EECS:3400 Electronics I s5ms_elct7.fm - Section Electronics I Midterm # Problems Points. 4 2. 5 3. 6 Total 5 Was the exam fair? yes no
EECS:3400 Electronics I s5ms_elct7.fm - 2 Problem 4 points For full credit, mark your answers yes, no, or not in all the given choices!. At room temperatures, the number of ionized donor/acceptor atoms in a semiconductor material doped for application in semiconductor devices, x depends on the concentration of donor/acceptor atoms. _ x is below 9% of all such atoms, x is close to 50% of all such atoms, x is above 99% of all such atoms, x depends on the intrinsic concentration of electron-hole pairs..2 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..3 Inside a semiconductor material: _x drift current is caused by the presence of an external electric field, x diffusion current is caused by the presence of the internal electric field, _x diffusion current is caused by the presence of a gradient in the concentration of minority charge carriers, x drift current is caused by the presence of the internal electric field,.4 PN-junction breakdown is caused by the: x avalanche multiplication of ionized donors inside the depletion region, x tunneling effect through thin potential barriers, x circuits where diodes are exposed to high reverse bias voltages, x avalanche multiplication of free charge carriers inside the depletion region, x tunneling effect through thick potential barriers.
EECS:3400 Electronics I s5ms_elct7.fm - 3 Problem 2 5 points Figure 2. shows the electrical circuit model of a pn-junction diode circuit in which all diodes are R i D V DD = 0V V DD + - V O v D V O =.5V Figure 2. A pn-junction diode circuit. 0.5 0.5 identical, and have their properties characterized by the following three statements: (a) they are "5mA devices", i.e. I DR (V DR =0.7V) =5mA, (b) for two values of the diodes current, i D and i D2, such that the relation i D2 =0i D holds, the corresponding diode voltages v D2 and v D are in the relation v D2 = v D + 0.058V. (c) diodes breakdown voltage value is V Z = 5.5V; Problem statement For the electric circuit model of Figure 2., demonstrate an ability to apply the known relation (b) and the Kirchhoff s voltage law, to determine: - the diode current i D at which the indicated voltage V o will be having the value shown in Figure 2., - the value of resistance R such that the indicated voltage drop V O will be having the value shown in Figure 2.. Solution Hint # For full credit, give answers to all questions, prepare all required circuit diagrams, write all equations for which the space is left, and show all symbolic and numerical expressions whose evaluation produces shown numerical results. An explicit demonstration of understanding the following solution steps is expected. 2. Indicate in Figure 2. the positive reference directions of the current i D and voltage v D of just one of the diodes. 2.2 Calculate the common value v D of the voltage drops across individual diodes in the circuit of Figure 2.; show your work in the space reserved for equation (2-). Since the diodes are identical, and the same current i D flows through both of them, the individual diode
EECS:3400 Electronics I s5ms_elct7.fm - 4 voltages are in the following relation with the output voltage V O, v D = V O.5 = 2 2 = 0.75V (2-) 2.3 Calculate the difference between the known diodes reference voltage V DR and the actual diodes voltage v D in the circuit of Figure 2.; show your work in the space reserved for equation (2-2). After selecting the notation: i D = I DR, i D2 = i D, v D = V DR, v D2 = v D, one has v D = v D2 - v D = v D - V DR = 0.75-0.7 = 0.05V (2-2) 2.4 Using two known diodes voltages, V DR and v D, and the diodes characterizing conditions (a) and (b), calculate the value of the actual current i D needed in the circuit model of Figure 2.; show your work in the space reserved for equation (2-3). With both, the calculated diodes voltage v D =0.825V > 0.2V and the reference voltage V DR = 0.7V > 0.2V, both current-voltage pairs, (i D,v D ) = (I DR,V DR ) and (i D2,v D2 ) = (i D,v D ), are well approximated by the forward-bias current-voltage relation, Therefore, i e v D -V DR D = V I T DR I DR I Se V DR V T and i D I S e v D V T v D i D = I DR e = 5 0 3 e 0.05 0.025 = 5 0 3 = 36.9mA V T e 2 (2-3) 2.5 Prepare the KVL equation for the circuit in Figure 2., and show it in the space reserved for equation (2-4) KVL: V DD - R i D - V O = 0 (2-4) 2.6 Solve the KVL equation (2-4) for the resistance R, and calculate the value of R. Show your work in the space reserved for equation (2-5). R = V DD - V O 0.5 = = 230Ω i D 0.0369 (2-5)
EECS:3400 Electronics I s5ms_elct7.fm - 5 Problem 3 6 points Given is a circuit with two ideal diodes shown in Figure 3.. R D 2 R 2 V M = 0V R = 7kΩ V M V D V +- + 2 - V N V N = 5V R 2 = 3kΩ Figure 3. A circuit with ideal diodes. Problem statement For the electric circuit model of Figure 3., demonstrate an ability to:. apply the piece-wise linear models of non linear circuit elements in the process of analysis of nonlinear circuits, 2. apply the large signal method of analysis to nonlinear electric circuits containing diodes in order to determine: - values of the voltages V and V 2 whose positive reference directions are indicated in the circuit model of Figure 3.. - values of the positive reference direction currents of diodes in the circuit model of Figure 3.. Hint # For full credit, give answers to all questions, prepare all required circuit diagrams, write all equations for which the space is reserved, and show all symbolic and numerical expressions whose evaluation produces the shown numerical results. Solution An 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.
EECS:3400 Electronics I s5ms_elct7.fm - 6 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., and draw the electrical model of the constructed 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 resistance of zero Ohms, and the internal resistance of a reverse biased diode is infinite). R C 2 A 2 I D2 R 2 V M V C V +- v + D 2 A - V N Figure 3.2 The circuit with ideal diodes replaced by their models for the biasing conditions guessed in Section 3.. 3.3 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 work in the space reserved for equations (3-). Hint #2 For a meaningful process of performing the analysis, the positive reference directions of diodes voltages/currents must be shown in the circuit of Figure 3.2. Failure to show those 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 2 is equal to 0V. R R V = V 2 = V N - 2 7 3 R + R V = 0.5V 2 M 5 - R + R 2 7+ 3 0 = 7+ 3 (3-) Which shows that the potential at C, being equal to V = 0.5V, is 0.5V above the potential of A, confirming that diode D is reverse biased in the circuit of Figure 3.2. To formally check the guess about the bias condition of the diode D 2, we ought to determine the direction of the current flowing through D 2 in the circuit of Figure 3.2. If the current of D 2 flows in the positive
EECS:3400 Electronics I s5ms_elct7.fm - 7 reference direction (anode to cathode), then D 2 is forward biased. Writing the KVL equation we obtain, V M +V N - I D2 R - I D2 R 2 = 0 solving the above equation for I D2, V M + V N 0 + 5 I D2 = = =.5mA R + R 2 (7 + 3) 0 3 (3.)bis 3.4 Compare the result of the analysis performed in Section.3 with the guesses made in Section., 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, x the biasing condition of both diodes has been guessed correctly, x the biasing 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. through.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 below the values of the voltages V and V which are indicated in the circuit of Figure.. Show your work in the space reserved for equations (3-2). Since now both guesses which led to the equivalent circuit of Figure 3.2 have been found correct, the results of the analysis performed on the circuit in Figure 3.2 are valid for the circuit of Figure 3.. Consequently, by equations (3-), (3.2) V = 0.5V V 2 = 0.5V 3.6 When the biasing conditions of all diodes have been guessed correctly, determine and write into the space reserved below the values of the currents flowing through diodes D and D 2 in the circuit of Figure 3.. Show your work in the space reserved for equations (3-3) Since the diode D is reverse biased, it does not conduct any current, so I D =0A. The current through diode D 2 has been determined by equations (3-)bis. Hence, the two current values, I D = 0A I D2 =.5mA (3.3)