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Power Electronics Test 1
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Power Electronics Test 1
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  • Question 1/10
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    In the circuit shown below, transistor Q is turned off at t = 0. The transistor turns off with its current falling to zero instantly. Diode Ds is ideal. The freewheeling diode Df does not turn on until the capacitor voltage reaches Vs).

    The value of capacitor Cs (in μs) so that the maximum dvdt across the transistor is 100 V/μs is

    Solutions

    After Q is turned off, the capacitor current becomes equal to I0

    ic=I0=Csdvdt

    dvdt=I0Cs=100vμsCs=0.2 μF
  • Question 2/10
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    A single-phase semi converter, supplied by on ac source, feeds power to an R-L-E load as shown in fig.

    Given that Vm = 90π volts and power loss in the battery is 1200 watt then the value of firing angle in degree is ______ (up to two decimal places)

    Solutions

    Concept:

    For Single-phase Simi converter

    Average output voltage

    Vo(av)=Vmπ(1+cosα)

    Power loss in battery = EI0

     Calculation:

    Power loss in battery = 1200 watt

    EI0 = 1200

    I0=120060=20A

    Average output power V0 = E + I0R

    = 60 + 20 × 3

    = 120 V

    Putting the value of output voltage

    120=90ππ(1+cosα)

    ⇒ cos α = 0.33

    ⇒ α = 70.73o
  • Question 3/10
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    In a boost chopper circuit if Vs, V0, L are input voltage, output voltage and inductance, respectively, and when the conducting switch is opened, the rate of change of inductive current is

    Solutions

    Voltage equation across inductance is given by

    vL=LdIdt

    Calculation:

    Circuit diagram for the boost chopper and its equivalent circuit with conducting switch open are shown below.

    Applying KVL in the above circuit

    Vs – VL – V0 = 0

    ⇒ Vs – V0 = VL

    Rate of change of inductive current  is 

    VL=LdILdt

    VsV0=LdILdt

    dILdt=VsV0L

  • Question 4/10
    1 / -0

    The single-phase full bridge inverter shown below is operated in the quasi-square wave mode at the frequency f = 100 Hz with a phase shift of β between the half bridge outputs Va0 and Vb0.

    With a purely resistive load R = 10 Ω, the value of β (in degrees) so that the average power supplied to the load is 2 kW is ________

    Solutions

    V0, rms=Vsβπ

    Average power =Vo, rms2R

    (Vsβπ)2R=2000

    Vs2×βR×π=2000

    (200)2×β10×π=2000

    β=π2=90
  • Question 5/10
    1 / -0

    In the single phase half wave rectifier shown below, assume that the load current i0 = ia is constant (load inductance L is very large) For R = 10 Ω, the rms value of the diode current iD for α = 45° is __________ (in A)

    Solutions

    Vdc=24022π (1+cos45)=92.21 V

    Ia=VdcR=9.22 A

    ID, rms=12π παiD2dθ

    =Iaα+π2π

    = 7.29 A
  • Question 6/10
    1 / -0

    The step-up dc-dc converter shown below is operated at a switching frequency of fs = 20 kHz.

    For the duty ratio δ = 0.7, the maximum value of the load resistance R so that the source current is is continuous is _______ (in ohms)

    Solutions

    When the switch Q is ON,

    L disdt=Vs, 0tδTs

    is(t)=I1+VsLt

    Where I1 = is (0)

    I2=I1+VsLδ Ts

    Where I2 = is (δ Ts)

    Power balance of the converter:

    pin = pout

    Vs Is=Vdc2R

    Is(av)=12 (I1+I2)

    =12 (2I1+VsL δ Ts)

    =I1+Vs2L δTs

    Vs (I1+Vs2L δTs)=Vs2R(1δ)2

    I1=VsR(1δ)2=Vs2LδTs

    For continuous source current,

    I1 > 0

    VsR(1δ)2>Vs2L δTs

    R<2fsLδ(1δ)2

    Maximum value of R=2fsLδ(1δ)2

    =2×20×103×500×106(0.7)(10.7)2

    = 317.46 Ω

  • Question 7/10
    1 / -0

    A three phase three pulse converter is operated from 3-phase, 230 V, 50 Hz supply with load resistance R = 10 Ω. An average output voltage of 50% of the maximum possible output voltage is required. Then the firing angle is ________ (in degrees)
    Solutions

    For R load, voltage is continuous when α ≤ 30° and average output voltage is given by

    Vo=3Vml2πcosα 

    Maximum possible output voltage =3Vml2π

    Required average output voltage =12(3Vml2π)

    Now,3Vml2πcosα=12(3Vml2π) 

    ⇒ α = 60°

    So, the required value can’t be obtained for α < 30°.

    For α > 30°,

    Average output voltage, Vo=3Vmp2π[1+cos(α+30)] 

    =3Vml23π[1+cos(α+30)] 

    3VmL23π[1+cos(α+30)]=(3Vml2π)(12) 

    ⇒ α = 67.69°

  • Question 8/10
    1 / -0

    The figure shows two choppers connected in parallel and the current through CH1 & CH2 is given below

    If α = 30o with some delay shown in figure. The load resistance is 1 Ω and supply is 100 V. What is the average value of current in A, if switching frequency is 1 kHz.

    Solutions

    From the given diagram we can say that two step down chopper is operating in parallel.

    VS = 100 V

    δ = 0.3

    For step down chopper

    V0 = δ VS

    δ = duty cycle = 0.3

    Load current I0=V0R=0.3×1001

    = 30 Amp

    For source current I, adding currents i1 & i2

    Frequency f = 1kHz

    T=1f=1msec

    δ = 0.3 (Ton = δ T)

    ∴ Ton = 0.3 m sec

    T2=0.5msec

    iav=I0×0.30.5=30×0.30.5

    = 18 A

  • Question 9/10
    1 / -0

    A single phase full bridge voltage source inverter feeds a load as shown in figure below. If the load as shown in figure below. If the load current is I0 = 200 sin (ωt – 30°), the power delivered to the load is ____ (kW)

    Solutions

    Power delivered to the load. P0 = V01 Ior cos ϕ

    V01 is rms value of fundamental output voltage

    I0r is rms value of load current

    V01=4Vdc2π

    Given that, Vdc = 400 V

    V01=4×4002×π=360.12V

    Power delivered, P0 = V01 Ior cosϕ

    =360.12×2002×cos30

    = 44.1 kW

  • Question 10/10
    1 / -0

    A 12-V battery feeds a single-phase full bridge inverter whose output is connected to an ideal single-phase transformer. Its primary has 10 turns and the load voltage is 230 V. For a load resistance of 100 Ω, the rms value of primary current is____

    Consider only the fundamental component of inverter out-put voltage.
    Solutions

    The output voltage for a signal phase full bridge inverter is

    Vo=n=1,3,54Vsnπsinnωt

    V01=4Vsπ2=4×12π×2=10.8 V

    Primary voltage of transformer (Ep) = 10.8 V

    Secondary voltage of transformer (Es) = 230 V

    Secondary current (Is)=230100=2.3 A

    EPEs=IsIP

    ⇒ Primary current (Ip)=Is×EsEp

    =2.3×23010.8=48.98 A

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