# Prob3 - Chng 3 KHUECH AI CONG SUAT AM TAN 3-16 V IC = 0,2A nen ay la iem Q bat ky a o doc cua ng tai AC V CC= 20V 1,58:1 R R L 20 b 1 1 1 1 =2 = =

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95 Chöông 3 : KHUEÁCH ÑAÏI COÂNG SUAÁT AÂM TAÀN ____________ Vì I C = 0,2A neân ñaây laø ñieåm Q baát kyø. a) Ñoä doác cuûa ñöôøng taûi AC : 50 1 20 . ) 58 , 1 ( 1 1 1 2 2 = = = L AC R N R = 0,02 b) Phöông trình ñöôøng taûi AC trong heä toïa ñoä toång quaùt : i C – I CQ = - AC R 1 (v CE – V CEQ ) + Cho v CE = 0 max C i = I CQ + 50 20 2 , 0 + = AC CEQ R V = 0,6A + Cho i C = 0 max CE v = V CEQ + I CQ. R AC = 20 + 0,2.50 = 30V c) Giaù trò ñ±nh cöïc ñaïi cuûa ñieän aùp collector khi khoâng bò saùi daïng : max Cm V = min[V CEQ , I CQ .R AC ] = min[20 , 0,2.50] = min[20,10] = 10V i C (A) v CE (V) 0 Q 0,2 i Cmax = 0,6 V CC = 20 v CEmax =30 DCLL( ) ACLL(-0,02) 20 b + CC = 1,58:1 L V 20V R R

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96 d) P L = 50 10 . 2 1 . 2 1 . 2 1 2 2 2 2 = = L Cm L Lm R N V R V = 1W e) P CC = I CQ .V CC = 0,2.20 = 4W η = 4 1 = CC L P P = 0,25 = 25% << η max = 50% Gioáng hình baøi 3-16 coù β = 40. a) Ñeå ñieän aùp ngoõ ra ñAnh – ñAnh cöïc ñaïi : A R V I I AC cc CQ cm 4 , 0 50 20 max = = = = A h I I fe CQ BQ 01 , 0 40 4 , 0 = = = b) max L P = AC Cm R I . 2 1 2 max = 2 1 (0,4) 2 .50 = 4W c) P CC = I CQ .V CC = 0,4.20 = 8W η max = 8 4 max = CC L P P = 0,5 = 50% max max max L CC L C P P P P - = = = 4W Cho I p = I Cm = 4±; V p = V Cm = 12V a) P L = 2 1 . 2 1 = Cm Cm I V .4.12= 24W b) P CC = I TB .V CC = CC Cm V I . 2 π L i L - + in in T1 T2 R 8 24V v v
97 = 24 .

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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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Prob3 - Chng 3 KHUECH AI CONG SUAT AM TAN 3-16 V IC = 0,2A nen ay la iem Q bat ky a o doc cua ng tai AC V CC= 20V 1,58:1 R R L 20 b 1 1 1 1 =2 = =

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