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Power Electronics Test 1
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Power Electronics Test 1
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  • Question 1/25
    2 / -0.33

    Due to low internal generation in GTO, the GTO has
    Solutions
    • GTO being a monolithic p-n-p-n structure just like a thyristor. In particular, the p-n-p-n structure of a GTO can be thought of consisting of one p-n-p and one n-p-n transistor connected in the regenerative configuration.
    • Due to low internal generation in GTO, the GTO has both holding and latching current of a GTO are considerably higher compared to a similarly rated thyristor.
    • Since the holding current of a GTO is considerably higher than that of a thyristor anode current variation can generate serious problem because the GTO might unlatch at an inappropriate moment.
    • To avoid this problem the gate drive unit of a GTO must feed the gate terminal with a continuous “back porch” current during the entire on period of the GTO.
    • This back-porch current must be larger than the gate trigger current.
  • Question 2/25
    2 / -0.33

    The reverse recovery time of a diode is trr = 3 μs and the rate of fall is didt=30A/μs. The stored charge of the diode is
    Solutions

    Important formulas:

    1) QRR=12×IRR×trr

    2) IRR=[2QRRdidt]1/2

    3) trr=[2QRRdidt]12

    Calculation:

    QRR=12trr2(didt)

    =12×(3×106)2×30Aμs

    QRR=135μC

  • Question 3/25
    2 / -0.33

    A single-phase semi converter shown below figure, is used to control the speed of the small separately excited dc motor at 4.5 kW, 220 V, 1500 rpm. The converter is connected to a single-phase 230 V, 50 Hz supply. The armature resistance is Ra = 0.50 Ω and the armature circuit inductance is La = 10 mH. The motor voltage constant is Kb = 0.1 V/rpm

    With the converter operates as a rectifier, the dc motor runs at 1200 rpm and carries an armature current of 16 A. Assume that the motor current is continuous and ripple-free. The firing angle is _______ (in degrees)
    Solutions

    Concept:

    The average output voltage of a single-phase semi-converter is given by

    V0=Vmπ(1+cosα)

    Vm is the maximum value of supply voltage

    α is the firing angle

    Calculation:

    The motor voltage constant is Kb = 0.1 V/rpm

    Speed of the motor = 1200 rpm

    Back emf (Eb) = 0.1 × 1200 = 120 V

    Armature current (Ia) = 16 A

    V0 = Eb + Ia Ra

    230×2π(1+cosα)=120+16(0.5)=128

    ⇒ α = 76.33°

  • Question 4/25
    2 / -0.33

    A single-phase fully controlled bridge circuit with R-L load is used to obtain the regulated dc output voltage. The RMS value of the ac input voltage 230 V and the firing angle at 60°, so that the load current is 4 A.

    If SCR T3 is damaged and gets open-circuited, the load current is _______ (in A)

    Solutions

    Concept:

    The average output voltage is given by, V1=2Vmπcosα

    The average output current, I1=V0R

    If SCR T3 is damaged and gets open circuited, the circuit acts as half wave rectifier.

    The average output voltage is given by, V2=Vmπcosα

    The average output current, I2=V0R

    I2I1=12

    Calculation:

    Given that, I1 = 4 A, α = 60°

    I2=I12=42=2A

  • Question 5/25
    2 / -0.33

    A motor load is connected to a single-phase full bridge converter type-controlled rectifier. The net energy is transferred from ac source to the motor (dc load) when
    Solutions

    During α to π, both vs and is are positive, power therefore flows from ac source to load. During the interval π to (π + α), vs is negative but is is positive, the load, therefore, returns some of its energy to the supply system. But the net power flow is from ac source to dc load because (π – α) > α.

    The average value of the output voltage is given by

    V0=2Vmπcosα

    Note:

    • If α > 90°, V0 is negative. If the load circuit emf E is reversed, this source E will feed power back to ac supply. This operation of full converter is known as the inverter operation of the converter.
    • The full converter with firing angle delay greater than 90° is called a line-commutated inverter. Such an operation is used in the regenerative braking mode of a dc motor in which case then E is counter emf of the dc motor.
    • During 0 to α, ac source voltage vs is positive, but ac source current is negative, power, therefore, flows from dc source to ac source.
    • From α to π, both vs and is are positive, power, therefore, flows from ac source to dc source. But the net power flow is from dc source to ac source because of (π – α) < α
  • Question 6/25
    2 / -0.33

    For a three-phase half-controlled rectifier with a freewheeling diode, the firing angle for the thyristor is 60o. The conduction angle for the freewheeling diode will be:
    Solutions

     

    For firing angle, greater than 60o the freewheeling diode starts to conduct.

    Therefore, the conduction angle for the freewheeling diode can be represented by the formula:

    θ = α - π3

    For firing angle α = π3; conduction angle for the freewheeling diode is 0o

  • Question 7/25
    2 / -0.33

    The thyristor in the figure turned on at t = 0. The inductor and capacitor are initially uncharged. The steady-state value of the capacitor voltage, VC is

    Solutions

    Taking Laplace at t = 0

    I(s)=20sLs+1cs

    =20Ls2+LC

    Taking inverse Laplace transform,

    i(t)=VsCLsintLC

    Vc(s)=I(s)×1Cs

    Vc(s)=20Ls2+LC×1Cs

    =20LCs[s2+1LC]

    Taking inverse Laplace transform,

    Vc(t)=Vs[1cosωt]

    Vc(t)=Vs[1costLC]

    The thyristor will allow current in positive direction only, i.e. till π.

    Vc(ωt=π)=Vs[1cosπ]=2Vs

    At steady state, V­c = 2 × 20 = 40 V

  • Question 8/25
    2 / -0.33

    For the simple chopper circuit shown, the average and rms value of currents for a duty cycle of 0.49, in amps, are (neglect the drop across chopper when ON)

    Solutions

    Concept:

    For a step-down chopper, average voltage and current is given by

    V0(avg) = δVs

    I0(avg)=δVsR

    rms voltage and current is given by

    V0(rms)=Vsδ

    I0(rms)=δVsR

    Where δ is duty cycle

    Calculation:

    We have given

    δ = 0.49

    Average load current I0(avg)=0.49×20010=9.8A

    Rms value of load current I0(rms)=0.49×20010=14A

  • Question 9/25
    2 / -0.33

    A single-phase fully controlled bridge circuit with R-L load is used to obtain the regulated dc output voltage. The RMS value of the ac input voltage Vs and the firing angle at 60°, so that the load current is I0.

    The ratio of the average output voltage of the converter when a freewheeling diode is connected across the load to the average output voltage of the converter without any freewheeling diode is ________ (up to two decimal places)
    Solutions

    Concept:

    In a single phase fully controlled bridge rectifier without freewheeling diode,

    The average output voltage, V0=2Vmπcosα

    The average output current, I0=V0R

    In a single phase fully controlled bridge rectifier with freewheeling diode,

    The average output voltage, V0,fd=Vmπ(1+cosα)

    The average output current, I0,fd=V0,fdR

    V0,fdV0=12(1+secα)

    I0,fdI0=V0,fdV0=12(1+secα)

    Calculation:

    Given that, firing angle (α) = 60°

    V0,fdV0=12(1+sec60)=1.5

  • Question 10/25
    2 / -0.33

    In a Buck converter, the input voltage is 30 V. The Duty cycle ‘D’ is varied from ‘zero’ to only progressively while feeding R load. The per-unit ripple in inductor would be maximum at a duty cycle is

    Solutions

    Input voltage = Vdc

    Output voltage = V0

    Inductance = L

    Capacitance = C

    When switch ‘sw’ closed (0 < t < DT)

    diLdt=VdcV0L 

    diL=VdcV0LdtiL(t)=VdcV0Ldt+ILmin 

    iL(t)=VdcV0Lt+ILmin

    When switch is open (DT < t < T)

    dLdt=V0L 

    diL=V0Ldt 

    iL=V0Ldt+ILmax 

    iL(t)=ILmaxV0L(tDT)

    Now ripple current in inductor

    ΔiL = ILmax – ILmin

    at t = DT iL(t) = Imax

    So ILmax=VdcV0LDT+ILmin

    Now ΔiL=ILmaxILmin=VdcV0LDTV0=DVdc

    ΔiL=VdcDVdcLDT=Vdc(1D)LDT 

    ΔiL α D (1 - D)

    To get maximum, dΔILdD=0

    dΔILdD=0VdcLT(12D)max=0 

    1 – 2Dmax = 0

    Dmax = 0.5

  • Question 11/25
    2 / -0.33

    A star-connected load of 20 Ω per phase is Fed from 450 V source through a 3-phase bridge inverter. For 180° mode determine load power in kW is ______ (correct up to 2 decimal places)
    Solutions

    Concept:

    Three-phase power =3(Iph)rms2×R

    (Iph)rm=(Vp)rmsRload 

    (VL)rms=Vdc23,(Vph)rms=Vdc3 

    (Vph)rms=Vdc23 

    Calculation:

    Load resistance, RL = 20 Ω

    Source voltage, Vs = 450 V

    Mode of operation = 180°  

    (Vph)rms=450×23 

    = 212.132 V

    (Iph)rms=212.13220 

    (Iph)rms=10.6A 

    Three-phase power is

    =3(Iph)rms2×R=3×10.62×20 

    = 6741.6 W = 6.74 kW

  • Question 12/25
    2 / -0.33

    A star connected load of 15 Ω per phase is fed from 420 V DC source through a 3-phase bridge inverter. If the inverter is operating in 120° conduction mode, the RMS value of thyristor current is _________ (in A)
    Solutions

    Concept:

    Parameter

    180° conduction mode

    120° conduction mode

    Phase voltage (Vph)

    23Vs

    Vs6

    Line voltage (VL)

    23Vs

    Vs2

    RMS load current (Ior)

    23RVs

    Vs6R

    RMS thyristor current (ITr)

    Vs3R

    Vs23R

     

    Calculation:

    The inverter is operating in 120° conduction mode.

    Supply voltage (Vs) = 420 V

    Load resistance (R) = 15 Ω

    The RMS value of thyristor current, ITr=42023×15=8.08A

  • Question 13/25
    2 / -0.33

    An IGBT chopper modulated power from a 200 V DC supply to a resistive load of 20 Ω. The switching frequency of the choppers is 1 kHz for VDS(sat) = 1.9 V. The chopper has a duty cycle, m = 0.8. The average power loss due to conduction is _______ (in W)
    Solutions

    Given that, Vs = 200 V

    VDS(sat) = 1.9 V

    Load resistance (R) = 20 Ω

    Duty cycle (m) = 0.8

    Steady on-state current, Il=VsVDS(sat)R=2001.920=9.9A

    Conduction losses, Pc=1T0tONVDS(sat)IDdt

    = VDC(sat) m ID = 1.9 × 0.8 × 9.9 = 15 W
  • Question 14/25
    2 / -0.33

    A 1000 V 25 A GTO controls power from a DC supply of 600 V to Rload of 30 Ω. The data sheet gives the following information: the on-state voltage drop, VGTO(ON) = 2.2 V; average gate power should not exceed 10 W; IG Turn-off = -25 A.
    Solutions

    Concept:

    The load current, Il=VsVGTO(ON)R

    The load power, P=Vl2R=(VsVGTO(ON))2R

    The total power gain =PPG

    Calculation:

    Given that, DC supply voltage (Vs) = 600 V

    Load resistance (R) = 30 Ω

    the on-state voltage drop, VGTO(ON) = 2.2 V

    The average gate power, PG = 10 W

    The load current, Il=6002.230=19.93A

    The load power, P=(6002.2)230=11912W

    The gate power, PG = 10 W

    The total power gain =PPG=1191210=1191.2
  • Question 15/25
    2 / -0.33

    The gate triggering circuit shown below has the specifications:

    Ig(min) = 100 mA

    Vg(min) = 2 V

    The Vtrig should be
    Solutions

    Given that,

    Ig(min) = 100 mA

    Vg(min) = 2 V

    Simplifying the above circuit diagram, we get below the circuit diagram where we assume SCR as a voltage device.

    Note: The diode in the given circuit is just for current reversal protection.

    Now applying KCL, we get

    VgVtrig30+Vg30+100mA=0

    ⇒ 2Vg – Vtrig + 3 = 0

    ⇒ 2Vg + 3 = Vtrig

    ⇒ Vtrig = 2 × 2 + 3 = 7 V
  • Question 16/25
    2 / -0.33

    A single-phase AC supply is connected to a full wave rectifier through a transformer. The rectifier is required to supply an average DC output voltage of Vdc = 400 V to a resistive load of R = 10 Ω. The kVA rating of the transformer is __________ (in kVA)
    Solutions

    Concept:

    In a full wave rectifier though a transformer,

    Average output voltage, V0=2Vmπ

    Average output current, I0=V0R

    RMS value of output voltage Vor = Vs

    RMS value of load current, Ior=VsR

    Average value of diode current, Id=Im2

    RMS value of diode current, Idr = Ior

    Peak value of diode current, Idm = √2 Ior

    Power delivered to the load = Vor Ior

    Input voltamperes = Vs Ior

    Calculation:

    Given that, DC output voltage (VDC) = 400 V

    Average output voltage of rectifier (V0) = 400 V

    2Vmπ=400

    2×2×Vsπ=400

    ⇒ Vs = 444.28 V

    Load resistance (R) = 10 Ω

    RMS value of load current, Ior=444.2810=44.428A

    kVA rating of transformer = 444.28 × 44.428 = 19.738 kVA
  • Question 17/25
    2 / -0.33

    For the three-phase circuit shown below, if the firing angle of each thyristor is 45°, then the average power consumed by the load is ________ W

    Solutions

    The output voltage waveform of the 3ϕ half-wave rectifier is given below.

    With respect to the above waveform:

    V0rms=Vmph22π3[[π(π6+α)]+12[sin(2α+π3)sin(2π)]]12

    V0rms=3Vmph2π[1.832+12[sin(π2+π3)]]12

    V0 rms = 126.67 V

    Power command by the load; P=(V0rms)2R=(126.67)220=0.802kW

  • Question 18/25
    2 / -0.33

    A single-phase semi-converter, connected to 230 V 50 Hz source, is feeding a load R = 10 Ω in series with a large inductance that makes the load ripple-free. The firing angle is 30o, then the current distortion factor is_________
    Solutions

    For a single-phase semi-converter, the peak value of the nth component of the current

    Isn=4I0nπcosnα2

    (Is1)rms=22πI0cosα2

    (Is)rms = RMS value of source current

    Isn=I0(παπ)1/2

    Current distortion factor, g=Is1Isn=22πI0cosα2I0(παπ)12

    g=22cosα2π(πα)

    Replacing the value of α=π6 we get,

    g=22cos(π12)π(ππ6)=0.9526

  • Question 19/25
    2 / -0.33

    A three-phase three pulse converter, fed from three-phase 400 V, 50Hz supply, has a load R = 2 Ω, E = 200 V, and large inductance so that load current is constant at 20A. If the source has an inductance of 2 mH, then the value of overlap angle for inverter operation is 
    Solutions

    Concept:

    Effect of source inductance: The presence of source inductance affects the rectifier output voltage

    V0 = Vd0 cos α – ΔVd0

    Where ΔVd0=Vd02[cosαcos(α+μ)]

    μ is the overlap angle

    Vd0=2Vmπ for 2 pulse

    =3VmL2π for 3 pulse

    =3Vmπ for 6 pulse

    Also, ΔVd0 = f Ls I0 (1 pulse)                                                      

    = 4 f Ls I0 (2 pulse)

    = 3 f Ls I0 (3 pulse)                                                                    

    = 6 f Ls I0 (6 pulse)

    Calculation:

    V0 = -E0 + I0 R

    Vd0 cos α – 3 f Ls I0 = -200 + (20)(2)

    Vd0=3Vm2π

    3Vm2πcosα3fLsI0=160

    3Vm2πcosα=154

    On solving we get cos α = - 0.57

    ⇒ α = 124.76°

    For three pulse converters:

    Vd02[cosαcos(α+μ)]=3fLsI0

    3Vm2π2[cos(127.76)cos(124.76+μ)]=3fLsI0

    On solving the above equation for overlap angle (μ) we get

    cos (α + μ) = -0.6145

    μ = 3.15°   

  • Question 20/25
    2 / -0.33

    In the circuit shown below if duty ratio D = 0.4 and switching frequency is 50 Hz, then the peak to peak ripple voltage of the capacitor is _________ V

    Solutions

    Concept:

    For a buck-boost converter

    V0=(α1α)Vs

    Capacitor voltage ripple:

    ΔVC=ΔQC=I0TonC

    ΔVC=I0αfC

    V0: output voltage

    Vs: source voltage

    α: duty ratio

    Calculation:

    The given circuit represents a buck-boost converter

    V0=(α1α)Vs=(0.40.6)×150=100V

    I0=V0R=10020=5A

    Ripple voltage of capacitor ΔV=I0αfC

    ΔV=5×0.450×200×106

    ⇒ ΔV = 200 V

  • Question 21/25
    2 / -0.33

    In a Buck-Boost Converter shown below, switching frequency is 100 kHz with duty cycle 0.4. Initial charges in the capacitor and initial current in the inductor are zero. Calculate the energy stored in the inductor in mJ after 10 cycles.

    Solutions

    Input voltage, Vdc = 50 V

    Output voltage, V0 = 150 V

    Inductance, L = 15 mH

    Capacitance, C = 1 μF

    Duty cycle, D = 0.4

    Frequency = 100 kHz

    When switch closed during 0 < t < DT, inductor will charge

    t = DT

    VL=Vdc=L=didt 

    di=VLDTL

    We know di become Imax at t = DT

    Imax=50×0.4×15×1031105=5×415×102 

    Imax=43×102A 

    From DT < t < T i.e. when switch open, the inductor will discharge.

    IL(t)=ImaxV0L(1D)T 

    =43×1015015×103×0.6×1105 

    =43×1020.610 

    IL(t) = -0.0466 A

    Important Observation:

    Current cannot become negative because of Diode D­1.

    So, at the end of each cycle inductor current is zero.

    So, inductor current becomes zero between DT < t < T

    So, energy stored after 10th cycle in the inductor

    =12×15×103×02 

    = 0 mJ

  • Question 22/25
    2 / -0.33

    A step-up chopper shown in figure is to deliver 3A

    In to the 10 Ω load. The battery voltage is 12 V, L = 20 μH, C = 100 μF and chopper frequency is 50 kHz. Determine the battery current variation in ampere

    Solutions

    Calculation:

    Vo = IoRL = 3 × 10 = 30 V

    Vb = Vo (1 – D)

    1D=1230=0.4

    D = 0.6

    Periodic time T=1f=150×103=20μs

    On time = DT = 0.6 × 20 × 10-6 = 12 μs 

    over an on-time the change in battery current will be

    Δi=VLL×Ton

    Δi=1220×106×20×106=7.2A

    Concept:

    When the transistor is switched on, current ramps up in the inductor and energy is stored.

    When the transistor is switched off, the inductor voltage reverses and acts together with the battery voltage to forward bias the diode, transferring energy to the capacitor.

    When the transistor is switched on again, load current is maintained by the capacitor, energy is stored in the inductor and the cycle can start again.

    The value of the load voltage is increased by increasing the duty cycle or the on-time of the transistor.

    During the transistor on-period, assuming an ideal inductor and transistor

    Vb = VL = Ldi/dt

    Vb/s = (sL) i(s)

    i(s) = Vb/s2L

    i(t) = (Vb/L)t = kt

    Over an on-time of TON, the change of battery current will be

    Δi = k TON

    During the off-period

    Vo = Vb + VL = Vb + L(di/dt) = Vb + L(Δi/Δt)

    = Vb + L(Vb/L)TON /Toff = Vb(1 + TON /Toff)

    Let the duty cycle

    D = TON T, and Toff = T – TON = T(1 - D). Now

    Vo = Vb (1 – DT/T(1 – D))

    which simplifies to

    Vo = Vb/(1 - D)

    In an ideal circuit, VbIb = V0I0. Therefore, from eq.

    Ib = (Vo/Vb)Ib = Io/(1 – D)

    Ib = Ib + Δi/2

    Ibo = Ib – Δi/2

    When the steady variation of battery current has been reached, the variation will be between lo and I1, as shown in.

  • Question 23/25
    2 / -0.33

    In a 3-phase bridge inverter feeds a star connected load for 180° conduction mode, RMS value of switch current is 15 A. For 120° mode, RMS value of switch current _______ (Correct up to 2 decimal places)
    Solutions

    Drawing output current waveform for 180° mode.

    (iT)rmsfor1802=12π[0π3Vdc3Rdωt+π32π323VdcRdωt+2π3πVdc3Rdωt] 

    =12π[2π3×(Vdc3R)2+π3(2Vdc3R)2] 

    iT2=(Vdc3R)2[13+4π3×2π] 

    iT2=(Vdc3R)2(iT)tms=Vdc3R 

    Vdc3R=15A

    Drawing output and switch current for 120°.

    (iT)rms2=(Vdc2R)2×2π3×π$

    (iT)rms=(Vdc2R)×13

    =452×13=12.99 

  • Question 24/25
    2 / -0.33

    The output voltage of an inverter is controlled by multiple pulse modulation technique. The peak value of fundamental voltage component is _________ V

    Solutions

    Concept: This method of pulse modulation is termed as multiple pulse modulation. r is an extension of single pulse modulation.

    The amplitude of the nth harmonic of two pulse waveform is given by:

    Vn=[8Vsnπsin(nγ)sin(nd2)]

    In general, γ=π2dN+1+dN where N is the number of pulses per half cycle (Here N = 2)

    Calculation:

    The amplitude of nth harmonic Vm=8Vsnπsin(nγ)sin(nd2)

    For peak value of fundamental voltage component

    V1=8Vsπsin(d2)sin(γ)

    γ=(π2dN+1)+dN

    γ = 56°, Vs = 200 V, N = 2

    (π180)56=(π2d3)+d2

    Solving we get d = 24°

    V1=8(200)πsin(242)sin(56)

    V1 = 87.78 V

  • Question 25/25
    2 / -0.33

    A single-phase full-bridge inverter has RLC load of R = 4Ω, L = 35 mH, C = 155 μF. The DC input voltage is 230 V and the output frequency is 50 Hz. The third harmonic component in load current is:
    Solutions

    Concept:

    General equation of nth harmonic voltage waveform for a single-phase full-bridge inverter is given by:

    VOn=4Vsnπsin(nωt)

    3rd harmonic waveform:V03=4Vs3πsin(3ωt)

    I03=(V03)rms|Z3|

    Calculation:

    Supply voltage (Vs) = 230 V

    (V03)rms=4Vs3π×12=4×23032π

    (V03)rns = 69.02 V

    Z3=R+j(3ωL+(1)3ωC)

    Z3=4+j(3ω(35×103)+(1)3ω(155×6))

    |Z3|=R2+(XLXC)2

    |Z3| = 26.44 Ω

    ∴ 3rd harmonic current component

    I03=(V03)rms|Z3|=69.0226.44=2.61A

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