Ch10s - Chapter 10 Problem Solutions 10.1 a. I1 = I 2 = 0 -...

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Chapter 10 Problem Solutions 10.1 a. () 12 22 3 2 3 2 31 2 02 2 1 BE C BE C CB E VV II RR VI RV I R R V V I R R IV V V V R γ γγ −− == + += + +− = + + ⎧⎫ ⎛⎞ ⎪⎪ =− + ⎨⎬ ⎜⎟ + ⎝⎠ ⎩⎭ b. 3 3 and 11 2 or 2 BE E C V V V R V I R + = c. ( ) 3 3 1 2 10 2 mA = 2.5 k 2 20 .7 10 2 mA 4.3 k 2.15 k C IR R = = = + 10.2 (a) 1 1 ln C BE T S I I = (i) 6 14 10 10 10 A, 0.026 ln 0.5388 V 10 10 A REF C BE O V I μ × = = = (ii) 6 14 100 10 100 A, 0.026 ln 0.5987 V 10 100 A REF C BE O V I × = = = (iii) 3 14 10 1 , 0.026 ln 0.6585 V 10 1 mA REF C BE O IIm A V I = = = (b) 2 1 REF O I I β = + (i) 6 14 10 9.615 A ln 2 1 50 9.615 10 0.026 ln 10 0.5378 V O O O BE BE T S I V V V I = = + × = = (ii) 6 1 14 100 96.15 10 96.15 A 0.026 ln 2 10 1 50 0.5977 V OO B E V × = + =
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(iii) () 3 1 14 1 0.9615 10 0.9615 mA 0.026 ln 2 10 1 50 0.6575 V OO B E II V ⎛⎞ × = == ⎜⎟ ⎝⎠ + = 10.3 11 1 12 on 3 0.7 3 0.250 21.2 K 0.250 0.2419 mA 22 60 4.03 A BE REF REF CC BB VV V I RR R I β μ +− −− = = = = ++ 10.4 1 on 5 0.7 5 18.3 0.5082 mA 0.5082 0.4958 mA 80 6.198 A BE REF REF REF V I R I I = = 10.5 (a) 1 15 0.7 15 or 58.6 0.5 BE REF VVo nV I k R (b) 00 . 7 1 5 28.6 0.5 BE REF k I Advantage: Requires smaller resistance. (c) For part (a): 29.3 max 0.526 58.6 0.95 29.3 min 0.476 58.6 1.05 0.526 0.476 0.05 5% O O O I mA I mA Im A Δ= = ± For part (b): 14.3 max 0.526 28.6 0.95 14.3 min 0.476 28.6 1.05 0.05 5% O O O I mA I mA A ± 10.6 a. 0 1 2 1 or 2.04 mA 100 15 0.7 7.01 k 2.04 REF REF I =+ = + = = =
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b. () 0 0 0 0 0 00 80 40 k 2 11 9.3 0.2325 mA 40 0.2325 11.6% 2 A CE V r I I I Vr II === Δ ⎛⎞ = Δ= = ⎜⎟ Δ ⎝⎠ ΔΔ = = 10.7 01 10 112 1 0 0 1 1 o r 1 1 C C REF C B B C REF C C REF In I I I I nn I I nI n I n n ββ β = =++=++ + =+ + + = + + 10.8 2 0.20 1 0.210 mA 2 40 1 50 . 7 4 . 3 20.5 K 0.21 REF OR E F REF I RR I = = + == = 10.9 a. . 7 0.239 mA 18 0.239 0.230 mA 2 1 50 REF I = = + b. 0 0 0 50 218 k 0.230 1.3 0.00597 mA 0.236 mA 217 A EC V r I IV I r = Δ= ⋅ Δ = = = c. 1 3.3 0.01516 mA 0.245 mA 217 = = 10.10 a. 1 1 . 7 5 1 9.3 k REF IR R −− = b. 22 m A REF I = = c. 2 . 7 For min 0.7 2.15 k 2 EC C C VR R = = = 10.11
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11 0.50 mA 0.25 mA 33 10 . 2 5 1 60 0.2625 mA 2.5 0.7 6.86 K 0.2625 OO A O B REF OA REF II I I RR β = == ⎛⎞ =+ = + ⎜⎟ ⎜⎟ ⎝⎠ = = = 10.12 1 10 0.7 37.2 K 0.25 R 10.13 21 31 2 a n d I 3 I (a) 23 1.0 , 1.5 I mA I mA (b) 13 0.25 0.75 I mA I mA (c) 12 0.167 0.333 I mA I mA 10.14 a.
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() 3 01 131 1 312 22 1 1 2 0 2 2 0 and 1 2 2 11 2 1 1 2 1 1 E CR E F C B C C BE BE EB B C BE REF C BE REF BE REF I II I I I I I VV I RR I V R V R V I R I β ββ == + = + + =++ = + =+ + ++ ⎛⎞ −= + ⎜⎟ ⎝⎠ + = + + b. 1 1 20 . 7 0.70 1 80 81 81 10 0.700216 0.000864 10 2 0.7 0.7011 mA 12.27 k REF REF REF I I IR R + = 10.15 a. 0 12 and 1 ... 1 1 ES i CR REF CR BS CR E SB RBB B N B R CR I I I I III I N I NI + = + + =+++ + = + + = 0 1 Then 1 or 1 1 1 CR REF CR REF i I I N + + = + + + b. 6 0.5 1 0.5012 mA 50 51 520 . 7 5 17.16 k 0.5012 REF I ⎡⎤ = ⎢⎥ ⎣⎦ −− = = 10.16
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() 0 11 22 1 0.5 1 0.5004 mA 15 0 5 1 520 . 7 5 17.19 k 0.5004 REF REF II I RR ββ ⎛⎞ =+ = + = ⎜⎟ ⎢⎥ + ⎝⎠ −− = = 10.17 0 0 1 2 1 2 For 0.8 mA 2 0.8 1 0.8024 mA 25 27 18 2 0.7 20.69 k 0.8024 REF REF REF I =⋅ + + = = = = 10.18
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The analysis is exactly the same as in the text. We have () 0 1 2 1 2 REF II ββ =⋅ ⎛⎞ + ⎜⎟ + ⎝⎠ 10.19 02 11 312 3 3 13 2 2 mA, 0.0267 mA 75 1 1 mA, 0.0133 mA 75 0.0133 0.0267 0.04 mA 0.04 0.000526 mA 17 6 1.000526 1 mA 10 2 0.7 8.6 8.6 k 1 B CB EB B E B REF C B REF REF I I I RR I β == = = =+ = + = = + =+ =≈ = 10.20 (a) ( ) 3 3 Assuming 2 100 400 K 0.25 100 400 20 M 2 o O AA O OR E F OO r R VV r = = = (b) 5 20 M 20 M 0.25 A O
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Ch10s - Chapter 10 Problem Solutions 10.1 a. I1 = I 2 = 0 -...

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