Benha University Faculty of Engineering Shoubra Electrical Engineering Department First Year Communications. Answer all the following questions Illustrate your answers with sketches when necessary. The exam consists of three pages. Exam Model Answer Page 1 of 12 2 st Semester Exam 25 May 2013 ECE 121: Electronics (I) Duration : 3 hours No. of questions: 5 Total Mark: 100 Marks Examiners: Dr. Ehsan Abaas Dr. Abdallah Hammad Question 1 (15 marks) Answer this question in the form of table. Choose the correct answer (only one answer is accepted). 1 The collector emitter voltage is usually (a) Less than the collector supply voltage (b) Equal to the collector supply voltage (c) More than the collector voltage (d) Cannot answer 2 A small collector current with zero base current is caused by the leakage current of the (a) Emitter diode (b) Collector diode (c) Base diode (d) transistor 3 For the base biased circuit, if the transistor operates at the middle of the load line, a decrease of the base resistance will move the Q point (a) Down (b) Up (c) Nowhere (d) Off the load line 4 When the Q point moves along the load line, V CE decreases when the collector current (a) Decreases (b) Stay the same (c) Increases (d) Non of the above 5 For the emitter biased circuit, when the current gain increases from 50 to 300, the collector current (a) Decreases by factor of 6 (b) Increases by a factor of 6 (c) Remains almost the same (d) Is zero 6 For the emitter biased circuit, if the emitter resistance increases, the collector voltage V C (a) Decreases (b) Stay the same (c) increases (d) Break down the transistor 7 The collector voltage of the voltage divider bias circuit is not sensitive to the change of (a) Emitter resistance (b) Supply voltage (c) Collector resistance (d) Current gain 8 If the emitter resistance doubles with the TSEB the collector current will (a) Stay the same (b) Increases (c) Doubles (d) Drop in half 9 A coupling capacitor is (a) An dc open and ac short (b) An dc short (c) A dc short and an ac open (d) An ac open 10 The output voltage of CE amplifier is (a) Amplified (b) Inverted (c) 180 o out of phase with the input (d) All of the above 11 When the emitter resistance R E doubles the ac emitter resistance r e (a) Remains the same (b) Decrease (c) Cannot be determined (d) Increases 12 The input impedance of the base decreases when (a) β decreases (b) β increases (c) Supply voltage increases (d) Into the base supply 13 If the load resistance increases in a zener regulator, the zener current (a) Decreases (b) Stays the same (c) Increases (d) = the source voltage/series resistance 14 The diode with a forward voltage drop of approximately 0.25 V is the (a) Step Recovery diode (b) Light emitting diode (c) Schottky diode (d) Photo diode 15 For the varactor diode, when the reverse voltage decreases, the capacitance (a) Stays the same (b) Decreases (c) Has more band with (d) Increases
Question 2 (20 marks) a (6 marks) Explain, how could you implement AND and OR gates using diodes. 2 inputs AND gate: Truth Table Assume a diode barrier voltage of V D = 0.7 V. There are four possible states, depending on the combination of input voltages. If at least one input is at zero volts, then at least one diode is conducting and V O = 0.7 V. If both V 1 = V 2 = 5 V 2 inputs OR gates Truth table The four conditions of operation of this circuit depend on various combinations of input voltages. If V 1 = V 2 = 0, there is no excitation to the circuit so both diodes are off and V O = 0. If at least one input goes to 5 V, for example, at least one diode turns on and V O = 4.3 V, assuming V D = 0.7 V. (The students may assume that the diode voltage drop = 0 V (ideal diode)) Page 2 of 12
b (7 marks) A certain voltage doubler has 20 V rms on its input. What is the output voltage? Draw the circuit, indicating the output terminals and PIV rating for the diode. V V rms m 20V 20 2 28.28V V 2V 56.56V out m Full wave voltage doubler PIV = 2Vm Half Wave voltage doubler PIV = 2 Vm (7 marks) Design a clamper to perform the function indicated in figure (1). The design should be the following circuit: with arbitrary values of the R and C values Page 3 of 12
Question 3 (20 marks) a (6 marks) For the zener circuit shown in figure (2). Given that, V Z1 = V Z2 = 5 V (assume that both zener diodes have a voltage drop of 0.7 V when they are forward bias. Explain the operation of the circuit, and then draw the o/p voltage waveform. Two back-to-back Zeners can also be used as an ac regulator. v ( V V ) i Z2 D For the positive half cycle Z 2 will be in the reverse bias region (open circuit) so Z 1 open as well v o v i For the negative half vi ( VZ1 VD) vi ( VZ1 VD) Z 1 will be in the reverse bias region (open circuit) so Z 2 open as well v v o i v ( V V ) i Z2 D Z 2 will operate in the reverse break down region (battery model V Z2 = 5), and Z1 is forward bias (0.7) Then v o ( V 0.7) 5.7 Z2 v ( V V ) i Z1 D Z 1 will operate in the reverse break down region (battery model V Z2 = 5), and Z1 is forward bias (0.7) Then v o ( V 0.7) 5.7 Z1 Page 4 of 12
b (6marks) Drive an expression for the ac resistance of the diode Page 5 of 12
c (8 marks) For each of the following special purpose diodes, Explain the physical construction [nott more than 4 lines], how could you bias them, state some applications, and finally draw the IV characteristics LED, photo diode, varactor and Schottky diode LED Photodiode LED is a PN junction fabricated from semiconductor materials such as GaAs and GaP LED is connected in forward bias to emit light When the diode is connected in forward bias, electrons and holes recombine and release energy in the form of light (Photons) The photon energy depends on the energy gap of the material that fabricate the LED The output light from LED is directly proportional to the forward current LED voltage drop is in the range 1.5 2.5 The photodiode is a PN junctionn that operates in reverse bias The photodiode has a small transparent window that allow the light to strike the PN junction Normal diode has a constant reverse saturation current, but photodiode has a reverse current that is directly proportional to the light intensity Applications: Photo detection, Optical switching applications and Burglar alarm system Applications: Indicating ON/OFF conditions, Optical switching applications, 7 Segments, Burglar alarm system, Remote control Schottky diode Varctor It is formed by joining a doped semiconductor region (usually N type) with a metal such as (Silver, Platinum or Gold) The Schottky diode has a very little junction capacitance and it can operate at much higher frequencies of 20 GHz or more The reduced junction capacitance also results in a much higher switching time A varactor diode is basically a reverse biased pn junction The capacitance parameters are controlled by the method of doping Doping Level Doping Level Distance from junction Distance from junction Abrupt doping profile Tunning range 4: 1 Hyper abrupt doping profile Tunning range 10:1 Application: Rectify high frequency signals, switching device in digital computers, and low voltage power supplies (V D = 0.25 V) li i di d i d i i i i
Question 4 (20 marks) a (10 marks) For the circuit shown in figure (3) Determine: (1) R C (2) R E (3) R B (4) V B The transistor operates in the active region. V V V 7.6 2.4 5.2 V CE C E CE 0.2 V 0.7 V V BE B E V V 0.7 2.4 0.7 3.1V B E V V I R R C CC C C C VCC VC 12 7.6 2.2 k I 2mA C 80 0.987 1 81 IC 2 IE 2.025mA 0.987 V I R R E E E E VE 2.4 1.185k I 2.025mA E Ic 2 IB 25A 80 V V I R CC B B B VCC VB 12 3.1 RB 356 k I 25A B Page 7 of 12
b (10 marks) The signal source switch of figure (4) is closed, and the transistor base current becomes The collector characteristic of the transistor is displayed in figure (5). If V CC = 14 V and R DC = 1 kω. Graphically Determine: (1) I CQ and V CEQ (2) i c and v ce (3) h FE at the Q point Page 8 of 12
Question 5 (25 marks) a (15 marks) Drive the expression for A v, r in, r in stage, r o and r o stage for (1) Common emitter amplifier (2) Common collector amplifier 1 Common emitter: Finding the input resistance Finding the voltage gain Finding the output resistance (The student may also use the T model It will give the same results) Page 9 of 12
2 Common collector 1 The students may also add the effect of Rs in calculation of r o and ro sta age. (It will be ok as well)
b (10 marks) Based on the derivations in part (a), For the multi stage amplifier circuit shown in fig (6), Calculate the numerical values for: (1) The overall voltage gain (2) The input impedance of this multistage amplifier. DC Analysis Stage 2 1 Stage 1 1 Solving using approximation Solving using approximation 7.58 V 4.58 ma 26 mv 4.58 ma 5.668.. 26 mv 0.983 ma. 2..7 25 6.2 2.7 7.58 0.7 1.5 56 25 56 56 12.5 0.7 12
AC Analysis Second Stage 680 120 or 119.9 5.668 Ω Stage 2 Stage 2 6.2 2.7 0.566 Stage 2 First Stage Stage 2 453 Ω 12000 453 26.4 12000 453 436.52 26.4 436.52. Total gain. Input Impedance Tr 1 Tr 1 100 26.44 12000 453 Tr 1 46.296 K Stage 1 Tr 1 Stage 1 56 56 46.296 Stage 1 15.366 K Page 12 of 12