EECE 310 Midterm, Summer 2010

EECE 310 – Midterm – Summer 2010 Page 1 of 5 American University of Beirut Department of Electrical and Computer Engineering EECE 310 – Electronics Midterm – July 16, 2010 Closed Book – No Programmable Calculators 90 minutes Profs. Chehab and Kayssi Penalty is 5 to 1 (1 to 4 wrong answers do not result in a penalty; 5 to 9 wrong answers cancel one correct answer; 10 to 14 wrong answers cancel two correct answers; and so on) Name: _________________________________ ID number: _________________ Unless otherwise specified, assume that: VT = 25 mV n = 1 ni = 1.5×1010 cm–3 q = 1.6×10-19 C p p dp J = −qD dx n n dn J = qD dx ( ) d p n J qp qn E μ μ = + I = J × A 0 2 ln A D T i N N V V n ⎛ ⎞ = ⎜ ⎟ ⎝ ⎠ EECE 310 – Midterm – Summer 2010 Page 2 of 5 1. The voltage gain of an amplifier is 100 V/V. Its power gain is 45 dB. Find the current gain in dB. a) 50 b) 40 c) 25 d) 45 e) 20 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2. A voltage amplifier is powered by a single 16 V DC supply. The amplifier is loaded by a 50 Ohm resistor and produces a peak-to-peak sinusoidal output voltage of 8 V across the load. Find the average DC supply current (in mA) if the efficiency of the amplifier is 20%. a) 200 b) 100 c) 250 d) 150 e) 50 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 3. A circuit consists of a 5 V DC source, a 12 kΩ resistor, and a forward-biased diode. The diode has IS = 10–15 A and n = 1.5. Starting with an initial guess for the diode current equal to the current that flows if the diode were ideal, find the value of the diode current (in mA) after completing two iterations. a) 0.386 b) 0.218 c) 0.471 d) 0.334 e) 0.417 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 4. The current-voltage characteristics of device X are shown on the ix–vx plot in the figure below. The x-axis division is 0.25 V, and the y-axis division is 200 μA. Using the load line graphical method, estimate the voltage vx (in V) when VA = 2 V and R = 1 kΩ. a) 0.64 b) 0.83 c) 1.37 d) 0.45 e) 1.12 + + - vX VA R iX vX iX 0 0 X - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - EECE 310 – Midterm – Summer 2010 Page 3 of 5 5. The diode in the circuit shown below is modeled by VD0 = 0.6 V and rD = 30 Ohms, when conducting, and by an open circuit when OFF. Determine the power dissipation in diode (in mW). a) 6.00 b) 7.50 c) 4.68 d) 2.40 e) 3.38 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 6. Assume that the diodes are ideal in the circuit shown below. Find the value of R (in kΩ) needed to obtain ID2 = 1 mA. a) 10 b) 4 c) 1 d) 5 e) 2 7. In the circuit of the previous problem, assume now that the diodes are modeled by a fixed 0.65 V drop when conducting, and an open circuit when OFF. Find the value of R (in kΩ) needed to obtain ID2 = 2 ID1. a) 1.46 b) 2.35 c) 1.29 d) 2.08 e) 6.71 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 8. Find the drift current density (in mA/cm2) that results due to the application of a 900 V/cm electric field to a volume of intrinsic semiconductor. The mobility of free electrons in the semiconductor is 1150 cm2/V.s, and it is three times larger than the mobility of holes. a) 3.3 b) 2.5 c) 0.83 d) 9.9 e) 4.9 9. The free electron density in a doped semiconductor is 500 cm–3 at room temperature. What is the type of impurity atoms? a) Acceptors b) Donors c) Silicon d) Not enough information to decide 10. The density of impurity atoms on the two sides of a PN junction is 10X × ni. The potential barrier is 0.921 V. Find X. a) 9 b) 5 c) 6 d) 7 e) 8 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - EECE 310 – Midterm – Summer 2010 Page 4 of 5 11. Find the power dissipation (in μW) in the 10 kΩ resistor in the circuit shown below when VA is 1.2 V DC. The diode is modeled by a fixed 0.7 V drop when conducting, and an open circuit when OFF. a) 9 b) 25 c) 16 d) 36 e) 4 12. Repeat the previous problem when VA is –1.2 V DC. a) 4 b) 25 c) 16 d) 36 e) 9 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 13. A GaAs forward-biased junction diode is conducting a current of 12 mA at a diode voltage of 1.1 V. Find the diode current (in mA) when the diode voltage is 1 V. a) 24 b) 1.2 c) 5.6 d) 7.8 e) 0.22 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - In the circuit shown below, the voltage V1 has a DC value of 5 V on which a sinusoidal signal equal to 0.1 sin(ωt) V is superimposed. Assume that the DC characteristics of the diode are modeled by a fixed voltage drop of 0.6 V when conducting, and an open circuit when OFF. 14. Find the incremental small-signal resistance rd for the diode (in Ohms). a) 14.9 b) 8.32 c) 2.84 d) 18.5 e) 1.52 15. Calculate iO(t) in mA. a) 4.4–0.199×sin(ωt) b) 8.8–0.199×sin(ωt) c) 8.8+0.199×sin(ωt) d) 4.4+0.1×sin(ωt) e) 4.4–0.1×sin(ωt) - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - EECE 31 The Zen vS is 20 V from 0 t 16. Find conditio a) 81 17. For t Zener vo a) 0.34 - - - - - - 18. A fu supply th transform modeled resistor a) 9.21 - - - - - - The diod remain c drop bel solutions 19. Find a) 521 20. Find a) 33.6 10 – Midterm ner diode sho V, but varies o 20 mA. d the largest v ons. b) 1 the value of R oltage VZ (in b) 0 - - - - - - - - ull-wave recti hrough step- mer turns rat d by a 0.75 V (in mA)? b) 8 - - - - - - - - de in the circ conducting w low 2 mA, an s are possibl d the value of b) 4 d the value of b) 1 – Summer 20 own below ha s by +/ –25% value of RX ( 52 RX calculated V)? .27 - - - - - - - - - ifier circuit w -down transfo tio for each s V drop for an .31 - - - - - - - - - cuit shown be when VA varie nd the power le] f R1 (in Ω). 07 f R2 (in Ω). 8.9 010 as VZ = 10 V %, i.e. betwee (in Ω) that en c) 267 d in the prev c) 0.56 - - - - - - - - - with a 3 kΩ l former having secondary is ny forward cu c) 4.91 - - - - - - - - - elow drops a es between 5 r dissipation + VA c) 372 c) 81.7 V at IZ = 25 m en 15 V and 2 nsures Zener d) 204 vious problem d) 0.11 - - - - - - - - - load operates g two second 15-to-1. The urrent. What d) 4.15 - - - - - - - - - a constant 0.7 5 V and 10 V of the diode R1 R2 d) 233 d) 24.4 mA, with rZ = 25 V. The lo region opera e) 5 m, what is the e) 0 - - - - - - - - - s from a 311 dary winding e rectifier use is the averag e) 2 - - - - - - - - - 7 V when co V. Moreover, should not e D e) 3 e) 6 = 5 Ω and IZK oad draws a c ation for the 33 e maximum 0.01 - - - - - - - - - V (peak) 50 gs and a cent es two diode ge (DC) curr .51 - - - - - - - - - onducting. Th the diode cu exceed 10 m 11 67.2 K = 5 mA. Th current that v diode under variation in t - - - - - - - - - 0 Hz sinusoid ter tap. The s each of wh rent in the loa - - - - - - - - - he diode shou urrent should W. [Note: m Page 5 of 5 he source varies r all the - - - - - dal wave hich is ad - - - - - uld d not multiple 5