Three-Phase, Half-Wave Controlled Bridge Rectifier Circuits
5.5 SERIES R-L LOAD PLUS A CONSTANT EMF WITH IDEAL SUPPLY
increase of load voltage
due to bypass diode =1+
(
+ 6)
3
cos /
cos α π
α (5.59)
At␣ ⳱ 60, for example, the ratio in Eq. (5.59) has the value 1.155. In other words, the effect of the bypass diode is to increase the average load voltage by 15.5%.
5.5 SERIES R-L LOAD PLUS A CONSTANT EMF WITH IDEAL SUPPLY
The most general form of burden on a three-phase, half-wave controlled bridge rectifier is shown inFig. 5.11.In addition to a series R-L impedance of arbitrary phase angle, the load current is opposed by an emf E, which is constant or which changes in a known manner, This might arise, for example, in battery charging or in the speed control of a separately excited dc motor. With the polarity shown in Fig. 5.11 the average load current is reduced because emf E is in opposition to the mean driving voltage Eavcreated by rectification:
I E E
R
E E
av R
av av
= − = −
0 cosα
(5.60)
FIG.11 Three-phase, half-wave controlled rectifier with series R-L load, incorporating a load-side emf.
Further accounts of the applications of bridge rectifiers to dc motor control are given in Refs. 10, 11, 15, 20, 25, and 27.
5.6 HIGHLY INDUCTIVE LOAD IN THE PRESENCE OF SUPPLY IMPEDANCE
Consider that a series inductance Lsnow exists in each phase of the supply. The terminal voltages eaN, ebN, and ecNat the bridge terminals (Fig. 5.12)are not sinusoi-dal while the bridge draws current from the supply but are given by Eqs.
(4.26)–(4.28). The supply voltages at the generation point are still given by Eqs.
(4.29)–(4.31).
In a controlled rectifier the effect of delayed triggering is to delay the start of conduction in the conductor containing the switch. Once conduction has begun the supply reactance delays the buildup of current for a period known as the overlap period. Much of the discussion of Sec. 4.4, relating to the three-phase, half-wave diode bridge rectifier is also relevant here. The effects of delayed triggering plus overlap are illustrated inFig. 5.13.At a firing angle␣ ⳱ 30 conduction com-mences. The phase current does not reach its final value Eav/R until the overlap period is completed, although the load current is smooth and continuous. The
FIG.12 Three-phase, half-wave controlled rectifier circuit, including supply inductance.
effect of overlap removes a portion (shown dotted) from the bridge terminal volt-ages that would be present with ideal supply. As a result, the area under the curve of the load voltage is reduced and so is its average value.
InFig. 5.13the average load voltage is given by
E e d t e e
When␣ ⳱ 0, Eq. (5.61) reduces to Eq. (4.33), which was derived for diode opera-tion. Since the time average value of the voltage on the load inductor is zero, then
I E
av R
= av (5.62)
If it is required to supply the load with constant current, then since Eavvaries with
␣and, resistor Rmostbe variedatthesame rate.Duringthefirstoverlap intervalin Fig. 5.13, thyristors Thaand Thcare conducting simultaneously. A complete circuit therefore exists such that, starting from point N in Fig. 5.12,
FIG.13 Waveforms of three-phase, half-wave controlled rectifier with highly inductive load, in the presence of supply inductance,␣ ⳱ 30, ⳱ 20: (a) load voltage, (b) supply current ia(t), (c) supply current ib(t), and (d) load current.
−e +L di − + =
dt L di dt e
AN s
a s
c
CN 0 (5.63)
Since thyristor Thbis in extinction, current ibis zero and
ia+ =ic Iav (5.64)
Eliminating icfrom Eqs. (5.63) and (5.64) gives
e e L di dt e
AN CN s
a
− =2 = AC (5.65)
Note that this voltage eAC(t), which falls across two line reactances in series, is not the same as the corresponding instantaneous load voltage (eANⳭ eCN)/2 (Fig.
5.13),which is also valid during the same period of overlap.
The instantaneous time variation of current ia(t) in the overlap period ␣ Ⳮ 30 t ␣ Ⳮ 30 Ⳮ may be obtained from Eq. (5.65). From condition (2)
K E The instantaneous time variation ia(t) may be written by combining Eqs. (5.66) and (5.67). The two expressions Eqs. (5.67) and (5.68) for constant of integration K may be equated to give
I E
av L
m s
= 2ω3
[
cosα−cos(α µ+ )]
(5.70)Note that the sign in the bracketed term of Eq. (5.70) is different from the corre-sponding sign in the similar expression Eq. (5.61) for the average load voltage.
Equations (5.61) and (5.70) may be combined to give
E E L
The first term of Eq. (5.71) is seen to be equal to Eq. (5.35) and represents the aver-age load voltaver-age with ideal supply. The second term in Eq. (5.71) represents the effect of voltage drop in the supply inductances.
Since Iav⳱ Eav/R, Eq. (5.71) can be rearranged to give
A discussion on the effects of delayed firing and of overlap on bridge opera-tion is given at the end of Chapter 7.
5.6.1 Worked Examples
Example 5.9 A three-phase, half-wave controlled rectifier contains induct-ance Lsin each supply line. The rectifier supplies a highly inductive load with a load current of average value Iav. Devise an equivalent circuit to show the effect on the average load voltage Eavof the switch firing angle␣ and the supply inductance.
From Eq. (5.71) it can be inferred that there is an effective open-circuit volt-age Eav0cos␣ on the load side. This constitutes a dc driving voltage that is reduced due to the effect of supply side inductance when current flows. A possible equiva-lent circuit is given inFig. 5.14,which also shows the effect of supply line resis-tance.
It is often permissible to assume that resistance Rs⳱ 0.
Example 5.10 A three-phase, half-wave rectifier uses silicon controlled rectifier switches and supplies a highly inductive load circuit in which the current is maintained constant at 50 A (by adjustment of the load resistance). The three-phase supply has a line-to-line voltage of 230 V at 50 Hz, and each supply line contains an effective series inductance Ls⳱ 1.3 mH plus a series resistance Rs⳱ 0.05. Calculate the average load voltage for firing angles ␣ ⳱ 0, 30, and 60.
FIG.14 Load-side equivalent circuit for a three-phase, half-wave controlled rectifier with supply impedance.
The average load-side voltage on open circuit is, from Eq. (5.35), Eav
0
3 3 2
230 2 3 155 3
= =
π . V
Without the effects of voltage drop in the supply lines the average load voltage would be
Eav⳱ Eav0cos␣ ⳱ 155.3 cos ␣
At a constant load current of 50 A there is a constant voltage drop 50⳯ 0.05 ⳱ 2.5 V in the supply line resistor. Also at a load current of 50 A the supply line induc-tor results in a voltage drop, Eq. (5.71), of
3 2
3 2 50 1 3 2 1000 50 9 75
ω π
π π Ls Iav
= × ×
=
.
. V
Because this calculation is performed in equivalent dc side terms, no question of phase relationship arises, and the voltage drops on the supply line resistor and induc-tor can be added algebraically. For all values of firing angle the voltage drop along a supply line is 2.5Ⳮ 9.75 ⳱ 12.25 V.
At␣ ⳱ 0,
Eav⳱ 155.3 ⳮ 12.25 ⳱ 143 V
At␣ ⳱ 30,
Eav⳱ 155.3 ⳯ 0.866 ⳮ 12.25 ⳱ 122.2 V At␣ ⳱ 0,
Eav⳱ 155.3 ⳯ 0.5 ⳮ 12.25 ⳱ 65.4 V
Example 5.11 A three-phase, half-wave controlled rectifier using SCR switches supplies power from a 415-V, 50-Hz bus with a short-circuit reactance of 0.415⍀/phase to a highly inductive load. The load current is 50 A when the load voltage is maximum. Calculate the communication time of the switches at␣ ⳱ 30.
The effect of commutation or simultaneous conduction between two phases is illustrated by the waveforms inFig. 5.13,for the case of␣ ⳱ 30. The per-phase supply reactance, (sometimes called the short-circuit reactance or commutation reactance) is given as
Xsc⳱ Ls⳱ 250Ls⳱ 0.415 ⍀
It can be assumed that the given value 415 V represents the rms line voltage.
Therefore,
兹3 Em⳱ peak line voltage ⳱ 兹2 ⳯ 415 ⳱ 587 V
Now maximum average load current occurs when␣ ⳱ 0. In Eq. (5.70)
50 587 At␣ ⳱ 0, the average load voltage is, from (5.71),
Eav= Em− × ×
Constant Load Resistance. If the load resistance is kept constant at the value R⳱ Eav/Iav⳱ 270/50 ⳱ 5.4 ⍀, from Eq. (5.62), then the average voltage with␣ ⳱ 30, from Eq. (5.72), is
Eav = × ×
The load current at␣ ⳱ 30 will then have fallen from Eq. (5.62), to
I E
av R
= av =234 1= 5 4. 43 35
. . A
Substituting into Eq. (5.70), 43 35 587
Constant Load Current. If the load current is kept constant at 50 A then from Eq. (5.70), at␣ ⳱ 30
With constant current of 50 A the average voltage at␣ ⳱ 30, from Eq. (5.61), is
Eav= ⋅ +
which is achieved by reducing the load resistor from 5.4⍀ to 232.8/50 ⳱ 4.660 ⍀.