EECE 311 Midterm, Fall 2010

American University of Beirut Faculty of Engineering and Architecture Department of Electrical and Computer Engineering EE 042 - Analog Electronics Instructors: A. Kayssi and M. Krisht Quiz 1 Saturday November 23, 1996 Closed Book Programmable Calculators Are Not Allowed. Time: 1.5 hours VERSION A The figures for this quiz are missing Name:_______________________________ ID#:__________________________ ! Provide your answer on the computer card only. ! Return the computer card attached to the question sheet. ! Use a pencil for marking your answers and ID number on the computer card. ! When using an eraser, make sure you erased well. ! On this sheet, write with a pen your name followed by your ID number. ! All questions are graded equally. 1. What is the version of your question sheet? (This question is not graded.) a) Version A b) Version B c) Version C d) Version D e) Version E Unless otherwise specified, assume VT = 25 mV, β is very large, and |VBE |= 0.7V 2. Determine RC and RE in the circuit of Figure 8 to have VO = 0V and VCE = 3V. Assume β = 200 (do not neglect IB). a) RC = 0.82 kΩ, RE = 1.65 kΩ b) RC = 1.65 kΩ, RE = 0.82 kΩ c) RC = 1.82 kΩ, RE = 0.65 kΩ d) RC = 0.65 kΩ, RE = 1.82 kΩ e) none of the above 3. Find the value of R required to satisfy the current relationships in the circuit of Figure 7. a) 100 kΩ b) 50 kΩ c) 200 kΩ d) 105 kΩ e) none of the above 4. For the circuit shown in Figure 13, find the voltage gain VO/Vi. Assume β = 40, and rπ = 1 kΩ. a) -33.7 b) -28.3 c) -14.6 d) -3.51 e) none of the above 5. Find the current gain IO/Ii directly, as defined in the circuit of Figure 12. The JFET parameters are IDSS = 4mA and VP= -2V and is biased at ID = 1mA. a) -340 b) -680 c) -170 d) -425 e) none of the above 6. A MOSFET source follower is biased at ID = 0.5mA. What is the value of VGS if the small signal output resistance is 500Ω and Vt = 1V. a) 2.2 V b) 2 V c) 1.5 V d) 1.2 V e) none of the above 7. Find the differential gain vO/(v1-v2) in the circuit of Figure 4. a) -100 b) -200 c) 200 d) 100 e) none of the above 8. Find the common-mode rejection ratio (CMRR) for the circuit of Figure 4. Assume VA = 100V. a) 73 dB b) 82 dB c) 66 dB d) 78 dB e) none of the above 9. Find the value of the emitter resistors, to be inserted between the emitters of Q1 and Q2 and node X in the circuit of Figure 4, to have a differential input resistance of 200KΩ. Assume β = 99. a) 1.5 kΩ b) 5 kΩ c) 150 Ω d) 35 Ω e) none of the above 10. Find the value of R in the circuit of Figure 3 in order to have VO = 2V. a) 7.17 kΩ b) 9.5 kΩ c) 14.3 kΩ d) 19 kΩ e) none of the above 11. All the BJTs in the circuit of Figure 9 are biased at IC=1mA and have β = 100. Find VO/V3 and V3/V2. (Hint: Include the loading effect of the last stage on the second stage). a) VO/V3 = -60 ,V3/V2 = 0.985 b) VO/V3 = -80 ,V3/V2 = 0.994 c) VO/V3 = -60 ,V3/V2 = 0.994 d) VO/V3 = -80 ,V3/V2 = 0.985 e) none of the above 12. Find VO/VS in the circuit of Figure 9. a) 540.5 b) 187.4 c) 932.6 d) 349.1 e) none of the above 13. Find the range of values of IO in the circuit of Figure 2, when β varies between 75 and 175. a) 0.918 to 0.925 mA b) 0.459 to 0.462 mA c) 0.468 to 0.471 mA d) 0.926 to 0.931 mA e) none of the above American University of Beirut Faculty of Engineering and Architecture Department of Electrical and Computer Engineering EE 042 - Analog Electronics Instructor: A. Kayssi Quiz 1 Saturday November 29, 1997 Closed Book Programmable Calculators Are Not Allowed. Time: 1.5 hours VERSION A Name:_______________________________ ID#:__________________________ ! Provide your answer on the computer card only. ! Return the computer card attached to the question sheet. ! Use a pencil for marking your answers and ID number on the computer card. ! When using an eraser, make sure you erased well. ! On this sheet, write with a pen your name followed by your ID number. ! All questions are graded equally. 1. What is the version of your question sheet? (This question is not graded.) a) Version A b) Version B c) Version C d) Version D e) Version E 2. Find the resistance ro for the MOSFET shown in Figure 1. The MOSFET parameters are K = 0.25 mA/V2, Vt = 1.1V, and VA = 1/λ = 60 V. Neglect λ in the DC analysis. a. 261.7 KΩ b. 250.8 KΩ c. 273.6 KΩ d. 286.5 KΩ e. none of the above 3. Calculate the voltage gain, vo/vs in the circuit of Figure 1. The MOSFET parameters are K = 0.25 mA/V2, Vt = 1.1V, and VA = 1/λ = 60 V. Neglect λ in the DC analysis. a. -6.45 b. -6.71 c. -6.84 d. -6.58 e. none of the above 4. For the differential amplifier shown in Figure 2, calculate the output voltage vo when v1 = 26 sin(ωt) mV and v2 = 24 sin(ωt) mV. The differential gain is 100 and the CMRR is 60 dB. a. 202.5 sin(ωt) mV b. 207.9 sin(ωt) mV c. 225 sin(ωt) mV d. 279.1 sin(ωt) mV e. none of the above 5. For the differential amplifier shown in Figure 3, calculate the collector current of transistor Q1. Assume VBE = 0.7 V, VT = 25 mV and RE = 10 KΩ. Neglect the DC base current in your calculations. a. 75.8 µA b. 37.9 µA c. 22.8 µA d. 113.8 µA e. none of the above 6. In the circuit of Figure 3, calculate the CMRR. Assume VBE = 0.7 V, VT = 25 mV, and RE = 10 KΩ. Neglect the DC base current in your calculations. The BJTs have β = 85 and VA = 70V. a. 93.3 dB b. 89.6 dB c. 94.3 dB d. 91.2 dB e. none of the above 7. For the circuit shown in Figure 4, find the voltage gain vo/vs. Assume VBE = 0.7V , VT = 25 mV, β = 100, VA = 100 V, and RS = 1 KΩ. Negelct the DC base current in your calculations. a. 0.922 b. 0.826 c. 0.963 d. 0.876 e. none of the above 8. Find ro for the three transistors shown in Figure 5. Transistor Q1 parameters are: K = 5 mA/V2, Vt = -2 V, VA = 1/λ = 50 V. Transistor Q2 parameters are β = 150 and VA = 80 V. Transistor Q3 parameters are β = 80 and VA = 60 V. Assume VBE = 0.7 V and VT = 25 mV. Neglect λ in the DC analysis of the MOSFET. Also, neglect the DC base currents of the BJTs. Assume VCC = 9 V. a. ro1 = 10 KΩ; ro2 = 100 KΩ; ro3 = 44.81 KΩ b. ro1 = 10 KΩ; ro2 = 49.23 KΩ; ro3 = 25.28 KΩ c. ro1 = 10 KΩ; ro2 = 65.98 KΩ; ro3 = 32.33 KΩ d. ro1 = 10 KΩ; ro2 = 39.26 KΩ; ro3 = 20.76 KΩ e. none of the above 9. Refer to the given of Problem 8. Find the gain of the last stage (vo/v3). a. 0.926 b. 0.945 c. 0.964 d. 0.957 e. none of the above 10. Refer to the given of Problem 8. Find the gain of the middle stage (v3/v2) taking into consideration the loading effect of stage 3 on stage 2. a. -214 b. -263.5 c. -109.8 d. -162.8 e. none of the above 11. Refer to the given of Problem 8. Find the overall gain vo/vs. a. 919.7 b. 1086.6 c. 727.8 d. 507.2 e. none of the above 12. What is the input resistance of the circuit in Figure 5? a. 200 Ω b. 9.9 KΩ c. 10 KΩ d. 1010 KΩ e. none of the above 13. Find the range of values of Io in the circuit of Figure 6 when β varies between 30 and 300. Assume that VBE = 0.7 V and R = 100 KΩ. a. 25 to 25.7 µA b. 30 to 30.9 µA c. 37.5 to 38.6 µA d. 45 to 46.3 µA e. none of the above American University of Beirut Faculty of Engineering and Architecture Department of Electrical and Computer Engineering EE 042 - Analog Electronics Instructors: L. Chaar and A. Kayssi Quiz 1 Saturday November 14, 1998 Closed Book Programmable Calculators Are Not Allowed. Time: 1.5 hours VERSION A Name:_______________________________ ID#:__________________________ ! Provide your answer on the computer card only. ! Return the computer card attached to the question sheet. ! Use a pencil for marking your answers and ID number on the computer card. ! When using an eraser, make sure you erased well. ! On this sheet, write with a pen your name followed by your ID number. ! All questions are graded equally. 1. What is the version of your question sheet? (This question is not graded.) a) Version A b) Version B c) Version C d) Version D e) Version E Assume in all problems that: VBE = 0.7 V, VT = 25 mV, and that all capacitors are large All problems are equally graded 2. Calculate the input resistance Ri of the amplifier shown in Figure 1. For the BJTs, β = 100, VA = 70V. Neglect base current during DC analysis. a) 112.5 KΩ b) 85.6 KΩ c) 73.5 KΩ d) 211.9 KΩ e) 180 KΩ 3. Calculate the output resistance Ro of the amplifier shown in Figure 1. For the BJTs, β = 100, VA = 70V. Neglect base current during DC analysis. a) 18.7 KΩ b) 20 KΩ c) 100 KΩ d) 16.67 KΩ e) 39.2 KΩ 4. Calculate the voltage gain vo/vs for the amplifier shown in Figure 1. For the BJTs, β = 100. Neglect base current during DC analysis. Also, neglect ro for the two transistors. a) 623.4 b) 586.2 c) 478.7 d) 345.9 e) none of the above 5. In the circuit of Figure 2, what should be the values of RE and RL to get a voltage gain vo/vs = -10, and an input resistance Ri = 500 KΩ? Assume β = 75 and re << RE. a) RE = 4.22 KΩ, RL = 100 KΩ b) RE = 5.39 KΩ, RL = 33.33 KΩ c) RE = 8.01 KΩ, RL = 220.2 KΩ d) RE = 6.58 KΩ, RL = 66.68 KΩ e) RE = 2.33 KΩ, RL = 51.42 KΩ 6. In the circuit of Figure 3, find the voltage gain vo1/vs. Assume β = 100. Neglect base current in the DC analysis. a) -0.984 b) -2.65 c) -18.3 d) 9.64 e) -322.5 7. In the circuit of Figure 3, find the voltage gain vo2/vs. Assume β = 100. Neglect base current in the DC analysis. a) 1 b) 0.923 c) 0.867 d) 0.954 e) 0.994 8. A differential amplifier employing 10 KΩ collector resistors, and for which the emitter bias current source is 400 µA, uses BJTs for which β = 200. It is driven differentially by signal sources (+vd/2 and - vd/2) whose source resistances are 10 KΩ each. The emitter current source has an output resistance of 500 KΩ. For an output taken single-endedly, find the magnitude of the voltage gain vo/vd and the CMRR. a) | vo/vd | = 28.5, CMRR = 69.1 dB b) | vo/vd | = 20.2, CMRR = 78.2 dB c) | vo/vd | = 33.5, CMRR = 82.3 dB d) | vo/vd | = 45.1, CMRR = 45.6 dB e) none of the above 9. Find the voltage gain vo/vd and the CMRR for the circuit of Problem 8 when two emitter resistors RE = 9 re are added to the circuit. a) | vo/vd | = 13.4, CMRR = 80.5 dB b) | vo/vd | = 3.83, CMRR = 51.7 dB c) | vo/vd | = 12.6, CMRR = 122.4 dB d) | vo/vd | = 6.63, CMRR = 44.9 dB e) none of the above 10. For the differential amplifier shown in Figure 4, find the magnitude of the differential gain vo/vd when vB1 = +vd/2 and vB2 = -vd/2. a) 23.6 b) 82.5 c) 94.1 d) 17.4 e) 41.2 11. For the circuit shown in Figure 4, calculate the CMRR. a) 67.2 dB b) 82.9 dB c) 75.6 dB d) 56.9 dB e) 49.1 dB 12. When the inputs to a differential amplifier are sin(t) - cos(t) mV and sin(t) + cos(t) mV, the output is 20000 cos(t) - sin(t) mV. Find the CMRR. a) 20 dB b) 40 dB c) 80 dB d) 100 dB e) 60 dB 13. A source follower amplifier uses MOSFETs for which Vt = 1V, VA is very large and body effect is negligible. If the MOSFET is biased at ID = 1 mA, VGS = 2V, what is the range of load resistances for which the voltage gain of the follower is greater than 0.9? a) > 500 Ω b) > 4500 Ω c) < 4500 Ω d) < 450 Ω e) > 450 Ω 14. The common source amplifier shown in Figure 5 uses MOSFETs for which k’(W/L) = 20 µA/V2 and VA = 100 V. How does the voltage gain vo/vs change when IREF is reduced from 100 µA to 1 µA? a) The magnitude of the gain decreases from 316.2 to 31.6 b) The magnitude of the gain decreases from 632 to 63.2 c) The magnitude of the gain increases from 63.2 to 632 d) The magnitude of the gain increases from 31.6 to 316.2 e) There is not enough data to solve this problem. 15. Which of the following statements is true? I. The voltage gain of an emitter-follower is much larger than 1 II. The input resistance of a common-base is very small III. The output resistance of a common-drain is very large IV. The current gain of common-emitter is always close to 1 a) only I is true b) only II is true c) only III is true d) only IV is true e) all are false 2K 180K 300K 300K 180K 20K 2K 18K 20K 20K 100K Vcc = +15V vS + Figure 1 RE 100 Ohm RL vS vo + - vo + - Figure 2 Ri Ri 1K + -5V 1K +5V vS 10K 20K 20K vo2 + - vo1 + - Figure 3 10K 10K +15V +15V 82K -15V vB1 vB2 v0 Figure 4 IREF vs vo Figure 5 + Ro ELEG 042 Quiz 1 – Fall 2000-2001 – Version A - Page 1 of 4 American University of Beirut Faculty of Engineering and Architecture Department of Electrical and Computer Engineering ELEG 042 - Analog Electronics Instructors: A. Kayssi and Z. Othman Quiz 1 Thursday November 23, 2000 Closed Book Programmable Calculators Are Not Allowed. Time: 1.5 hours VERSION A Name:_______________________________ ID#:__________________________ ► Provide your answer on the computer card only. ► Return the computer card attached to the question sheet. ► Use a pencil for marking your answers and ID number on the computer card. ► When using an eraser, make sure you erased well. ► On this sheet, write with a pen your name followed by your ID number. ► All questions are equally graded. ► Only the computer card will be considered in grading. PENALTY is FOUR to ONE 1. What is the version of your question sheet? (This question is not graded.) a) Version A b) Version B c) Version C d) Version D e) Version E ELEG 042 Quiz 1 – Fall 2000-2001 – Version A - Page 2 of 4 Assume in all problems that: |VBE| = 0.7 V VT = 25 mV β = 100 unless otherwise specified Capacitors are very large Early effect is negligible unless otherwise specified In the circuit of Figure 1, RB = 300K. 2. Find the voltage gain vo/vi. a) –0.81 b) –0.97 c) –1.59 d) –2.34 e) –1.21 3. Find the input resistance Ri in KΩ. a) 380.5 b) 175.8 c) 292.7 d) 339.6 e) 238.7 4. Find the current gain io/ii. a) –41.2 b) –31.0 c) –35.4 d) –38.1 e) –33.0 In the circuit of Figure 2, assume RE = 0.5K. 5. Find the voltage gain vo/vi. a) 0.900 b) 0.928 c) 0.952 d) 0.825 e) 0.943 6. Find Rib2 at the base of Q2 (in KΩ.) a) 203.2 b) 253.9 c) 152.4 d) 101.6 e) 50.8 7. Find Rib1 at the base of Q1 (in MΩ.) a) 16.39 b) 11.36 c) 21.74 d) 27.03 e) 6.17 8. Find the current gain io/ii. a) 396.6 b) 220.1 c) 152.3 d) 94.3 e) 116.5 In the circuit of Figure 3, assume VCC = 12V. 9. Find the voltage gain vo/vi. a) 95.2 b) 75.2 c) 35.3 d) 55.2 e) 115.2 10. Find the input resistance vi/ii in Ohms. a) 8.5 b) 27.3 c) 13.0 d) 17.6 e) 10.3 In the circuit of Figure 4, assume I = 2.5mA. 11. Find the differential voltage gain (vc1 – vc2)/vd, when vB1 = +vd/2 and vB2 = -vd/2. a) –9.9 b) –11.3 c) –12.4 d) –13.2 e) –11.9 ELEG 042 Quiz 1 – Fall 2000-2001 – Version A - Page 3 of 4 12. Find the differential input resistance Rid in KΩ. a) 26.5 b) 30.3 c) 22.7 d) 25.3 e) 24.2 13. Find the input bias current in microAmps when β1=80 and β2=120. a) 12.88 b) 5.15 c) 10.31 d) 20.61 e) 8.24 14. Find the input offset current in microAmps when β1=80 and β2=120. a) 4.08 b) 8.16 c) 5.10 d) 2.04 e) 3.26 15. A differential amplifier has a CMRR of 70 dB. The output voltage is 100 mV when the inputs are v1 = 21.5 mV and v2 = 20 mV. Find the differential gain of this amplifier. a) 66.4 b) 53.5 c) 58.6 d) 46.4 e) 63.9 In the circuit of Figure 5, the MOSFETs have Vt = 2.2V and ½ k’n(W/L) = 1 mA/V2. Neglect body effect and channel length modulation. 16. Find ID in mA. a) 0.090 b) 0.0563 c) 0.116 d) 0.0730 e) 0.160 17. Find gm for the MOSFETs in mA/V. a) 0.600 b) 0.540 c) 0.474 d) 0.800 e) 0.681 18. Find vo1/vs. a) –5.12 b) –5.68 c) –7.58 d) –4.49 e) –6.45 19. Find vo2/vs. a) 18.4 b) 29.3 c) 22.7 d) 40.4 e) 14.2 20. Find ro in KΩ for the MOSFETs given that channel length modulation for AC signals is not negligible and is modeled by λ = 0.1 V-1. a) 62.5 b) 177.7 c) 137.0 d) 111.1 e) 86.2 21. Find Ro in KΩ, given that channel length modulation for AC signals is not negligible and is modeled by λ = 0.1 V-1. a) 16.9 b) 17.5 c) 16.2 d) 15.2 e) 18.0 ELEG 042 Quiz 1 – Fall 2000-2001 – Version A - Page 4 of 4 1000K vi vo 0.1K 10K 10K +10V vi vo 500K 500K 10K RE +12V vi vo Vcc 1K 1K +10V 1K 2K 0.1K 0.1K I Q1 Q2 RB Ri io ii Q2 Q1 Rib2 ii Rib1 io 500K io ii vB1 vB2 vC1 vC2 +12V 300K 300K 100K 20K 20K 10K 10K 50K 10K vo2 vs + 100K vo1 ID Ro Figure 1 Figure 2 Figure 3 Figure 4 Figure 5 ELEG 042 Quiz 1 – Fall 2000-2001 – All Versions - Page 4 of 4 ELEG 042 Quiz 1 – Fall 2001-2002 – Page 1 of 4 American University of Beirut Faculty of Engineering and Architecture Department of Electrical and Computer Engineering ELEG 042 - Analog Electronics Instructors: A. Kayssi and Z. Othman Quiz 1 November 2001 Closed Book Programmable Calculators Are Not Allowed. Time: 1.5 hours Name:_______________________________ ID#:__________________________ ► Provide your answer on the computer card only. ► Return the computer card attached to the question sheet. ► Use a pencil for marking your answers and ID number on the computer card. ► When using an eraser, make sure you erased well. ► On this sheet, write with a pen your name followed by your ID number. ► All questions are equally graded. ► Only the computer card will be considered in grading. PENALTY is FOUR to ONE Penalty is calculated as follows: One to three wrong answers result in no penalty; four to seven wrong answers result in canceling one correct answer; eight to eleven wrong answers result in canceling two correct answers; and so on. ELEG 042 Quiz 1 – Fall 2001-2002 – Page 2 of 4 Assume in all problems that: |VBE| = 0.7 V VT = 25 mV β = 100 unless otherwise specified Capacitors are very large Early effect and body effect are negligible unless otherwise specified In the circuit of Figure 1, I2 = I3 = 0.24 mA. 1. Find the collector current of transistor Q2 in mA. a) 1.21 b) 0.667 c) 0.242 d) 0.828 e) 0.354 2. Find the collector current of transistor Q1 in µA. a) 2.4 b) 12 c) 6.6 d) 8.2 e) 3.5 3. Find the input resistance at the base of Q2 (Ri2) in KΩ. a) 3.75 b) 2.06 c) 10.3 d) 3.02 e) 7.07 4. Find the input resistance at the base of Q1 (Ri) in MΩ. a) 0.758 b) 0.610 c) 0.417 d) 1.43 e) 2.08 5. Find the voltage gain vc2/vb2 (vc2 and vb2 are the small-signal collector and base voltages of transistor Q2, respectively.) a) –138.7 b) –475.4 c) –261.5 d) –95.1 e) –324.9 6. Find the voltage gain vo/vi a) –68.4 b) –236.7 c) –46.6 d) –161.5 e) –129.8 7. Find the output resistance Ro in Ohms. a) 116.8 b) 125.9 c) 164.2 d) 194.3 e) 132.9 In the circuit of Figure 2, assume the DC value of the output voltage to be VO = -2 V. The MOSFET parameters are Vtn = 0.8 V, ½ k’n(W/L)n = 0.5 mA/V2, λn = 0.01 V-1, and Vtp = - 0.8 V, ½ k’p(W/L)p = 0.2 mA/V2, λp = - 0.01 V-1. 8. Find VGS for transistor Q2 in Volts. a) 1.86 b) 1.66 c) 2.03 d) 1.41 e) 2.18 9. Find ID of Q1 in mA. a) 0.139 b) 0.205 c) 0.105 d) 0.0687 e) 0.172 10. Find the voltage gain vo/vg2 (vg2 is the small-signal voltage at the gate of Q2.) a) 0.862 b) 0.846 c) 0.820 d) 0.873 e) 0.766 11. Find the voltage gain vo/vi a) –9.65 b) –8.26 c) –6.37 d) –10.8 e) –11.7 In the circuit of Figure 3, assume I = 250 µA and RB = 30 K. 12. Find the value of RE (in KΩ) to get the specified I when v1 = v2 = 0. a) 37.1 b) 30.9 c) 92.9 d) 61.9 e) 46.4 13. Find Ad = vo2 / vd when v1 = +vd/2 and v2 = – vd/2 a) 49.8 b) 31.1 c) 39.3 d) 45.2 e) 53.4 ELEG 042 Quiz 1 – Fall 2001-2002 – Page 3 of 4 14. Find Acm = vo2/vcm when v1 = v2 = vcm a) –0.664 b) –0.531 c) –0.266 d) –0.796 e) –0.398 15. Find the CMRR in dB, corresponding to the values of the gains calculated in the previous two questions. a) 41.4 b) 37.5 c) 36.5 d) 38.6 e) 39.9 16. Find the differential input resistance vd / ib1 in KΩ, when v1 = +vd/2 and v2 = – vd/2. a) 127.3 b) 100.4 c) 161 d) 110.5 e) 93.7 17. Find the common-mode input resistance vcm/(2ib1) in MΩ, when v1 = v2 = vcm a) 3.14 b) 9.42 c) 4.71 d) 6.28 e) 3.77 18. Find the input bias current in µA. a) 0.743 b) 1.238 c) 0.495 d) 0.990 e) 1.485 19. Find the input offset current (in nA) when β is matched to within +/- 1.5%. a) 22.3 b) 37.1 c) 29.7 d) 14.9 e) 44.6 A differential amplifier has a differential voltage gain of 350 and a CMRR of 85 dB. A differential input signal of vd = 2 sin(ωt) mV is applied, along with a common-mode voltage of vcm = 2 sin(ωt) V. Assume that the common-mode gain is positive. 20. Determine the amplitude of the ideal output voltage in Volts. a) 0.7 b) 0.6 c) 0.3 d) 0.5 e) 0.4 21. Determine the amplitude of the actual output voltage in Volts. a) 0.528 b) 0.638 c) 0.422 d) 0.739 e) 0.317 In the circuit of Figure 4, assume Vcc = 7 V. 22. Find the collector current of the BJT in mA. a) 1.71 b) 1.49 c) 1.26 d) 0.811 e) 1.04 23. Find the value of re for the BJT in Ohms. a) 23.9 b) 30.5 c) 16.7 d) 19.6 e) 14.5 24. Find the input resistance Ri in KΩ. a) 148.4 b) 142.4 c) 145.8 d) 137.9 e) 131.4 25. Find minimum value of the total output voltage in Volts when vs = sin(ωt) Volt. a) 2.076 b) 1.624 c) 2.529 d) 0.725 e) 1.174 ELEG 042 Quiz 1 – Fall 2001-2002 – Page 4 of 4 Figure 3 Figure 2 vi vo 10K Vcc Q2 Q1 Ri Figure 1 5K + - Q3 Ro I2 I3 Ri2 vi vo 35K 5K -5V +5V Q1 Q2 +10V 50K 50K Q1 Q2 v1 v2 vO1 vO2 -10V RE RB RB + + I Figure 4 vo Vcc 10K 10K 20K 2K Ri 20K Vcc + vs American University of Beirut Department of Electrical and Computer Engineering EECE 311 – Electronics II Instructor: Ayman Kayssi Final Exam January 21, 2006 Open Book Time: 180 minutes Name: ID Number: Problem 1 [10 points] a) Refer to the circuit shown in Figure 1. What should be the value of R in order to have class AB operation? Assume |VBE(on)| = 0.65 V and neglect base current. Note that the bias circuit of Q1 has negligible effect and is not shown. b) The average DC current drawn from each of the two power supplies is 20 mA. What is the efficiency of the amplifier if the peak-to-peak output voltage is 10 V? vs + R 0.2 mA 100 Ω -15 V +15 V 10K C Q1 Figure 1 Problem 2 [20 points] a) For the amplifier circuit shown in Figure 2, assume VD = 0.7 V for a conducting diode, VBE = 0.7 V, and VCE(edge of sat) = 0.3 V. The BJTs are identical and have negligible base current. The FETs are identical with k’n (W/L) = 2.5 mA/V2 and Vt = 1 V. Find VOUT when V1 = V2 = 0. b) Find the total circuit power dissipation when V1 = V2 = 0. c) Find the input common-mode range. -10V +10V -10V +10V -10V 10K Vout 5K 14.3K 2.6K -10V 5K 14.3K +10V +10V 10K 10K +10V 10K 10K +10V 15.3K -6V -10V 4K V V 1 2 V s Figure 2 Problem 3 [20 points] Consider the amplifier shown in Figure 3. For the MOSFET, gm = 10 mA/V, and Cgs = Cgd = 0.1 pF. For the BJT, β = 100, gm = 50 mA/V, Cπ = 1 pF and Cμ = 0.5 pF. a) Find the midband gain vo/vs. b) Find the upper 3-dB frequency using Miller’s theorem and the open circuit time constants method. Figure 3 Problem 4 [20 points] Design a low-pass Butterworth filter to meet the following specs: Amax = 2.5 dB Amin = 20 dB fP = 15 KHz fS = 25 KHz Show the op-amp implementation of the filter. Minimize the number of op-amps in your implementation, and use 1 nF capacitors whenever possible. Q 1 10K V S + 10K V o Vbias Vcc Q 2 Problem 5 [15 points] 5V - 5V Q3 Q4 Q2 Q1 Q5 Q6 Q10 Q7 Q8 Q9 IREF v2 v1 Q11 Figure 4 Consider the circuit shown in Figure 4. a) What is the inverting input for this op-amp? Explain. b) Find the output resistance of the op-amp. Assume that k’n = 25 μA/V2, k’p = 10 μA/V2, Vtn = 0.75 V, Vtp = -0.75 V, and IREF = 100 μA. The sizes of the MOSFETs are as follows: (W/L)1 = (W/L)2 = (W/L)9 = (W/L)10 = 20 (W/L)3 = (W/L)4 = 50 (W/L)6 = (W/L)8 = 10 (W/L)5 = 100 (W/L)7 = 25 (W/L)11 = 5 Problem 6 [15 points] The op-amp shown in Figure 5 is modeled by an infinite input resistance, an open- loop open-circuit differential voltage gain function of 107/(s+50), and an output resistance of 500 Ω. What is the maximum load capacitance CL that can be connected at the output of a voltage follower configuration (as shown in the figure), if the phase margin of the amplifier is 60 degrees? Vo Vi + - + CL Figure 5 American University of Beirut Department of Electrical and Computer Engineering EECE 311 – Electronics II Fall 2005 – 2006 (Section 2) Quiz 1 – Solution Problem 1: v i Q 1 V I I v O a) I = ½ k’n (W/L) (VGS – Vtn)2 => 40 u = 0.5 (0.2m)(5)(VI – 0.5)2 => VI = 0.5 + sqrt(40u/0.5m) = 0.783 V b) vO = vDS => vOmin = vDSmin = vGS – Vtn Assuming vGS = VGS (DC value) => vOmin = 0.783 – 0.5 = 0.283 V c) gm = ID/(VOV/2) = 40u/(0.283/2) = 0.283 mA/V ro = VA/ID = (1/0.1) / 40u = 250 K gain = vo/vi = –gm(ro // Ro of current source) = –0.283(100//250) = –20.2 V/V Q3 IREF VSG + - Q2 d) ID(Q2) = ½ k’p (W/L) (VOV2)2 This current is equal to 40u => 40u = 0.5(0.08m)(W/L)(0.2)2 => (W/L) = 25 for Q2 (W/L) = 25 for Q3, since Q2 and Q3 are matched. e) VSG = |Vtp| + |VOV2| = 0.54 + 0.2 = 0.74 V f) Output resistance is ro2 = VAP / ID2 = (1/0.12) / 40u = 208.33 K g) For Q2: |VDS| ≥ |VGS| – |Vtp| => VDD – VO ≥ 0.2 V => VOmax = VDD – 0.2 V Problem 2: In this problem, there is an implicit assumption that the signal source (not shown) has a large source resistance, and that the source does not affect the DC bias. I vi + - vo RB 300K RL 10K CL 20pF a) DC analysis: current in 300K resistor = IB KCL at collector: 0.5 mA = IB + IC + VO/10K But VO = 300K×IB + VBE = 300K×IB + 0.7 V => 0.5 mA = IB + β IB + (300K×IB + 0.7 V)/10K => 0.5 mA – 0.07 mA = 91 IB => IB = 4.725 uA IC = 60 IB = 0.2835 mA VO = 300K×4.725 uA + 0.7 V = 2.118 V b) The Miller constant is the gain from base to collector without RB, and is given by –gm R’L with R’L = ro (BJT) // Ro (current source) // RL ro (BJT) = VA/IC = 80/0.2835m = 282.2 K R’L = 282.2 K // 100 K // 10K = 8.807 K gm = IC/VT = 0.2835m/25m = 11.34 mA/V Miller’s K = –gm R’L = –11.34×8.807 = –99.87 Equivalent resistance from base to ground due to RB: R1 = RB/(1 – K) = 300K/(1+99.87) = 2.974 K The input resistance is Rin = rπ // R1 rπ = VT/IB = 25m/4.725u = 5.291 K Rin = 5.291 // 2.974 = 1.904 K c) The resistance seen by Cπ is Rin (we can use the approximation from part (b)) = 1.904 K. The more accurate estimate is obtained as follows: iRB = (vi – vo)/RB = (1 – gain)vi/RB => vi/iRB = RB/(1 – gain) = 300K/(1 + 97) = 3.061 K [note: the value of the voltage gain is found in part (d)] Resistance seen by Cπ is equal to: rπ // (vi/iRB) = 5.291 K // 3.061 K = 1.939 K rpi RB gmvi R’L vi vo Resistance seen by Cpi Resistance seen by Cmu Resistance seen by CL The resistance seen by CL is equal to: R’L // (resistance due to controlled source) // (RB + rπ): vπ = vi = vx rπ/(rπ+RB) = vx×5.291/(5.291+300) = vx/57.7 The controlled current source is therefore equivalent to a resistance equal to 57.7/gm = 5.088 K The resistance seen by CL is therefore: 8.807 // 5.088 // 305.29 = 3.191 K The resistance seen by Cμ is equal to: RB // (R’L + rπ + gm×rπ×R’L) = 300 K // (8.807 K + 5.291 K + 11.34×8.807×5.291 K) = 193.2 K. τH = 1.939 K×1 pF + 3.191 K×20 pF + 193.2 K×1 pF = 258.96 nsec fH = 1/(2π τH) = 614.6 KHz d) The low-frequency voltage gain is determined without using Miller’s theorem. The current in RB is (vi – vo)/RB. This current is equal to gm vi + vo/R’L. Therefore: (1/RB – gm) vi = vo(1/R’L + 1/RB) => The voltage gain is vo/vi = (1/RB – gm)/ (1/R’L + 1/RB) = (1 – gm RB)/(RB/R’L + 1) = (1 – 11.34×300)/(300/8.807+1) = –97 V/V or 39.74 dB e) Asymptotic Bode Plot Gain in dB 39.74 614.6 KHz f (log scale) -20 dB/dec Problem 3 IREF R1 10K R2 10K IO 2V VO a) IREF = (2 – VGS1)/10K = ½ k’n (W/L) (VGS1 – Vtn)2 = (0.5)(0.2m)(10)(VGS1 – 0.4)2. Solving the quadratic equation we get: VGS = 0.753 V (the other root is negative). IREF is therefore (2 – 0.753)/10K = 0.1247 mA b) VGS2 = VGS1 – IO×R2 and IO = ½ k’n (W/L) (VGS2 – Vtn)2 = ½ k’n (W/L)(VGS1 – IO×R2 – Vtn)2 Therefore IO = (0.5)(0.2m)(10)(0.753 – IO×10K – 0.4)2 => IO = 20.86 uA The value of VGS2 is 0.753 – 20.86u×10K = 0.5444 V c) The minimum output voltage corresponds to VDS2 = VGS2 – Vtn => VO – IOR2 = VGS2 – Vtn => VOmin = 20.86u×10K + 0.5444 – 0.4 = 0.353 V d) The output resistance is given by ro2 + R2 + gm2 ro2 R2 ro2 = VA/IO = (1/0.08)/20.86u = 599.23 K gm2 = IO/(VOV/2) = 20.86u/((0.5444-0.4)/2) = 0.289 mA/V RO is therefore equal to: 10K + 599.23K + 599.23K×0.289m×10K = 2340.5 K Problem 4 In this problem, there is an implicit assumption that the signal source (not shown) has a large source resistance. vo vi vo vi VDC a) Using Miller’s theorem, the time constant at the input of the circuit is given by τH = rπ (Cπ + Cμ (1 + gm ro)) since the only “load” is the output resistance of the BJT itself. τH = rπ (Cπ + Cμ (1 + gm ro)) = rπCπ + rπCμ + rπCμgm ro = rπCπ + rπCμ + βroCμ The third term is much larger than the first two, and therefore τH is approximately βroCμ. fH = 1/(2π τH) = 1/(2π βroCμ) b) The current source output resistance is approximately ro. The output resistance of the amplifier becomes Ro ≈ ro//ro = ro/2. The open-circuit voltage gain becomes –gm ro/2 = –A0/2. The 3-dB frequency becomes 1/(2π τH) = 1/(2π β(ro/2)Cμ) = 2 fH c) The current source output resistance is approximately βro. The output resistance of the amplifier becomes Ro ≈ βro//ro ≈ ro. The open-circuit voltage gain becomes –gmro = –A0. The 3-dB frequency becomes 1/(2π τH) = 1/(2π βroCμ) = fH d) For an unloaded cascode amplifier, the output resistance of the unloaded amplifier is given by βro. The output resistance of the amplifier becomes Ro ≈ βro//ro ≈ ro. The open-circuit voltage gain becomes (see Fig. 6.41 in textbook) –βA0 ro/(ro+βro) ≈ –A0. The 3-dB frequency becomes 1/(2π τH) ≈ 1/(2π roCμ) = βfH (see Fig. 6.42 in textbook) e) The output resistance of the amplifier becomes Ro ≈ βro//βro = βro/2. The open- circuit voltage gain becomes –βA0 βro/(βro+βro) = –βA0/2. The 3-dB frequency becomes 1/(2π τH) ≈ 1/(2π (βro/2)Cμ) = 2fH Problem 5 R1 R2 R3 vi2 vo R4 vi1 For the difference amplifier, the output voltage is given by: 2 4 2 1 3 4 1 1 2 (1 ) R R R o i i R R R R v v v − + = + + => 1 2 4.7 4.674 o i i v v v = − + = Av1 vi1 + Av2 vi2 = Ad (vi1 – vi2) + Acm (vi1+vi2)/2 => Ad = (Av1 – Av2)/2 = –4.687 V/V and Acm = Av1 + Av2 = –0.026 V/V => CMRR = 20 log|4.687/0.026| = 45.12 dB EECE 311 Quiz 2 Page 1 of 7 American University of Beirut Departmen