Ch13p - Chapter 13 Exercise Solutions EX13.1 I C1 = I C 2...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 13 Exercise Solutions EX13.1 12 9.5 A 9.5 A 0.0475 A 47.5 nA 200 CC BB II μ =≅ == = EX13.2 17 13 16 16 50 . 60 . 6(5 ) 0.22 mA 40 0.75 (0.75)(0.22) 0.165 mA 0.165 (0.165)(0.1) 0.6 200 50 0.000825 0.01233 13.2 A REF B R E F C C I I I I −−− = = + =+ = EX13.3 31 4 0.18 10 10 exp D T V V −− ⎛⎞ ×= ⎜⎟ ⎝⎠ 3 14 3 14 0.18 10 ln 10 0.18 10 (0.026)ln 10 0.6140 DT D VV V × = × = = 21 . 2 2 8 V BB DD 14 20 14 2 exp 0.6140 0 e x p 0.026 BB CCS T V III V / 14 20 0.541 mA EX13.4 6 100 10.5 M 0.0095 o r = Ω Then, using results from Example 13.4
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
[ ] [ ] () 16 6 26 4 4 1 ( || ) 10.5 1 (0.365)(1|| 547) 14.3 M 100 10.5 M 0.0095 9.5 10.5 ||14.3 || 4.07 889 0.026 act o m A o C d Rr g R r V r I A π =+ = + == Ω ⎛⎞ =− ⎜⎟ ⎝⎠ EX13.5 19 13 32 2 1 9 2 0 2 17 2 2 100 556 K 0.18 (1 )[ || ] 7.22 (51)(556 ||111) 4.73 M 100 185 K 0.54 100 185 K 0.54 (200)(201)(50)(185 || 4730 ||185) 4070[50 [9.63 (201)(0.1)]] 182358786.9 32450.1 5 oA iP act o v v RR R R A A β = + Ω = ++ = 62 EX13.6 [] 17 13 17 17 22 19 100 100 185 K 185 K 0.54 0.54 185 1 (20.8)(0.1|| 9.63) 566 K 7.22 566 ||185 2.88 K 51 100 556 K 0.18 oo B C C e C rr R R R R = + 20 0.65 2.88 || 556 0.0689 51 68.9 22 68.9 90.9 e O R R + = Ω EX13.7 12 2 11 4 1 | | 30(1 562) 16890 pF 4.07 M || 14.3||10.5 6.05 M i i oa c t o CC A R r = = Ω Then || 6.05 || 4.07 2.43 M eq o i R = Ω Then 61 2 2 2 (2.43 10 )(16890 10 ) 3.88 Hz PD eq i f RC ππ ×× =
Background image of page 2
EX13.8 For 5 , M 6 M 2 2 5 5 2 55 5 2 5 5 40 (12.5) 250 A / V 2 0.25( 0.5) 225 56.25( 0.25) 10 56.25 55.25 4.0625 0 55.25 3052.5625 914.0625 2(56.25) 0.902 V 10 0.902 40.4 A 225 P SG SG SG SG SG SG SG SG SG set Q K V V VV V V V II μ ⎛⎞ == ⎜⎟ ⎝⎠ +− −= −+ = = ±− = = Current in 14 20.2 A MM EX13.9 2 12 40 (12.5) 250 A/V 2 PP KK = From Exercise Ex 13.8, 20.2 A DD 42 4 2 7 77 78 7 27 11 2.48 M (0.02)(0.0202) 2 ( || ) 2(0.25)(0.0404)(2480 || 2480) 176 80 (12.5) 500 A / V 2 2 2 (0.50)(0.0404) 0.284 mA/V 1.24 M (0.02)(0.0404) ( oo D dP Q o o d n mn Q D vm rr I AK I r r A K gK I I Ag λ = = Ω = = = Ω = 2 || ) (0.284)(1240 ||1240) 176 (176)(176) 30,976 Vd V AA A =⋅ = = EX13.10 (a) 1 1 1 0.08 0.25 2 2 (20) 22 0.6325 mA/V Q n m I k W g L ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ ⎝ ⎠ = 2 4 1 800 K (0.01)(0.25 2) 1 533.3 K (0.015)(0.25 2) o o r r / / 2 4 1 ( || ) (0.6325)(800 || 533.3) 202.4 dm o o d r r A =
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
52 5 0.04 22 ( 8 0 ) ( 0 . 2 5 ) 1.265 mA/V n mQ k W gI L ⎛⎞ ⎛ ⎞ == ⎜⎟ ⎜ ⎟ ⎝⎠ ⎝ ⎠ = 5 9 1 266.7 K (0.015)(0.25) 1 400 K (0.01)(0.25) o o r r 25 5 9 2 ( || ) (1,265)(266.7 || 400) 202.4 mo o Ag r r A =− Overall gain (assuming 3 1) A = 12 (202.4)( 202.4) 40,966 d AAA =⋅ = = EX13.11 18 66 1681 0 4 4 8 25 40 2 2 (25)(25) 224 / 80 2 2 (25)(25) 316 / 11 2 (0.02)(25) 1 (0.02)(50) DD p mm D
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 11

Ch13p - Chapter 13 Exercise Solutions EX13.1 I C1 = I C 2...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online