Solution9 - UNIVERSITY OF CALIFORNIA AT BERKELEY EECS...

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Unformatted text preview: UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department Page 1 of 13 EECS 40/42/100, Spring 2007 Prof. Chang-Hasnain Homework #9 (Note: EE42/100 is out of 85 pts, EE40 is out of 100 pts) Due at 6 pm in 240 Cory on Wednesday, 04/04/07 Total Points: 100 Put (1) your name and (2) discussion section number on your homework. You need to put down all the derivation steps to obtain full credits of the problems. Numerical answers alone will at best receive low percentage partial credits. No late submission will be accepted expect those with prior approval from Prof. Chang-Hasnain. *Note: Power gain is defined as the ratio between power to a load and power from an input source. 1. Hambley, P14.33 [5 points] Because of the variation requirements, we must use 1% resistors. To satisfy the 1k input impedance requirement we place a 1k resistor at the input. To satisfy a gain of 10, we require a ratio of 9 in the feedback line resistances. In this configuration, maximum output variation in the lower direction is: ( ) ( ) 82 . 9 01 . 1 1 99 . 9 1 Gain = + = k k The maximum in the upper direction is: ( ) ( ) 18 . 10 99 . 1 01 . 1 9 1 Gain = + = k k Both of these values satisfy the 3% variation requirement. *Students may have different amplifier structures (such as cascaded inverting amplifiers) *1pt for drawing your amplifier structure labeled with values *2pts for satisfying the gain relationship *1pt for satisfying the gain variation requirement *1pt for satisfying the input resistance variation requirement UNIVERSITY OF CALIFORNIA AT BERKELEY EECS Department Page 2 of 13 2. Hambley, P14.39 [10 pts] a) Applying nodal analysis at the output node, and KVL around the outer loop: [5 pts] o s o o ol in o s V V V R V V A R V V = = + 1 1 , o o ol o o in o o s ol in s R V A R V R V R V A R V + + = + ( ) + + = + o in ol in o o o in in ol o s R R A R R V R R R A R V 1 ( ) ( ) ( ) 1 99999 . 10 1 1 25 10 1 25 1 5 5 = + + + = + + + = M M A R R A R R V V ol in o ol in o s o b) By Ohms Law: [2 pts] ( ) = = = in CL s in CL s s in o s s R A V R A V V R V V i 1 , where A CL is the closed loop gain computed in (a) = = = = G M A R i V Z CL in s s in 100 99999 . 1 1 1 This is similar to the infinite input impedance assumed for ideal op-amps....
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This homework help was uploaded on 04/02/2008 for the course EE 40 taught by Professor Chang-hasnain during the Spring '07 term at University of California, Berkeley.

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Solution9 - UNIVERSITY OF CALIFORNIA AT BERKELEY EECS...

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