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SCRs with a peak forward rating of 2 kV and an average onstate current rating of 50 A are used in a singlephase midpoint converter. If the factor of safety is 2.5, the power that can be handled by this converter is ______kW
The maximum voltage across SCRs in mod point connection is 2 Vm. So the maximum voltage of SCR = = 400 V
The maximum average power that the midpoint converter can handle is
= 12.732 kW
In a center tap full wave rectifier, if the peak voltage applied between the center tap and one of the secondary 200 V, then the reverse biased diode is applied with a maximum voltage of:
Concept:
Peak inverse voltage: It is a voltage across the diode when it is reversed bias.
The following circuit is the center tapped fullwave rectifier.
Consider for positive half cycle, D_{1} is FB and D_{2} is reversed biased.
In loop 1, Apply KVL
⇒ V_{0} = V_{in}
Now, apply KVL in loop 2
V_{in} + V_{0} = V_{D2}
⇒ V_{D2 }= V_{in} + V_{in} = 2V_{in}
PIV of the centre tapped full wave rectifier = 2V_{in}
Calculation:
Given that, the centertapped fullwave rectifier and V_{in} = 200 V (Peak) between the centre tap and one end of secondary.
Peak inverse voltage = 2V_{in} = 400 V
A fullwave rectifier uses 2 diodes. The internal resistance of each diode is 20 Ω. The transformer RMS secondary voltage from centre tap to each end of the secondary is 50 V and the load resistance is 980 Ω. Mean load current will be
Concept:
Center tapped full wave rectifier:
Analysis: The DC output voltage or average output voltage can be calculated as follows,
V_{0} = 2V_{m} / π
Now we can calculate the average or mean current of load by dividing the average load voltage by load resistance RL. Therefore mean load current is given by
I_{0} = V_{0} / RL
If the internal resistance of the diode is given in that case mean load current I_{0} = V_{0} / (R_{L }+ r)
Where r = internal resistance of the diode.
Calculation:
Given that
Rms value of supply voltage V = 50 V
The internal resistance of diode r = 20 Ω
The load resistance R_{L} = 980 Ω
Maximum voltage on the secondary side Vm = √2 V = √2 × 50 = 70.7 V
Average or DC output voltage V0 = (2 × 70.7) / π = 45 V
Average or mean load current is
I_{0} = V_{0} / (R_{L} + r) = 45 /(980 + 20) = 45 mA
In a twodiode full wave rectifier, with a load current requirement of 4.2 A, what should be the current ratings of the diodes used?
Concept:
Center tapped full wave rectifier:
Analysis:
The DC output voltage or average output voltage can be calculated as follows,
V_{0 }= 2V_{m} / π
Now we can calculate the average or mean current of load by dividing the average load voltage by load resistance RL. Therefore mean load current is given by
I_{0} = V_{0} / RL
Current through each diode will be half of the load current, as each diode conducts only for the half cycle.
Calculation:
Load current = I_{0} = 4.2 A
Current through each diode = I_{0} / 2 = 4.2 / 2 = 2.1 A
∴The current ratings of the each diode is 2.1 A.
A singlephase midpoint fullwave SCR converter with maximum midpoint voltage of V_{m} volts develops an average output voltage across a resistive load at firing delay angles of 0 and π/2 rad., respectively, as
A singlephase midpoint fullwave SCR converter:
In this circuit, two thyristors are used and are connected as shown below
Where,
V_{s }is the source voltage
V_{m} is the maximum voltage of the midpoint
Let α be the firing angle
V_{0} is the output voltage
When the thyristors are triggered with a firing angle 0 and π/2
Then the output waveform will be as follows
Let the average output voltage of the circuit be V_{a}
The value of source voltage is V_{m} sin ωt
Calculation:
Given,
α_{1} = 0, α_{2} = π/2
For the first thyristor, the average output voltage is
V_{a} = 2V_{m} / π
For the second thyristor, the average output voltage is
V_{a} = V_{m} / π
In a 12 phase full wave rectifier if source frequency is 30 Hz, then the ripple frequency will be_____Hz
We know that
Ripple frequency = 2nf = 2 × 12 × 30 = 720 Hz
A singlephase AC supply is connected to a full wave rectifier through a transformer. The rectifier is required to supply an average DC output voltage of V_{dc} = 400 V to a resistive load of R = 10 Ω. The kVA rating of the transformer is __________ (in kVA)
Concept:
In a full wave rectifier though a transformer,
Average output voltage, V_{0} = 2V_{m}/π
Average output current, I_{0} = V_{0}/R
RMS value of output voltage V_{or} = V_{s}
RMS value of load current, I_{or} = V_{s}/R
Average value of diode current, I_{d} = I_{m}/2
RMS value of diode current, I_{dr} = I_{or}
Peak value of diode current, I_{dm} = √2 I_{or}
Power delivered to the load = V_{or} I_{or}
Input voltamperes = V_{s} I_{or}
Calculation:
Given that, DC output voltage (V_{DC}) = 400 V
Average output voltage of rectifier (V_{0}) = 400 V
⇒ V_{s} = 444.28 V
Load resistance (R) = 10 Ω
RMS value of load current, I_{or} = 444.2810 = 44.428A
kVA rating of transformer = 444.28 × 44.428 = 19.738 kVA
A singlephase fullwave controlled rectifier, operating at 120 V rms and 60 Hz ac supply, has a firing angle of 60°. The average value of its output voltage is
Concept:
The average output voltage of the singlephase fullwave rectifier with a resistive load is given by
V_{0} = 2V_{m}/π cos α
Where,
V_{m} = peak value of source voltage
α = firing angle
Calculation:
Given that:
Source voltage Vs = 120 V rms
firing angle α = 60^{0}
From the source voltage peak value V_{m} = √2 × 120 V
The average output value of a singlephase fullwave rectifier is given by
A 1ϕ full wave converter with RLE load wit R = 10Ω, L = 8mH and E = 150V. The AC voltage source is 230V, 50Hz for continuous conduction. Find the average value of load current for firing angle delay of 120°.
Concept:
The average output voltage of a 1ϕ full wave converter is given by:
Case 1: α ≤ 90°
V_{o(avg)} = I_{o}R + E
This is a motoring mode of operation of a rectifier.
Case 2: α ≥ 90°
V_{o(avg)} = I_{o}R  E
This is a generating mode of operation of a rectifier.
In generating mode, the polarity of battery voltage gets reversed.
Calculation:
Given, α = 120°
I_{o} = 4.64 A
An AC voltage of maximum value equal to 100V is applied to a singlephase fully controlled bridge circuit. The peak inverse voltage rating of each SCR used will be ______.
Concept:
Singlephase fully controlled bridge circuit:
Calculation:
Given that
Vm = 100 V (AC signal)
PIV of bridge Rectifier = PIV of SCR T_{1} and T_{3} = Vm = 100 V
Similarly, PIV of SCR T_{2 }and T_{2} is Vm = 100 V.
i.e. PIV of each SCR T_{1}, T_{2}, T_{3}, T_{4} is Vm = 100 V
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