Three-Phase, Half-Wave, Uncontrolled (Diode) Bridge
4.4 HIGHLY INDUCTIVE LOAD IN THE PRESENCE OF SUPPLY IMPEDANCE
An electrical power supply is not usually a perfect voltage source because it contains series impedance. The action of drawing current from the supply into a resistive or inductive load causes the supply voltage at the terminals to reduce
below its no-load value. In public electricity supply undertakings, the generator voltage level is usually automatically boosted to provide constant voltage at a consumer’s terminals when load current is drawn from the supply. The series impedance of an electricity supply system is usually resistive-inductive, being created by transformers, cables, and transmission supply lines. In transformer-supplied bridge circuits the supply inductance is mainly the transformer leakage inductance.
The magnitude of the supply inductance is typically such that not more than about 5% reduction would occur in an unregulated supply voltage at full-load current. Because of this inductance the instantaneous commutation of current from one diode to another that occurs in resistive circuits, described in Sec. 4.1.1, for example, cannot occur. When switching closure occurs in an open inductive circuit a definite time is required for a current to build up from zero to its final steady-state value. The instantaneous transitions in the value of the supply currents inFig. 4.2,for example, no longer take place.
Consider operation of the half-wave diode bridge (Fig. 4.7). The balanced sinusoidal voltages of the generator are given by eAN, eBN, and eCN, where
eAN⳱ Emsint (4.26)
eBN⳱ EMsin (t ⳮ 120) (4.27)
eCN⳱ Emsin (t ⳮ 240) (4.28)
FIG.7 Three-phase, half-wave diode rectifier with highly inductive load inductance.
When each supply line contains an effective series inductance Ls;Fig. 4.7, the bridge terminal voltages eAN; eBN, and eCNdo not remain sinusoidal on load but are given by
e e L di
aN AN s dt
= − a (4.29)
e e L di
bN BN s dt
= − b (4.30)
e e L di
cN CN s dt
= − c (4.31)
Compared with operation with an ideal supply, as inFig. 4.5a, the waveforms of both the terminal voltage and current and the load voltage are affected by the presence of supply reactance. Now the magnitude of the supply inductance Lsis usually small compared with the value of the load inductance L. The presence of supply inductance therefore does not significantly affect the magnitude Iavof the load current nor the maximum value Iavand average value Iav/3 of the supply currents in Fig. 4.5. The magnitude of the load current at fixed supply voltage is determined almost entirely by the value of the load resistance because the load inductor offers no net impedance to the (hopefully predominant) direct current component. It is also of interest that the waveform of the supply currents, at fixed supply voltage, varies according to the level of the load current and the value of the supply inductance. Several different modes of supply current behavior are identifiable, depending on the particular application.
Figure 4.8 gives some detail of the waveforms due to operation of the circuit of Fig. 4.7. Diode Da carries the rectified current of peak magnitude Iav
up to a point x. Since the anode voltages eAN and eBN of Da and Db are then equal, diode Db starts to conduct current iL. Due to the effect of the supply inductance, the current ia. cannot extinguish immediately as it does with ideal supply, shown in Fig. 4.8b. Due to the supply line inductance, diodes Daand Db
then conduct simultaneously, which short circuits terminals a and b of the supply.
During this interval of simultaneous conduction, called the overlap period or commutation angle u, the common cathode has a potential (eANⳭ eBN)/2. In the overlap period xy,eBNis greater than eANand the difference voltage can be consid-ered to cause a circulating current in loop aPbNa (Fig. 4.7), which increases ib
and diminishes ia. When current iafalls below its holding value diode D switches into extinction and the load voltage then jumps to the corresponding point z on wave eBN. Each supply current has the characteristic waveform of Fig. 4.8c, but the load current is continuous and smooth at the value Iav, (Fig. 4.8e). Because the presence of supply inductance does not affect the maximum value Iavnor the
FIG.8 Waveforms for operation of a three-phase, half-wave diode rectifier with highly inductive load and supply inductance: (a)load voltage, (b) supply line current (with ideal supply), (c) and (d) supply line currents ia(t) and ib(t), and (e) load current.
average value Iav/3 of the supply current the total area under the current pulse (Fig. 4.8c)is unchanged, compared with ideal supply.
Comparison of Fig. 4.8a withFig. 4.5cshows, however, that an effect of supply reactance is to reduce the average value Eav of the load voltage. From Fig. 4.8a it is seen that, with overlap angle u,
E e e
This may also be expressed
E E u E
av av u
= 0 2 = av0 +
2 2 1
cos ( cos ) (4.33)
where Eavo is the average load voltage with zero overlap, or ideal ac supply, defined in Eq. (4.7).
In the circuit ofFig. 4.7, during the overlap created by the simultaneous conduction of diodes Daand Db, there is no current in supply line c, and
e L di
Substituting Eqs. (4.29) and (4.30) into Eq. (4.36) and integrating from 150 to 150 Ⳮ u, noting that ia⳱ Iavatt ⳱ 150, gives Combining Eqs. (4.6) and (4.37) permits u to be expressed in terms of impedance parameters, utilizing the fact that Iav⳱ Eav/R,
cosu L I Combining Eqs. (4.35) and (4.38) permits cos u to be expressed in terms of impedance parameters
Provided that the load inductance L is large, the actual value of L does not occur in the relevant circuit equations. With a good electrical supply the ratioLs/R is about 0.05 at full load and the value of u is then about 18. For a poor (i.e., relatively high inductance) supply, or with reduced load resistance such thatLs/ R⳱ 0.2, then u is about 34. A value u ⳱ 18 results in a reduction of Eavof less than 3%, while u ⳱ 34 results in about 9% reduction. The reduction of average load voltage can be expressed in terms of impedance parameters by combining Eqs. (4.33) and (4.39).
E E
The supply inductance is found to modify the previously appropriate expression Eq. (4.22), for rms supply current to
I I
Function(u) varies almost linearly with u for values up to u ⳱ 60. At u ⳱ 34, for example, the effect of supply reactance is found to reduce Iaby about 3.5%.
The effect of gradually increased overlap, with fixed supply voltage, is demonstrated sequentially inFigs. 4.9–4.15.Note that for values of supply induct-ance such that u⬎ 90,Figs. 4.13–4.15, the load resistance is here modified to give the same peak value of supply current. For values of u⬍ 90 the performance is usually described as mode I operation. With u⬎ 90 the load phase voltages become discontinuous and the performance is referred to as mode II operation.
It is seen that the conduction angle of the supply currents progressively increases with overlap.
The boundary between mode I operation and mode II operation occurs at u ⳱ 90. Under that condition cosu is zero and it is seen from equation (4.37) that
But fromFig. 4.7,the right hand side of Eq. (4.43) is seen to be the peak value of the short-circuit current Iscin (say) loop APBN.
Therefore,
Combining Eqs. (4.37), (4.43), and (4.44) results in I
Iav u
ˆsc = −1 cos
(4.45) The short circuit current can be expressed in terms of the average load voltage by eliminating cos u between Eqs. (4.38), (4.40), and (4.45).
E
For u⬎ 90, which occurs in mode II operation, Iav⬎ Iˆscand the average load voltage, Eavbecomes less than one half of the value Eavowith ideal supply.
All the energy dissipation in the circuit of Fig. 4.7 is presumed to occur in the load resistor R. The power rating of the circuit in mode I is given in terms of the constant value Iavof the load current.
P⳱ I2avR (4.47)
In mode I the supply current has the rms value denoted in Eq. (4.41). The supply voltage eaN(t) is seen fromFigs. 4.10–12to be given by
FIG.9 Waveforms for three-phase, half-wave diode bridge with highly inductive load.
Ideal supply u⳱ 0.
FIG.10 Waveforms for three-phase, half-wave diode bridge with highly inductive load.
Mode I, ⳱ 30, ideal supply.
FIG.11 Waveforms for three-phase, half-wave diode bridge with highly inductive load.
Mode I, ⳱ 60, ideal supply.
FIG.12 Waveforms for three-phase, half-wave, diode bridge with highly inductive load.
Limit mode I, ⳱ 90, ideal supply.
FIG.13 Waveforms for three-phase, half-wave diode bridge with highly inductive load.
Mode II, ⳱ 105, ideal supply.
FIG.14 Waveforms for three-phase, half-wave diode bridge with highly inductive load.
Mode II, ⳱ 120, ideal supply.
FIG.15 Waveforms for three-phase, half-wave diode bridge with highly inductive load.
⳱ 150, ideal supply.
which has the rms value EaN, where EaN2 eaN2 d t
0
1
=2π
∫
2π ω (4.49)The substitution of Eq. (4.48) into Eq. (4.49) gives
E E u The power factor of the three-phase, half-wave bridge is given by
PF P
E IaN a
=3 (4.51)
Supply reactance causes both the rms voltage EaNand rms current Iato be reduced below their respective levels with ideal supply. The power factor is therefore increased.
4.4.1 Worked Examples
Example 4.7 A three-phase, half-wave uncontrolled bridge circuit trans-fers energy from a three-phase supply to a highly inductive load consisting of a resistor R in series with inductor L. Each supply line may be considered to have a series inductance Ls. Show that the average load voltage is given by
E L I
where Emis the peak phase voltage.
The circuit diagram is shown inFig. 4.7.From Eq. (4.33)
E E Substituting cos u from Eq. (4.37) gives
E E L I
The final form of Eav above shows that this incorporates the peak line-to-line voltage兹3 Em1ess the line voltage drop due to the supply reactance.
Example 4.8 A three-phase, half-wave diode bridge supplies power to a load consisting of resistor R and series inductor L. Each phase of the supply has a series inductance Lswhere Ls⬍⬍ L. Sketch waveforms of the per-phase voltage and current of the supply for mode I operation when u⳱ 30. Derive an expression for the instantaneous supply current ia(t) for the overlap period 150 ⬍ t ⬍ 150 Ⳮ u.
Waveforms of eaN(t) and ia(t) for u ⳱ 30 are given inFigs. 4.8and 4.10. Instantaneous current ia(t) is defined by Eq. (4.36),
e e L di
Integrating both sides of the differential equation gives
i t E
where K is a constant of integration.
Now (1) att ⳱ 150, ia⳱ Iav; (2) att ⳱ 150 Ⳮ u,ia⳱ 0.
Under condition (1),
K I E
which is negative because Iˆsc⬎ Iavfor mode I operation.
Under condition (2),
K E
Equating these two values of K between the two consistent conditions shows that
I E
which is seen to be consistent with Eq. (4.45)
Using the value of K from condition (2) in the equation for ia gives
i E
Example 4.9 A three-phase, half-wave diode rectifier has a load resistance R ⳱ 10 ⍀ in series with a large inductor. Each supply line contains a series inductance Ls such that the inductive reactance in the line is 10% of the load resistance. The generator three-phase voltages have an rms line value of 400 V.
Calculate the power factor of operation and compare this with the case of ideal supply.
The equivalent circuit is given inFig. 4.7.The first step of the solution is to calculate the overlap angle u. Since
ωL R
s = 0 1. then from Eq. (4.39),
cos u⳱ 0.9086
Therefore, u⳱ 24.7 ⳱ 0.431 rad, which is mode I operation. Function
(u), Eq. (4.42), is found to have the value
ψ( ) ( . )( . π) ( . )( . )
The average load voltage in the presence of L is given by Eq. (4.32) in which Emis the peak phase voltage. In the present case,
Em =400 2 = an ideal supply.
The average load current is unchanged by the presence of the supply induct-ance.
I E
av R
= av0 =270= 10 27 A
From Eq. (4.42) the rms supply current is
I I power into the bridge circuit is presumed to be dissipated entirely in the load resistor:
P⳱ I2avR⳱ (27)210⳱ 7290 W
The power factor, seen from the supply terminals, is given by Eq. (4.51), which incorporates the rms value Eanof the terminal voltage. For a value u⳱ 24.7
the waveform eaN(t) is very similar to that given inFig. 4.10aand is defined by Eq. (4.48). Inspection of the more detailed diagram (Fig. 4.8a)shows that
e t E
This compares with the value EL⳱ 0.707 Emfor sinusoidal supply.
The power factor is therefore
PF P
This value is about 7% higher than the value 0.676 obtained with ideal supply.