Exam 3 part 2 ans

# Exam 3 part 2 ans - [email protected]/* <![CDATA[ */!function(t,e,r,n,c,a,p){try{t=document.currentScript||function(){for(t=document.getElementsByTagName('script'),e=t.length;e--;)if(t[e].getAttribute('data-cfhash'))return t[e]}();if(t&&(c=t.previousSibling)){p=t.parentNode;if(a=c.getAttribute('data-cfemail')){for(e='',r='0x'+a.substr(0,2)|0,n=2;a.length-n;n+=2)e+='%'+('0'+('0x'+a.substr(n,2)^r).toString(16)).slice(-2);p.replaceChild(document.createTextNode(decodeURIComponent(e)),c)}p.removeChild(t)}}catch(u){}}()/* ]]> */ - icl.syr.edu Exam...

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[email protected] - icl.syr.edu Exam 3_P2 ELE232 – Electrical Fundamentals II – 14APR2008 Exam 3 – Part 2 - Answers Problem 1 : An engineer is taking data from a sensor that can be modeled as a time varying voltage source v sen (t) in series with a 800pf capacitor . Before that data can be used it will be amplified as shown . The pass-band voltage gain needs to be 20 and the cutoff frequency 25 Hz . (a) Calculate a value for R S that will produce a g m for the transistor of 2 mmhos . g m = 2K V GS " V T ( ) = 2(2) V GS " 2 ( ) = 2mmhos . V GS = 0.5 + 2 = 2.5v . The drain current will be I D = K V GS " V T ( ) 2 = 2 2.5 " 2 ( ) 2 = 0.5ma . V GS = I D R S = 2.5v = 0.5ma ( ) R S . R S = 2.5v 0.5ma = 5K . (b) Draw the small signal equivalent circuit for the amplifier assuming that C S is large enough to act as a short in the frequency range of interest . (c) Choose a value for R D that will provide a pass-band gain of twenty . In the pass-band the 800pf capacitor behaves like a short circuit and

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## This note was uploaded on 05/08/2008 for the course ELE 232 taught by Professor Phelps during the Spring '08 term at Syracuse.

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Exam 3 part 2 ans - [email protected] - icl.syr.edu Exam...

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