Ch08s - Chapter 8 Problem Solutions 8.1 a. b. i. VD D = 80...

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Chapter 8 Problem Solutions 8.1 a. b. i. 80 V DD V = Maximum power at 40 V 2 25 0.625 A 40 80 40 64 0.625 DS T D DS V V P I V RR == = = ii. 50 V V = Maximum power at 25 V 2 25 1 A 25 50 25 25 1 T D DS V V P I V = = 8.2 a.
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(max) 2 2( m a x ) 2(20) So 1.67 A 24 (/ 2 ) 24 12 7.2 1.67 1.67 20.8 mA 80 24 0.7 1.12 k 20.8 CC QC Q Q CQ CC CC CC LL CQ CQ B BB V PI P I V VV RR I I I β =⋅ == = = b. () ( ) 0 0 1.67 7.2 462 0.026 (max) 12 (max) 12 V 26 mV 462 CQ L vm L T PP v IR Ag R V V V A = = = 8.3 a. For maximum power delivered to the load, set 2 CEQ V V = sus Set 25 V CE ( ) ,max D,max 25 Then 0.1 250 mA 25 12.5 125 mA 0.1 max 0.125 12.5 2 1.56 W P 125 1.25 mA 100 25 0.7 19.4 k 1.25 CC Cm L Cm C CQ CC Q BQ V I R II I V I =< = = b. ( ) ( ) ( ) 2 2 11 max 0.125 100 max 0.781 W rms 22 LC Q L L R P ⋅ = = 8.4 Point (b): Maximum power delivered to load. Point (a): Will obtain maximum signal current output. Point (c): Will obtain maximum signal voltage output.
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8.5 a. b. () 2 2 2 2 5 V, 0.25 5 4 0.25 A, 37.5 V, 9.375 W 6 V, 0.25 6 4 1.0 A, 30 V, 30 W 7 V, 0.25 7 4 2.25 A, 17.5 V, 39.375 W 8 V, 0.25 2 8 4 40 2.92 10 3. GG D D S GG D D S GG D D S GG D D S D S DS D VI V P V P V P V V V V I == = = = = = = = = = ⎡⎤ ==− ⎣⎦ = = = 2 71 A, 10.8 W 9 V, 0.25 2 9 4 40 1.88 V 10 3.81 A, 7.16 W GG D D S D S D P V V V V IP = = = c. Yes, at ,max 7 V, 39.375 W > 35 W GG D VP P = 8.6 a. 2 2 12 2 2 1 Set 25 V 2 50 25 1.25 A 20 1.25 4 6.5 V 0.2 Let 100 k 6.5 50 13 k 100 87 k DD DSQ DQ DQ n GS TN GS GS DD V V I IK V V V R VV RR R R R =− += = ⎛⎞ = ⎜⎟ + ⎝⎠ = = = b. ( ) 1.25 25 31.25 W DD Q D S Q D PIV P = c. ,max ,max ,max ,max ,max 22 . 5 A 50 V 31.25 W Q D DS DD DS D II I VVV P = = = = = d.
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() ( ) 0 0 2 2 0 22 0 . 2 1 . 2 5 1 / 1 20 0.5 10 V 10 11 2.5 W 2 0 31.25 2.5 28.75 W mL i mn D Q LL L QQ V gR V g KI AV V V PP R = == = =⋅ = =− = 8.7 (a) (b) ,max 25 DD j P P Slope T ,max ,max 60 At 0, 25 145 C 0.5 Dj j PT T + (c) ,max ,max j case D dev amb TT P θ = or 145 25 2 60 dev amb dev amb C/W θθ −− = 8.8 ,max ,rated dev case ,max amb dev case ,rated or 150 25 2.5 C/W 50 = = ° ja m b D j D P P dev amb dev case case amb case amb case amb Then 150 25 2.5 125 2.5 −= + + =+ D P 8.9 dev amb dev case case snk snk amb dev dev dev case dev case case dev case case sink case snk sink c 45 2 0 W 25 20 1.75 0.8 3 111 136 C 20 1.75 35 35 136 35 101 C 20 0.8 16 C θθθ =⋅ = = + + + += = = = = ° = D S D D D PIV P P T P ase sink 16 101 16 85 C −= − T 8.10
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() dev amb dev case case amb case amb case amb 200 25 25 3 4 C/W θθ −− −= + + D TT P 8.11 ,max amb dev case ,rated ,max amb dev case case snk snk amb 175 25 10 C/W 15 175 25 10 W 10 1 4 θ θθθ == = ° = ++ = = j D j D D P P P 8.12 2 1 2 50% L S SC C Q CC LP P Q CC Q CC Q P P PVI V I VI η ηη = =⋅ ⎛⎞ =⋅= ⎜⎟ ⎝⎠ ⋅⋅ = = 8.13 32 S max 4.8 V 0.7 5 4.3 mA 1 min 0.7 max 4.3 mA 1 so 3.6 5.5 V min 4.3 V o CC Io L v ii v vv i = = =+ = = −≤ = 8.14
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() () () ()( ) 2 3 33 2 2 3 2 3 32 2 22 0 20 05 K 12 0.5 5 21 1 2 0 11 11 4 12 2 212 1.072 V 12 1.072 0.5 3.93 mA sat 1.072 0.5 0.572 V min min : max 3.93 min 3.93 V 1 mi GS DG S T N GS GS GS GS GS GS GS DD DS GS TN o I V IV V R VV V II V V vi V v −− =− = −= = ±+ = == = = = = ( ) 1 2 11 1 1 n min 3.93 0.5 min 3.43 V max 5 sat 5 0.572 max 4.43 V 4.43 max 3.93 8.36 mA 1 8.36 12 0.5 1.33 V max 4.43 1.33 max 5.76 V oT N I oD S o D S G S Io G S I vV v v I V vv V v =+ = + = = = = + = 8.15 a. Neglect base currents. 0
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This homework help was uploaded on 02/10/2008 for the course ECE 138 taught by Professor N/a during the Spring '08 term at CSU Fresno.

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Ch08s - Chapter 8 Problem Solutions 8.1 a. b. i. VD D = 80...

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