hwk7_sol

hwk7_sol - #$#$ I"E-0;8*58'95%0-2'4!< 1 Hambley 14.9 Figure...

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EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 7 SOLUTIONS 1. Hambley, 14.9 Figure P14.9 is an inverting amplifier with gain given by: A v = - R 2 /R 1 = - 30 / 10 = - 3 Thus v o ( t ) is just v in amplified by A v . Sketches shown below: =# 755:) 2’&28’* ,-,:)4’4 5&’-2’5:%4 ’-2:80’-. >’&2++;((?4 ,-0 @+9?4 :,A4 *; A&’*% 2’&28’* %B8,*’;-4# C+%- 4;:/% (;& *+% B8,-*’*’%4 ;( ’-*%&%4*# "#$%& 722;&0’-. *; *+% 4899’-.D5;’-* 2;-4*&,’-*E *+% ;8*58* /;:*,.% ;( ,- ;5 ,95 ,4489%4 *+% /,:8% &%B8’&%0 *; 5&;082% <%&; 0’((%&%-*’,: ’-58* /;:*,.% ,-0 <%&; 28&&%-* ’-*; *+% ;5D,95 ’-58* *%&9’-,:4# C+’4 5&’-2’5:% ,55:’%4 A+%- -%.,*’/% (%%01,23 ’4 5&%4%-* 18* -;* A+%- 5;4’*’/% (%%01,23 ’4 5&%4%-*# "#$%’ C+% ’-/%&*’-. ,95:’(’%& 2;-(’.8&,*’;- ’4 4+;A- ’- F’.8&% "G#G ’- *+% *%H*# C+% /;:*,.% .,’- ’4 .’/%- 1) " 6 I 7 / ! " ( *+% ’-58* ’95%0,-2% ’4 %B8,: *; " E ,-0 *+% ;8*58* ’95%0,-2% ’4 <%&;# "#$%) C+’4 ’4 ,- ’-/%&*’-. ,95:’(’%& +,/’-. , /;:*,.% .,’- .’/%- 1) = " 6 ! " ! " # C+84E A% +,/% # $ # $ % & * ! ; 6JJJ 2;4 6 = ! " K3%*2+%4 ;( ’- L M ,-0 L M ,&% GN= Figure 1. Sketch of v in ( t ) and v o ( t ) Replacing R L by a 10kΩ resistor does not change anything. This is because the Op- Amp is an active device and can provide current as needed. 2. Hambley, 14.11 The negative feedback allows us to use the summing-point constraint. There are two 1
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2 EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 7 SOLUTIONS ways to attack this problem. The first way is to use a Thevenin Equivalent Circuit on the left-half of the circuit to convert it to the Inverting amplifier. Doing this gives us V th = V in / 3 R eq = R + 2 R k R = 5 * R/ 3 From here we can calculate A v , or closed-loop gain. A v = V o / V th * V th / V in = - 15R / R * ( 5 / 3 ) * ( 1 / 3 ) = - 3 The second option is to analyze the circuit directly. One quick trick is to use the summing-point constraint that makes v - equal to v + , or ground. From there we can find the currents easily as shown: This gives v in = 2 R (2 i ) + Ri and v o = - 15 Ri . Solving gives A v = - 3 3. Hambley, 14.12 Using the summing point constraint gives: i D = v in /R i D = I s * exp( v D /nV T ) v D = - v o Solving gives: v in /R = I s * exp( - v o /nV T ) v o = - nV T ln( v in RI s ) 4. Hambley, 14.20 (a) v - = v + = ground = 0 V KVL gives:
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EE 40, FALL 2008 PROF. CHANG-HASNAIN HOMEWORK 7 SOLUTIONS 3 !"#$") !"# 6.6&60#’)&6, +-71&/&#’ %.6/&,(’+)&.6 &3 3".26 &6 A&,(’# BC;BB &6 )"#
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This note was uploaded on 10/24/2009 for the course EE 40 taught by Professor Chang-hasnain during the Fall '07 term at Berkeley.

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hwk7_sol - #$#$ I"E-0;8*58'95%0-2'4!< 1 Hambley 14.9 Figure...

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