# Prob5 - BAI TAP CHNG 5 5_15 4V i1 < 10k 0,6V o io > i2 30k...

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114 Ta coù o o K K K v v v 6 , 0 30 20 30 = + = + o v v v 6 , 0 = = + - Ta coù : a) v - = 4 K .1mA + v o = v o + 4 (V) ->v + = v - = v o + 4 (V) maø v + = -1 K .1mA = -1V BAØI TAÄP CHÖÔNG 5 5_15 0,6V 0,6V > io < < < i1 i2 i3 Vo 20k 30k 10k + - 4V suy ra ) ( 02 , 0 20 6 , 0 ) ( 02 , 0 30 6 , 0 ) ( 06 , 0 10 6 , 0 3 2 1 mA v v v i mA v v i mA v v i o K o o o K o o K o = - = = = = = vaø 4 V = v o -0,6v 0 =0,4v o -> v o = 10(V) do ñoù: i 1 = 0,6 (mA) i 2 = 0,2 (mA) i 3 = 0,2 (mA) b i o = i 1 + i 3 = 0,8 (mA) Coâng suaát phaùt ra töø nguoàn : P ng = 4 V .i 1 = 4.0,6 = 2,40 (mV) 5.16 V - V + 1mA 4k Vo 1k 2k 3k neân : v o + 4 = -1 -> v 0 = -5 V

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115 Neáu coù ñieän trôû noái giöõa AB laø 5 K thì V AB =10 -3 .5.10 3 =5V Ta coù : v - = 4 K .i +v o = v + v - =-5 K .i + 5 + v + a) Ta coù : 5 50 10 o V Vo k k v v = = = - + (1) maø : K o K s v v v v i 20 50 - = - = - - suy ra : suy ra : v + = v - = -1 ( V) b) V+ v- > < i i + - 5V 4k Vo 1k 3k b i = 5/5k = 1 (mA) do v + = v - suy ra : v + = v - = -i.1 K = -1 (V) 5.17 V - V+ > i + - Vs 9V Vo 40k 20k 10k 50k hay 5 2 7 s o v v v - = - (2) Töø (1) vaø (2) : - - = - v v v s 5 5 2 7 hay 18v - =-2v s + - = - = - = v V v v s ) ( 1 9 neân v o = -5(V) b)Tìm giaù trò R ñeå v o = -10(V) (taêng gaáp ñoâi)
116 Taïi ñaàu nuùt (1) : K o K s v v R v v v 20 50 - + = - - - - (2) Khöû v - ôû (1) & (2) ta coù : K o o K K K s v v o R v 20 5 ) 20 1 5 1 1 ( 50 - + + = a) Ta coù : ; 3 . 3 1 A A v v K K v = = + Maët khaùc v + =v - neân : v A = 3v + = 3v - suy ra taïi nuùt A : K A K A S v v v v i 2 3 + - - + - = + + - + = + = v v v K K K K ) 2 2 3 2 ( 2 2 3 2 Ta vaãn coù : 5 o v v v = = - + (1 V - V + > i < i 1 > ie R + - Vs 9V Vo 40k 20k 10k 50k Thay soá : K K V V K K K V R R 11 , 11 20 10 5 10 ) 20 1 50 1 1 ( 50 9 = + - + + = 5_19 v + v- A is Vo 1k 4k 2k 3k vôùi i s = 1mA thì v + = 0,6 V K A K o v v v v 3 . 4 - - - - = hay ) ( 1 3 6 , 0 . 2 4 6 , 0 V v K V K V o - = - = b) Neáu coù ñieän trôû 3 K maéc // nguoàn doøng Ta vaãn coù :

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117 K A K A K A V v v v v v 2 3 3 3 + - - + - = - ) ( 125 , 1 ) ( 375 , 0 V v v V v A = = = - + x Khi K=0: v + = v - = 0 neân maïch laø khueách ñaïi ñaûo : 1 2 R R A - = a) v A = 3v + = 3v - Taïi A : v + v- A 3k + - 3V Vo 1k 4k 2k 3k neân : K K K A K o V v v v v 3 375 , 0 . 2 . 4 ) ( 375 , 0 3 4 - = - - = - - v o = -0,625(V) 5_22 R2 R3 kR3 Vo R4 R1 + - Vi x Khi K = 1 v + = v - = v i ->taïo maïch ñeäm (do khoâng coù doøng qua R 1 &R 2 ) ->A=1 Vaäy 1 1 2 - A R R b) Khi coù noái theâm R 4 Ta coù v + = v - = kv i (1) Taïi nuùt 1 :
118 2 4 1 R v v R v R v v o i - + = - + + + (2) Khöû v + ôû (1) & (2) ta coù : 2 2 4 1 1 1 1 1 R v R R R kv R v o i i - + + = neân 2 1 2 4 1 1 1 1 1 R v R R R R k v o i = - + + -> - + + = 1 2 4 1 2 1 1 1 1 R R R R k R A c) Ñeå coù : 5 5 - A

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Prob5 - BAI TAP CHNG 5 5_15 4V i1 < 10k 0,6V o io > i2 30k...

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