nht.

d. dng gi nh

3.5. Mch khuych i bng transistor. 19. Khi mch khuych i bng BJT c phn cc ng hot ng ch A, th . . . . . . . a. tip gip base - emitter c phn cc thun v tip gip base - collector c phn cc ngc; b. tip gip base - emitter c phn cc ngc v tip gip base - collector c phn cc ngc; c. tip gip base - emitter c phn cc thun v tip gip base - collector c phn cc thun; d. tip gip base - emitter c phn cc ngc v tip gip base - collector c phn cc thun. 20. mch khuych i hot ng ch A, th tip gip base - collector ca BJT cn phi . . . .. a. h mch; b. kn mch; c. c phn cc thun; d. c phn cc ngc. 21. H s khuych i in p ca mch khuych i bng BJT bng . . . . . . a. VB/VE; b. Vin / Vout; c. Vout / Vin;

d. VCC / VC.

22. in p phn cc ti collector (VC) ca mch khuych i hot ng ch A xp x bng . . . ... a. VCC; b. mt na VCC; c. 0V; d. 0,2V. 3.6. Phn tch tn hiu mch khuych i phn cc base. 23. Tr khng vo ca mch khuych i phn cc base s bng . . . . . . . . a. 1k ; b. t l nghch vi beta; c. t l thun vi beta; d. khng phi cc trng hp trn. 24. Tr khng ra ca mch khuych i phn cc base s bng . . . . . . . . a. Rc; b. t l nghch vi beta; c. t l thun vi beta; d. 1k . 25. lch pha gia hai tn hiu vo v ra ca mch khuych i phn cc base bng . . . . . . . . o o o o a. 0 ; b. 90 ; c. 180 ; d. 270 . 26. Cng thc chung tnh h s khuych i in p ca mch khuych i phn cc base l . . . ... a. Av = VCC / Vc; b. Av = VB / VE; c. Av = rc / re; d. Av = RL x .



3.7. Php o tr khng vo v ra. 27. Tr khng vo ca mt mch khuych i bng transistor c th o c bng cch s dng . . .. . a. ng h o in tr [ohmmeter]; b. ng h o tr khng; c. my v c tuyn; d. in th k mc ni tip vi my to sng. 28. Tr khng ra ca mt mch khuych i bng transistor c th o c bng cch s dng . . . . .. a. ng h o in tr [ohmmeter]; b. ng h o tr khng; c. my v c tuyn; d. in th k t vo v tr ca in tr ti.

3.8. H c tuyn ra ca BJT. 29. H c tuyn ra ca BJT l th ca . . . . . . a. dng base theo in p collector - emitter; emitter; c. dng collector theo in p collector - emitter;

b. dng collector theo in p base d. dng emitter theo in p base - emitter.

3.9. Sai hng trong mch BJT. 30. Khi kim tra mt BJT tt bng ng h o in tr, th BJT s biu hin . . . . . . a. s biu hin t s in tr thun - nghch cao trn c hai tip gip; b. s biu hin t s in tr thun - nghch cao trn tip gip collector - base; c. s biu hin t s in tr thun - nghch cao trn tip gip emitter - base; d. khng phi cc trn. 31. Khi u que dng ca mt ng h o in tr [ohmmeter] c ni n base, cn u que m c ni n collector ca mt transistor NPN, th gi tr in tr o c l bao nhiu ? a. 0 ; b. in tr thp; c. 5k ; d. in tr cao. 32. Khi u que m ca mt ohmmeter c ni n cc base v u que dng c ni n cc emitter ca mt transistor NPN, th gi tr in tr o c l bao nhiu ? a. 0 ; b. in tr thp; c. 5k ; d. in tr cao. 33. in tr o c gia hai cc collector v emitter ca mt transistor tt l bao nhiu ? a. 0 ; b. in tr thp; c. 5k ; d. in tr cao. 34. Gi tr in p trn collector ca transistor hnh 3.40a, l bao nhiu ? a. 0,2V; b. 0,7V; c. 7,5V; 35. in p trn collector ca transistor mch hnh 3.40b l bao nhiu ? a. 0,2V; b. 0,7V; c. 7,5V;

d. 15V. d. 15V.



36. Mc in p DC trn collector ca transistor mch hnh 3.41, l bao nhiu ? a. 0,2V; b. 0,7V; c. 7,5V; d. 15V. 37. in p DC trn cc base ca transistor mch hnh 3.41, l bao nhiu ? a. 0,2V; b. 0,7V; c. 7,5V; d. 15V. 38. in p tn hiu trn collector ca transistor mch hnh 3.41, l bao nhiu ? a. 50mVpp; b. 0,2Vpp; c. 7,5Vpp; d. 15Vpp. 39. Nu t u ra (C2) hnh 3.41, h mch, th mc in p tn hiu trn collector ca transistor l bao nhiu ? a. 50mVpp; b. 0,2Vpp; c. 7,5Vpp; d. 15Vpp.

B. Bi tp. 1. Nu = 80 mch hnh 3.42, th cc tr s c c trn cc ng h IB, IE, IC, v VCE l bao nhiu ?

VRb = V1 VBE = 1.7 0.7 = 1 VI B = VRb / Rb = 0.01 mA IC = IB = 0.8 mA IE = IB + IC = 0.81 mA

V E ni t nn VE = 0

VCE = VC = V2 VR2 = V2 ICR2 = 10 0.8*6 = 5.2 V



2. Nu dng base l 30 A v dng collector l 4mA, th gi tr ca dng emitter l bao nhiu? IE = IB + IC = 0.03 + 4 = 0.43 mA 3. Nu dng base l 20 A v dng collector l 4mA, th gi tr ca beta l bao nhiu ? = IC / IB = 4/0.02 = 200 4. Nu transistor mch hnh 3.43, c dng nh mt chuyn mch. Hy tnh in p ti collector khi transistor dn. Khi trans dn th VC = 0.2 V

5. Tnh in p collector khi transistor mch hnh 3.43, ngng dn. Khi trans ngng dn th VC = VCC = 20 V 6. Nu ca transistor trong mch hnh 3.43, bng 50, th tr s in p nh nht cn thit ti u vo transistor bo ho l bao nhiu ? Ta c : I C V CC C 20/(5*103 ) 4 mA = /R = = I B I C = 4/50 = 0.08 = / mA ( thc ra trong trng thi bo ha th VC gim xung 0.2V v VRc = VCC 0.2, nhng nh nn ta khng xt) VRb = IB*Rb = 0.08*10 = 0.8 V VBB = VRb + VBE = 0.8 + 0.7 = 1.5 V 7. Khi tn hiu vo mch hnh 3.43, l 5V, th gi tr cn thit lm cho transistor bo ho l bao nhiu ? IB = ( VBB VBE )/Rb = (5 0.7)/104 = 0.43 mA IC = VCC/RC = 20/5000 = 4 mA = IC/IB = 4/0.43 = 9.3 8. Trong mch hnh 3.43, gi s dng collector l 4mA, v dng vo l 0,5mA, nu tng dng vo ln 1mA, th dng collector s bng bao nhiu ? Khi tn hiu dng vo bng 0.5mA th mch hot ng trng thi dn bo ha, nn khi tng dng vo ln 1mA th cng khng lm dng IC thay i nn IC = 4mA 9. Nu in p "m" [on] ti u vo c cho l 3,7V, th tr s in tr vo ln nht c th s dng nhn c s bo ho mch hnh 3.43 l bao nhiu ? ( min = 50). Ta c : IC = VCC/RC = 20/5000 = 4 mA IB = IC/ = 4/50 = 0.08 mA Suy ra gi tr in tr c th s dng l Rb = (VBB VBE)/IB = (3.7 0.7 )/(0.08*103) = 37.5 k

10. Hy tnh cc tr s yu cu di y i vi mch hnh 3.44JC. ( = 80). VB = VBE + VE = 0.7 V VC = 13.4 V

VC = VCC VRc = 20 ICRc = 20 IBRc = 20 80*(20 0.7)*3300 / 775000 = 13.4 V



11. Cn phi thay i gi tr in tr Rb (775k ) mch hnh 3.44JC to ra in p collector bng 10V ( = 80). Ta c : VC = 10V VRc = VCC VC = 20 10 = 10 V IC = VRc /Rc = 10/3300 = 3.03 mA IB = IC/ = 3.03/80 = 0.0379 mA Vy gi tr Rb l : Rb = (15 0.7 )/(0.0379*10-3)= 509 k 12. mch hnh 3.45JC, nu Rc = 3,3k , v beta l 120, hy tnh tr s ca Rb mch cho phn cc im gia (Vc = 7,5V) Ta c : VRc = VCC VC = 15 7.5 = 7.5 V IC = VRc / Rc = 7.5/3300 = 2.272 mA IB = IC/ = 2.272/120 = 0.0189 mA Rb = (VCC VB)/IB = (15 - 0.7)/(0.0189*10-3) = 755 k

13. in p trn collector trong mch hnh 3.45JC s bng bao nhiu nu Rb = 755k , Rc = 3,3k , v = 240 ?. Ta c : IB = (15 0.7)/755000 = 0.0189 mA IC = IB = 0.0189*240 = 4.536 mA Vy : VRc = ICRc = 4.536*10-3*3.3*10-3 = 14.97 V VC = VCC VRc = 15 - 14.97 =0.03 V 14. Trong mch hnh 3.45JC, nu Rb = 600k v beta bng 200, th gi tr no ca Rc s thch hp cho phn cc im gia (Vc = 7,5V) ? Ta c : IB = (15 0.7)/600000 = 0.0238 mA IC = I B = 200*0.0238 = 4.76 mA Vy gi tr Rc l : Rc = (VCC VC)/IC = (15 7.5)/4.76*10-3 = 1.57 k 15. Trong mch hnh 3.45JC, nu Rb l 800k v Rc l 4,7k , th beta cn thit cho phn cc im gia (Vc = 7,5V) l bao nhiu ? Ta c : IB = (15 0.7)/800000 = 0.0179 mA , IC = (15 7.5)/4700 = 1.596 mA = IC/IB = 1.596/0.0179 = 89 16. Tr s ca r'e, zin, v zout ca mch hnh 3.46JC, l bao nhiu ? r'e = 16.6 zin = 2486 zout = 5 k ta c : IB = (VCC VB)/Rb = (15 0.7)/1.43*106 = 0.01 mA IC = IB = 150*0.01 = 1.5 mA IE = IB + IC = 1.5 + 0.01 = 1.51 mA Vy : re' = 25mV/IE = 25/1.51 = 16.6 Zin = Rb// re = (1.43*106*150*16.6) / (1.43*106 + 150*16.6) = 2486 Zout = Rc = 5 k



17. H s khuych i in p ca mch hnh 3.46JC, l bao nhiu ? Av = 200 H s khuych i : AV = rc/re , ta c : rc = Rc//RL = 10*5/(10 + 5) = 3.33 k AV = 3.33*103/16.6 = 200 18. Nu t in C2 h mch, th h s khuych i in p ca mch hnh 3.46JC, bng bao nhiu ? Av = 300 C2 h mch th rc = Rc = 5 k AV = 5*103 / 16.6 = 300

19. i vi mch hnh 3.47JC, hy tnh: Vc = 6.42 V; VB = 0.7 V; VE = 0 ; VCE = 6.42 V; zin = 2096 ; zout = 6 k; Vout = 4.58 V; Av = 9 Ta c : IB = (VCC VB)/Rb = (15 0.7)/1.2*103 = 0.01192 mA, IC = IC = 0.01192*120 = 1.43 mV VRc = ICRC = 1.43*10-3*6*103 = 8.58 V Vy : VC = VCC VRc = 15 8.58 = 6.42 V VE = 0 V VCE = VC VE = 6.42 V VB = VBE + VE = 0.7 V re = 25mV/(IB + IC) = 25/(1.43 + 0.01192) = 17.4 rc = Rc//RL = 6*12/(6 + 12) = 4 k Vy : Zin = Rb// re = (1.2*106*120*17.4) / (1.2*106 + 150*17.4) = 2096 Zout = Rc = 6 k AV = rc/re = 4000/17.4 = 229 Vout = Vin AV = 0.02*229 = 4.58 V 20. i vi mch hnh 3.48JC, hy xc nh: Vc = 7V ; VB = 0.7V ; VE = 0V ; VCE = 7V ; zin = 1596; zout =4 k; Vout = 4.8V ; Av =120



Gii tng t bi 19 C. Cu hi phn mch BJT. 4.1. Gii thiu. 1. Ti sao cn phi n nh mch khuych i bng BJT chng li s thay i beta ? a. do beta thay i theo nhit , b. do beta thay i theo s thay i cc t ghp tng; c. do beta khc nhau trong cc BJT cng loi; d. c a v c. 2. Gi tr beta in hnh ca mt transistor c th xem xt l . . . . . . . . . a. + 50% v - 50%; b. +50% v - 100%; c. + 100% v - 50%; 100%.

d. + 100% v -

3. Nu beta thay i, th s thiu n nh im phn cc trong mch khuych i nh th no ? a. in p collector s thay i; b. dng collector s thay i; c. dng emitter s thay i; d. tt c cc trn. 4.2. Phn cc phn p. 4. Trong mch phn cc phn p, ti sao in p ti im ni ca R b1 v Rb2 c xem l c lp vi dng base ca transistor ? a. dng base khng chy qua Rb1 hoc Rb2; b. dng base nh so vi dng chy qua Rb1 v Rb2; c. ch c dng emitter nh hng n dng chy qua Rb1 v Rb2; d. t ni tng (t ghp tng) chn dng base chy qua mch phn p. 5. Trong cc mch khuych i phn cc phn p, s chnh lch in p gia emitter v base lun lun bng . . . . . a. 0V; b. 0,2V; c. 0,7V; d. 2V. 6. Trong cc mch khuych i phn cc phn p, khi tnh c in p emitter DC, th dng collector ti im tnh c th tnh gn ng bng cch chia in p emitter cho . . . . . a. in tr nhnh base; b. in tr nhnh emitter; c. in tr nhnh collector; d. in tr ca ti. 7. Trong cc mch khuych i phn cc phn p, in p collector c tnh bng cch . . . . . . . a. nhn dng collector vi in tr collector; b. nhn dng collector vi in tr ti; c. cng in p base vi in p emitter; d. tr st p trn in tr collector khi in p ngun. 4.3. Cc tham s tn hiu ca mch phn cc phn p. 8. Mch phn cc phn p c lp vi beta, nhng phi tr gi cho s khng ph thuc vi beta l g ? a. lm gim n nh; b. tr khng ra thp; c. suy gim h s khuych i in p; d. c b v c.





9. Khi tnh tr khng vo, hai in tr base (Rb1 v Rb2) xut hin di dng . . . . . vi cc linh kin khc. a. ni tip; b. ni tip / song song; c. song song; d. ni tip ngc chiu nhau. 10. in tr ng ca tip gip base - emitter l c mc . . . . . . a. ni tip vi in tr tn hiu nhnh base; b. song song vi in tr tn hiu nhnh base; c. song song vi in tr tn hiu nhnh emitter; d. ni tip vi in tr tn hiu nhnh emitter. 11. Tr khng ra ca mch khuych i emitter chung s bng . . . . . . a. in tr collector; b. in tr ti; c. in tr collector mc song song vi in tr ti; d. beta ln in tr collector. 4.4. Cc thay i mch khuych i phn cc phn p. 12. Kiu mch khuych i phn cc phn p no c tr khng vo cao nht ? a. c r mch t ton b; b. phn tch in tr emitter; c. khng c r mch t; d. tt c cc trn. 13. Kiu mch khuych i phn cc phn p no c tr khng ra cao nht ? a. c r mch t ton b; b. phn tch in tr emitter; c. khng c r mch t; d. tt c cc trn.

14. Kiu mch khuych i phn cc phn p no c h s khuych i cao nht ? a. c r mch t ton b; b. phn tch in tr emitter; c. khng c r mch t; d. tt c cc trn. 15. Kiu mch khuych i phn cc phn p no c mo dng t nht ? a. c r mch t ton b; b. phn tch in tr emitter; c. khng c r mch t; d. tt c cc trn. 4.5. Mch khuych i phn cc emitter. 16. in p base ti im tnh ca mch khuych i phn cc emitter thng bng . . . . . . a. 0V; b. 0,7V; c. 2V; d. Vcc. 17. Nhc im ca mch khuych i phn cc emitter khi so vi mch khuych i phn cc phn p l mch khuych i phn cc emitter yu cu . . . . . a. cc transistor c beta cao hn; b. hai ngun cung cp; c. gi tr Vcc cao hn; d. khng phi cc trn. 18. H s khuych i in p ca mch khuych i phn cc emitter l . . . . . a. ph thuc vo beta; b. c tnh bng cng thc chung nh i vi mch khuych i phn cc phn p; c. bng vi x re. d. lun lun cao hn so vi h s khuych i in p ca mch kh. i phn cc phn p. 4.6. Mch khuych i phn cc hi tip kiu in p. 19. Cc mch khuych i phn cc hi tip kiu in p thc t thch hp cho lm vic vi . . . . a. cc tn hiu tn s cao; b. cc ngun cung cp in p thp; c. cc mch cn tr khng vo rt cao; d. cc mch cn tr khng ra rt thp.



20. Tr khng vo ca mch khuych i phn cc hi tip kiu in p b nh hng bi . . . . a. gi tr cng sut trn in tr collector; b. h s khuych i in p ca b khuych i; c. gi tr in tr ca in tr hi tip; d. c b v c. 4.7. Mch khuych i nhiu tng ghp RC. 21. Ti sao cn phi bit tr khng vo ca mi tng trong mt b khuych i nhiu tng ? a. do tr khng vo ton mch l tch ca tr khng vo ca mi tng; b. do h s khuych i in p ca mt tng b tc ng bi tr khng vo ca tng tip theo ; c. do tr khng vo ca mt tng l in tr ti ca tng trc; d. c b v c. 22. Mt trong nhng u im chnh ca vic s dng cc t ghp gia cc tng l g ? a. cc t ghp cho php mch khuych i nhiu tng truyn cc tn hiu DC; b. cc t ghp tng cho php cc mch phn cc trong mi tng c lp nhau; c. cc t ghp tng r mch in tr emitter nn lm tng h s khuych i; d. c b v c. 23. H s khuych i in p ton b ca mch khuych i nhiu tng s bng vi . . . . . . . . a. tng tr khng vo ca mi tng; b. tch h s khuych i in p ca mi tng; c. h s khuych i in p ca tng u tin; d. h s khuych i in p ca tng cui cng.

24. Tr khng vo ca ton b mch khuych i nhiu tng s bng vi . . . . . . . . a. tng tr khng vo ca mi tng; b. tch tr khng vo ca mi tng; c. tr khng vo ca tng u tin; d. tr khng vo ca tng cui cng. 25. Tr khng ra ca ton b mch khuych i nhiu tng s bng vi . . . . . . . . a. tng tr khng ra ca mi tng; b. tch tr khng ra ca mi tng; c. tr khng ra ca tng u tin; d. tr khng ra ca tng cui cng.

4.8. Cc t in ghp tng v r mch. 26. Tr s in dung ca cc t r mch v ghp tng l mt trong nhng yu t chnh khi xc nh . ..... a. tn s ct thp; b. h s khuych i in p; c. h s khuych i dng in; d. tn s ct cao. 27. Nu tr khng vo ca tng th hai l 1k , th t ghp ni gia tng th nht v tng th hai s c Xc vo khong . . . . . i vi tn s thp nht s c khuych i. a. 1 ; b. 10 ; c. 100 ; d. 1k ; e. 10k . 4.9. Cc b khuych i ghp trc tip. 28. Cc mch khuych i ghp trc tip c u im hn cc mch khuych i ghp RC l ch chng c th khuych i . . . . . . a. cc mc tn hiu ln hn; b. cc tn hiu tn s cao; c. cc mc tn hiu nh hn; d. cc tn hiu tn s thp. 29. Trong thc t cc b khuych i ghp trc tip c th d b nh hng vi vn . . . . . . a. v h s khuych i; b. v bo ho; c. v tri DC; d. v tr khng. 4.10. Sai hng ca cc mch bng transistor.

on n



30. in p o c trn collector ca Q1 trong mch hnh 4.24JC, vo khong 20VDC,



a. mch ng chc nng; b. t C2 b ngn mch; c. t C2 b h mch; d. in tr R1 b h mch. 31. in p o c trn collector ca Q2 mch hnh 4.24JC, l 13,8VDC, a. mch ang lm vic ng chc nng; b. transistor Q2 b h mch gia collector v emitter; c. t C5 b ngn mch; d. in tr R8 b ngn mch. 32. in p DC ti im ni ca hai in tr R4 v R5 mch hnh 4.24JC, bng 0V. a. mch ang lm vic ng chc nng; b. transistor Q1 b ngn mch gia collector v emitter; c. t C4 h mch; d. in tr R2 b ngn mch. 33. H s khuych i in p tn hiu ca Q 2 mch hnh 4.24JC, gn bng hai ln h s khuych i tnh c,

a. mch ang ng chc nng;

b. t C3 h mch;

34. H s khuych i in p tn hiu ca Q1 gn bng 3, c. t C3 b ngn mch; a. mch ng chc nng; d. t C5 b h mch. c. t C4 b ngn mch;

b. t C4 h mch;

d. t C2 h mch.

D. Bi tp. 1. Cho mch hnh 4.25JC, hy tnh:



Vc = 10.25 V; VB = 3,8 V; VE = 3.1 V; VCE = 7,15 V; zin = 5,85 k; zout = 2 k; Vout = 30mVp-p; Av = 1,5 Ta c : VB = VCCRb2/(Rb1 + Rb2) = 18*8/(30 + 8) = 3,8 V VE = VB VBE = 3.8 0.7 = 3.1 V IE = VE / Re = 3,1/800 = 3.875 mV IC VRc = ICRc = 3.875*2 = 7,75 V VC = VCC VRc = 18 7,75 = 10.25 V VCE = VC VE = 7,15 V Ta c : r b = Rb1 // Rb2 = 30*8 /(30 + 8) = 6.316 k = 6316 in tr ni tip gip : r'e = 25mV/IE = 25/3,875 = 6,45 Vy re = r'e + Re = 6.45 + 800 = 806.45 Zin = r b//(re) = 6316*100*806,45 / (6316 + 100*806,45) = 5,85 k ( CHO = 100 ) Zout = Rc = 2 k Ta c : r c = Rc//RL = 3*2/(3 + 2) = 1,2 k AV = rc/re = 1200 / 806.45 = 1,5 Vout = AV Vin = 1,5*20mVp-p = 30mVp-p 2. Cho mch hnh 4.26JC, hy tnh Vc =10.25 V; VB = 3,8 V; VE = 3.1 V; VCE = 7,15 V; zin = 584 ; zout = 2 k; Vout = 3,72 Vp-p; Av = 186

Tnh tng t nh trn, cc thng s VC,VB,VE,VCE,Zout gi nguyn Nhng re thay i re = re = 6,45 v RE b ni tt bi t C Zin = r b//(re) = 6316*100*6,45 / (6316 + 100*6,45) = 584 AV = rc/re = 1200 / 6.45 = 186 Vout = AV Vin = 186*20mVp-p = 3,72 Vp-p

3. Cho mch hnh 4.27JC, hy tnh, Vc =10.25 V; VB = 3,8 V; VE = 3.1 V; VCE = 7,15 V; zin = 3,9 k; zout = 2 k; Vout = 3,72 Vp-p; Av = 186 Tnh tng t nh trn, cc thng s VC,VB,VE,VCE,Zout gi nguyn Nhng re thay i re = re + Re = 100 + 6,45 = 106,45 v RE b ni tt bi t C Zin = r b//(re) = 6316*100*106,45 / (6316 + 100*106,45) = 3,9 k AV = rc/re = 1200 / 106.45 = 11,3 Vout = AV Vin = 11,3*20mVp-p = 226 mVp-p



4. Mc tn hiu ln nht c th cung cp ti u vo ca mch hnh 4.27JC, trc khi tn hiu ra bt u b xn l bao nhiu ?

5. Nu tn s lm vic thp nht l 100Hz, hy chn tr s cho C1, C2, v C3 trong mch hnh 4.27JC, C1 = 4,08 F; C2 = 5,3 F; C3 = 159 F; ta c : XCmax = 1/10 gi tr in tr ni tip vi t , t suy ra C= 1/( 2f XC ) Vy C1 ni tip Zin XC1 = 390 C1 = 1/( 200*390 ) = 4,08 F C2 ni tip RL XC2 = 300 C2 = 1/( 200*300 ) = 5,3 F C3 ni tip Re XC1 = 10 C3 = 1/( 200*10 ) = 159 F 6. Cho mch hnh 4.28JC, hy tnh: Vc = . . . . ; VB = . . . . ; VE = . . . . ; VCE = . . . . ; zin = . . . . ; zout = . . . . . ; Vout = . . . . ; Av = . . . .. B ni t nn VB = 0 VE = VB VBE = 0 0,7 = - 0,7 V IE = (VEE VE) / (RE + Re) = (- 12 + 0,7)/(300 + 5300) = 2mA IC IE = 2 mA VRc = ICRc = 2*3,3 = 6,6 V VC = VCC VRc = 12 6,6 = 5,4 V VCE = VC VE = 5,4 (-0,7) = 6,1 V Ta c : r b = Rb1 = 2 k in tr ni tip gip : r'e = 25mV/IE = 25/2= 12,5 Vy re = r'e + Re = 12,5 + 300 = 312,5 Zin = r b//(re) = 2000*100*312,5 / (2000 + 100*312,5) = 1,9 k ( CHO = 100 ) Zout = Rc = 3,3 k Ta c : rc = Rc//RL = 3,3*6/(3,3 + 6) = 2,13 k AV = rc/re = 2130 / 312,5 = 6,8 Vout = AV Vin = 6,8*50mVp-p = 340mVp-p 7. Hy xc nh cc tr s yu cu di y cho mch hnh 4.29JC, tng 1: VB = 4,8 V ; VE = 4,1 V; VC = 8,8 V ; Av =6,1; zin = 3,8 k; zout = 1 k tng 2: VB = 3,2 V; VE = 2,5 V; VC = 8V ; Av = 12,7; zin = 1k; zout = 390 Ton b mch: Av = 77,4 ; zin = 3,8 k ; zout = 390



Tnh tng t ring r cho tng tng Nhng : - RL ca tng th 1 bng Zin ca tng 2

8. Nu tn s lm vic thp nht l 1kHz, hy chn cc gi tr cho C1, C2, C3, C4, v C5 trong mch hnh 4.29JC, C1 = 0,42 F; C2 = 1,6 F; C3 = 1,6 F; C4 = 17,7 F; C5 = 65,2 F; tng t nh bi s 5 trong : C1 ni tip Zin1 XC1 = 380 C1 = 1/( 2000*380 ) = 0,42 F C2 ni tip Zin2 XC2 = 100 C2 = 1/( 2000*100 ) = 1,6 F C3 ni tip RL2 XC3 = 100 C3 = 1/( 2000*100 ) = 1,6 F C4 ni tip re1 XC4 = 8,2 C4 = 1/( 2000*8,98 ) = 17,7 F C5 ni tip re2 XC5 = 2,2 C5 = 1/( 2000*2,4 ) = 65,2 F

9. Hy tnh cc tr s yu cu di y i vi mch hnh 4.30JC, tng 1: VB = 0V ; VE = - 0,7V ; VC = 3,8V ; Av = 1,7 ;

zin = 360 k ; zout = 6 k VE = 4,5V Av = 4,6;

tng 2:VB = 3,8V ; ; VC = 0,3V ; zin =740 k; zout = 7,4 k Ton mch: zin = 360 k; = 7,8

zout = 7,4 k;

Av

tnh tng t ring r cho tng tng

E. Cu hi phn mch BJT khc. 5.1. Mch khuych i collector chung. 1. Trong mch khuych i collector chung, cc collector . . . . . a. c ni vi mc t ca ngun DC; b. c ni vi mc t ca tn hiu; c. ni vi Vcc; d. c b v c. 2. Mch khuych i collector chung c th s dng kiu phn cc . . . . . . . . a. emitter; b. hi tip in p collector ; c. phn p ; d. c a v c. 3. Cu hnh collector chung c . . . . . . a. tr khng vo cao v tr khng ra thp;

b. tr khng vo thp v tr khng ra bng

Email: [email protected] Rc ; c. tr khng vo cao v tr khng ra bng Rc;


d. ngoi cc trng hp trn.

4. H s khuych i in p ca mch collector chung l . . . . . a. ln hn 1; b. ng bng 1; c. hi nh hn 1; d. bng vi . 5.2. H s khuych i dng in v cng sut. 5. H s khuych i cng sut ca mt mch c th tnh c bng php nhn. . . . . a. beta vi h s khuych i in p; b. bnh phng dng ti vi in p ti; c. h s khuych i in p vi h s khuych i dng; d. cng sut vo v cng sut ra. 6. Cu hnh BJT no cho s khuych i cng sut ? a. emitter chung; b. base chung; trn.

c. collector chung;

d. tt c cc

5.3. Cp Darlinhton. 7. Cp Darlington thay cho transistor thng thng trong mch collector chung s . . . . . . . a. lm tng tr khng vo ca mch; b. cho php mch iu khin ti in tr thp hn; c. thay i in p phn cc emitter bng 0,7V; d. tt c cc trn. 8. Nu cp Darlington c lp bng hai transistor vi mi transistor bng 40, th mch Darlington c beta bng . . . . . a. 40; b. 80; c. 60; d. 1600. 9. Mch Darlington khi hot ng khuych i ch A, in p gia cc base v emitter l bao nhiu ? a. 0,2V; b. 0,7V; c. 1,4V; d. 2V. 5.4. Tng collector chung trong mch khuych i nhiu tng. 10. Thng thng, tng collector chung l tng cui cng trc ti. Chc nng chnh ca tng collector chung l . . . . . . a. cho khuych i in p; b. lm b m cho cc b khuych i in p khi in tr ti thp; c. cho s o pha tn hiu; d. cho ng dn tn s cao ci thin p ng tn s tng cui cao. 11. Mt b khuych i in p c ba tng c cc h s khuych i l 32, 16, v 1. H s khuych i in p ca ton mch l bao nhiu ? a. 49V; b. 49; c. 512V; d. 512. 5.5. Cc mch khuych i base chung. 12. Cu hnh base chung c . . . . a. tr khng vo cao v tr khng ra thp; c. tr khng vo cao v tr khng ra bng Rc;

b. tr khng vo thp v tr khng ra bng Rc; d. ngoi cc trn.

13. H s khuych i dng ca mch base chung l . . . . a. ln hn 1; b. ng bng 1; c. hi nh hn so vi 1;

d. bng .

5.6. So snh cc cu hnh ca mch khuych i.

Email: [email protected] 14. Cu hnh emitter chung c . . . . . a. tr khng vo cao v tr khng ra thp; Rc ; c. tr khng vo trung bnh v tr khng ra bng Rc;


b. tr khng vo thp v tr khng ra bng d. ngoi cc trng hp trn.

15. Khi BJT c mc theo cu hnh base chung, th mch c kh nng cung cp . . . . . a. h s khuych i in p v cng sut; sut; c. h s khuych i dng in, in p v cng sut; b. h s khuych i dng in v cng d. ch c h s khuych i in p.

16. Khi BJT c mc theo cu hnh collector chung, th mch c kh nng cung cp . . . . . . a. h s khuych i in p v cng sut; b. h s khuych i dng in v cng sut; c. h s khuych i dng in, in p v cng sut; d. ch c h s khuych i in p. 17. Khi BJT c mc theo cu hnh emitter chung, th mch c kh nng cung cp . . . . . a. h s khuych i in p v cng sut; b. h s khuych i dng in v cng sut; c. h s khuych i dng in, in p v cng sut; d. ch c h s khuych i in p. 5.7. Ngun dng in. 18. K hiu no sau y l k hiu ca ngun dng in ? d

19. Mt ngun dng hng 10mA cung cp cho ti 1k , mc in p trn ti l bao nhiu ?

20. Mt ngun dng hng 10mA cung cp cho ti 1k . Nu thay i in tr ti l 2k , th dng chy qua ti l bao nhiu ? a. 1mA; b. 5mA; c. 10mA; d. 40mA. 5.8. Mch khuych i vi sai. 21. Mt mch khuych i vi sai c thit k vi ngun dng hng 10mA, ngun cung cp 12V, v in tr nhnh collector (Rc) cho c hai BJT l 1k . C hai u vo c thit lp l 100mV. Mc chnh lch in p o c gia hai hai u ra collector ca transistor l bao nhiu ? a. 0V; b. 100mV; c. 5V; d. 7V. 22. Mt mch khuych i vi sai c thit k vi ngun dng hng 10mA, ngun cung cp 12V, v in tr nhnh collector (Rc) cho c hai BJT l 1k . C hai u vo c thit lp l 100mV. Mc in p o c trn collector ca mt trong hai u ra ca transistor so vi t l bao nhiu ? a. 0V; b. 100mV; c. 5V; d. 7V. 5.9. Sai hng trong cc mch BJT. 23. in p tn hiu trn ti trong mch hnh 5.21JC, o c l 1Vpp. Vn c nhiu kh nng xy ra nht i vi mch l g ? a. beta ca transistor thp; b. t in ghp u ra b ngn mch; c. transistor b ngn mch ti tip gip base - emitter; d. mch ang lm vic ng chc nng.



24. in p trn base mch hnh 5.21JC, o c l 1,2VDC. Vn c nhiu kh nng xy ra nht i vi mch l g ? a. beta ca transistor thp; b. t in ghp u ra b ngn mch; c. transistor b ngn mch ti tip gip base - emitter; d. mch ang lm vic ng chc nng. 25. Tn hiu ra mch hnh 5.21JC, dao ng trn mc 6VDC. Vn c nhiu kh nng xy ra nht i vi mch l g ? a. beta ca transistor thp; b. t in ghp u ra b ngn mch; c. transistor b ngn mch ti tip gip base - emitter; d. mch ang lm vic ng chc nng.

26. Mc tn hiu vo mch hnh 5.22JC, c t mc 0. in p trn collector ca Q1 v Q2 o c l 12V. Vn c nhiu kh nng xy ra nht i vi mch l g ? a. transistor Q1 h mch; b. transistor Q2 h mch; c. ngun dng b hng; d. mch lm vic bnh thng. 27. Mc tn hiu vo mch hnh 5.22JC, c t mc 0. in p trn collector ca Q1 o c l 12V v in p trn Q2 o c l 0,2V. Vn c nhiu kh nng xy ra nht i vi mch l g ? a. transistor Q1 h mch; b. transistor Q2 h mch; c. ngun dng b hng; d. mch lm vic bnh thng 28. Mc tn hiu vo mch hnh 5.22JC, c t mc 0. in p trn collector ca Q1 v Q2 o c l 6V. Vn c nhiu kh nng xy ra nht i vi mch l g ? a. transistor Q1 h mch; b. transistor Q2 h mch; c. ngun dng b hng; d. mch lm vic bnh thng

F. Bi tp. 1. Cho mch hnh 5.23JC, hy tnh cc tr s ti im lm vic tnh Q: VB = 6,7 V ; VE = 6 V; VC = 12 V ; VCE = 6 V; IE = 60 mA; Mch khuych i C chung V B = VCCR 2/(R 1 + R 2) = 12*1/(1 + 0,8) = 6,7 V VE = VB VBE = 6,7 0,7 = 6 V VC = VCC = 12 V VCE = VC VE = 12 6 = 6 V IE = VE/Re = 6/100 = 60 mA

2. Tnh cc gi tr zin, zout, v Av ca mch hnh 5.23JC. CHO = 100



zin = . . . . ; zout = . . . . . ; Av = . . . . . ta c : R1//R2 = 1*0,8/(1 + 0,8) = 0,444 k re = R3//RL = 100*50/(100 + 50) = 37,5 rb = R1//R2//Rint = 364 vy : Zin = (R1//R2)//(re) = 444*37500/(444 + 37500) = 396 Zout = R3//(rb/) = 100*3,64/(100 + 3,64) = 3,5 Av = re / (re + r'e) = 1 3. i vi mch hnh 5.23JC, hy tnh h s khuych i dng in (Ai), v h s khuych i cng sut (Ap). ta c : iout = Vout/RL iin = Vin/Zin Ai = iout/iin = AvZin/RL = 1*396/50 = 7,9 Ap = Voutiout/VinZin = AvAi = 7,9 4. Cho beta ca Q1 l 150 v beta ca Q2 bng 30. ca cp Darlington hnh 5.24JC, l bao nhiu ? = Q1*Q2 = 150*30 = 4500 5. Cho mch hnh 5.25JC, hy tnh cc gi tr ti im lm vic tnh Q: VBQ1, VEQ1, VCQ1, VBQ2, VEQ2, VCQ2, v IEQ2. (gi s Mch C chung dng cp DARLINTON VBQ1 = VCCR2/(R1 + R2) = 18*2,2/(2 + 2,2) = 9,4 V VEQ1 = VBQ1 VBEQ1 = 9,4 0,7 = 8,7 V VCQ1 = VCC = 18 V VBQ2 = VEQ1 = 8,7 V VEQ2 = VBQ2 VBEQ2 = 8,7 0,7 = 8 V VCQ2 = VCQ1 = 18 V IE = VEQ2/R3 = 8/25 = 320 mA

Q1

= 90 v

Q2

= 40)

6. Hy tnh cc gi tr ca zin, zout, v Av cho mch hnh 5.25JC. (cho Q1 = 90 v Q2 = 40). Tnh tng t bi 2 nhng = Q1*Q2 = 90*40 = 3600 Zin = 1,024 k Zout = 0,19 Av = 1 7. Cho mch hnh 5.25JC, hy xc nh h s khuych i dng in (Ai) v h s khuych i cng sut (Ap). (gi s Q1 = 90, v Q2 = 40). Tng t bi 3 Ai = iout/iin = AvZin/RL = 1*1024/25 = 41 A p = Voutiout/VinZin = AvAi = 41 8. Hy tnh h s khuych i in p, h s khuych i dng in, v h s khuych i cng sut cho h thng hnh 5.26JC.

Av = 8*22*1 = 176 Ai = AvZin/RL = 176*2000/25 = 14080

eated by Mondeo Email: [email protected]


Ap = AvAi = 176*14080 = 2478080

9. Tnh cng sut tn hiu c phn b trn ti trong c hai mch hnh 5.27JC. Mi mch c cung cp bi ngun tn hiu 2Vpp (khng ti) c tr khng ni l 2k . (cho Q1 = 90 v Q2 = 40). Ta c : Vin rms = Vmax/ 2 = 1/2 = 0,707 V = Q1*Q2 = 90*40 = 3600 Mch (a) : Vout = VrmsRL/(Rint + RL) = 8,7 mV P = Vout2/RL = 3,05 W Mch (b) : Zint = (R1//R2)//(re) = 2,42 k Vin = VrmsZin/(Zin + Rint) = 0,387 V Vout = AvVin = 0,387 V ( mch C chung) P = Vout2/RL = 6 mW

10. Hy tnh cc mc in p phn cc DC: VB, VE, Vc, v VCE cho mch hnh 5.28JC. Mch khuych i B chung VB = VCCRb2/(Rb1 + Rb2) = 15*18/(90 + 18) = 2,5 V VE = VB VBE = 2,5 0,7 = 1,8 V IE = VE/Re = 1,8/1800 = 1 mA VC = VCC VRc = 15 1*6,8 = 8,2 V VCE = VC VE = 8,2 1,8 = 6,4 V

11. Hy tnh h s khuych i in p tn hiu (Av), v tr khng vo (zin), tr khng ra (zout), v dao ng ca in p ra ln nht gn ng cho mch hnh 5.28JC. rc = RL//Rc = 4048 , re = r'e = 25mV/IE = 25 Av = rc/re = 162 Zin = r'e = 25 , Zout = Rc = 6,8 k dao ng ca in p ra ln nht gn ng bng 2VCE = 2*6,4 = 12,8 V 12. Hy tnh in p trn collector, base, v emitter ca c ba transistor trong mch hnh 5.29JC. Mch khuych i vi sai Q1 : st p trn in tr 10k l V = (15 (- 15))*10/(10 + 18) = 10,7 V VB1 = V + VEE = 10,7 15 = - 4,3 V V E1 = VB1 VBE = - 4,3 0,7 = - 5 V V C1 = VE2 = VE3 V B2 = VB3 = 0 V ( ni t )



VC1 = VE2 = VE3 = 0 0,7 = - 0,7 V IC1 = IE1 = (VE1 VEE)/Re1 = ( - 5 + 15 )/5 = 2 mA IE2 = IE3 = IC1 = 1 mA VRc2 = VRc3 = 1*3,3 = 3,3 V V C2 = VC3 = 15 3,3 = 11,7 V 13. Hy tnh h s khuych i in p ca u ra n v h s khuych i in p vi sai cho mch hnh 5.29JC. Ta xem nh mch khuych i gm 2 tng rc = Rc = 3,3 k re = 25 mV/IE = 25 Av = rc/re = 132 ( i vi u ra n ) i vi u ra cn bng th : Vout ca u ra cn bng bng 2 ln V out ca u ra n nn A'v = 2Av = 264

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