374k HW5 Solution - 2.14 me(2.26 me cxarnplc 2.4_1_24{1 IE E = —{11 v u 1 9660 38” I 2.1 T 1‘ swam Ina:1 fBSiStflt‘S:13 shown by the

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Unformatted text preview: 2.14 me (2.26}, me cxarnplc 2.4 _1_24{1 _ IE _ E - = — {11 v u 1 9660 38” I'.'\ 2.1? T 1‘}! swam] Ina-:1 fBSiStflt‘S :13. shown by the dashed load lines fd'llnwing. The maximum power i5 2.5 “W. The load resiamr, R = WI = ([15 WHILEHA} = ifllfll kfl. (a) shows the pi'ob1em—the RC product is toe- high. {b} shows the simplest selutien—the transistor input resistance is much lower than R. {C} shows that an op amp provides a virtual ground that prevides a low input resistance. {d} shows that if R is divided h}; 10, the gain may be achieved by a noninverting ampiifier. Active cempenents must have adequate speed. tar R l kfl .9 3:9 I! C — C x _ {'11} W10 1 id} 1".""7.7;"Elsififlzgg; a; _______ __ ____ ._ ____.3an i4 : 3143 "d _ (Door a (q (’7‘ 1 End \ \ and + 1. (VoVrarJC 4”“ng 394C, 2 (ii “W EL 3WP—INC ’r TL H’UJRHUC, +— CL L+ flu)? K: ’TJT— {INC-“(H.Hqgjfil‘ggh) :: (9.1 C “A: Ki J gram/L g C, C‘XWfT— him (At '31 : (zze’iine‘) C I " 22,313“?— 1 (100} L 101 5‘6} : 1” '“V/(W‘Isi) '________________,___._ 1.6 H1 -31.348 ...
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This note was uploaded on 03/19/2008 for the course EE 374K taught by Professor Pearce during the Fall '07 term at University of Texas at Austin.

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374k HW5 Solution - 2.14 me(2.26 me cxarnplc 2.4_1_24{1 IE E = —{11 v u 1 9660 38” I 2.1 T 1‘ swam Ina:1 fBSiStflt‘S:13 shown by the

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