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Material Type: Exam; Professor: Xu; Class: Electronics; Subject: Electrical & Computer Engineer; University: Virginia Polytechnic Institute And State University; Term: Fall 2008;
Typology: Exams
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Electronics I, ECE 2204, CRN 11883 Exam 1 Chapter 1&2. Diodes Use proper UNITS and ONLY 3 significant figures in all calculations. NAME: _______________________________________ Signature: _____________________________________ By signing this exam, you agree that you have followed the Honor Code. Part 1: Please circle the right answers to each question: (5% for each question)
DQ T
10V D (a)
(b) 10V D (c) 10V Z (d) 10V Z (e) 10V Z (f) 10V
right: (a) VL= 5V when Vps= 10V (b) VL= 5V when Vps= 3V (c) VL= 0V when Vps= 3V (d) VL= 3V when Vps= 3V
100 20V 12V
rZ A R r
i Z PS zO Z^0.^0792 100 1
Thus, PZ^ IZ VZ IZ (^ IZ rZ VZO )^0.^0792 (^0.^0792 ^1 ^12 )^0.^958 W Step 2: The critical condition for Iz = 0 is when Vz just reaches its breakdown voltage, 12V, and its current has yet to ramp up from zero, as shown in below: 100 20V (^) 0A 12V 100 20V (^) 0A 12V Therefore, the 12V is generated by the voltage divider between RI and RL. Thus, V R R
i L L L^12 ^20 RL 150 ohm As long as RL 150 ohm, there will be no enough voltage shared by RL to breakdown the zener diode. Under such condition, the power dissipation in zener diode is zero due to zero IZ.
Vac Answer: Step 1: Do the DC analysis to find the Q-point of the diode: ( 1 ) 2 R 1
Thus, ) (^2 )
Based on the piecewise linear model of diode in Figure 1.30 (a), forward-biased diode can be treated as the Vr in series with rf. Therefore, diode itself have the following I-V relationship: VD ID rf Vr ID 0. 7 ( 3 ) From (2) and (3), we can get: I^ D ^4.^3 /^101 ^0.^0425 ( A ) V (^) D ID 0. 7 0. 7425 ( V ) (THIS IS THE Q-POINT OF DIODE UNDER 10V DC INPUT VOLTAGE) Step 2: Draw the load line : 0.7V 5V 0.05A Q-point VD ID Step 3: