Three-Phase, Half-Wave Controlled Bridge Rectifier Circuits
5.2 RESISTIVE LOAD WITH IDEAL SUPPLY AND SHUNT CAPACITOR COMPENSATION
5.2.1 Worked Example
Example 5.4 A resistive load R⳱ 100 ⍀ is supplied from an ideal three-phase 50-Hz supply via a three-three-phase, half-wave controlled bridge rectifier. Equal capacitors C are connected across each phase of the supply. Calculate the value of capacitance C that would result in maximum power factor at␣ ⳱ 30. What is the percentage improvement of power factor then realized by capacitance compensa-tion?
At␣ ⳱ 30, the optimum value of the compensation capacitance is given by Eq. (5.26)
The new value of the rms supply current is given by Eq. (5.27)
Isamin / /
In the absence of the capacitors only the first term of Eq. (5.27) is valid, so that Isa =240 =
100 0 402. 1 522. A
The ratio of compensated to uncompensated power factor is the inverse ratio of the respective rms currents
PF PF
I I
c s
s a
a
= = =
min
.
. .
1 522 1 495 1 018
The improvement of power factor at␣ ⳱ 30 is therefore 1.8%
5.3 HIGHLY INDUCTIVE LOAD AND IDEAL SUPPLY
A three-phase, half-wave controlled bridge rectifier with a highly inductive load is shown in Fig. 5.5. The purpose of the load inductor is to smooth the load current to, as nearly as possible, an ideal direct current with no ripple at all.
The supply voltages defined by Eqs. (5.1)–(5.3) are still valid. Circuit power dissipation is presumed to take place only in the load resistor.
If the load inductance is sufficiently high the load current becomes very smooth as shown inFig. 5.6c–g. The instantaneous load voltage retains the same form as for resistive load (Fig. 5.2)until␣ ⳱ 30. For ␣ ⳱ 30, the load voltage now goes negative for part of its cycle as in, for example, Fig. 5.6f. At␣ ⳱ 90 ⳱
FIG.5 Three-phase, half-wave controlled rectifier with highly inductive load.
FIG.6 Waveforms of three-phase, half-wave controlled rectifier with highly inductive load: (a) supply phase voltages, (b) load voltage (␣ ⳱ 0), (c) load current (␣ ⳱ 0), (d) load voltage (␣ ⳱ 30), (e) load current (␣ ⳱ 30), (f) load voltage (␣ ⳱ 60), (g) load current (␣ ⳱ 60), and (h) load current (␣ ⳱ 90).
/2, the negative and positive alternations (not shown) are equal so that the average load voltage and current become zero (Fig. 5.6h).
The load voltage eL(t) in Fig. 5.6 is now described by Eq. (5.4) for both ␣
⬍ 30 and ␣ ⬎ 30. Unlike the case with resistive load (Fig. 5.2),only one mode of operation now occurs and this is equal to the mode 0 ␣ /6 that occurred with resistive load. The average load voltage with highly inductive load is therefore given by Eq. (5.6), repeated below.
For 0 ␣ /2,
The average load voltage is given inFig. 5.4and is seen to be less than the corre-sponding value with resistive load for all values of␣ in the working range. The average load current is, once again, given by
I E
With highly inductive load the smooth load current waveform satisfies (very nearly) the current relationship
iL(ω =t) Iav =IL =Im (5.37)
The load current ripple factor is therefore zero.
The supply current now consists of rectangular pulses of conduction angle 2/3. Each phase carries identical pulses with a phase difference of 2/3 or 120
from the other two phases. In phase a,Fig. 5.7,the supply current, in the first supply voltage cycle, is given by
ia( t) Iav The rms value of this supply current is therefore
I i t d t Combining Eqs. (5.36) and (5.39) gives
I E
a R
= av0
3 cosα
(5.40)
FIG.7 Supply currents of three-phase, half-wave controlled rectifier with highly inductive load
The average value of the supply current is seen, by inspection ofFig. 5.6,to be one-third of the average load current. From Eq. (5.36), therefore,
I I E
a av R
av
av =1 =
3 3
0 cosα
(5.41) The total load power dissipation may be obtained from Eqs. (5.36) and (5.37),
P I R E
L L R
= 2 = av 2
0 2
cos α (5.42)
It may be assumed that the power is transferred from the supply to the load equally by the three phases, so that
P E
R P P
a av
b c
= 0 = =
2 2
3 cos α (5.43)
The power factor may be obtained by the substitution of Eqs. (5.40) and (5.43) into Eq. (5.19)
PF P
When␣ ⳱ 0, Eq. (5.44) reduces to Eq. (4.25) which was derived for the uncon-trolled (diode), half-wave bridge. Some properties of the three-phase, half-wave bridge circuit with highly inductive load are given inTable 5.2.
If the bridge is compensated by equal capacitors at the supply point in the manner ofFig. 5.4,the supply current in phase a becomes
i t E The substitution of Eq. (5.45) into the defining integral gives an expression for the rms current
It is seen that (5.46) reduces to (5.39) when the bridge is uncompensated, and, in effect, Xc⳱ ⴥ. The rms supply current may be expressed in terms of the rms thyris-tor current Iav/兹3 by combining Eqs. (5.39) and (5.46).
I E The terminal voltage and load power are not affected by the connection of shunt capacitance at the supply. The power factor is therefore improved, by reduction of the rms supply current, if
3 2
TABLE5.2 Some Properties of the Three-Phase, Half-Wave Controlled Bridge Rectifier with Highly Inductive Load and Ideal Supply (0 90°)
Circuit property Expression
Instantaneous supply current
RMS supply current (Is)
Average load current
Average load power
Power factor
RMS load current (IL)
a1
b1
Displacement factor (cos 1) Distortion factor
Rearranging Eq. (5.48) gives as the required criterion
X E The minimum value of rms supply current obtainable by capacitance compensation may be obtained, as before, by differentiating Eq. (5.47) with respect to C and equat-ing to zero. The appropriate optimum values of Xcand C then prove to be
X E
With this value of capacitance, the value of the rms supply current minimum value is
Isamin =Ia 1− 9 cos 2 2 22
π α (5.51)
Both the power and the supply phase voltage are unaffected by the connection of good quality capacitors across the terminals. The power factor with optimum capac-itor compensation PFcis therefore obtained by substituting Eq. (5.51) into Eq.
(5.44).
5.3.1 Worked Examples
Example 5.5 A three-phase, half-wave controlled rectifier supplies a highly inductive load from an ideal three-phase supply. Derive an expression for the aver-age load voltaver-age and compare this with the corresponding case for resistive load at
␣ ⳱ 60.
The circuit diagram is shown inFig. 5.5.AlthoughL ⬎⬎ R at supply fre-quency, the inductor offers negligible impedance at the load frequency (ideally zero). Resistor R may well represent the only load device that is being powered.
The load voltage eL(t) shown inFig. 5.6is a continuous function for all values of␣ and is seen to be described by Eq. (5.4), which has an average value
Eav =3 3Em =Eav
2π cosα 0cosα (Ex. 5.5a)
This is precisely the same expression that is pertinent to resistive loads in the mode when 0 ␣ /6. For resistive loads where ␣ ⬎ /6, the current becomes discon-tinuous, the modal behavior changes, and the average load voltage is represented by Eq. (5.7), reproduced below.
E E
At␣ ⳱ 60, the average voltage with inductive load, from (Ex. 5.5a), is Eav(inductive load) 1Eav
=2 0
Similarly, at␣ ⳱ 60, the average load voltage with resistive load, from (Ex. 5.5b), is
(inductive load) . resistive load)
3
= 2 = 0 866
This result is seen to be consistent with the data ofFig. 5.4.
Example 5.6 In the three-phase rectifier bridge ofFig. 5.5a three-phase, 415-V, 50-Hz, supply transfers power to a 100-⍀ load resistor. What is the effect on the power transferred at (a)␣ ⳱ 30, (b) ␣ ⳱ 60, if a large inductance is con-nected in series with the load resistor.
With resistive load the input power per phase is given by Eqs. (5.16) and (5.17). Assuming that each phase provides one-third of the load power, then PL⳱ 3Pa. At␣ ⳱ 30, from Eq. (5.16), assuming that 415 V is the line voltage,
At␣ ⳱ 60, from Eq. (5.17),
With highly inductive load the load power is given by Eq. (5.42) for all␣.
P E
The effect of the load inductor is therefore to reduce the rms load current and thereby the load power dissipation below the value obtained with the load resistor acting alone. If the same load power dissipation is required in the presence of a high induc-tor filter, then the value of the load resisinduc-tor must be reduced to permit more load current to flow.
Example 5.7 A three-phase, half-wave controlled rectifier has ideal thyris-tor elements and is fed from a lossless supply. The rectifier feeds a load consisting of a resistor in series with a large filter inductor. Sketch the waveform of the reverse voltage across a thyristor at␣ ⳱ 60 and specify its maximum value.
The circuit is shown inFig. 5.5with some typical load voltage and current waveforms given inFig. 5.6.Let the reverse voltage on thyristor Thabe designated eTaas shown in Fig. 5.5. When thyristor Thcis conducting, for example, the instanta-neous value of this reverse voltage can be written
−eT t =ecN−eaN
a(ω ) (Ex. 5.7a)
From Eqs. (5.1) and (5.3) it is seen that
Similarly, if thyristor Thbis conducting, the instantaneous reverse voltage across Thais But it can be seen from the three-phase voltage waves ofFig. 5.8that
3 7
The reverse voltageⳮeTacan therefore be defined in terms of the line-to-line supply voltages, which is obvious from the basic expressions in Eqs. (Ex. 5.7a) and (Ex.
5.7b).
The waveform of the reverse blocking voltage at thyristor Tha, for␣ ⳱ 60, is given in Fig. 5.8. Its peak value is clearly equal to兹3Em, being the peak value of the line-to-line supply voltage.
Example 5.8 Equal capacitors C are to be used across the supply to neutral terminals to compensate the power factor of a three-phase, half-wave, thyristor con-trolled bridge rectifier with a highly inductive load in which the load resistor R⳱ 100⍀. Calculate the value of capacitance that will give maximum power factor and the degree of power factor improvement at (1)␣ ⳱ 30 and (2) ␣ ⳱ 60 if the supply voltage is 240 V/phase at 50 Hz.
FIG.8 Voltage waveforms of three-phase, half-wave controlled rectifier with highly induc-tive load,␣ ⳱ 60: (a) load voltage and (b) voltage across thyristor Tha.
The appropriate circuit diagram is given inFig. 5.3.For both of the specified values of thyristor firing angle the optimum capacitance is given by Eq. (5.50). The rms supply current Iain the absence of capacitance is, from Eq. (5.40),
I E
R E
R
a av
m
=
=
0
3 3 2
cos
cos α
π α
At␣ ⳱ 30,
Ia =3 2 240
( )
=2 100 3 2 1 4
π . A
At␣ ⳱ 60,
Ia =
( )
Substituting into Eq. (5.50) gives at␣ ⳱ 30
The supply current reduction is given in Eq. (5.51).
I
The connection of the capacitors makes no difference to the transfer of power or to the supply voltages. The power factor is therefore improved by the inverse ratio of the supply current reduction. At␣ ⳱ 30,
PF
The actual values of power factor after compensation may be obtained from Eq.
(5.52). At␣ ⳱ 30, PFc⳱ 0.621. At ␣ ⳱ 60, PFc⳱ 0.36.