Midterm Ece 431 531 s16 Form1

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    ECE 431/531 Midterm (Form 1)

    Name:

    Each question has ONE correct answer. Only choose one answer.

    May 4 2016

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    A boost converter (with unidirectional power pole) is shown below. It is operating in CCM steadystate. The input voltage is 10 V. The FET duty ratio is 60%. The load resistance R L is 3 Ohms.The inductor is 25 µH. The switching frequency is 250 kHz. Assume the output capacitor is largeenough such that the ripple in vo is small.

    + −vin

    L iL

    idiode

    C

    iC

    R L

    +

    vo

    1. What is the output voltage vo?

    A) 16.7 VB) 25.0 V

    C) 5.4 V

    D) 6.0 V

    E) 45.0 V

    2. What is the inductor average current īL ?

    A) 7.50 A

    B) 13.89 A

    C) 20.83 A

    D) 5.00 A

    E) 3.33 A

    3. What is the inductor ripple current ∆ iL ?

    A) 0.48 A

    B) 0.64 A

    C) 1.06 A

    D) 0.75 AE) 0.96 A

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    4. What is the diode average current īdiode ?

    A) 2.0 A

    B) 18.8 A

    C) 9.4 A

    D) 8.3 A

    E) 5.6 A

    5. What is the diode ripple current ∆ idiode ? (Where ∆ idiode is the peak-peak of the currentwaveform.)

    A) 14.1 A

    B) 21.3 A

    C) 7.5 A

    D) 3.7 A

    E) 18.8 A

    6. What is the capacitor average current īC ?

    A) -3.0 A

    B) 19.2 A

    C) 0 A

    D) 8.3 A

    E) 20.8 A

    7. What is the capacitor ripple current ∆ iC ? (Where ∆ iC is the peak-peak of the currentwaveform.)

    A) 21.3 A

    B) 3.6 A

    C) 0.9 A

    D) 14.1 A

    E) 20.7 A

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    8. Which of following is true (choose one)?

    A) Discontinuous conduction mode is not possible in this boost converter due to the uni-directional power pole.

    B) Discontinuous conduction mode in a boost converter causes vo to be greater than vin1 − D

    and approach innity as RL approaches innity (i.e., the load decreases).C) This boost converter will go into discontinuous conduction mode if the average inductor

    current becomes sufficiently large.

    D) A low-side diode will prevent the possibility of discontinuous conduction mode in thisboost converter.

    E) This boost converter cannot enter discontinuous conduction mode as the inductor currentcannot be maintained at zero for any part of the switching period.

    A buck converter is used to drive a DC motor an in electric vehicle as shown below. The available

    DC bus voltage, V d , is 350 V, supplied by batteries. At a cruising speed of 40 mph, the voltageproduce by the motor, em , is 170 V. The motor winding resistance R m is 1.5 Ω. The motor currentim required to hold this speed is 30 A. The switching frequency is 25 kHz. The current and voltagefall and rise times are trv = 10 ns, tf i = 15 ns, tri = 13 ns, tf v = 17 ns. The on-state drain-sourceFET resistance R DS is 0.05 Ohms. The converter inductance is 1 mH.

    9. What is the buck converter duty ratio? (Assume duty ratio in this case to be dened as thepercentage on-time for the high-side FET.)

    A) 0.36

    B) 0.71

    C) 0

    D) 0.49

    E) 0.61

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    10. What is the average buck converter inductor current īL ?

    A) 18 A

    B) 30 A

    C) 49 A

    D) 0 A

    E) 107 A

    11. What is ∆ iL ?

    A) 2.8 A

    B) 6.2 A

    C) 13.9 A

    D) 3.3 A

    E) 36.8 A12. Neglecting any power loss in the buck converter, what is the average current drawn from the

    batteries, īd?

    A) -30.0 A

    B) 30.0 A

    C) 18.4 A

    D) 0 A

    E) 48.8 A

    13. What is the switching loss in the high-side FET?A) 7.2 W

    B) 6.6 W

    C) 73.3 W

    D) 116.7 W

    E) 14.4 W

    14. What is the conduction loss in the low-side FET? (Assume the anti-parallel diode conductionloss, which is during the switching events, is small and negligible. Also assume the ripple

    current is small.)A) 73.3 W

    B) 116.7 W

    C) 17.4 W

    D) 219.4 W

    E) 27.6 W

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    Assume now that the vehicle driver would like to brake the vehicle regeneratively and the requiredmotor current to do so is − 15 A. The vehicle speed is the same and em is still 170 V.

    15. What now is the duty ratio?

    A) 0.47

    B) 0.44C) 0.55

    D) 0.42

    E) 0.00

    16. What now is the īL ?

    A) -4.6 A

    B) -15 A

    C) 0 A

    D) -48.8 A

    E) -30 A

    17. What now is the ∆ iL ?

    A) 47.2 A

    B) 8.4 A

    C) 3.4 A

    D) 4.4 A

    E) 2.1 A18. What now is īd?

    A) -10.0 A

    B) 0 A

    C) -6.3 A

    D) -35.6 A

    E) 10.0 A

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    A single phase rectier is connected to 240 V RMS at 60 Hz. The equivalent load resistance R L onthe DC side is 10 Ω.

    19. Approximately what is vo?

    A) 228 V

    B) 120 VC) 3224 V

    D) 370 V

    E) 320 V

    20. Approximately what is the average current īdr ?

    A) 0 A

    B) 3224 A

    C) 32 A

    D) 322 A

    E) 23 A

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