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final+version+C+solutions - EECS 40 Fall 2008 Prof...

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EECS 40, Fall 2008 Prof. Chang-Hasnain Final Exam Solutions Version C 8:00 am – 11:00 am, Saturday December 13, 2008 Total Time Allotted: 180 minutes Total Points: 300 1. This is a closed book exam. However, you are allowed to bring 4 pages (8.5” x 11”), double-sided notes. 2. No electronic devices, i.e. calculators, cell phones, computers, etc. 3. SHOW all the steps on the exam. Answers without steps will be given only a small percentage of credits. Partial credits will be given if you have proper steps but no final answers. 4. Remember to put down units. Points will be taken off for missed unit. Last (Family) Name:_____________________________________________________ First Name: ____________________________________________________________ Student ID: ___________________________ Discussion Session: ________________ Signature: _____________________________________________________________ Score: Problem 1 (116 pts) Problem 2 (36 pts) Problem 3 (58 pts) Problem 4 (90 pts) Total (300 pts)
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Page 2 of 17 Problem 1: Transistor Amplifier (116 pts) Note: The sub-parts are somewhat independent. Read through the entire problem even if you are not able to solve some parts. Good luck! In this problem you are about to analyze the amplifier circuit shown in the figure below. The threshold voltage V t0 = 1 V and the capacitor impedance at signal frequency is negligible. D D 2.15 kOhm 1 kOhm C C 600 Ohm v s *sin(wt) v in v out 5 V D D Figure 1: Amplifier circuit for Problem 1 The diodes D in figure 1 have the following I-V characteristic: I diode [mA] V diode [V] 0.65 0.95 1.25 1 2 Figure 2: I-V characteristics of diodes in circuit shown in figure 1.
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Page 3 of 17 a) Derive an expression for V diode [V] as a function of I diode [A] for the “on-region” of the diode and give an equivalent circuit that has the same I-V characteristic in that region (10 pts) V diode = 0.65 + 300 * I diode Equivalent circuit: A 0.65 V voltage source in series with a 300 Ohm resistor b) Calculate V G , the DC voltage at the gate of the MOS transistor (15 pts). We basically have a resistive voltage divider between 5 V and 3 * 0.65 V, with resistors of 2150 Ohm and 3 x 300 Ohm V G = 1.95 + (5-1.95)*900/(2150 + 900) = 1.95 + 3.05/3050 * 900 = 2.85 V Alternatively once can solve the following equation for the current through the 3 diodes in series I diodes =(5 – 3 * V diode )/2150 = (5 – 3*(0.65 + 300 I diodes ))/2150 to get I diodes = 1 mA V G = 5 – 2150*0.001 = 2.85 V c) Express V GS , the DC voltage between the gate and the source of the transistor, as a function of I DS . (4 pts) V GS = V G – V diode = 2.85 – 0.65 – 300 * I DS = 2.2 – 300 * I DS
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Page 4 of 17 d) Assume that for a similar circuit you found out that V GS = 3 - I DS * 500. Using the I-V characteristic for the transistor in saturation (given below) and Load-Line analysis find V GSQ and I DSQ (12 pts) (Note that the given V GS vs. I DS relation is not necessarily equal to what you found in part c). This is a new starting point for the problem in case you encountered some problems in parts a) – c)) Figure 3: I DS vs. V GS for the transistor used in circuit in Figure 1 (in saturation) From the load-line analysis above we found I DS = 2mA and V GS = 2 V 3000/500 3
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Page 5 of 17 e) For the values found in d) and given in the problem, in Figure 1 and the I-V characteristic of the diode: Is the transistor in saturation? Why or why not? (12 pts)
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