Ch08p - Chapter 8 Exercise Solutions EX8.1 a VCC = 30 V VCE...

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Chapter 8 Exercise Solutions EX8.1 a. 30 V, 30 , 10 CC CE C C C CE VV I R I V == = 1 Maximum power at 15 2 10 10 2 15 3 2 So 15 30 22.5 Maximum Power 10 W 3 CE CC C CE LL I V RR = =− = b. () ( ) ,max 15 V, 2 A 15 0 15 2 7.5 Maximum Power 1 7.5 7.5 W CC C CE C L VI R EX8.2 (a) 82 . 4 19.2 C TP T θ ° Δ= = (b) 85 3.7 23.0 W T P P Δ = EX8.3 Power 1 12 12 watts DD S iv =⋅ = = c. ( ) sink amb snk amb sink sink 25 12 4 73 C TTP TT =+ b. ( ) case sink case snk case case 73 12 1 85 C a. ( ) dev case dev case dev dev 85 12 3 121 C EX8.4
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() ( ) ,max amb dev case D,rated ,max amb ,max dev case case snk snk amb ,max case amb ,max case snk snk amb case 200 25 3.5 C/W 50 200 25 29.2 W 3.5 0.5 2 25 29.2 0.5 2 98 C J J D D D TT P P P TTP T θ θθθ θθ −− == = ° = ++ = = =+ + + EX8.5 a. 10 4 60 mA 0.1 DQ DQ II = = b. ( ) ( ) 9 60 0.050 2.7 V min 4 2.7 1.3 V 10 ds DS vv ⎛⎞ =− = ⎜⎟ ⎝⎠ So maximum swing is determined by drain-to-source voltage. 22 . 5 5 . 0 V PP V = c. ( ) 2 2 2.5 11 31.25 mW 0 . 1 10 60 600 mW 31.25 5.2% 600 P LL L SD D D Q L S V PP R PV I P P ηη =⋅ = =⋅= = = EX8.6 Computer Analysis EX8.7 No Exercise Problem EX8.8 a. 0 00 2 2 1 BB IG S n GSn I Dn n GSn TN Dn GSn TN n GSn GSn Dn oD n o V vvv dv dv dv dv iK vV i K dv dv di dv di dv
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111 So 2 GSn Dn nD n dv di K i =⋅ At 0 0, 0.050 A Dn vi == So 11 1 5 2 0.2 0.050 GSn Dn dv di = D nLD p ii i =+ For a small change in () 0 L Dn Dp i i →Δ − −Δ So 1 2 D nL Δ=Δ or 00 11 1 1 1 0.025 22 2 2 0 Dn L L di di dv dv R = Then ( ) 0 5 0.025 0.125 GSn dv dv Then 0 1 0.125 1.125
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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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Ch08p - Chapter 8 Exercise Solutions EX8.1 a VCC = 30 V VCE...

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