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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
After Q is turned off, the capacitor current becomes equal to I0
ic=I0=Csdvdt
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)
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
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
Voltage equation across inductance is given by
vL=LdIdt
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
Vs−V0=LdILdt
dILdt=Vs−V0L
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 ________
V0, rms=Vsβπ
Average power =Vo, rms2R
⇒(Vsβπ)2R=2000
⇒Vs2×βR×π=2000
⇒(200)2×β10×π=2000
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)
Vdc=24022π (1+cos45∘)=92.21 V
Ia=VdcR=9.22 A
ID, rms=12π ∫παiD2dθ
=Iaα+π2π
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)
When the switch Q is ON,
L disdt=Vs, 0≤t≤δTs
⇒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(1−δ)2
⇒I1=VsR(1−δ)2=−Vs2LδTs
For continuous source current,
I1 > 0
⇒VsR(1−δ)2>Vs2L δTs
⇒R<2fsLδ(1−δ)2
Maximum value of R=2fsLδ(1−δ)2
=2×20×103×500×10−6(0.7)(1−0.7)2
= 317.46 Ω
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°
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.
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
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)
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
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____
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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