Ch05s - Chapter 5 Problem Solutions 5.1 (a) iC 510 = = 85 6...

Info iconThis preview shows pages 1–7. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 5 Problem Solutions 5.1 (a) () ( ) ( ) 510 85 6 85 0.9884 18 6 6 6 5 1 6 C B EB E i i ii i A ββ β αα βμ == = = + =+ = = (b) ( ) ( ) 2.65 53 0.050 53 0.9815 54 1 54 0.050 2.70 E i m A = = = = = = 5.2 (a) For 110: = 110 0.99099 11 1 1 α = + For 180: = 180 0.99448 181 0.99099 0.99448 ≤≤ (b) 110 50 A 5.50 CB C I II m A = or 180 50 A 9.00 CC I Im A μ = = so 5.50 9.0 C I mA 5.3 (a) 1.12 9.33 A 120 121 1.12 1.13 mA 120 120 0.9917 121 B E i i = ⎛⎞ ⎜⎟ ⎝⎠ (b) 50 2.5 mA 20 21 50 52.5 mA 20 20 0.9524 21 B E i i 5.4 (a) 1 = 0.9 9 0.95 19 0.98 49 0.99 99 0.995 199 0.999 999 (b)
Background image of page 1

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

View Full DocumentRight Arrow Icon
β 1 α = + 20 0.9524 50 0.9804 100 0.9901 150 0.9934 220 0.9955 400 0.9975 5.5 (a) () ( ) 1.2 14.8 A 81 80 1.2 1.185 mA 81 80 0.9877 81 5 1.185 2 2.63 V B C C I I V μ = ⎛⎞ == ⎜⎟ ⎝⎠ =− = (b) ( ) 0.80 9.88 A 81 80 0.80 0.790 mA 81 80 0.9877 81 5 0.790 2 3.42 V B C C I I V = = (c) Yes, CB VV > so B-C junction is reverse biased in both areas. 5.6 For 5 0, 2.5 mA 2 CC VI = 2.5 2.546 mA 0.982 C EE I II = 5.7 (a) ( ) 0.75 12.3 A 61 60 0.75 0.738 mA 61 60 0.9836 61 10 0.738 5 10 6.31 V B C C C I I R V = = (b) ( ) 1.5 24.6 A 61 60 1.5 1.475 mA 61 60 0.9836 61 1.475 5 10 2.625 V B C I I = (c) Yes, 0 C V < in both cases so that B-C junction is reverse biased.
Background image of page 2
5.8 () 10 10 1.2 1.76 mA 5 1.76 1.774 mA 0.992 C C C C EE V I R I II α −− == = = 5.9 / 13 0.685 10 exp 27.67 mA 0.026 19 1 27.67 90 27.98 mA 27.67 0.307 mA 90 RE T VV CS C EC D C BB e I I I β = ⎛⎞ = = ⎜⎟ ⎝⎠ + = = 5.10 Device 1: / 3 0.650/ 0.026 11 0.5 10 EB T vV o E o iI e I e = ×= So that 15 1 6.94 10 EO I A Device 2: 3 0.650/ 0.026 2 12.2 10 Eo Ie Or 13 2 1.69 10 Eo I A 13 2 15 1 1.69 10 Ratio of areas Ratio 24.4 6.94 10 Eo Eo I I × = × 5.11 (a) 250 250 1 A oo C V rr k I (b) 250 2.50 0.1 A C V M I 5.12 0 0 3 3 0 60 100 12.9 V CB CE BV BV BV = 5.13 0 0 3 3 3 220 220 56 3.93 56 60.6 BV BV = = = 5.14
Background image of page 3

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

View Full DocumentRight Arrow Icon
() 0 0 3 3 3 00 0 50 50 184 V CB CE BV BV BV BV BV β = == = 5.15 (a) 0.7 10 1.86 mA 5 75 1.86 1.836 mA 76 0.7 4 3.3 V 10 3.3 3.65 K 1.836 E C C CC I I V RR −− ⎛⎞ ⎜⎟ ⎝⎠ =− + = = = (b) ( ) 0.5 0.00658 mA 76 0.00658 25 0.164 V 75 0.5 0.493 mA 76 15 8.11 K 0.493 B BB B B C I VI R V I = −−− = = (c) () ( ) ( ) () ( ) 10 0.7 4 8 76 7.3 4 0.132 1.767 mA 75 1.767 1.744 mA 76 8 1.744 4 1.767 4 8 16 6.972 7.068 1.96 V E E EE C CE CE I OI II I V V =+ + = ⎡⎤ ⎣⎦ = (d) ) ( ) ( ) 5 10 20 0.7 2 10 0.263 2 0.7 76 0.3506 mA 4.61 A 5 0.3506 10 1.49 V E E EB C C I I IIV V μ + + = + + + = = 5.16 For Fig. 5.15 (a) 5 5% 5.25 K E R = 0.7 10 1.77 mA 5.25 1.75 mA 10 3.3 3.83 K 1.75 55 % 4 . 7 5 K 0.7 10 1.96 mA 4.75 1.93 mA 10 3.3 3.47 K 1.93 E C C E E C C I I R R I I R = = = So 1.75 1.93 mA 3.47 3.83 K IR ≤≤ ≤ ≤ For Fig. 5.15(c) 4 5% 4.2 K E R =
Background image of page 4
() ( ) ( ) ( ) ( ) ( ) 80 . 7 0.0222 mA 1.66 mA 10 76 4.2 1.69 mA 16 1.66 4 1.69 4.2 16 6.64 7.098 2.26 V 45 % 3 . 8 K . 7 0.0244 1.83 mA 10 76 3.8 1.86 mA 16 1.83 4 1.86 3.8 16 7.32 7.068 1.61 V BC E CE CE E E CE CE II I V V R I V V == = + = =− = = = + = = So 1.66 1.83 mA 1.61 2.26 V CC E IV ≤≤ ≤ ≤ 5.17 2.5 0.7 120 0.015 70 15 1.05 52 . 5 2.38 1.05 BB EB BB B CQ CC ECQ CQ VV RR k I IA m A k I μ = 5.18 (a) 1 500 2.0 A 10 . 7 1 . 7 V 3 1.7 3 0.2708 mA 4.8 0.2708 1 135.4 134.4 0.002 0.9926 1 0.269 mA 3 3 1.7 4.7 V B B B E E E E E B CB C CE E CE V VI RI R I V V I R I I I V ββ β αα −− = =− − −+ = =+ = = = = = + = = =−− = (b) ( ) 54 0.5 mA 2 40 . 7 5 4 0.7 100 0.5 8 5 0.5 0.043 1 11.63 0.043 10.63, 0.9140 1 EE B C C BC E B E B B IR I I R III I I I I βα α = = =+ ++ += + = = = + 5.19
Background image of page 5

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

View Full DocumentRight Arrow Icon
() [] 0.7 5 4.3 33 4.3 50 50 51 3 51 4.3 50 55 1 0 35 1 Now , so 1 3.27 5 14.1 9.05 2.12 V 2.12 4.3 0.727 mA 3 B B E B CE B CC C BC B B EE V V I V II V VI R VV V V −− + == + ⎛⎞ ⎜⎟ ⎝⎠ + =− =+ = = −+ = = 5.20 10 10 2 0.80 mA 10 10 0.7 2 0.7 1.3 V 1.3 0.026 mA 50 0.80 0.026 0.774 mA E BE B BB B CEB C V VV V R III I = =− = − = = =−= = ( ) 0.774 29.77 0.026 29.77 0.9675 1 30.77 10 2 0.774 10 10 4.26 V C B EC E C E C C EC I I V R V ββ β αα = = + =−=− ⎡⎤ ⎣⎦ = Load line developed assuming the V B
Background image of page 6
Image of page 7
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 / 33

Ch05s - Chapter 5 Problem Solutions 5.1 (a) iC 510 = = 85 6...

This preview shows document pages 1 - 7. Sign up to view the full document.

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