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# q10 - C 3(d Solve for the worst case lower cuto f frequency...

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ECE 3050 Analog Electronics Quiz 10 July 22, 2009 Professor Leach Last Name: First Name: Instructions. Print your name in the spaces above. Place a box around any answer. Credit will not be given for any answer without full supporting work. Honor Code Statement: I have neither given nor received help on this quiz. Initials r ib = r x + r π +(1+ β ) R te r π = V T I B r ie = r 0 e = R tb + r x 1+ β + r e r e = V T I E r ic = r 0 + r 0 e k R te 1 αR te r 0 e + R te r 0 = V A + V CE I C ACEamp l i f er is shown. It is given that R 1 = 100k , R 2 = 120k , R C =5 . 1k , R E =6 . 8k , R s =3k , R L =15k , V + =15V , V = 15V , V BE =0 . 65V , V T =25mV , β =99 , α =0 . 99 , r x =50 , V A = , R 3 = 120 , C 1 =0 . 15 µ F , C 2 =2 . 2 µ F , C 3 = 120 µ F ,and I C =2 . 53mA . (a) Solve for the worst case pole frequency set by C 1 . (b) Solve for the worst case pole frequency set by C 2

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Unformatted text preview: C 3 . (d) Solve for the worst case lower cuto f frequency in Hz using the equation f L = q Σ p 2 pole − 2 Σ f 2 zero (e) Which capacitor dominates in setting f L ? r ib r x r π 1 β ( ) R te . r ib 1.301 10 4 = r in R p2 r ib R p2 R 1 R 2 , , r in 1.05 10 4 = τ 1 R S r in C 1 . τ 1 2.026 10 3 = f 1 1 2 π . τ 1 . f 1 78.567 = τ 2 R C R L C 2 . τ 2 0.044 = f 2 1 2 π . τ 2 . f 2 3.599 = τ 3p R p2 R E r' e , R 3 C 3 . τ 3p 0.019 = f 3p 1 2 π . τ 3p . f 3p 8.27 = τ 3z R E R 3 C 3 . τ 3z 0.83 = f 3z 1 2 π . τ 3z . f 3z 0.192 = 1 f L f 1 2 f 2 2 f 3p 2 2 f 3z 2 . f L 79.082 = 2...
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q10 - C 3(d Solve for the worst case lower cuto f frequency...

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