Electronics I. Midterm #1
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1 The University of Toledo Section f6ms_elct7.fm - Electronics I Midterm # Problems Points Total 5 Was the exam fair? yes no
2 The University of Toledo f6ms_elct7.fm - 2 Problem 4 points For full credit, mark your answers yes, no, or not in all given choices!. When a pn-junction diode is biased by the voltage v D = v AC = 0.2V, the total current flow through the diode is the consequence of the cumulative effect of: x the transition of electrons through the depletion region from the n-side to the p- side of the pn-junction, x the transition of holes through the depletion region from the n-side to the p-side of the pn-junction, x the transition of electrons through the depletion region from the p-side to the n- side of the pn-junction, x the transition of holes through the depletion region from the p-side to the n-side of the pn-junction..2 The magnitude of the depletion region capacitance of a pn-junction diode: x depends on the concentration of donor atoms in the n-region, x depends on the magnitude of the reverse bias voltage applied to the diode, x is always greater than the diffusion capacitance of the diode, x depends on the concentration of acceptor atoms in the p-region..3 The height of the potential barrier V bo of the pn-junction in thermal equilibrium, at room temperature, x depends on the temperature of the pn-junction, x depends on the atmospheric pressure at the surface of the piece of semiconductor material which contains the pn-junction, x depends on the magnitude of the external electric field, x depends on the concentration of acceptor atoms on the p-side of the junction...4 In a pn-junction diode: x the terminal called the anode is connected to the semiconductor region which is doped by the toms of a three-valent element. x the terminal called the cathode is connected to the semiconductor region which is doped by the atoms of a five-valent element, x the terminal called the cathode is connected to the semiconductor region which is doped by the atoms of a three-valent element, x the terminal called the anode is connected to the semiconductor region which is doped by the atoms of a five-valent element.
3 The University of Toledo f6ms_elct7.fm - 3 Problem 2 5 points Given is the electric circuit model shown in Figure 2.. i D V DD = 30V T o =293K V DD R V R R = MΩ V Z = 30V T =333K V Ro = V R (T o ) =.5V Figure 2. Electric circuit model containing a pn-junction diode. The pn-junction diode in the circuit of Figure 2. has the following properties: a) the diode is "00mA devices", i.e. I DR (V DR =0.7V) =00mA, b) the diode s breakdown voltage is V Z, c) temperature dependence of diode s reverse leakage current I L is described by: I L doubles for every 0 o K increase in diode s temperature, d) the value of diode s reverse leakage current at room temperature is not known, but the voltage V R has been measured at room temperature T o, and its value is known to be V Ro. Problem Statement Demonstrate an ability to determine:the value of the voltage V R, to which the voltage V R in the circuit of Figure 2. will be equal at temperature T. 0.5 Hint # For full credit, give answers to all questions, prepare all required circuit diagrams, write all equations for which the space has been reserved, and show all symbolic and numerical expressions whose evaluation produces shown numerical results. Problem Solution For full credit, explicit demonstration of understanding the following solution steps is expected. 2. From a consideration of the information that is presented in Figure 2, make a determination regarding the operating region of the diode at temperature T o, and check correspondingly the conditions on all three lines below, x the diode is forward biased, x the diode is reverse biased, x the diode is in its breakdown region. That the diode in the circuit of Figure 2. is reverse biased should be obvious after five weeks spent studying pn-junctions and diodes. That the diode does not operate in the breakdown region follows from the voltage division between the diode and the resistor. Therefore, the current which
4 The University of Toledo f6ms_elct7.fm creates the voltage drop V R across the resistor R in the circuit of Figure 2. is the reverse leakage current of the diode. Ohm s law provides the reverse leakage current I Lo at temperature T o. 2.2 Calculate the value I Lo of diode s reverse leakage current at temperature T o. Show your work in the space reserved for equation (2-). By Ohm s law, V Ro.5 I Lo = = R 0 6 = =.5µA (2-) Calculate the increase in temperature T between the temperature T o at which the reverse leakage current of the diode has been calculated under 2., and the temperature T at which the value of the voltage V R ought to be determined. Show your work in the space reserved for equation (2-2). T = T - T o = = 40K (2-2) 2.4 Determine the number of decades of degrees Kelvin n 0K in the temperature increase T. Show your work in the space reserved for equation (2-3). This calculation is straight forward, T n 0K = 0C 40 = 0 = 4 (2-3) 2.5 Determine the ratio of the diode s reverse leakage currents at temperatures T and T o.show your work in the space reserved for equation (2-4) By the diode s property b), and by the calculated number of decades of degrees Kelvin of temperature increase, I L n = 2 0C = 2 4 = 6 I Lo (2-4) Determine the reverse leakage current of the diode at temperature T. Show your work in the space reserved for equation (2-5) Solving the equation (2-4) for I L, I L = 6I L0 = = 24µA (2-5) 2.7 Determine the value of the voltage V R at temperature T. Show your work in the space reserved
5 The University of Toledo f6ms_elct7.fm - 5 for equation (2-6) By Ohm s law, V R = R I L = = 24V (2-6) I Lo =.5µA V R = 24V
6 The University of Toledo f6ms_elct7.fm - 6 Problem 3 6 points Given is a circuit with two ideal diodes shown in Figure 3.. R D R V M V D 2 V 2 Figure 3. A circuit with ideal diodes. V N V M = 9V V N =5V R = 4kΩ R 2 = 8kΩ Problem Statement Demonstrate an ability to:. apply the Hint # For full credit: all equations, all answers to questions, all circuit models and other graphical representations are expected to be entered into the space designated for them; all shown numerical results must be preceded by the symbolic and numeric expressions whose evaluation produces the shown results. Problem Solution For full credit, explicit demonstration of understanding the following solution steps is expected. 3. Make an educated guess as to the bias conditions of the two diodes in the circuit of Figure 3., and show your guess by checking the conditions on all four lines below, x the diode D is forward biased, x the diode D is reverse biased, x the diode D 2 is forward biased, x the diode D 2 is reverse biased. 3.2 Construct the linear circuit which results when the ideal diodes in the circuit of Figure 3. are replaced by their models for the biasing condition guessed in Section 3.. Draw the electrical model of the constructed linearized circuit in the space reserved for Figure 3.2 Substituting the ideal diodes D and D 2 by their equivalent circuits for the states guessed in Section 3., gives the circuit of Figure 3.2 (by the definition of an ideal diode, a forward biased diode has an internal
7 The University of Toledo f6ms_elct7.fm - 7 resistance of zero Ohms, and the internal resistance of a reverse biased diode is infinite). R C I D A R 2 V M -+ V v D2 V 2 -+ C 2 A 2 V N Figure 3.2 The circuit with ideal diodes replaced by their models for the biasing conditions guessed in Section To check the validity of the guesses made in Section 3., perform an analysis of the circuit of Figure 3.2 to determine the voltage across the diodes which were guessed reverse biased, and to determine the current through the diodes which were guessed forward biased. Show your calculations, and state your findings in the space reserved for equations (3-). Hint #2For a meaningful process of performing the analysis, the positive reference directions of these voltages/currents must be shown in the circuit of Figure 3.2. Failure to show these positive reference directions reduces the credit for this part to 0.. In the circuit of Figure 3.2, the voltages V and V 2 are equal since the voltage drop across the forward biased ideal diode D is equal to 0V. R 2 R V = V 2 = V M R + R V = 9 V 2 N + R + R = 8+ 4 (3.) This shows that the potential at C 2, being equal to V 2 =V, is V above the potential of A 2 ; which confirms the guess that diode D 2 is reverse biased in the circuit of Figure 3.. To check the guess about the forward bias condition of the diode D, the direction of the current flowing through D in the circuit of Figure 3.2 ought to be determined; then if that current flows in the positive reference direction (anode to cathode), the guessed forward bias has been confirmed; otherwise the guess was incorrect. Summing the voltage rises in the direction of current I D one obtains, KVL: V N -V M - I D R - I D R 2 = 0 when solved for I D, I D = V N - V M 5-9 = = 0.5mA R + R 2 (8 + 4) Compare the result of the analysis performed in Section 3.3 with the guesses made in Section 3., to make a conclusion as to whether the bias conditions of both diodes were guessed correctly. Indicate your conclusion by appropriate checks on both lines below, The obtained positive value of the current I D means that the current through diode D flows in the positive reference direction, which confirms the guess that diode D is forward biased in the circuit of Figure 3..
8 The University of Toledo f6ms_elct7.fm - 8 x the bias conditions of both diodes has been guessed correctly, x the bias condition of one, or more diodes has been guessed incorrectly. If the biasing condition of at least one diode is incorrect, repeat the steps of Sections 3. through 3.4 using the free space on the opposite page. 3.5 When the biasing conditions of all diodes have been guessed correctly, determine and write into the space reserved for equations (3-2) the values of the voltages V and V 2 which are indicated in the circuit of Figure 3.. Since both guesses which led to the equivalent circuit of Figure 3.2 have been found correct, the results of analysis performed on the circuit in Figure 3.2 are also valid for the circuit of Figure 3.. Consequently, by equations (3-), (3-2) V = V V 2 = V 3.6 When the biasing conditions of all diodes have been guessed correctly, determine and write into the space reserved for equations (3-3) the values of the currents flowing through diodes D and D 2 in the circuit of Figure 3.. Since the diode D 2 is reverse biased, it does not conduct any current, so I D2 =0A. The current through diode D has been determined by equations (3-); hence, the two currents, I D = 0.5A I D2 = 0A (3-3)
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