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W09_314_HW08_solution

# W09_314_HW08_solution - EECS 314 Student’s name...

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EECS 314 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 200 9 Alexander Ganago Page 1 of 4 Problem 3 Half-Wave Rectifier circuits with semiconductor diodes The Big Picture DC power supplies (such as in your computer) include rectifiers, which are built of semiconductor diodes. This problem introduces the simplest rectifier circuit called the half-wave rectifier. The goal of this problem is to help you develop the intuition of what the rectifier should and shouldn’t/can’t do so that you are well prepared for the exam problems such as Part 6 of this problem. In each part of this problem, calculations are minimal, but attention to detail is very important. At frequencies as low as 60 Hz, we can safely neglect all RC effects due to the effective capacitance of the diode (note that many diodes include several charged layers, which look – and act as capacitors, but their effects are significant at higher frequencies). Consider each circuit at two moments of time, labeled T 1 and T 2 : for each moment, calculate the output voltage as you did for DC circuits (problem 1 of this assignment). The Assignment Part 1 (6 points) Your answers: V OUT (T 1 ) = ______________ V V OUT (T 2 ) = ______________ V Continued on the next page Part 1 2 3 4 5 6 Total If we assume the diode is off, we can replace with an open circuit model. In this case, no current flows and Vout = 0, so V = 5 volts at T1. Since V = 5 is greater than V_D0, that implies our assumption was wrong. Replacing with the constant voltage source (=V_D0) model, we get Vin = Vout = 5V at T1. Repeating the assumption, at T2, we get V = -5V, this is less than V_D0, so our assumption was correct and the open circuit model remains and Vout = 0V. general approach: 1. Assume diode is off and replace with open. 2. solve for V across the diode. 3. if V< V_D0, open circuit can stay, otherwise replace diode with a voltage source = V_D0.
EECS 314 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above © 200 9 Alexander Ganago Page 2 of 4 Problem 3, continued Part 2 (6 points) Your answers: V OUT (T 1 ) = ______________ V V OUT (T 2 ) = ______________ V Part 3 (6 points) Your answers: V OUT (T 1 ) = ______________ V V OUT (T 2 ) = ______________ V Continued on the next page If we assume the diode is off at T1, we can replace with an open. this would give V = -5 which is less than V_D0, so the "off" (open circuit) assumption was correct and the open circuit stays and Vout = 0 When V is greater than V_D0 (@T1, where Vin = 5V will do), the diode can be replaced with the constant voltage source model (=0.7V); that would make Vout = Vin - 0.7V = 4.3V At T2, V < V_D0 and the diode acts as an open, so no current flows and Vout = 0V.

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EECS 314 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above
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W09_314_HW08_solution - EECS 314 Student’s name...

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