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How many thyristors are there in a threephase full wave controller?
ThreePhase Full Converters:
If all diode is replaced by thyristor, a threephase full converter bridge formed as shown below,
Waveform:
Voltage and Current Equation:
The average output voltage (V_{0}) is given by,
Where V_{m} is the maximum value of line voltage.
Voltage and current waveform for α = 30° and for constant current can be drawn as,
From waveform RMS value of Thyristor current (I_{T} = i_{T1}) will be,
Where I0 is constant DC load current.
In a single Phase full converter with resistive load and firing angle α. The load current is nonzero and zero, respectively for
Concept:
Phasecontrolled converters:
Single phase full converter with resistive load:
Working:
Current flowing through a resistive load will be
I = V_{o}/R
We can see from the above graph that the current will be zero from '0 to α' and the current will be nonzero from α to π.
So, the current will be nonzero and zero from 'π  α' and 'α' respectively.
3 phase, 6 pulse converter shown in the figure, the load is taking ripple free constant current of 10 A. The average output voltage of the converter is 150 V at a firing angle α = 30º. Find the value of line resistance R in ohms. (Up to two decimal places)
Concept:
The average output voltage of 3 phase 6 pulse converter is given by
here V_{ml} is the maximum value of linetoline voltage
The voltage drop in the resistance is given as
2RI0, here 2 resistance will come in to picture in every 60º conduction of two phases
so this voltage drop is subtracted from the final output of this converter.
Solution:
The average output voltage is equal to
R = 4.19 Ω
In a dual converter, if the firing angle of one bridge is 30°, then the firing angle of second bridge is ______.
Concept:
Modes of operation of Dual converter:
Noncirculating current mode:
Circulating current Mode:
Application of dual converter:
Calculation:
Given that, firing angle of one bridge is 30° (i.e. α_{1} = 30°)
We know that firing angles can never be greater than 180°.
i.e. α_{1} + α_{2} = 180°
∴ 30° + α_{2} = 180°
α_{2} = 180°  30°
α_{2} = 150°
A delayed fullwave rectified sinusoidal current has an average value equal to half its maximum value. Find the delay angle θ.
Concept:
The average output voltage of a fullwave controlled rectifier with R load is given by:
Where V_{m} is the maximum value of supply voltage
α is the firing angle or delay angle
Average load current:
Calculation:
Given that, the average value of the average load current is equal to half of its maximum value.
A delayed fullwave rectified sinusoidal current has an average value equal to one – third its maximum value. Find the delay angle.
Concept:
Considered Vm is the maximum value of AC input voltage of converter and V0 is the average output voltage converter and α is delay angle.
For singlephase semi converter or delayed fullwave rectifier,
Calculation:
Given that,
V_{0} = V_{m}/3
Hence, the equation becomes,
or,
π/3 = (1 + cosα)
cos α = 0.047
α = cos^{1}(0.047)
Distortion factor (DF) and total harmonic distortion (THD) are related by.
Total harmonic distortion:
Where,
V_{or} = Total rms value
V_{01} = Fundamental rms value
Distortion factor (DF): It is the ratio of the fundamental component to r.m.s value of the waveform.
Distortion factor (DF) and total harmonic distortion (THD) are related as
A freewheeling diode in a phasecontrolled rectifier:
Freewheeling diode:
A singlephase, fullbridge diode rectifier fed from a 230 V, 50 Hz sinusoidal source supplies a series combination of finite resistance, R, and a very large inductance, L. The two most dominant frequency components in the source current are:
Concept:
Fourier series of source current in a full bridge rectifier is given by
For n = 1, 3, 5, ......
The most dominant frequency components in the above expression are f, 3f
Application:
A singlephase, fullbridge diode rectifier fed from a 230 V, 50 Hz sinusoidal source
Dominant frequencies = f, 3f
= 50 and 150 Hz
A sixpulse thyristor bridge rectifier is connected to a balanced threephase, 50 Hz AC source. Assuming that the DC output current of the rectifier is constant, the lowest harmonic component in the AC input current is
In a sixpulse thyristor bridge rectifier, the harmonics present are = 6 k ± 1
So, the harmonics are = 5, 7, 11, 13, ...
lowest harmonic component = 5th harmonic supply frequency = 50 Hz
5th harmonic frequency = 5f = 250 Hz
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