Bài tập mạch điện tử

10.5 50501000500 100.500 h. R)h1(hR//R R//R i i . i i i i 4 1fe 2E2fe2ie2b1C 2b1C 1b 1C 1C 2b 1b 2b −=−= ++ −= +++ −== (8) ( ) ( ) 5,01010 10 hR//R R//R i i 33 3 1ie1bi 1bi i 1b = + = + = (9) Thay (7), (8), (9) vaøo (2) ta ñöôïc: iC2 hie1 1K Ri 100K ii hie2 1K 100ib1 RC1 1K Rb2 1K hfe2RE2 5050 ib2 ib1 iC1 Rb1 1K Vo2 100ib2 hie3 1K RC2 1K hfe3RE3 5050 ib3 Khoa Ñieän - Ñieän töû Vieãn thoâng Maïch Ñieän Töû I Moät soá baøi taäp maãu 30 ( ) ( ) mA V27,195,0.63,7.5050 i V i 01 ≈−−= Zi = Ri//Rb1//hie1 ≈ 500Ω [ ] Ω≈++=      ++= 5,471093010//50 h R h Rh//RZ 3fe 2C 3fe 3ib3Eo II. Transistor maéc vi sai vaø Darlingtôn 1) Baøi 6-23: E.C – E.C. a- Cheá ñoä DC V25,29. 10.310 10V. RR R VV 33 3 CC 2111 11 2BB1BB = + = + == mA55,1 42,710 7,025,2 h R R2 7,0V II 3 fe b E 1BB 2EQ1EQ = + − = + − == IE = 2IE1 = 3,1mA VCEQ1 = VCC – 2RE.ICQ1 = 9 – 2.500.1,55.10-3 = 7,45V VCEQ2 = VCC – 2RE.ICQ1 – RC2.ICQ2 = 9 – 103.1,55.10-3 – 2,5.103.1,55.10-3 = 3,575V RE3 50Ω Zo hib3 10 R/hfe3 930 Rc2 hfe3 =10 +VCC =9V ii R21 3K iL ib3 R11 K RE1 500 T1 T2 RC2 2,5K R22 3K R12 1K T3 T4 RC4=RL 60 RE4 60 hfe=100 iC4 hie1 hfe2ib2 100ib2 RL (1+hfe)RE1 (1+hfe)ib4 ii hie3 RC2 2,5K Rb2 750 h2feRE4 ib1 ib2 hfe3hie4 Rb1 750 hie2 Khoa Ñieän - Ñieän töû Vieãn thoâng Maïch Ñieän Töû I Moät soá baøi taäp maãu 31 V725,34,1875,39VVI.RVV 4BE3BE2CQ2CCCR 4E =−−=−−−= mA62 60 725,3 R V II 4E R 4EQ4CQ 4E ≈=== mA62,0 10 10.62 h I III 2 3 fe 4CQ 4BQ3EQ3CQ ≈=≈== − VCEQ4 = VCC – ICQ4(RC4 + RE4) = 9 – 62.10-3.120 = 9 – 7,44 = 1,56V VCEQ3 = VCEQ4 – VBE4 = 1,56 – 0,7 = 0,86V Ω=== − −− 2258 10.55,1 10.25.100.4,1 I 10.25.h4,1h 3 3 1EQ 3 1fe1ie Ω=== − −− 5645 10.62,0 10.25.100.4,1 I 10.25.h4,1h 3 3 1EQ 3 3fe3ie Ω=== − −− 45,56 10.62 10.25.100.4,1 I 10.25.h4,1h 3 3 1EQ 3 4fe4ie b- Cheá ñoä AC ( ) 50500101.500h1RR fe1E'E ==+= ( )[ ]( ) [ ] Ω=+=+++= K76,617101.606045,56h1Rh1hR 3fe4E4fe4ie' 4E i 1b 1b 2c 2c 3b 3b L i L i i i . i i . i i . i i i i A == (1) ( )( ) 10201101.101h1h1 i i 4fe3fe 3b L ==++= (2) ( ) 3 33 3 ' 4E4iefe3ie2C 2C 2c 3b 10,4 605,631 5,2 10.76,6175700564510.5,2 10.5,2 Rhh1hR R i i −≈ − = +++ − = ++++ − = (3) ( ) 1001.h i i i i i i 2fe 1b 2b 2b 2c 1b 2c −=−== (4) (Vì R’E raát lôùn neân coi ib2 ≈ ib1) 3'' E1ieb b i 1b 10.4,128 28322258750 750 RhR R i i − = ++ = ++ = (5) iC2 hie1 2258 100ib2 RC 60Ω R’E 50,5K Zo (1+hfe)2ib3 10201ib3 ii hie3 5645 RC2 2,5K Rb 750 R’E4 618K ib1 ib2 hie4(1+hfe) 5700 Rb1 750 hie2 2258 ib3 iL Zo’ Khoa Ñieän - Ñieän töû Vieãn thoâng Maïch Ñieän Töû I Moät soá baøi taäp maãu 32 vôùi [ ] Ω≈=+= 28323008//5050Rh//RR b2ie'E''E Thay (2), (3), (4), (5) vaøo (1) ta coù: Ai = 10201.(-4.10-3).(-100).128,4.10-3 = 522,3 (laàn) Zi = Rb1//(hie1 +R’’E) ≈ 750//(2258 + 2832) = 654Ω Zo = ∞ ⇒ Z’o = Zo//RC = RC = 60Ω 2) Baøi 6-24: E.C – C.C a- Cheá ñoä DC AÙp duïng ñònh luaät K.II ΣVkín = 0 cho voøng 2 ta coù: VBE3 + IEQ3RE3 – VEE = 0 (1) mA3,2 10 7,03 R VV I 3 3E 3BEEE 3EQ = − = − = mA15,1 2 I II 3EQ2EQ1EQ === VCE1 = VCE2 = VCC – RC1ICQ1 – VE1 (2) Maët khaùc aùp duïng ñònh luaät K.II ΣVkín = 0 cho voøng 1 ta coù: -VBB1 + RbIBQ1 + VBE1 + VE1 = 0 (3) ⇒ VE = VBB1 – RbIBQ1 – VBE1 = 1 – 104.1,15.10-5 – 0,7 =0,185V Thay vaøo (2) ta ñöôïc: VCE1 = VCE2 = 6 – 103.1,15. 10-3 – 0,185 = 4,665V ≈ 4,67V Ta coù VE1 = VCE3 + RE3.IEQ3 - VEE (4) Suy ra VCE3 = VEE + VE1 – RE3IEQ3 = 3 + 0,185 – 103.2,3.10-3 = 0,885V VRE6 = VCC – RC2ICQ2 – VBE4 - VBE5 - VBE6 = 6 – 103.1,15.10-3 – 2,1 = 2,75V mA275 10 75,2 R V I 6E 6RE 6EQ === VCE6 = VCC – VRE6 = 6 – 2,75 = 3,25V VCE5 = VCE6 – VBE6 = 3,25 – 0,7 = 2,55V VL Zo Zi ii -3V Rb1 10K RC1 1K Rb2 10K RE 1K VBB2 1V T4 T5 RC2 1K VBB1 1V T3 RE6 10 T6 Khoa Ñieän - Ñieän töû Vieãn thoâng Maïch Ñieän Töû I Moät soá baøi taäp maãu 33 VCE4 = VCE5 – VBE5 = 2,55 – 0,7 = 1,85V Ω=== − − − − 3043 10.15,1 10.25.100.4,1 10.15,1 10.25.h4,1h 3 3 3 3 1fe1ie mA75,2 h I I 6fe 6EQ 5EQ == ; A10.75,2h I I 5 5fe 5EQ 4EQ − == Ω=== − −− 72,12 10.275 10.25.100.4,1 I 10.25.h4,1h 3 3 6EQ 3 6fe6ie hie5 = 1272Ω; hie4 = 127.200Ω b- Cheá ñoä AC ( ) Ω≈≈+= 766E3fe6E' 6E 1010.Rh1RR i 2b 2b 4b 4b L i L T i i . i i . i V i V A == (1) Ω≈= 7' 6E 4b L 10R i V (2) 4 4 2 733 23 ' 6E4ie2C 2fe2C 2b 2C 2C 4b 2b 4b 10.3,96 106,3811 10 1010.6,38110 10.10 Rh3R h.R i i . i i i i − −= ++ −= ++ −= ++ −== (3) 485,0 1010.086,610 10 Rh2R R i i i i 434 4 2b1ie1b 1b i 1b i 2b −= −+ −= ++ −=−= (4) Thay (2), (3), (4) vaøo (1) ta coù AT = 107.(-96,3).10-4.(-0,485) = 46728V/A = 46,7V/mA Zi = Rb1//[2hie1 + Rb2] ≈ 6,15KΩ ( ) Ω≈=      += 37,0382,0//10 h h3 h R //RZ 3 fe 4ie 3 fe 2C 6Eo iC2 hie1 3043 VL hfe2ib2 100ib2 Zo Zi ii hie4 127,2K RC2 1K Rb2 10K R’E6 ib1 Rb1 10K hie2 3043 ib4 ib2 hie5.hfe4 127,2K hie5.hfe4.hfe5 127,2K Zo RE6 3hie4 h3fe =0,381 Rc2 h3fe ≈10 Khoa Ñieän - Ñieän töû Vieãn thoâng Maïch Ñieän Töû I Moät soá baøi taäp maãu 34 Chöông VII: MAÏCH KHUEÁCH ÑAÏI HOÀI TIEÁP. I. Hoài tieáp aùp, sai leäch doøng. 1) Baøi 7-4    = Ω= 40h K1h GT fe ie ;     == = ? i i A ?T KL i L i Ñaây laø daïng hoài tieáp aùp, sai leäch doøng. a- Tính ñoä lôïi doøng T: cho ii = 0 ' 1 1b 1b 2b 2b 1 i ' 1 1 V i . i i . i V 0iV V T = = = (1) ( ) 32fe22E 2b 1 10.41h1R i V =+= (2) ( ) ( ) 941,0 85 80 10.4110.411010.2 40.10.2 h1Rh1RhR h.R i i . i i i i 3333 3 2fe22E2fe21E2ie1C 1fe1C 1b 1C 1C 2b 1b 2b −=−= +++ −= +++++ −== (3) 634 1ief1ief ' 1 ' 1 ' 1 1b 10.91 1010 1 hR 1 hR V . V 1 V i − = + = + = + = (4) V1 ii RE22 1K Rf 10K RC1 2K RC2 2K iL RL 100 +VCC RE21 1K T1 VL iC2 hie1 hfe1ib1 40ib1 RE21(hfe2+1) 41K Zo Zi ii RC1 2K RC2 2K ib1 ic1 40ib2 Rf V1 hie2 1K ib2 V’1 + - RE22(hfe2+1) 41K RL 100 Khoa Ñieän - Ñieän töû Vieãn thoâng Maïch Ñieän Töû I Moät soá baøi taäp maãu 35 Thay (2), (3), (4) vaøo (1) ta coù: T = 41.103.(-0,941).91.10-6 = -3,51 b- Tính i L i i iA = cho V’1 = 0 i 1b 1b 2b 2b L i L i i i . i i . i i i i A == (1) 6,3940. 1010.2 10.2h. RR R i i . i i i i 3 3 2fe L2C 2C 2b 2C 2C L 2b L −= + −= + −== (2) 941,0 i i 1b 2b −= (nhö (3) ôû phaàn treân) (3) 234 4 1ief f i 1b 10.91 1010 10 hR R i i − = + = + = (4) Thay (2), (3), (4) vaøo (1) ta coù: Ai = (-39,6).(-0,941).91.10-2 = 33,9 ≈ 34 c- Tính Aif, Zif, Zof. 54,7 51,31 34 T1 A A iif =+ = − = Zi = Rf//hie1 = 104//103 ≈ 910Ω Ω= + = − = 202 51,31 910 T1 Z Z iif Zo = RC2 = 2KΩ Ω= + = − = 443 51,31 10.2 T1 Z Z 3 o of 2) Baøi 7-11      ∞→ Ω= = C 10h 100h GT ib fe ;        = == ?T Z;Z ? i i A KL oi i L i ii Rf = Rb =10K RC 2K C iL RL +VCC RE 100 C Khoa Ñieän - Ñieän töû Vieãn thoâng Maïch Ñieän Töû I Moät soá baøi taäp maãu 36 hie = hib.hfe = 10.100 = 1KΩ a- Ñoä lôïi voøng T: cho ii = 0 ' L b b L i ' L L V i . i V 0iV V T = = = (1) 433 233 fe LC C L b C C L b L 10.5 1010 10.10.10h. RR R .R i i . i V i V −= + −= + −== (2) 634 iefief ' L ' L ' L b 10.91 1010 1 hR 1 hR V . V 1 V i − = + = + = + = (3) Thay (2), (3) vaøo (1) ta coù: T =(-5.104).91.10-6 = -4,55 b- Tính Ai, Zi, Zo. i b b L ' Li L i i i . i i 0Vi i A = = = (1) 50100. 1010 10h. RR R i i . i i i i 3 3 fe LC C b C C L b L −= + −= + −== (2) 234 4 ief f i b 10.91 1010 10 hR R i i − = + = + = (3) Thay (2), (3) vaøo (1) ta coù: Ai = (-50).91.10-2 = -45,45 Zi = Rf//hie = 104//103 = 910Ω Zo = RC = 103Ω = 1KΩ c- Tính Aif, Zif, Zof. 2,8 55,41 50 T1 A A iif −=+ −= − = Ω= + = − = 164 55,41 910 T1 Z Z iif Ω= + = − = 180 55,41 10 T1 Z Z 4 o of hie1 iL VL hfeib 100ib Zo ii RC 1K ib ic Rf Rf 10K V’L + - RL Khoa Ñieän - Ñieän töû Vieãn thoâng Maïch Ñieän Töû I Moät soá baøi taäp maãu 37 II. Hoài tieáp aùp, sai leäch aùp: 1) Baøi 7-10      ∞→ Ω= = C 50h 50h GT ib fe ;        = == ?T Z;Z ? i i A KL oi i L i a- Tính ñoä lôïi voøng T (cho ii = 0) ' L 1b 1b 2b 2b L i ' L L V i . i i . i V 0iV V T = = = (1) 33 34 34 2fe 2Cf 2Cf 2b 2C 2C L 2b L 10.5,8350.10.67,150. 10.210 10.2.10 h. RR R.R i i . i V i V −=−= + −= + −== (2) ( ) 46,12 10.5,210.83,0 50.10.83,0 h hR//R R//R i i . i i i i 33 3 1fe 2ie2b1C 2b1C 1b 1C 1C 2b 1b 2b −= + −= + −== (3) 6 3533 3 1fef1E1fe1ie1b 1fe1E ' L 1b 10.2,1 10.210.5 1. 10.510.5,2890 10.5 'Rh.R 1. R.hhR h.R V i − −= +++ − = +++ − = (4) VL ii Rf =10K C +VCC R21 8K C R11 1K RC1 1K RE1 100 R22 10K R12 10K RC2 2K RE2 1K iC1 hfe1ib1 50ib1 hie1 2,5K hie2 2,5K RC2 2K Zo ii RC1 1K Rb2 5K Rf 10K ib2 ib1 iC2 hfe2ib2 50ib2 Rf.hfe1 5.105 RE1hfe1 5K V ’ L + - Rb1 890 Khoa Ñieän - Ñieän töû Vieãn thoâng Maïch Ñieän Töû I Moät soá baøi taäp maãu 38 Thay (2), (3), (4) vaøo (1) ta coù: T = (-83,5.103).(-12,46).(1,2.10-6)= -1,25 b- Tính AT, Zi, Zo. i 1b 1b 2b 2b L i L T i i . i i . i V i V A == (1) 2b L i i vaø 1b 2b i i tính nhö treân theo coâng thöùc (2), (3) ( ) 107,010.95,410.5,2890 890 hRRhR R i i 33 1fef1E1ie1b 1b i 1b = ++ = +++ = (4) Thay (2), (3), (4) vaøo (1) ta coù: Ai = (-83,5.103).(-12,46).(0,107) = 111.103V/A = 111V/mA Zi = Rb1//[hie1 + (RE11//Rf)(1 + hfe)] = 890 //[2500 + 4950] = 795Ω Zo = Rf = 10KΩ c- Tính AVf, Zof, Zif. mA V49 A V10.49 25,11 10.111 T1 A A 3 3 T Tf ==+ = − = ( ) Ω=+=−= 1788)25,11(795T1ZZ iif Ω= + = − = 4444 25,11 10 T1 Z Z 4 o of 2) Baøi 7-12      ∞→ Ω= = C 50h 20h GT ib fe ;      = = ?T Z;Z ?A KL oi V VL Vi + - ri 1K Rf =1K C +VCC R21 10K C R11 1K RC1 500 RE12 82 R22 10K R12 1K RC2 500 RE22 82 C RE11 22 C RE21 22 Khoa Ñieän - Ñieän töû Vieãn thoâng Maïch Ñieän Töû I Moät soá baøi taäp maãu 39 a- Tính ñoä lôïi voøng T (cho Vi = 0) ' L 1b 1b 2b 2b L i ' L L V i . i i . i V 0iV V T = = = (1) ( ) 666720.500//10h. RR R.R i i . i V i V 3 2fe 2Cf 2Cf 2b 2C 2C L 2b L −=−= + −== (2) ( ) ( ) 52,3 1835 6460 4621050323 20.323 h h1RhR//R R//R i i . i i i i 1fe 2fe21E2ie2b1C 2b1C 1b 1C 1C 2b 1b 2b −=−= ++ −= +++ −== (3) ( ) ( )( ) ( )( ) ( ) 6 3 3 1bi1ie1fef11E 1fef11E 1fef , L , L ' L 1b 10.88,10 4,41546 10.452 4,4761050452 452. 10.21 1 R//rhh1R//R h1R//R . h1R V . V 1 V i − − −=−= ++ −= +++ + + −= (4) Thay (2), (3), (4) vaøo (1) ta coù: T = (-6667).(-3,52).(-10,88.10-6)= -0,255 b- Tính AV, Zi, Zo. i 1b 1b 2b 2b L i L V V i . i i . i V V V A == (1) 2b L i V vaø 1b 2b i i tính nhö treân theo coâng thöùc (2), (3) ( ) ( )( ) 5 3 3 1fef11E1ie1bi 1bi i i ii 1b 10.24 4,1978 10.4,476 45210504,476 4,476. 10 1 h1R//RhR//r R//r . r V . V 1 V i − − == ++ = +++ = (4) hfe1ib1 20ib1 Rf(hfe1+1) 21K hie2 1050 Vi + - Zo Zi Rb1 910 RCb2 910 ib1 iC2 RE11(1+hfe1) 462 iC1 V’L + - ri 1K RC1 500 RC2 500 ib2 ii hie1 1050 RE21(1+hfe2)=462Ω 20ib2 Rf 1050 Rf(hfe1+1) 21K Rb1 910 ib1 RE11(1+hfe1) 462 ri 1K hie1 1050 V’L Rf(1+hfe1) Khoa Ñieän - Ñieän töû Vieãn thoâng Maïch Ñieän Töû I Moät soá baøi taäp maãu 40 Thay (2), (3), (4) vaøo (1) ta coù: AV = (-6667)(-3,52).24.10-5 = 5,63 Zi = Rb//[hie1 + (RE11//Rf)(1 + hfe1)] = 103 //1502 ≈ 600Ω Zo = Rf = 1000Ω c- Tính AVf, Zif, Zof. 486,4 255,01 63,5 T1 A A VVf =+ = − = ( ) Ω=+=−= 753)255,01(600T1ZZ iif Ω= + = − = 797 255,01 1000 T1 Z Z oof