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Unformatted text preview: Rd Vdd Rss R2 Vdd R1 T1 2 4 6 8 10 12 14 16 18 2054321 1 2 3 4 5 6 GatetoSource Voltage (V) Drain Current (mA) ECE 255 ELECTRONIC ANALYSIS AND DESIGN Fall 2009 Homework 5 Solutions Problem 1 Select element values for the circuit shown below to maintain D 4 I 6 m A for transfer functions ranging from 2 GS D V I 20 1 m A 4 = + to 2 GS D V I 7.2 1 m A 1.44 = + assuming that V DD = 16 V. Shown below are the two limiting transfer functions. Since the only requirement was D 4 I 6 m A there are many possible solutions. My first guess is represented by the solid red line where V G = 4 V and R SS = 1 k . Another easy solution is given by the broken orange line which goes through the point I D = 5 mA and V GS =  1 V and has V G = 6 V. For the orange line GS SS D V 7 V R 1.4 k I 5m A = = = . The limiting load line, shown as the broken blue line , can be determined from the extreme operating points which are V GS =  1.8091 V @ 6 mA and V GS = .3667 V @ 4 mA. The load line running through these points has an intercept V G = 2.518 V and slope corresponding to R SS = 721.2 . To satisfy the design requirements, V G and R SS must be at least as large (or larger) than the limiting values. Completing the design using the red load line : given that V DD = 16 V we can let R 2 = 100 k . Then R 1 = 300 k to provide a gate voltage of V G = 4 V. R SS was already assumed to be 1 k . The only restriction on choosing R D is that it can not be so large as to put the FET into the ohmic region of operation. Let R D = 1 k , then V DG 6 V and either transistor will be safely in the B.P.O. region. + V i + V o Problem 2 Show your calculation to determine the largest value of R D that can be used in the circuit that you biased in Problem 1 that will guarantee operation in the BeyondPinchOff region. The limiting case is determined by the operating point with the highest current (assuming that this operating point is also associated with the device with the largest V P , which is almost always the case). The gatetosource voltage at the highest possible current is determined from 2 GS V 6 20 1 4 = + , therefore, GS 6 V 4 1 20 = V GS = 1.809 or 6.191 V. Clearly the first value is correct....
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This note was uploaded on 01/20/2010 for the course ECE 255 taught by Professor Staff during the Spring '08 term at Purdue UniversityWest Lafayette.
 Spring '08
 Staff

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