Prob4 - 103 f o =30Mhz r b , e =1 K r bb' =0 h fe =100 f T...

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Unformatted text preview: 103 f o =30Mhz r b , e =1 K r bb' =0 h fe =100 f T =500Mhz C b'c =2p T e b fe e b r h C ω 1 ' ' = → C b'e =31,8pF 4_3 Cho maïch khueách ñaïi coäng höôûng nhö sau : R 2 R 1 100 L->oo->oo->oo i Vcc i L Ce Cc1 RFC Cc2 R 1k Re 1k 10k Sô ñoà thay theá R b =1 K //10 K ~1 K ,q m =h fe /r b'e = 1 , mho 1K 1K 1k ' B b ' e ' b L L m ' b g v i i L Rb r C R V Ta coù R=R b //r b'e =0,5K C b' = C b'e + C M = 31,8p + (1+g m R L )C b'c = 31,8p + (1+0,1.10 3 )2p = 234pF Coäng höôûng taïi f o = 30 Mhz thì ) ( 30 2 1 ' Mhz LC f b o = = π suy ra : H f C L o b μ π π 12 , ) 10 . 30 ( 4 . 10 . 234 1 ) 2 ( 1 2 6 2 12 2 ' = = =- ♦ Baêng thoâng : ) ( 36 , 1 10 . 234 . 500 . 2 1 2 1 12 ' Mhz RC BW b = = =- π π ♦ Ñoä lôïi doøng : 104 ) ( 1 . ' ' ω ω ω ω o o i m i b b L i L i jQ R g i v v i i i A- +- = = = Vôùi : 22 10 . 234 . 10 . 5 . 10 . 3 . 28 , 6 12 2 7 ' ≈ = =- b o i RC Q ω neân : A im = -g m R = -0,1.500 = -50 ) 10 . 3 10 . 3 ( 22 1 1 50 7 7 ω ω- +- = ⇒ j A i 4_4 Thieát keá maïch coâng höôûng ñôn coù : A im = 10db = 3,16 r i = 1 K f o = 40 Mhz R L = 1 K BW = 1Mhz Vcc = 10 V Cuoän daây coù Q = 50; Sô ñoà caàn thieát keá coù daïng nhö sau : i C i L 1k 1k e ' b L L m ' b g v i i r L i r Rp C R V ôû ñaây ta choïn transistor coù : g m =0,01 mho ; r bb' =0 ; h fe = 100 C b'c = 10p, C b'e =1000p 2 2 ' 10 10- = = m fe e b g h r Ta coù : 6 10 2 1 = = RC BW π vôùi R = r i //Rp//r b'e C = C b'e + (1+g m R L )C b'c + C' : C' laø tuï gheùp ngoaøi 105 vaø c o c o e b i Q C Q C r Rp r R ω ω + = + + = + + =- 3 ' 10 . 4 , 1 2500 1 1000 1 1 1 1 1 vôùi c c r L Q ω = : heä soá toån hao cuûa cuoän daây Do ñoù : ) 50 ) ' 1110 ( 10 . 4 , 1 .( ) ' 1110 ( 2 1 10 3 6 C pF...
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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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Prob4 - 103 f o =30Mhz r b , e =1 K r bb' =0 h fe =100 f T...

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