q01 - R 2 1100 v s 5 g m 1 20 v 1 v s R 1 R 1 R 2 . g m v 1...

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ECE3050Ana logE lectron icsQu iz1 August 26, 2009 Professor Leach Last Name: First Name: Instructions. Print your name in the spaces above. Place a box around any answer. Honor Code Statement: I have neither given nor received help on this quiz. Initials 1o f2 . For v s =5V , R 1 =11k , R 2 =1 . 1k ,and g m =1 / 20 : (a) Use superposition, voltage division, current division, and Ohm’s Law to solve for v 1 . (b) Use the values of v s and v 1 to solve for
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Unformatted text preview: R 2 1100 v s 5 g m 1 20 v 1 v s R 1 R 1 R 2 . g m v 1 . R p2 R 1 R 2 , . v 1 v s R 1 R 1 R 2 1 g m R p2 R 1 R 2 , . . v 1 0.089 = v 2 v s v 1 v 2 4.911 = 2 of 2. (a) With the aid of a graph, illustrate how the value of the diode small-signal resistance is de f ned. [It is the reciprocal of the slope of the i D versus v D graph.] (b) Draw and label the hybrid- model of the BJT. 1...
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This note was uploaded on 05/26/2011 for the course ECE 3050 taught by Professor Hollis during the Summer '08 term at Georgia Institute of Technology.

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