Ch17p - Chapter 17 Exercise Solutions EX17.1 a-0.7-5 iE = = 1 mA RE = 4.3 k RE iC1 = iC 2 = 0.5 mA = 5 3.5 RC1 = RC 2 = 3 k RC1 b v1 = 1 V i(1

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Chapter 17 Exercise Solutions EX17.1 a. 12 1 2 1 0.7 ( 5) 1 mA 4.3 k 53 . 5 0.5 mA 3 k EE E CC C C C iR R ii RR R −− == = Ω b. i. 1 1 V v = 10 1 0 1 02 (1 0.7) ( 5) 1.23 mA 4.3 5 (1.23)(3) 1.31 V 5 V CE v v v = = == =− = = ii. 1 v =− 02 02 01 1 mA 5 (1)(3) 2 V 5 V E iv v v = = = EX17.2 34 () ( 5 . 2 ) CXY CR Pi i i i +++ a. 3 4 logic 1 3.22 mA 0 0.7 5.2 3 mA 1.5 1.4 5.2 2.53 mA 1.5 (3.22 0 3 2.53)(5.2) 45.5 mW XY C X Y CR vv i i i i PP = = −+ =+ + + = b. 3 4 logic 0 0 2.92 mA 2.53 mA 3 mA (0 2.92 2.53 3)(5.2) 43.9 mW C X Y CR i i i i = = = = + + = EX17.3 1 2 5 3.5 (2.45 0.7) 1.4 K 0.25 3.5 2(0.7) 8.4 k 7 K 0.25 2.45 9.8 K 0.25 R R R += = = EX17.4
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34 3 51 3 4 0.7 ( 5.2) 14 . 5 K 0.7 0.7 ( 5.2) 3.22 mA 1.18 1.40 mA 1.0 mA 1.4 ( 5.2) 0.844 mA 4.5 (3.22 1.4 1.4 1.0 0.844)(5.2) 40.8 mW MAX CXY IR R R i ii i i P −− == −− − = =+ + + + = EX17.5 3 0.7 ( 5.2) (10) (1.18)(51) (5 . 2 ) 1.5 0( 0 . 7 ) 5 . 2 4 . 5 (51) (10) 0.24 1.5 (1.18)(51) 51 1 1 (6) (0.7)(51) 5.2 4.5(10) 0.24 1.5 (1.18)(51) 0.24 1.5 (1.18)(51) oR L oR oR oR oR oR v i v i Vv v v ⎛⎞ = ⎜⎟ ⎝⎠ = −+ + + ⎡⎤ ⎢⎥ ⎣⎦ + = + + [212.50 0.6667 0.166168] 148.75 3.4666 0.747757 [213.3328] 152.9644 0.7170 oR oR oR v v v ++ =++ −= =− EX17.6 3 3 (100)(0.026) 2.6 K 1 1 38.46 mA / V 0.026 m r g π 3 23 3 3 (1 ) 0.24 2.6 (101)(4.5) 457.34 nn b C n b VV I Rr R V I β +++ + + = 32 3 33 ( ) (0.24 2.6) 457.34 0.00621 ) (101) (4.5) 457.34 0.9938 n Ob C On n V VI R r V R + + = ′ = EX17.7 0 0.2 (1.7) 117.6 A 0.4 on 1.7 1.7 0.4 3.40 k 0.1176 1.7 1.3 1.5 V 2 QC C Q Q RQ C C C RR PIV I I QvI R R R μ =⋅ = = = + = =
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EX17.8 (a) 1 1 1 5 ( ) 2 0.8 2(0.7) 2.2 52 . 2 0.70 4 () 50 . 1 1.225 4 ( ) (0.70 1.225)(5) XY BE Y CC
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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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Ch17p - Chapter 17 Exercise Solutions EX17.1 a-0.7-5 iE = = 1 mA RE = 4.3 k RE iC1 = iC 2 = 0.5 mA = 5 3.5 RC1 = RC 2 = 3 k RC1 b v1 = 1 V i(1

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