Ch11p - Chapter 11 Exercise Solutions EX11.1 vE = -VBE ( on...

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Chapter 11 Exercise Solutions EX11.1 () ( ) 12 on 0.7 V 0.5 mA 10 0.5 10 5 V EB E E CC vV v II vv =− == EX11.2 2 1 0.99 1e x p 1 x p 0.99 1 exp 1 0.99 1 ln 1 119.5 mV 0.99 C Q d T d T d T dT d i I v V v V v V v ⎛⎞ + ⎜⎟ ⎝⎠ += ⎡⎤ ⎢⎥ ⎣⎦ EX11.3 a. ( ) 1 0 0.7 V 0.25 8 2 V 3 V 3.7 V E RC C C EC v Vv v v = Δ= = = b. 1 2.5 V 3.2 V 6.2 V EE C v v = = c. 1 2.5 V 1.8 V 1.2 V C v v = EX11.4 Let 1 , Q I mA = then 0.5 CQ CQ I Im A 0.5 19.23 / 0.026 mm g gm A V = At 2 22 1 , 2 c Cd m C d v vA g R v So, 1 150 19.23 15.6 2 RR k = At 1 11 1 2 c m C d v g R v So, 1 100 19.23 10.4 2 k −= If 10 and 10 , VV +− =+ dc biasing is OK. EX11.5 ( ) ( ) ( ) ( ) ( ) ( ) 0.8 12 20 .026 2 0.0594 1 101 0.8 100 1 1 0.026 100 0.0594 1sin 59.4sin QC T cm QO T oc m c m IR V A V v t m V t V β ωω μ = + + + = EX11.6
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(a) () 40 , 0 50 40 2.0 dc m o vV v vm V μ == = (b) 40 200 m v V μμ Assuming 0 cm A > ()( ) ( ) 10 50 60 20log 0.05 50 40 0.05 200 2.010 cm cm o o A A v V = = =+ = EX11.7 a. Diff. Gain 4 QC d T I R A V = For 12 5 V vv Minimum collector voltage 2 5 V 15 5 10 V 2 Q CC I vR = ⋅= = or 20 V IR = for max. d A Then 20 max 192 20 .026 dd AA = = b. If 0.5 mA, 40 k ( ) 0 10 2 1 1 Then 20 0.199 201 0.5 100 1 0.026 200 192 and 20log 59.7 dB 0.199 T cm Q T cm cm dB dB V A V CMRR β ⎛⎞ ⎜⎟ ⎝⎠ = + ⎡⎤ + ⎢⎥ ⎣⎦ = =− + = = EX11.8 ( ) 21 100 0.026 10.4 K 0.25 2 10.4 101 0.5 122 K id E T CQ id Rr R V r I R π + = = EX11.9 ( ) 9.62 10 10 2 1 9.62 mC d mE E gR A R = + = + 1 9.62 4.81 0.396 K EE RR += = ( ) ( ) 2 1 2 10.4 101 0.396 100.8 K id E id R R + = + =
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EX11.10 () ( ) ( ) 2 4 13 4 1 2 44 2 444 2 2 4 10 10 0.1 80 0.8 10 8 1.6 0.64 81 1 . 84 . 8 8 0 11.8 11.8 4 8 4.88 1.81 V 28 GS nG S T N GS GS GS GS GS GS GS GS V IK V V R VV VVV V ==− −= + −− = ±+ == ( ) 1 12 2 11 2 01 02 10 1.81 0.102 mA 80 0.102 0.0512 mA 2 0.0512 0.050 0.8 1.81 V 5 0.0512 40 2.95 V Q DD N GS GS II KV V vv = = =− = = ( ) ( ) 01 1 1 14 : 1.81 0.8 1.01 V max sat 2.95 1.01 1.81 max 3.75 V : 1.81 0.8 1.01 V min sat 5 1.81 1.01 5 min 2.18 V 2.18 3.75 V cm DS GS TN cm DS GS cm cm DS GS TN cm GS DS cm cm Max v V sat V V V V v Min v V sat V V vV V v v = + + = = =+ =+− −≤ EX11.11 max max 1 / 22 15 5 nQ ff df D d KI g gm A V Ag R A = = EX11.12 2 1 1 1 2 nn D dd QQ Q KK i I ⎛⎞ ⎜⎟ ⎝⎠ Using the parameters in Example 11.11, 2 0.5 / , 1 , K mA V I mA then 2 1 10 . 5 0 . 5 0.90 1 1 2 1 D Q i I + ⋅− By trial and error, 0.894 d = EX11.13 a.
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() 2 02 2 4 04 4 02 04 0.5 9.615 mA/V 2 2 0.026 125 500 k 0.25 85 340 k 0.25 9.615 500 340 1946 Q m T A C A C dm d I g V V r I V r I Agrr A == = = = = b. 02 04 9.615 500 340 100 644 L dd AgrrR AA = ⎡⎤ = = ⎣⎦ c. ( ) 150 0.026 15.6 k 0.25 23 1 . 2 k T CQ id id V r I Rr R π β = = = d.
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Ch11p - Chapter 11 Exercise Solutions EX11.1 vE = -VBE ( on...

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