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

This preview shows pages 1–2. Sign up to view the full content.

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 1 of 2. The fi gure shows a precision recti fi er circuit. The input signal is a sine wave. For R 1 = 10 k , R 2 = 10 k , R 3 = 5 k , R 4 = 10 k , and R 5 = 20 k , sketch the time domain waveforms for the v 1 and the v O nodes. The output signal is a negative going full wave recti fi ed sine wave with a peak voltage of 2 v I . 2 of 2. R S = 1 k , R 1 = R 2 = 30 k , R 3 = 3 k , and R 4 = 1 k , 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 + s/ 20 1 + s/

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern