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

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Chapter 10 Problem Solutions 10.1 a. ( ) ( ) 1 2 1 2 2 2 3 2 3 1 2 2 3 1 2 0 2 2 2 2 1 2 2 BE C BE C C BE V V I I R R V I R V I R R V V V V I R R R R I V V V V R R R γ γ γ γ γ γ = = + + = + + = + + = + + b. ( ) 1 2 3 3 and 1 1 2 2 2 or 2 BE C BE C V V R R I V V V V R V I R γ γ γ = = = + = c. ( ) ( ) ( ) 3 3 1 2 1 2 1 2 1 2 10 2 mA = 2.5 k 2 2 0.7 10 2 mA 4.3 k 2.15 k C I R R I I R R R R R R − − = = − − = = = + = = = + 10.2 (a) 1 1 ln C BE T S I V V I = (i) ( ) 6 1 1 14 10 10 10 A, 0.026 ln 0.5388 V 10 10 A REF C BE O I I V I μ μ × = = = = = (ii) ( ) 6 1 1 14 100 10 100 A, 0.026 ln 0.5987 V 10 100 A REF C BE O I I V I μ μ × = = = = = (iii) ( ) 3 1 1 14 10 1 , 0.026 ln 0.6585 V 10 1 mA REF C BE O I I mA V I = = = = = (b) 2 1 REF O I I β = + (i) ( ) 1 2 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 I I V V V I μ = = = = + × = = (ii) ( ) 6 1 14 100 96.15 10 96.15 A 0.026 ln 2 10 1 50 0.5977 V O O BE I I V μ × = = = + =

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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 O O BE I I V × = = = + = 10.3 ( ) ( ) 1 1 1 1 2 1 2 1 2 on 3 0.7 3 0.250 21.2 K 0.250 0.2419 mA 2 2 1 1 60 4.03 A BE REF REF C C C C B B V V V I R R R I I I I I I I β μ + − − = = = = = = = = + + = = 10.4 ( ) ( ) ( ) 1 1 2 1 2 1 2 on 5 0.7 5 18.3 0.5082 mA 0.5082 0.4958 mA 2 2 1 1 80 6.198 A BE REF REF REF C C C C B B V V V I R I I I I I I I I β μ + − − = = = = = = = = + + = = 10.5 (a) ( ) ( ) 1 1 1 15 0.7 15 or 58.6 0.5 BE REF V V on V I R R k R + − − = = = Ω (b) ( ) ( ) 1 1 0 0.7 15 28.6 0.5 BE REF V V on V R R k I + − − = = = Ω 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 I mA = = = = 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 I mA = = = = Δ = ± 10.6 a. 0 1 1 2 2 1 2 1 or 2.04 mA 100 15 0.7 7.01 k 2.04 REF REF I I I R R β = + = + = = =