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Unformatted text preview: ECE 442 HW2 Solutions 1. a) * Graphical analysis: the i v relations of the diode and the resistor are as follows: i D = I S e v D nV T i R = 1 v D 250 Since the diode and the resistor are connected in series, their currents must be equal. Plotting the two relations on the same plot gives us the solution for this system of equations, as shown in Fig 1a. Figure 1a. The i v relations of the diode (solid curve) and the resistor (dashed) The bias current and voltage are read from the intersecting point, as follows: I D = 1 . 1105 mA , V D = . 721 V * Iterative analysis: to update the current at each iteration step, we will use the resistor current equation I D = 1 V D . 25 k . To update the voltage, we will use the following voltage relation: V D = nV T ln ( I D I S ) We start with V 1 = V D = . 7 V , and I 1 = 1 V D 1 . 25 k = 1 . 2 mA . V D 2 = 1 × 25 m × ln ( 1 . 2 m 1 15 ) = . 695 V I D 2 = 1 . 695 . 25 k = 1 . 22 mA V D 3 = 1 × 25 m × ln ( 1 . 22 m 1 15 ) = . 696 V 1 I D 2 = 1 . 696 . 25 k = 1 . 22 mA Iterating further does not change the values of V D and (more importantly) I D significantly, so we can con clude that the iteration proces converges after three iterations, to result in the bias current and voltage as follows:...
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This note was uploaded on 02/21/2010 for the course ECE 442 taught by Professor Staff during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Staff

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