EECE 310 Midterm, Fall 2013

EECE 310 – Midterm Exam – Fall 2013 - Page 1 of 6 American University of Beirut Department of Electrical and Computer Engineering EECE 310 – Electronics Midterm – October 24, 2013 Closed Book – No Programmable Calculators 90 minutes There are 21 problems and 6 pages. All problems are equally graded. 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 J p = −qDp dp dx J n = qDn dn dx 𝑫𝒏 𝝁𝒏 = 𝑫𝒑 𝝁𝒑 = 𝑽𝑻 J d = qpµ p + qnµn ( )E I = J × A EECE 310 – Midterm Exam – Fall 2013 - Page 2 of 6 =================================================================== The efficiency of an amplifier loaded by a 1 kΩ resistor is 43%. The amplifier is powered from a single 8 V DC supply. The supply current is 10 mA. The voltage gain of the amplifier is 45 dB. The input and output voltages are sine waves with zero average. 1. Find the power dissipated (lost) in the amplifier (in mW). a) 65.6 b) 53.6 c) 61.6 d) 45.6 e) 69.6 2. Find the peak value of the input voltage in (in mV). a) 34.11 b) 30.18 c) 25.65 d) 46.64 e) 40.86 =================================================================== 3. A large forward current is flowing in a Silicon diode. When the diode voltage increases by Δ = 27 mV (from VD to VD + Δ), this current doubles (from ID to 2ID). Find the value of n for this diode. a) 1.21 b) 1.33 c) 1.44 d) 1.56 e) 1.67 =================================================================== A Zener voltage regulator shown in the figure below uses a diode with VZ0 = 5.6 V, rZ = 25 Ω, and IZK = 2.5 mA. The supply VA is an unregulated 13.8 V voltage source. Assume R1 = 1 kΩ. VA R1 vO + - 4. If the supply voltage VA changes by 0.9 V, what is the corresponding change in the output voltage vO (in mV)? a) 41.46 b) 26.83 c) 31.71 d) 36.59 e) 21.95 =================================================================== EECE 310 – Midterm Exam – Fall 2013 - Page 3 of 6 The rectifier in the figure shown below is fed with a voltage Vin having a periodic triangular waveform with a peak value Vp = 12 V, zero average, and a period of 20 ms. Assume that the diode is ideal. 5. Find the average (DC) value of the output voltage Vo (in V). a) 3 b) 2.75 c) 2.25 d) 2.5 e) 2 6. An 800 µF capacitor is connected in parallel with RL (= 1 kΩ). Find the output ripple voltage (in mV). a) 300 b) 250 c) 275 d) 200 e) 225 =================================================================== 7. A full-wave bridge rectifier circuit with a purely resistive load operates from a sinusoidal supply through a step-down transformer having a single secondary winding. The RMS voltage at the transformer secondary is 14 V, and the bridge uses four diodes each of which can be modeled to have a 0.7 V drop for any forward current. Find the average voltage across the load (in V). a) 16.6 b) 9.4 c) 11.2 d) 13.0 e) 14.8 =================================================================== A diode with n = 1 is operating such that its voltage is vD = 725 + 3 cos(100πt) mV, and its current is iD = ID + 0.3 cos(100πt) mA. 8. Find the small-signal resistance rd of this diode (in Ohms). a) 15 b) 12 c) 10 d) 20 e) 30 9. Estimate the value of the reverse saturation current IS for this diode (×10–16 A). a) 5.3 b) 6.4 c) 4.2 d) 2.1 e) 3.2 =================================================================== EECE 310 – Midterm Exam – Fall 2013 - Page 4 of 6 The diodes in the circuit shown below are ideal. The voltage Vin is sine wave with a 10 V peak value and zero DC component, and VB = 6 V. 10. Find the maximum forward current in diode D2 (in mA). a) 106.7 b) 100 c) 93.3 d) 80 e) 86.7 11. Find the Peak Inverse Voltage (PIV) of diode D2 (in V). a) 4 b) 5 c) 8 d) 7 e) 6 12. Find the fraction of each cycle during which diode D1 conducts. a) 0.436 b) 0.403 c) 0.369 d) 0.333 e) 0.295 =================================================================== In the circuit shown below, the Zener diode has VZ = 6 V at IZ = 5 mA, with rZ = 20 Ω and IZK = 2 mA. Assume R1 = 1 kΩ, R2 = 50 Ω, and RL = 3.2 kΩ. The source voltage is vS(t) = 13 + A sin(ωt) V. R1 R2 RL vS + - 13. Find the voltage across RL at time t = 0 (in V). a) 5.32 b) 5.78 c) 6.24 d) 6.69 e) 7.15 EECE 310 – Midterm Exam – Fall 2013 - Page 5 of 6 14. Find the maximum value of A (in V) to keep the Zener diode operating in the breakdown region. a) 3.07 b) 2.42 c) 1.76 d) 3.73 e) 4.39 =================================================================== The doping concentration (density) profile of acceptor atoms in a piece of semiconductor material of length L = 1 mm, is shown in the figure below. NMIN is much larger than ni, and NMAX = 7NMIN. SEMICONDUCTOR MATERIAL x L L/2 L/4 3L/4 0 L Doping Concentration (Density) NMIN NMAX NMIN+NMAX 2 15. The hole concentration (density) is largest at x = x1, and the free electron concentration (density) is largest at x = x2. Find x1 and x2. a) x1 = L, x2 = 0 b) x1 = 0, x2 = L c) x1 = L/2, x2 = L/2 d) x1 = L, x2 = L 16. What is the direction of flow of the hole diffusion current? a) Negative y (down) b) Positive y (up) c) Unknown d) Positive x e) Negative x 17. Under equilibrium conditions, the total current should be zero. Find the magnitude of the electric field E (in V/m) at x = L/2 that must exist to maintain the zero-current condition. Consider majority carriers only. a) 40.0 b) 38.9 c) 37.5 d) 33.3 e) 35.7 =================================================================== In the circuit shown below, the diodes are modeled by a fixed 0.75 V drop when conducting, and by an open circuit when OFF. The voltage vS is a square wave with zero average, and symmetrical levels at ±8 V. Assume RS = 5 kΩ. EECE 310 – Midterm Exam – Fall 2013 - Page 6 of 6 vS –1 V RS vO + – vS(t) time 18. Find the diode current (in mA), when the diodes are conducting. a) 1.7 b) 1.9 c) 1.1 d) 1.3 e) 1.5 19. Find the average (DC) value of the output voltage vO (in V). a) 4.75 b) 4.25 c) 3.75 d) 3.25 e) 2.75 =================================================================== A nonlinear amplifier is characterized by the equation vOUT = a + b×vIN 3, where vIN is the input voltage, and vOUT is the output voltage. vIN(t) and vOUT(t) vary as shown in the figure below, where vIN(t) is a sine wave that varies between 2.9 V and 3.1 V, VMIN = 1 V, and VMAX = 4.5 V. 20. Find the value of b (in V–2). a) –0.741 b) –0.648 c) –0.370 d) –0.463 e) –0.556 21. Find the value of a (in V). a) 23.00 b) 12.00 c) 14.75 d) 17.50 e) 20.25 =================================================================== time vIN(t) 3.1 2.9 time vOUT(t) VMAX VMIN