Ch14p - Chapter 14 Exercise Solutions EX14.1 a ACL =-50 ACL =-49.949 1 1 5 104 51 dA 51 = 10 CL = 0.0102 ACL 5 104 b dACL ACL ACL =-50 ACL =-49.943

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Chapter 14 Exercise Solutions EX14.1 a. () 4 50 49.949 1 15 1 51 0 CL CL AA = =− ⎡⎤ ⎛⎞ + ⎜⎟ ⎢⎥ × ⎝⎠ ⎣⎦ b. 4 4 51 10 0.0102% 0 50 49.943 51 1 4.5 10 CL CL CL CL CL CL dA dA = × = + × EX14.2 a. For 0 0 R = 43 3 11 1 0 0 . 0 10 10 10 k 1 if R R =+ + = + = Ω b. For 0 10 k R 44 3 1 1 1 0 1 1 0 0.1 10 10 1 1 1 3 10 31 0 k 3 if i f R RR ++ =+× ≅ + Ω EX14.3 4 53 3 40 40 1 10 99 1 1 99 1 1 41 0 4 . 0 5 91 0 100 4.04 10 k 4.04 M R + = + ×+ × =× Ω EX14.4 2 1 0 0 R R += a. 5 0 0 1 0 10 0.1 100 100 f f R R == b. 5 2 0 1 0 10 10 100 f R 2 00 10 k 10 ff EX14.5 From Equation (14.43)
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() 0 00 4 4 1 25 25 11 21 0 50 10 25 CL CL PD CL A Af f j fA A f f jj = +⋅ / == × / a. 2 kHz f = 0 0 25 peak 1.25 mV I v v v = = b. 20 kHz f = 0 0 1 25 peak 0.884 mV 2 I v v v =⋅ = c. 100 kHz f = 0 2 0 25 25 4.90 5.099 0 0 / 2 0 0.245 mV I v v v = + = EX14.6 Full-scale response 15 5V =× = 5 2.5 s 2 tt μ = = EX14.7 a. 6 0 5 0.63 10 2m a x 2 1 1.0 10 100 kHz SR FPBW V FPBW FPBW ππ × = b. 6 4 0.63 10 1.0 10 0 10 kHz FPBW FPBW π × × = EX14.8 14 2 0 14 1 0 1.85 10 ln (0.026)ln 0 2.03 mV S ST S S I VV I V ⎛⎞ × ⎜⎟ × ⎝⎠ = EX14.9 We need 12 3 4 1 2 ,0 . 6 V , a n d 1 0 V C C EC EC CE CE ii v v v v = = = By Equation (14.60(a)) 1 3 3 10 exp 1 50 0.6 exp 1 50 BE CS T EB S T v iI V v I V ⎡⎤ =+ ⎢⎥ ⎣⎦ By Equation (14.60(b))
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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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Ch14p - Chapter 14 Exercise Solutions EX14.1 a ACL =-50 ACL =-49.949 1 1 5 104 51 dA 51 = 10 CL = 0.0102 ACL 5 104 b dACL ACL ACL =-50 ACL =-49.943

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