Test: Single Phase FW AC-DC - 1
30 Questions MCQ Test Power Electronics | Test: Single Phase FW AC-DC - 1
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The below shown single phase controlled rectifier configuration is a
It is a two pulse i.e full wave M-2 connection.
In the below shown circuit, considering the load to be purely R load, the thyristor T2 would conduct from ωt =
The bottom winding forward biases the SCR T2 at π, conduction starts after the pulse is given at α & stops when T2 is naturally commutated at 2π.
For a single pulse M-2 rectifier with RL load and firing angle α >0°, the value of the inductance L is such that the load current is continuous.
Choose the correct statement.
Choose the correct statement.
As mentioned, L is such that the load current is continuous. Hence, it never falls below a predefined positive value.
For the circuit below, the transformer ratio is 1:1 and Vs = Vm sinωt. When T1 is forward biased and is triggered at α than T2 is
When T1 is gated, the other SCR T2 is reversed biased due to the windings so connected.
The circuit turn-off time of M-2 type connection is
When T1 is gated, SCR T2 is off from ωt = α to π
Thus t = π-α(offtime)/ω
Same can be said when T2 is gated and T2 is off.
Find the expression for the average value of the output voltage for the below given circuit. Consider the load current to be continuous, firing angle = α, transformer ration 1:1 and Vs = Vm sinωt.
As the load current is continues, the voltage is positive from α to π and negative from π to π+α and so on. Thus,
Vavg = 1/Period ∫ Vm sinωt d(ωt)
Where period = π
And the integral runs from α to α+π.
Fully controlled implys that all the elements are “fully controlled” hence, it uses SCRs only except the FD.
A single phase full-converter using R load is a _________ quadrant converter and that using an RL load without FD is a __________ quadrant converter
In R load both V and I are positive, in RL load the voltage can be negative but current is always positive.
A single phase full controlled bridge converter (B-2) uses
4 SCR’s are connected in a bridge fashion.
B-2 is the bridge type controller, in which 2 devices conduct at a time. One acting as the current supplying path and other acts as a return path.
T1-T2 are gated together at a time and T3-T4 are gated together at a time.
With respect to the below given circuit, from ωt = 0 to α, ____ thyristor(s) block the forward voltage
T1 and T2 are not yet gated, hence they block the forward voltage.
With respect to the below given circuit, from ωt = 0 to α,
T1 and T2 are not yet gated, hence they block the forward voltage where as no voltage exists across T3 and T4.
For the below given circuit, when the devices T1 and T2 are conducting from ωt = α till they are naturally commutated at π voltage across the other two pair of devices is
When T1 and T2 are conducting they act like S.C’s and the other pair of devices i.e. T3 and T4 has to take care of the reverse voltage.
A single-phase full converter B-2 type connection has a RLE type of motor load connected. The minimum requirement to turn-on the device is
Device will not turn on until the supply voltage exceeds the motor back emf E.
For the below shown circuit with continuous load current.
Each thyristor pair conducts for
As the load current is continuous each pair conducts from π to π+α i.e π radians.
T1-T2 are gated together at a time and T3-T4 are gated together at a time.
For the below shown circuit has dis-continuous load current waveform.
Each thyristor pair conducts for
Load current is discontinues than SCR pair conducts for less than 90° depending on the value of the inductor L.
A single-phase full converter with B-2 type of connection has a continuous load current waveform. The thyristor pairs T3, T4 is triggered at ωt =
First, at α T1-T2 conduct, then after π+α the next pair is excited i.e. T3-T4 and likewise.
The PIV for each diode in a single-phase, full converter with B-2 type of controlled rectifier is
Consider supply voltage as Vs = Vm sinωt.
The peak inverse voltage handled by each device is Vm in case of B-2 type of configuration.
A motor load is connected to a single-phase full converter B-2 type controlled rectifier. The net energy is transferred from ac source to the motor (dc load) when
Converter will work as a line commuted inverter when π-α > α.
The below shown rectifier configuration has continues load current, find the expression for the average value of output voltage when the supply Vs = Vm sinωt is connected.
Vo = 1/π x [∫ Vm sinωt d(ωt)] Where the integration runs from α to π+α.
Find the expression of the rms value of output voltage for a single-phase M-2 type rectifier with RL load and continues load current. Transformer ratio is 1:1 with supply voltage Vs = Vm sinωt
Vor2 = 1/π x [∫ Vm2 sin2ωt d(ωt)], Where the integration runs from α to π+α
Vor = Vm2/2 = Vs.
A single-phase full controlled converted with RLE load will act like a line-commutated inverter when the firing angle α
It will act like a inverter i.e. most of the current would flow from the battery or back emf E to the source. For α=90° it will not act like a converter nor an inverter.
In converter operation, with output voltage = Vo and RLE load.
The output voltage obtained by a converter is always greater than the counter or back emf E.
Vo = IR + E.
In inverter operation, with output voltage = Vo and a RLE load connected
In inverter operation Vo < E.
A motor load is connected to a single-phase full converter B-2 type controlled rectifier, the net energy is transferred from ac source to the motor (dc load) when
It works as a converter when π-α < α.
The below shown rectifier configuration has continues load current, find the expression of the RMS value of output voltage when the supply Vs = Vm sinωt
Vor2 = 1/π x [∫Vm2 sin2ωt d(ωt)] Where the integration runs from α to π+α
Vor = Vm2/2 = Vs.
The PIV of diodes in M-2 is 2Vm, whereas that in B-2 type of connection is Vm.
An SCR has the peak forward voltage = 1000 V. Find the maximum voltage that the SCR can handle if employed in a M-2 type full controlled converter circuit. Use factor of safety (FOS) = 2.5
In M-2 type configuration maximum voltage handled is 2Vm
Therefore, 1000/2×2.5 = 200 V.
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