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Power Electronics Test 2

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Power Electronics Test 2
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Weekly Quiz Competition
  • Question 1
    1 / -0
    Which of the following statement is true regarding power MOSFET and BJT?
    Solution
    • Power MOSFET has lower switching losses but its on-resistance and conduction losses are more. A BJT has higher switching losses but lower conduction losses.
    • MOSFET is a voltage-controlled device whereas BJT is a current controlled device.
    • MOSFET has positive temperature coefficient for resistance. This makes parallel operation of MOSFETs easy. If a MOSFET shares increased current initially, it heats up faster, its resistance rises, and this increased resistance causes this current to shift to other devices in parallel.
    • A BJT has negative temperature coefficient, so current-sharing resistors are necessary during parallel operation of BJTs.
    • In MOSFET, secondary breakdown does not occur, because it has positive temperature coefficient.
    • As BJT has negative temperature coefficient, secondary breakdown occurs. With decrease in resistance, the current increases. This increased current over the same area results in hot spots and breakdown of the BJT.
  • Question 2
    1 / -0
    In a power MOSFET, as the gate to source voltage is increased beyond the threshold voltage, the thickness of depletion layer is
    Solution
    In a power MOSFET, as the gate to source voltage is increased beyond the threshold voltage, the thickness of depletion layer is remains constant.
  • Question 3
    1 / -0

    Consider the following statements with regard to a GTO:

    1. The turn-off gain of the GTO is large.

    2. Large negative gate current pulses are required to turn off the GTO.

    3. GTO has a large reverse blocking capability.

    Which of the above statements is/are correct?
    Solution
    • GTO is a four-layer, three junction semiconductor device with three external terminals, namely, the anode, the cathode and the gate.
    • The main advantage of the GTO over the SCR is that it can be turned on or off by applying the proper pulse to the gate.
    • The capability of the GTO to be turned on by a gate signal and also turned-off by a gate signal of negative polarity give it a unique capability within the thyristor family of devices.
  • Question 4
    1 / -0
    The reverse recovery time of a diode is trr = 3 μs and the rate of fall of the diode current is \(\frac{{di}}{{dt}} = 30\;A/\mu s\). The storage charge QRR and the peak inverse current IRR will be respectively
    Solution

    Important formulas:

    1) \({Q_{RR}} = \frac{1}{2} \times {I_{RR}} \times {t_{rr}}\)

    2) \({I_{RR}} = {\left[ {2\;{Q_{RR}}\frac{{di}}{{dt}}} \right]^{1/2}}\)

    3) \({t_{rr}} = {\left[ {\frac{{2{Q_{RR}}}}{{\frac{{di}}{{dt}}}}} \right]^{\frac{1}{2}}}\)

    Calculation:

    \({Q_{RR}} = \frac{1}{2}t_{rr}^2\;\left( {\frac{{di}}{{dt}}} \right)\)

    \( = \frac{1}{2} \times {\left( {3 \times {{10}^{ - 6}}} \right)^2} \times 30\frac{A}{{\mu s}}\;\)

    \({Q_{RR}} = 135\;\mu C\)

    \({I_{RR}} = {\left( {2\;{Q_{RR}}\frac{{di}}{{dt}}} \right)^{\frac{1}{2}}} = 90\;A\)

  • Question 5
    1 / -0

    Consider the following statements:

    (1) IGBT has low input impedance compared to PMOSFET

    (2) IGBT and PMOSFET are both voltage controlled devices.

    (3) IGBT can be designed for higher voltages compared to PMOSFETS.

    (4) IGBT converters are more costly and bigger in size compared to BJT converters.

    The correct statements are:

    Solution

    IGBT has low input impedance compared to PMOSFET

    IGBT and PMOSFET are both voltage controlled devices.

    IGBT can be designed for higher voltages compared to PMOSFETS.

  • Question 6
    1 / -0
    A MOSFET rated for 20 A, carries a periodic current expressed as \(i\left( t \right) = \left\{ {\begin{array}{*{20}{c}}{15,\;\;0 < t < \frac{T}{4}}\\{0,\;\;\frac{T}{4} < t < T}\end{array}} \right.\).  The on-state resistance of the MOSFET is 0.25 Ω. The average on-state power loss of device for four cycles is ________ (in W)
    Solution

    Concept:

    The average on-state power loss per cycle is given by

    \({P_{avg}} = I_{rms}^2{R_{on}}\)

    Where Irms is the rms value of current

    Ron is the on-state resistance

    Calculation:

    \(i\left( t \right) = \left\{ {\begin{array}{*{20}{c}}{15,\;\;0 < t < \frac{T}{4}}\\{0,\;\;\frac{T}{4} < t < T}\end{array}} \right.\)

    \({i_{rms}} = \sqrt {\frac{1}{T}\mathop \smallint \limits_0^T {{\left( {i\left( t \right)} \right)}^2}dt} \)

    \( = \sqrt {\frac{1}{T}\mathop \smallint \limits_0^{\frac{T}{4}} {{\left( {15} \right)}^2}dt} \)

    \( = \sqrt {\frac{1}{T}\left[ {225t} \right]_0^{\frac{T}{4}}} = \sqrt {\frac{{225}}{4}} = 7.5\;A\)

    The average on-state power loss per cycle, Pavg = (7.5)2 × 0.25 = 14.0625 W

    For four cycles, the power loss = 4 × 14.0625 = 56.25 W
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