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# Ch06p - Chapter 6 Exercise Problems EX6.1(a I BQ = VBB VBE...

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Chapter 6 Exercise Problems EX6.1 (a) () 20 . 7 2 650 BB BE BQ B VV o n I A R μ == Then ( ) ( ) 100 2 0.20 50 . 21 5 2 CQ BQ CEQ CC CQ C I IA m A I R V βμ = =− = = (b) Now 0.20 7.69 / 0.026 CQ m T I g mA V V = and ( ) 100 0.026 13 0.20 T CQ V rk I π β = Ω (c) We find ( ) 13 7.69 15 2.26 13 650 vm C B r Ag R rR ⎛⎞ ⎜⎟ + ⎝⎠ + EX6.2 0.92 0.7 100 BB BE BQ B o n I R or 0.0022 BQ I mA = and ( ) 150 0.0022 0.33 CQ BQ I Im A = (a) ( ) 0.33 12.7 / 0.026 150 0.026 11.8 0.33 200 606 0.33 CQ m T T CQ A o CQ I gm A V V V I V I = = Ω = Ω (b) and om o C s B r vg v r R v v ππ = + & so 11.8 12.7 606 15 11.8 100 which yields 19.6 o o C B v vr r R R A + + & & EX6.3 (a) 1.145 0.7 50 BB EB BQ B o n I R or 0.0089 BQ I mA = ( ) Then 90 0.0089 0.801 CQ BQ I A =

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Now () ( ) 0.801 30.8 / 0.026 90 0.026 2.92 0.801 120 150 0.801 CQ m T T CQ A o CQ I gm A V V V rk I V I π β == = = Ω = Ω (b) We have omo C Vg V rR = & and s B r VV ⎛⎞ =− ⎜⎟ + ⎝⎠ so 2.92 30.8 150 2.5 2.92 50 which yields 4.18 o vm o C sB v Vr Ag r R R A + + & & EX6.4 Using Figure 6.23 (a) For 0.2 , CQ Im A = 7.8 15 , 60 125, ie fe hk h << Ω < < 44 6 . 21 0 5 01 0, re h −− ×< < × 5 13 oe hm h o s μ << (b) For 5 , CQ I mA = 0.7 1.1 , 140 210, ie fe h Ω < < 1.05 10 1.6 10 , re h < × 22 35 oe h o s EX6.5 12 2 35.2 || 5.83 5 5.83 5 5.83 35.2 TH TH CC RR R k R = Ω =⋅ = ++ & or 0.7105 TH = Then 0.7105 0.7 5 TH BE BQ TH o n I R or 2.1 BQ I A = and ( ) ( ) 100 2.1 0.21 50 . 2 1 1 0 CQ BQ CEQ CC CQ C I IA m A I R βμ = = and 2.9 CEQ = Now 0.21 8.08 0.026 100 476 0.21 CQ m T A o CQ I g mA V V I = = Ω And ( ) ( ) 8.08 476 10 o C r R && so 79.1 v A EX6.6
() 12 2 250 75 57.7 75 5 75 250 TH TH CC RR R k R VV == = Ω ⎛⎞ ⎜⎟ ++ ⎝⎠ && or 1.154 1.154 0.7 1 57.7 121 0.6 TH TH BE BQ TH E o n I β = + or ( ) 3.48 120 3.38 0.418 BQ CQ BQ IA I m A μ βμ = = (a) Now ( ) 0418 16 08 0026 120 0 026 746 0 418 CQ m T T CQ I . g. m A / V V. . V r. k I. π = β = Ω We have om C Vg V R π =− We find ( ) ( ) 1 7.46 121 0.6 ib E Rr R =++ = + or 80.1 ib Rk Also 250 75 57.7 57.7 80.1 33.54 ib k RRR k Ω Ω & We find 33.54 33.54 0.5 ib s ss ib S V RRR R =⋅ = or 0.985 s s ′ = Now 11 2 1 0 . 6 7.46 sE R V r ππ ⎡⎤ + =+ ⎢⎥ ⎣⎦ or ( ) 0.0932 0.0932 0.985 s s V So ( ) ( ) ( ) 16.08 0.0932 0.985 5.6 o v s V A V or 8.27 v A EX6.7 2 15 85 12.75 85 12 15 85 TH TH CC R k R ===Ω = ⎜⎟⎜⎟ or 10.2 TH = Now ( ) 1 CC BQ E EB BQ TH TH VI R V o n I R V + + +

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so () () ( ) 12 101 0.5 0.7 12.75 10.2 BQ BQ II =+ + + which yields 0.0174 BQ Im A = and () ( ) ( ) ( ) ( ) 100 0.0174 1.74 101 0.0174 1.76 12 1.76 0.5 1.74 4 CQ BQ EQ ECQ CC EQ E CQ C I A A VV I R I R β == = =− or 4.16 ECQ = Now ( ) 100 0.026 1.49 1.74 T CQ V rk I π = Ω We have ππ = ⎛⎞ + ⎜⎟ ⎝⎠ om CL s mE Vg VR | | R V VVg V R r Solving for V and noting that , m g r = we find ( ) 1 100 4 2 1.49 101 0.5 o v sE RR V A Vr R ++ = + & & or 2.56 v A EX6.8 Dc analysis 10 0.7 0.00439 100 101 20 0.439 , 0.443 BQ CQ EQ I mA A A + Now ( ) 100 0.026 5.92 0.439 0.439 16.88 / 0.026 100 228 0.439 m o gm A V Ω Ω (a) Now 100 5.92 5.59 o C B s BS B V r R Rr RrR k =⋅ + Ω & & & && 5.59 Then 0.918 5.59 0.5 s s V = + ( ) Then 16.88 0.918 228 10 o v s V A V & or 148 v A (b) 0.5 100 5.92 6.09 in S B R r k = + = Ω
and 10 228 9.58 oC o RRr k == = Ω && EX6.9 (a) () ( ) 0.25 9.615 / 0.026 100 400 0.25 9.615 400 100 769 CQ m T A o CQ vm o c I gm A V V V rk I Ag r r = = Ω =− (b) ( ) ( ) 9.615 400 100 100 427 o c L v r r r A & & EX6.10 ( ) ( ) ( ) ( ) 50 . 7 0.00672 10 126 5 0.84 , 0.847 10 0.84 2.3 0.847 5 BQ CQ EQ CEQ I mA Im A I m A V + or 3.83 CEQ VV = dc load line CE C C E V I R R +− ≅−− + or 10 7.3 CE C VI ac load line (neglecting r o ) ( ) ( ) 2.3 5 1.58 ce c C L c c vi R R i i 1.37 0.84 3.83 10 I C (mA) AC load line DC load line Q -point V CE (V) EX6.11 (a) dc load line ( ) 12 4.5 EC CC C C E C VVI R R I ≅− + = ac load line ec c E C L R R R + & or ( ) 0.5 4 2 1.83 ec c c i + & Q-point values 12 0.7 10.2 0.0150 12.75 121 0.5 1.80 1.82 3.89 BQ CQ EQ ECQ I mA A A −− + =

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2.67 1.80 3.89 12 AC load line DC load line I C (mA) Q -point V EC (V) (b) () ( ) 1.80 1.8 1.83 3.29 min 3.89 3.29 0.6 CC Q EC EC iI m A vV
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Ch06p - Chapter 6 Exercise Problems EX6.1(a I BQ = VBB VBE...

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