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Mathematical Analysis
Analysis of this three-phase controlled rectifier is in many
ways similar to the analysis of single-phase bridge rectifier
circuit. We are interested in output voltage and the source current.
The average output voltage, the rms output voltage, the ripple
content in output voltage, the total rms line current, the fundamental
rms current, THD in line current, the displacement power factor
and the apparent power factor are to be determined. In this section,
the analysis is carried out assuming that the load current is
a steady dc value.
AVERAGE OUTPUT VOLTAGE
Before getting an expression for the output voltage, it is preferable
to find out how the output voltage waveform varies as the firing
angle is varied. In one cycle of source voltage, six pairs conduct,
each pair for 60o. This means that the period for output
waveform is one-sixth of the period of line voltage. The output
waveform repeats itself six times in one cycle of input voltage.
The waveform of output voltage can be determined by considering
one pair. It is seen that when vR(q)
= E* Sin (q), SCR S1 and S6 conduct
when q varies from 30o +
a to 90o + a
, where a is the firing angle. Then
The waveform of output can be plotted for different firing angles.
The applet below takes in the firing angle as an input and plots
the output. The peak line-to-line voltage is marked as 'U' and
the applet starts with the instant an SCR is fired and displays
the output waveform for one input cycle period.
The average output voltage of the bridge circuit is calculated
as follows, with a change in variable, where q
= a + 60o.
In the expression above, U is the peak line-to-line voltage,
whereas E is the amplitude of phase voltage of 3-phase supply.
RMS OUTPUT VOLTAGE
The rms output voltage is calculated as follows:
The ripple factor of the output voltage is then:
The applet below displays the average output voltage, the rms
output voltage and the ripple factor for the case of continuous
conduction through the load.
It is seen that the average output voltage is negative when firing
angle exceeds 90o. It means that power flow is from
the dc side to the ac source. When the firing angle is kept in
the region 0o < a <
90o, this circuit is said to be operating in the rectifier
region. When the firing angle is kept in the region 90o
< a < 180o, this circuit
is said to be operating in the inverter region.
When the circuit operates in the rectifier region, the net power
flow is from the ac source to the dc link. In the inverter region,
the net power flow is in the reverse direction. To operate in
the inverter region, it is necessary to have a dc source present
in the dc link which can provide the power that is fed back to
the ac source.
RMS LINE CURRENT
The rms line current is relatively easy to find out if the dc
current is ripple-free and steady. The load current is ripple-free
if the inductance in the dc link is relatively large. To maintain
load current at any firing angle, it is necessary that the dc
link should contain a voltage source. Given that the resistance
of the load circuit is zero, the voltage source should equal the
average output voltage of the bridge circuit. The waveforms shown
below are based on the assumption that these conditions are met.
It has been shown that if vR(q)
= E*Sin (q), SCR S1 conducts
when q varies from a
+ 30o to a + 90o
and that SCR S4 conducts when q
varies from a + 210o to
a + 270o. If the amplitude
of dc load current is assigned to be unity, the line current waveform
is then a rectangular pulse, remaining at + 1 from a
+ 30o to a + 150o,
at - 1 from a + 210o to
a + 330o, and zero elsewhere.
The amplitude of the fundamental in line current is then 3.464/p
( which evaluates to
nearly 0.78) and the amplitude of other odd harmonics is
3.464/np,
where n is the odd harmonic number. When the dc load current is
steady and has a magnitude of unity, the rms line current is obtained
as shown in equation (5). The rms value of the fundamental
is obtained as shown in equation (6). Equation (6) is based on
how trigonometric Fourier coefficients are defined for waveforms
with quarter-wave symmetry. When the line current is a rectangular
and symmetric, the phase current is the same as the line current
and the fundamental component of the phase current lags the phase
voltage by an angle equal to the firing angle. Hence the displacement
power factor is expressed as shown by equation (7). Since the
line current is not sinusoidal, the apparent power factor, usually
referred to just as the power factor in most of the texts, is
less than DPF and is represented by equation (8). Since the line
current is not sinusoidal, the distortion component in the line
current has to be computed. This component, called the THD( Total
Harmonic Distortion ), is calculated as shown in equation (9).
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