# q14 - (a For the lower cuto f frequency solve for the worse...

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ECE 3050 Analog Electronics Quiz 14 April 22, 2009 Professor Leach Name Instructions. Print your name in the space above. Honor Code: I have neither given nor received help on this quiz. Initials 1o f2 . The f gure shows a precision recti f er circuit. The input signal is a sine wave. For R 1 =1 0k , R 2 =10k , R 3 =5k , R 4 =10k ,and R 5 =20k , sketch the time domain waveforms for the v 1 and the v O nodes. The output signal is a negative going full wave recti f ed sine wave with a peak voltage of 2 v I . 2o f2 . R S =1k , R 1 = R 2 =30k , R 3 =3k ,and R 4 =1k , C 1 =1 µ F ,and C 2 =50 µ F . The impedances seen looking into the base and into the emitter are z ib =10 4 1+ s/ 100 1+ s/ 10 z ie = 500 1+

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Unformatted text preview: (a) For the lower cuto f frequency, solve for the worse case pole frequency for C 1 . (b) For the lower cuto f frequency, solve for the worse case pole frequency for C 2 . (c) Which pole dominates in calculating the lower cuto f frequency f L ? 1 R S 1000 R 1 30000 R 2 30000 R 3 3000 R 4 1000 r ib 1000 r ie 50 worst case values which are the high frequency limits C 1 1 10 6 . C 2 50 10 6 . τ 1 R S R p3 R 1 R 2 , r ib , C 1 . 1 2 π . τ 1 . 82.144 = this frequency dominates because it is the highest τ 2 R p2 r ie R 3 , R 4 C 2 . 1 2 π . τ 2 . 3.034 = 2...
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## This note was uploaded on 05/26/2011 for the course ECE 3050 taught by Professor Hollis during the Summer '08 term at Georgia Tech.

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q14 - (a For the lower cuto f frequency solve for the worse...

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