EECE 310 Exam 1, Fall 2007

EECE 310 Electronics – Section 1 Quiz 1 – November 3, 2007 CLOSED BOOK – One and One-Half Hours Penalty 5:1 on Questions 1 - 7 (7%) 1. In the given circuit, VSRC = 5 V dc and the diode is an ideal diode switch. Determine the dc current I so that the reverse voltage across the diode is 2 V. A. 5 mA B. -5 mA C. -7 mA D. 6 mA E. None of the above Solution: When the ideal diode switch is reverse biased, it behaves as an open circuit. From KVL: -VR = VSRC + I1 = 0; -2 = 5 + I, or I = -7 mA. (7%) 2. In the given circuit, vSRC = V. The diode may be modeled as having VD0 = 0.6 V and rD = 5 . Determine t at which the diode will just begin to conduct in the forward direction. A. 1.25 ms B. 1.67 ms C. 1.84 ms D. 2.12 ms E. None of the above Solution: At the instant of conduction, the forward voltage across the diode is 0.6 V and the current through the diode is zero. It follows that vSRC = = 50.6 = 3 V. This gives 1 – + – + VSRC 1 k I + – VR + – I1 – + – + vSRC 1 k + – 0.6 V I = 0 0.6 mA 4 k – + – vSRC 1 k 4 k – + – vSRC 1 k I ; ms. (7%) 3. In the given circuit, the zener diode may be modeled as having VZ0 = 5 V, rZ = 0, and in the forward direction as having VD0 = 1 V, rD = 0. vSRC is the periodic waveform shown. Determine the average value of vL. A. 5 V B. 1 V C. 3 V D. 2 V E. None of the above Solution: During the positive half-cycle, the diode behaves as a zener diode, and vL = VZ0 = 5 V. During the negative half-cycle, the diode conducts in the forward direction, and vL = 1 V. The average value is V. (7%) 4. Determine the average value of vR in the preceding problem. A. 0.5 V B. -0.5 V C. 1 V D. 3 V E. None of the above Solution: During the positive half-cycle, vR = 10 – 5 = 5 V. During the negative half- cycle, vL = -4 V. The average value is V. Alternatively, it can be argued that: average vSRC = average vR + average vL. Since average vSRC = 2.5, it follows that average vR = 2.5 – 2 = 0.5 V. 2 – + – vSRC 5 k + – vL + – vR t, s vSRC, V 10 -5 1 2 (7%) 5. In the circuit shown, the diodes can be modeled in the forward direction by VD0 = 0.7 V, rD = 0. At what value of vI (not vO) will limiting just begin? A. 0.7 V B. 1.2 V C. 1.4 V D. 2.4 V E. None of the above Solution: When limiting just begins, the upper diode will present a voltage of 0.7 V. Current can flow in the 1 k resistor, which at the beginning of limiting will be very small, so that the lower diode does not conduct. Hence, limiting will just begin when vO = 0.7 V. At this value of vO the current in the 1.4 k resistor is 0.5 mA, which causes a voltage drop of 0.5 V across the series 1 k resistor. It follows that vI = 0.5 + 0.7 = 1.2 V. (7%) 6. A p-type semiconductor having a majority carrier concentration of 1017/cm3 is in contact with an intrinsic semiconductor in the same crystal. Determine the electric potential of the intrinsic semiconductor with respect to the p-type semiconductor, assuming ni = 1010/cm3 and VT = 0.026 V. A. 0.23 V B. 0.38 V C. 0.42 V D. 0.57 V E. None of the above Solution: At equilibrium, the electrochemical of holes must be the same in both types of semiconductor, where . In the p-type semiconductor, , whereas in the intrinsic semiconductor, . Equating hp to hi, Cr, Vr, and r cancel out, giving V. 3 1 k + – vO 1.4 k 1 k + – vI (7%) 7. An n-type semiconductor has ND = 1018/cm3. At the temperature under consideration, the donor impurities are 90% ionized, ni being 108/cm3. Determine the equilibrium concentration of majority carriers. A. 108/cm3 B. 9108/cm3 C. 91017/cm3 D. 1018/cm3 E. None of the above Solution: = 0.91018. (17%) 8. An x-ray machine requires a dc voltage of 150 kV at a dc current of 100 mA, with peak-to-peak ripple of 500 V. A half-wave (not full-wave) rectifier with a capacitor-input filter is to be used. (a) If at this voltage, a 0.01 F capacitor is practicable, what should be the supply frequency? (b) If the transformer used has a primary voltage of 100 V rms, what should be the primary-to-secondary turns ratio. Solution: (a) Assuming that the approximate analysis holds, then for the half-wave rectifier, , or kHz. To check the validity of the approximate analysis, 600 . (b) . Since vr << VDC, kV. The diode voltage drop is also small compare to , so that = 150 kV. The rms secondary voltage is . The primary-to-secondary turns ratio is , or , or . 4 (17%) 9. Given a zener diode regulator using a zener diode of VZ0 = 9 V and IZmin = 2 mA. The input varies between VImax and VImin, where VImax – VImin = 5 V. A variation of the load current between zero and 20 mA, at constant input voltage, changes the load voltage by 0.2 V. A change in the input voltage between VImax and VImin changes the load voltage by 0.1 V, at a constant load current of 20 mA. Determine: (a) RS; (b) rZ; (c) VZ at IZ = 10 mA; (d) VImin; and (e) minimum power rating of zener diode. Solution: At constant input voltage, , or, . At constant load current: . Dividing gives: (a) . (b) It follows that , so that . (c) 9.10 V; (d) At IZ = 2 mA, V; V; (e) , which gives mA, 9.32 V; minimum power rating of zener diode = 9.320.0314 = 0.293 W. (17%) The waveform of Problem 3 is applied to the circuit shown. Determine VC assuming a very large capacitor Solution: C is discharged during the positive half cycle, the discharge current being mA. The charge lost is . During the negative half cycle, C is charged, the charging current being mA. The charge gained is . Equating these: , so that VC = 10.5 V. 5 vSRC VC 3 k _ + + _ 1 k 4 V 6 V