Prob2 - 2-1 Vcc R4 1k rbb' V Ii iL L R3 1k + Vbb r1 R2 ii...

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72 i B=B 1 b //r b'e b'e m v c i L b i g C R R r bb' b i b'e b'e M b' v m L i L L R g V C C1 i r1 R2 R4 1k R3 1k Vcc + Vbb Ii Vôùi giaû thieát C b'c = 0 thì C M = 0; r bb' = 0 : ngaén maïch B - B'. Sô ñoà chA coøn laïi nhö sau : Theo giaû thieát ta coù : I CQ = 2 m± suy ra : h ie = 2 25 . fe h = 12,5.h fe maø : h ie = r bb' + r b'e = r b'e (r bb' = 0) Do vaäy : 12,5.h fe = r b'e (1) g m = 5 , 12 1 ' = e b fe r h = 0.08 mho Taàn soá cao 3 dB : f h = e b e b M e b e b C R C C R ' ' ' ' 2 1 ) ( 2 1 π = + vôùi : R b'e = (r i // R b + r bb' ) // r b'e = R b // r b'e (r bb' = 0; r i = ) neân : 2-1
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73 f h = e b e b b C r R ' ' . // 2 1 π (2) Ñoä lôïi doøng taàn giöõa : A im = i b b L i L i v v i i i ' ' . = = -g m .R b // r b'e Theo giaû thieát : im A = 32 dB = 40 Suy ra : R b // r b'e = Ω = = 500 08 , 0 40 m im g A vôùi : R b = 10 3 Ω r b'e = = - 500 10 10 . 500 3 3 10 3 Ω Töø (2) ta coù : C b'e = 500 . 10 . 800 . 2 1 // . 2 1 3 ' = e b b h r R f = 400 pF Töø (1) suy ra : h fe = 5 , 12 10 5 , 12 3 ' = e b r = 80 Vaäy : h fe = 80; r b'e = 1K; C b'e Cho sô ñoà maïch nhö sau : Caùc thoâng soá : - ω r = 10 9 rad/s - h fe = 100 - C b'c = 5pF - r bb' = 0 - I EQ = 10 mA i CC L i L Q1 NPN +V i 20uF 20uF 20uF R 1k 100 1k 10k 1k 10k
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74 = = = = = = pF g C mho I g K K K R T m e b EQ m b 400 / 4 , 0 40 9 , 0 10 // 1 ' ω i L i L i c1 b b'e b'e e e c2 m v b'e +C M c B' E g R C R r i C C C R R r a) Tính ñoä lôïi taàn giöõa A im : cho ngaén maïch caùc tuï C c1 , C c2 , C e vaø boû qua caùc phaàn töû C b'e , C b'c (cho hôû maïch hai ñaàu caùc phaàn töû aáy). Khi ñoù : A im = i e b e b L i L i v v i i i ' ' . = vôùi : * L c c m e b L R R R g v i + - = . ' * e b b i b i e b i e b r R r R r r i v ' ' ' // // . + = Vaäy : 38 ) // ( ). // ( . . ' ' = + + - = e b b i e b b i L c c m im r R r r R r R R R g A b) Tìm taàn soá 3 dB f h : Ta coù : C b'e = 400 10 4 , 0 9 = = T m g pF
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75 Xeùt ôû taàn soá cao ta seõ thaáy raèng caùc tuï gheùp ngoaøi C c1 , C c2 , C e coù trôû khaùng raát beù do ω raát lôùn ngaén maïch caùc tuï gheùp ngoaøi. Sô ñoà chA coøn : b i b'e b'e r i //r B' L i c v m +C M L b'c C R R g C R i Vaø : C M = [1 + g m .(R c //R L )]C b'c C M = (1 + 0,4.500).5 pF C M = 1000 (pF) Taàn soá cao 3 dB : ω h = ) ( 1 ' ' M e b e b C C R + (R b'e = r i // R b // r b'e = 196 Ω ) ω h = 12 10 ). 1000 400 .( 196 1 - + = 3,64 (Mrad/s) << c b c L C R R ' ) // ( 1 tính toaùn laø hôïp lyù. Vaäy : h Cho sô ñoà maïch nhö sau vôùi caùc thoâng soá : ω T = 10 9 rad/s, C b'e = 6 pF, r bb' = 0; I EQ = 1 m±, h fe = 20 g m = 0,04 mho. i c1 i L EE - i c2 r i bb' e' R E //R L C' = +r R' b'c Yh Yh R C + - v + - v V Vcc Q1 NPN C C 1k 1k 0.5k
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76 R’ = r b’e + h fe R e ’, C’ = ' ' 1 e m e b R g C + * Ñoä lôïi taàn giöõa : hôû maïch C b'c , C' : A vm = e b i L e fe L e fe r r R R h R R h ' ) // ( ) // ( + + = 0,9 * Taàn soá cao 3 dB : R' = r b'e + h fe R e ' >> R e ' : boû qua R e '. Do ñoù taàn soá xaûy ra ñieåm cöïc : ω 1 = ) ' )( ( 1 ) ' ( ' 1 ' ' ' ' C C R h r C C R c b e fe e b c b + + = + Thay soá : R' = h fe . CQ I 25 + h fe .R E //R L = 10,5 K C' = ' ' ' 1 / 1 e m T m e m e b R g g R g C + = + ω = 2 pF Suy ra : ω 1 = 12 Mrad/s : cho Z i Ñeå tính taàn soá ω h ta xeùt hai tröôøng hôïp sau : ω β = 12 ' ' 10 . 40 . 500 1 .
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This note was uploaded on 09/08/2011 for the course ELCTRONIC 555 taught by Professor Kaimebenmahdi during the Spring '10 term at Abu Dhabi University.

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Prob2 - 2-1 Vcc R4 1k rbb' V Ii iL L R3 1k + Vbb r1 R2 ii...

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