HW 7 ans - AC part , called v ds , of drain to source...

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fwphelps@syr.edu - icl.syr.edu ELE232 – Electrical Fundamentals II – 21MAR2008 Homework 7 – Due: Not later than Monday, 31MAR2008 Problem 1 : The input voltage source produces a 200 mv peak to peak sinewave with no DC component. (a) With the signal source turned off, the voltage drop from drain to source V DS should equal 8v DC . Determine the size of R D that will make this happen. V GS = 0v . I D = I DSS = 5 ma . V DS = 8 v = 25 v ! 5 ma R d ( ) . R d = 25 v ! 8 v 5 ma = 3 . 4 K (b) Calculate a value for g m , the trans-conductance of the JFET . g m = 2 I DSS |V P | = 2 ( 5 ) 4 = 2 . 5 mmhos (c) Turn the signal source back on. Calculate the peak to peak size of the
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Unformatted text preview: AC part , called v ds , of drain to source voltage . v ds = ! g m R d v in = ! 2 . 5 ( 3 . 4 ) 200 mv = 1 . 7 v . Problem 2 : (a) Determine the quiescent values for V DS and I D . I D = 12 1 ! 1 . 75 5 " # $ % & 2 = 5 . 07 ma . V DS = 25 v ! 5 . 07 ma( 3 . 3 K) = 8 . 27 v (b) Determine the small signal voltage gain for the amplifier. K V = ! g m R d g m = 2 I DSS |V P | 1 ! V GS V P " # $ % & = 2 ( 12 ) 5 1 ! 1 . 75 5 " # $ % & = 3 . 12 mmhos . K V = ! 3 . 12 mmhos( 3 . 3 K) = 10 . 3...
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This note was uploaded on 05/02/2008 for the course ELE 232 taught by Professor Phelps during the Spring '08 term at Syracuse.

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