Naturally commutating converters
222
During diode conduction the circuit is defined by the Kirchhoff voltage equation
di
L
+ Ri = vR + vL = v = 2 V sin ωt
(V)
(11.1)
dt
where V is the rms ac supply voltage. Solving equation (11.1) yields the load (and diode) current
11
i (ωt ) =
2V
Z
{sin
(ωt - φ ) + sin φ e-ωt / tan φ }
0 ≤ ωt ≤ β ≥ π
Naturally Commutating Converters
(A)
(11.2)
(rad)
R
The converter circuits considered in this chapter have in common an ac voltage supply input and a dc
load output. The function of the converter circuit is to convert the ac source energy into controlled dc
load power, mainly for highly inductive loads. Turn-off of converter semiconductor devices is brought
about by the ac supply voltage reversal, a process called line commutation or natural commutation.
Converter circuits employing only diodes are termed uncontrolled (or rectifiers) while the incorporation of
only thyristors results in a (fully) controlled converter. The functional difference is that the diode conducts
when forward-biased whereas the turn-on of the forward-biased thyristor can be controlled from its gate.
An uncontrolled converter provides a fixed output voltage for a given ac supply and load.
Converters employing a combination of both diodes and thyristors are generally termed half-controlled
(or semi-controlled). Both fully controlled and half-controlled converters allow an adjustable output
voltage by controlling the phase angle at which the forward biased thyristors are turned on. The polarity
of the output (load) voltage of a fully controlled converter can reverse (but the current flow direction is
not reversible), allowing power flow into the supply, a process called inversion. Thus a fully controlled
converter can be described as a bidirectional converter as it facilitates power flow in either direction.
The half-controlled converter, as well as the uncontrolled converter, contains diodes which prevent the
output voltage from going negative. Such converters only allow power flow from the ac supply to the dc
load, termed rectification, and can therefore be described as unidirectional converters.
Although all these converter types provide a dc output, they differ in characteristics such as output ripple
and mean voltage as well as efficiency and ac supply harmonics. An important converter characteristic
is that of pulse number, which is defined as the repetition rate in the direct output voltage during one
complete cycle of the input ac supply.
A useful way to judge the quality of the required dc output, is by the contribution of its superimposed
ac harmonics. The harmonic or ripple factor RF is defined by
RFv =
q =1 r =1 s =1
p=qxrxs
p=1
Φ
vL = 0 = Ldi/dt
∴ current slope = 0
i
2
V rms
−Vdc2
V2
= rms2 − 1 = FF 2 − 1
2
Vdc
Vdc
Single-phase uncontrolled converter circuits – ac rectifiers
11.1.1 Half-wave rectifier circuit with an R-L load
A simple half-wave diode rectifying circuit is shown in figure 11.1a, while various circuit electrical waveforms are shown in figure 11.1b. Load current commences when the supply voltage goes positive at ωt
= 0. It will be seen that load current flows not only during the positive half of the ac supply voltage, 0 ≤
ωt ≤ π, but also during a portion of the negative supply voltage, π ≤ ωt ≤ β. The load inductor stored
energy maintains the load current and the inductor’s terminal voltage reverses and is able to overcome
the negative supply and keep the diode forward-biased and conducting. This current continues until all
the inductor energy, ½Li2, is released (i = 0) at the current extinction angle (or cut-off angle), ωt = β.
BWW
X=
ωL
R
where FF is termed the form factor. RFv is a measure of the voltage harmonics in the output voltage
while if currents are used in the equation, RFi gives a measure of the current harmonics in the output
current. Both FF and RF are applicable to the input and output, and are fully defined in section 11.10.
The general analysis in this chapter is concerned with single and three phase ac supplies mainly
feeding inductive dc loads. A load dc back emf is used in modelling the dc machine. Generally,
uncontrolled rectifier equations can be derived from the corresponding controlled converter circuit
equations by setting the controlled delay angle α to zero. Also purely resistive load equations generally
can be derived by setting inductance L to zero in the L-R load equations and R-L load equations can be
derived from R-L-E equations by setting E, the load back emf, to zero.
11.1
Z=
√R2+ω2L2
current extinction
angle
β
π
-VR
Figure 11.1. Half-wave rectifier with an R-L load:
(a) circuit diagram and (b) waveforms, illustrating the equal area and zero current slope criteria.
Z = √(R2 + ω2 L2) (ohms)
tan φ = ω L / R = Q and R = Z cos φ
i (ωt ) = 0
(A)
(11.3)
β ≤ ωt ≤ 2π
(rad)
The current extinction angle β is determined solely by the load impedance Z and can be solved from
equation (11.2) when the current, i = 0 with ωt = β, such that β > π, that is
where
Power Electronics
223
Naturally commutating converters
sin( β - φ ) + sin φ e- β / tan φ = 0
(11.4)
This is a transcendental equation which can be solved by iterative techniques. Figure 11.2a can be used
−1
to determine the extinction angle β, given any load impedance (power factor) angle φ = tan ω L / R .
The mean value of the rectified current, the output current, I o , is given by integration of equation (11.2)
∫
Io =
1
2π
Io =
2V
2π R
β
0
i (ωt ) d ωt
(A)
(1 − cos β )
(11.5)
(A)
while the mean output voltage Vo is given by
Vo =
1
2π
∫
β
2 V sin ωt d ωt = Io R =
0
2V
2π
(1 − cos β )
(11.6)
(V)
Since the mean voltage across the load inductance is zero, Vo = I o R (see the equal area criterion below).
Figure 11.2b shows the normalised output voltage Vo / V as a function of ωL / R.
The rms output (load) voltage and current are given by
Vrms =  12π

irms =
∫ (
β
0
V cos φ
R
2V
 1

 2π
) sin ωt dωt 
2
2
½
= V  12π {β − ½ sin 2 β }
½
sin β cos ( β + φ ) 
V  1


β −
 =
cos φ
Z  2π



½
sin β cos ( β + φ ) 

β −

cos φ



½
(11.7)
224
From equations (11.6) and (11.7) the harmonic content in the output voltage is indicated by the voltage
form factor.
FFv =
V rms π {β − ½ sin 2 β }
=
Vo
1 − cos β )
½
(11.8)
For a resistive load, when β = π , the form factor reduces to a value of 1.57. The ripple factor is therefore
FFv 2 − 1 = 1.21. For a purely resistive load the voltage and current form factors are equal.
The power delivered to the load, which is the power delivered to the load resistance R, is
2
PL = i rms
R
(11.9)
The supply power factor, using the rms current in equation (11.7), is
power, PL
pf =
apparent power
(11.10)
½
sin β cos ( β + φ ) 
i 2 R i R V R rms  1 

cos
cos
= rms = rms =
=
−
=
×
β
φ
µ
φ


2π 
cos φ
i rmsV
V
V


The characteristics for an R-L-E load can be determined by using α=0 in the case of the half-wave
controlled converter in section 11.3.1iii.
For a purely inductive load, L, β = 2π is substituted into the appropriate equations. The average output
voltage tends to zero and the current is given by
i (ωt ) =
2V
ωL
{1 − cos ωt }
(A)
which has a mean current value of √2 V/ωL.
11.1.1i – Inductor equal voltage area criterion
The average output voltage Vo, given by equation (11.6), is based on the fact that the average voltage
across the load inductance, in steady state, is zero. The inductor voltage is given by
vL = L di / dt
(V)
which for the circuit in figure 11.1a can be expressed as
∫
β /ω
0
vL ( t ) dt =
∫
iβ
io
L di = L (iβ − i0 )
(11.11)
If the load current is in steady state then iβ = io , which is zero here, and in general
∫v
√2/π
= 0.45
Figure 11.2. Single-phase half-wave converter characteristics: (a) load impedance angle ø versus
current extinction angle β and (b) variation in normalised mean output voltage Vo / V versus ωL/R.
L
dt = 0
(Vs)
(11.12)
The inductor voltage waveform for the circuit in figure 11.1a is shown in the last plot in figure 11.1b. The
inductor equal voltage area criterion implies that the shaded positive area must equal the shaded
negative area, in order to satisfy equation (11.12). The net inductor energy at the end of the cycle is zero
(specifically, unchanged since i o = i β ), that is, the energy in equals the energy out. This is a useful aid
in predicting and drawing the load current waveform.
It is useful to superimpose the supply voltage v, the load voltage vo, and the resistor voltage vR
waveforms on the same time axis, ωt. The load resistor voltage, vR = Ri, is directly related to the load
current, i. The inductor voltage vL will be the difference between the load voltage and the resistor
voltage, and this bounded net area must be zero. Thus the average output voltage is Vo = I o R . The
equal voltage areas associated with the load inductance are shown shaded in two plots in figure 11.1b.
11.1.1ii - Load current zero slope criterion
The load inductance voltage polarity changes from positive to negative as energy initially transferred into
the inductor, is released. The stored energy in the inductor allows current to continue to flow after the
input ac voltage has reversed. At the instant when the inductor voltage reverses, its terminal voltage is
zero, and
vL = Ldi / dt = 0
(11.13)
that is di / dt = 0
The current slope changes from positive to negative, whence the voltage across the load resistance
ceases to increase and starts to decrease, as shown in figure 11.1b. That is, the Ri waveform crosses
the supply voltage waveform with zero slope, whence when the inductor voltage is zero, the current
begins to decrease. The fact that the resistor voltage slope is zero when vL = 0, aids prediction and
sketching of the various circuit waveforms in figure 11.1b, and subsequent waveforms in this chapter.
Power Electronics
225
Example 11.1:
Naturally commutating converters
Alternatively, as would be expected, the average diode current is
V
209.95V
= 21.0A
I D = Io = o =
10Ω
RL
The peak diode current is
∧
2V s − V o
2 × 230V − 210V
ID =
=
= 115.3A
Ri
1Ω
Half-wave rectifier – constant current
In the dc supply half-wave rectifier circuit of figure Example 11.1, the source voltage is 230√2 sin(2π 50t)
V with an internal resistance Ri = 1 ohm, RL = 10 ohms, and the filter capacitor C is very large. Calculate
i. the mean value of the load voltage, Vo
∧
ii. the diode average and peak currents, ID, I D
iii. the capacitor peak charging and discharging current
iv. the diode reverse blocking voltage, VDR
θ
Ri
D
The capacitor peak charging current is the difference between the peak diode current and the
load current, viz., 115A - 21A = 94A, while the peak discharging current is the average load
current of 21A.
iv.
The diode reverse voltage is the difference between the instantaneous supply voltage and the
output voltage 210V. This is a maximum at the negative peak of the ac supply, when the diode
voltage is √2×230V + 210V = 535.3V. During any period when the load is disrupted, the output
capacitor can charge to √2×230V, hence the diode can experience, worse case, 2×√2×230V =
650.5V.
♣
Vo
is
iC
C
Vs
iii.
√2Vs
Vs
io
ωt
o
π
Vo
RL
α
2π
β
iD
∧
ID
ID
226
ωt
11.1.2 Half-wave circuit with an R-L load and freewheel diode
Figure Example 11.1. Single-phase half-wave rectifier: (a) circuit and (b) waveforms.
Solution
i.
Because the load filter capacitor is large, it is assumed that the dc output voltage is ripple free
and constant. The capacitor provides the load current when the ac supply level is less that the
dc output. The load current and peak diode (hence supply) current are therefore
∧
V
2V s − V o
Io = o
ID =
RL
Ri
The ac supply provides current, through the rectifying diode, during the period
1
2V s sin ωt −Vo
is =
α ≤ ωt ≤ β
Ri
(
)
If the capacitor voltage is to be maintained constant, the charge into the capacitor must equal
the charge delivered by the capacitor when the diode is not conducting, that is
β
∫ (i
s
− i o ) d ωt =
α
α + 2π
∫
i o d ωt
β
The circuit in figure 11.1a, which has an R-L load, is characterised by discontinuous current (i = 0) and
high ripple current. Continuous load current can result when a diode Df is added across the load as
shown in figure 11.3a. This freewheel diode prevents the voltage across the load from reversing during
the negative half-cycle of the ac supply voltage. The inductor energy is not returned to the ac supply,
rather is retained in the load circuit. The stored energy in the inductor cannot reduce to zero
instantaneously, so the current is forced to find an alternative path whilst decreasing towards zero.
When the rectifier diode D1 ceases to conduct at zero volts it blocks, and diode Df provides an alternative
load current freewheeling path, as indicated by the waveforms in figure 11.3b.
The output voltage is the positive half of the sinusoidal input voltage. The mean output voltage (thence
mean output current) is
Vo = Io R =
Vo =
2V
2
2
Manipulation yields
R
1Ω
tan½θ − ½θ = π i = π
= 0.1π
10Ω
RL
An iterative solution yields θ = 99.6º, that is, the diode conducts for a period of 5.53ms, every
cycle of the ac supply. The capacitor voltage is
π −θ
Vo = 2V s sin α = 2V s sin
2
180° − 99.6°
= 2 × 230V × sin
= 209.95V
2
ii.
The average diode current is given by
β
1 1
1 
π −θ

ID =
2V s sin ωt −Vo d ωt =
2V s × 2 × cos
−V o × θ 
2π α∫ R i
2π R i 
2

(
=
)
1 
180° − 99.6°
99.6° 
2 × 230V × 2 × cos
− 209.95V × π ×
= 21.0A
2π 1Ωi 
2
180° 
∫
π
0
2V
sin ωt d ωt
(11.14)
= 0.45 ×V = Io R
π
(V)
The rms value of the load circuit voltage v0 is given by
Vrms =
Also
Vo = 2V s sin α
π −θ
π +θ
α=
β=
1
2π
1
2π
∫ (
π
0
= V
)
2
2V sin ωt d ωt
= 0.71× V
(11.15)
(V)
2
The output ripple (ac) voltage is defined as
2
Vrms
− Vo2
VRi
=
 2V 

2 

2
− 
2V


π 
2
= 0.545 ×V
(11.16)
hence the load voltage form and ripple factors are defined as
FFv = Vrms / Vo
= ½π = 1.57
RFv
VRi / Vo =
( )
Vrms 2
Vo
−1
(11.17)
= FFv − 1 = ¼π 2 − 1 = 1.211
2
After a large number of ac supply cycles, steady-state load current conditions are established, and from
Kirchhoff’s voltage law, the load current is defined by
di
L + Ri = 2 V sin ωt
(A)
0 ≤ ωt ≤ π
(11.18)
dt
and when the freewheel diode conducts
di
L + Ri = 0
(A)
π ≤ ωt ≤ 2π
(11.19)
dt
Power Electronics
227
Naturally commutating converters
I D1 =
V
2πR
( 2 − (1 + e
V
I Df = I o − I D1 =
− π / tan φ
228
) × sin φ )
2
(1 + e
− π / tan φ
) × sin
(11.22)
2
φ
2πR
In figure 11.3b it will be seen that although the load current can be continuous, the supply current is
discontinuous and therefore has a high harmonic content.
The output voltage Fourier series (Vo + V1 + Vn = 2, 4, 6..) is
q =1 r =1 s =1
p=qxrxs
p=1
vo ( t ) =
2V
.
π
+
2V
sin ωt −
2
∞
2V
.
.
π
n
∑ (n
= 2,4,6
2
2
cos nωt
− 1)
(11.23)
Dividing each harmonic output voltage component by the corresponding load impedance at that
frequency gives the harmonic output current, whence rms current. That is
V
Vn
Vn
(11.24)
In = n =
=
2
Zn
R + jnω L
R 2 + ( nω L )
and
I rms = I o2 +
Example 11.2:
Figure 11.3. Half-wave rectifier with a load freewheel diode and an R-L load:
(a) circuit diagram and parameters and (b) circuit waveforms.
During the period 0 ≤ ωt ≤ π, when the freewheel diode current is given by iDf = 0, the supply current,
which is the load current, are given by
i (ωt ) = io (ωt ) = 2 V sin(ωt − φ ) + ( I o 2π + 2 V sin φ )e −ωt / tanφ
(A)
Z
Z
(11.20)
0 ≤ ωt ≤ π
for
2V
1 + e −π / tan φ
I o 2π =
sin φ π / tan φ
(A)
Z
e
− e −π / tan φ
where Z =
R 2 + (ω L) 2
(ohms)
tan φ = ω L / R
During the period π ≤ ωt ≤ 2π, when the supply current i = 0, the freewheel diode current and hence
load current are given by
io (ω t ) = iDf (ω t ) = I o1π e − (ωt −π ) / tan φ
(A)
π ≤ ω t ≤ 2π
(11.21)
for
I o1π = I o 2π eπ / tan φ
(A)
For discontinuous load current (the freewheel diode current iDf falls to zero before the rectifying diode D1
recommences conduction), the appropriate integration gives the average diode currents as
∑
n = 1, 2, 4, 6..
½ I n2
(11.25)
Half-wave rectifier
In the circuit of figure 11.3, the source voltage is 240√2 sin(2π 50t) V, R = 10 ohms, and L = 50 mH.
Calculate
i. the mean and rms values of the load voltage, Vo and Vrms
ii. the mean value of the load current, Io
iii. the current boundary conditions, namely Io1π and Io2π
iv. the average freewheel diode current, hence average rectifier diode current
v. the rms load current, hence load power and supply rms current
vi. the supply power factor
If the freewheel diode is removed from across the load, determine
vii. an expression for the current hence the current extinction angle
viii. the average load voltage hence average load current
ix. the rms load voltage and current
x. the power delivered to the load and supply power factor
From the rms and average output voltages and currents, determine the load form and ripple factors.
Solution
i.
From equation (11.14), the mean output voltage is given by
V = 2 V = 2 × 240V = 108V
o
π
π
From equation (11.15) the load rms voltage is
Vrms = V / 2 = 240V / 2 = 169.7V
ii. The mean output current, equation (11.5), is
V
2 ×240V
Io = o
= 2V
= 10.8A
R
π×10 Ω
πR =
iii. The load impedance is characterised by
Z =
R 2 + (ω L) 2
= 102 + (2π × 50Hz × 0.05) 2
= 18.62 Ω
tan φ = ω L / R
= 2π × 50Hz × 0.05H /10Ω = 1.57 or φ = 57.5° ≡ 1rad
From section 11.1.2, equation (11.20)
Power Electronics
229
I o 2π = 2 V
I o 2π = 2 × 240V
× sin(tan −1 1.57) ×
i (ωt ) =
1 + e−π /1.57
= 3.41A
eπ /1.57 − e−π /1.57
18.62 Ω
Hence, from equation (11.21)
I o1π = I o 2π eπ / tan φ = 3.41× eπ /1.57 = 25.22A
Since I o 2π = 3.41A > 0 , continuous load current flows.
=
∫
2 × 240V
18.62Ω
(ωt - φ ) + sin φ e-ωt / tan φ }
{sin
(ωt -1.0) + 0.841× e-ωt /1.57 }
0 ≤ ωt ≤ β
(A)
(rad)
The current extinction angle β is found by setting i = 0 and solving iteratively for β. Figure 11.2a
gives an initial estimate of 240° (4.19 rad) when φ = 57.5° (1 rad). That is
0 = sin ( β -1.0) + 0.841× e- β /1.57
gives β = 4.08 rad or 233.8°, after iteration.
Vo =
2V
2π
(1 − cos β ) =
2 ×240V
2π
(1 − cos 4.08) = 86.0V
The average load current is
I o = Vo / R = 86.0V/10Ω = 8.60A
v. The load voltage harmonics given by equation (11.24) can be used to evaluate the load current at
the load impedance for that frequency harmonic.
∞
2V
2V
2 2V
v (t ) =
+
sin ωt − ∑
cos ( nωt )
2
2
π
n = 2,4,6 ( n − 1) π
.
{sin
viii. The average load voltage from equation (11.6) is
π
π
−
1
25.22A


25.22A × e−ωt /1.57rad d ωt =
× 1.57rad × 1 − e 1.57  = 5.46A
2π 0
2π


The average input current, which is the rectifying diode mean current, is given by
I s = I D1 = I o − I Df = 10.8A − 5.46A = 5.34A
.
2V
Z
= 18.23 × {sin (ωt -1.0) + 0.841× e-ωt /1.57 }
iv. Integration of the diode current given in equation (11.21) yields the average freewheel diode current.
π
π
1
1
I Df =
iDf (ω t ) d ωt =
I o1π e −ωt / tan φ d ωt
2π ∫0
2π ∫0
=
230
vii. If the freewheel diode Df is removed, the current is given by equation (11.2), that is
1 + e−π / tan φ
eπ / tan φ − e−π / tan φ
sin φ
Z
Naturally commutating converters
ix.
The load rms voltage is 169.7V with the freewheel diode and increases without the diode to, as
given by equation (11.7),
Vrms = V  12π {β − ½ sin 2 β }
.
½
= 240V  12π {4.08 − ½ sin 2 × 4.08} = 181.6V
½
The following table shows the calculations for each frequency component.
The rms load current from equation (11.7) is decreased to
harmonic
Vn =
n
Z n = R 2 + ( nω L )
2 2V
( n2 − 1) π
(V)
.
½ I n2
(Ω)
Vn
Zn
(A)
10.00
10.80
(116.72)
2
In =
0
(108.04)
1
(169.71)
18.62
9.11
41.53
2
72.03
32.97
2.18
2.39
4
14.41
63.62
0.23
0.03
6
6.17
94.78
0.07
0.00
8
3.43
126.06
0.03
I o2 +
∑½I
0.00
2
n
=
160.67
irms =
V
Z
 1 
sin β cos (α + β + φ ) 

β −

2
π
cos φ




½
½

sin 4.08cos ( 4.08 + 1.57 ) 
4.08 −
 = 9.68A
cos1.57



Removal of the freewheel diode decreases the rms load current from 12.68A to 9.68A.
=
x.
240V  1
×
18.62Ω  2π
The load power is reduced without a load freewheel diode, from 1606.7W with a load freewheel
diode, to
2
P10 Ω = irms
R = 9.682 × 10Ω = 937W
The supply power factor is also reduced, from 0.74 to
Pout
937W
pf =
=
= 0.40
Vrms I rms 240V × 9.68A
The rms load current is
I rms = I o2 +
∑ ½I
2
n
=
.
160.7 = 12.68A
n=1, 2, 4..
Load factor
The power dissipated in the load resistance is therefore
2
P10 Ω = I rms
R = 12.68A 2 × 10Ω = 1606.7W
The freewheel diode rms current is
π
2
1
I Df =
I o1π e − (ωt ) / tan φ d ωt
2π ∫0
π
2
1
25.22A × e − (ωt ) /1.57rad d ωt = 8.83A
=
2π ∫0
Thus the input (and rectifying diode) rms current is given by
.
.
(
)
(
)
2
− I Df2 rms
I D1rms = I srms = I rms
= 12.68 − 8.83 = 9.09A
2
2
vi. The input ac supply power factor is
Pout
1606.7W
=
= 0.74
pf =
Vrms I rms 240V × 9.09A
circuit with freewheel diode
circuit without freewheel diode
form factor
ripple factor
form factor
ripple factor
FF = rms/ave
RF = √FF2-1
FF = rms/ave
RF = √FF2-1
Voltage factor
169.7V/108V = 1.57
1.21
181.6V/86V = 2.1
1.86
Current factor
12.68A/10.8A = 1.17
0.615
9.68A/8.60A = 1.12
0.517
♣
11.1.3 Single-phase, full-wave bridge rectifier circuit
Single-phase uncontrolled full-wave bridge circuits are shown in figures 11.4a and 11.4b. Both circuits
appear identical as far as the load and supply are concerned. It will be seen in part b that two fewer
diodes can be employed but this circuit requires a centre-tapped secondary transformer where each
secondary has only a 50% copper utilisation factor. For the same output voltage, each of the secondary
windings in figure 11.4b must have the same rms voltage rating as the single secondary winding of the
transformer in figure 11.4a. The rectifying diodes in figure 11.4b experience twice the reverse voltage,
(2√2 V), as that experienced by each of the four diodes in the circuit of figure 11.4a, (√2 V).
Power Electronics
231
Naturally commutating converters
Figure 11.4c shows bridge circuit voltage and current waveforms. With an inductive passive load, (no
back emf) continuous load current flows, which is given by
2V 
2sin φ

io (ωt ) =
sin (ωt − φ ) +
0 ≤ ωt ≤ π
(11.26)
× e −ωt / tan φ 
Z 
1 − e −π / tan φ

Appropriate integration of the load current squared, gives the rms load (and ac supply) current:
½
V
I rms = 1 + 4sin 2 φ tan φ × (1 + e −π / tan φ )  = I s
(11.27)
Z
The load experiences the transformer secondary rectified voltage which has a mean voltage (thence
mean load current) of
1 π
2 2V
2 V sin ω t d ω t = I R =
(V)
V =
= 0.90 V
(11.28)
o
π
∫
o
0
π
Since the average inductor voltage is zero, the average resistor voltage equals the average R-L voltage.
1
2π
∫ (
2π
0
)
2
2 V sin ω t d ω t = V
From the load voltage definitions in section 11.10, the load voltage form factor is
V
V
FFv = rms =
= 1.11
Vo
2 2V
(11.30)
π
The load ripple voltage is
2
Vrms
− Vo2
VRI
= V 2 − (2
2
)V
2
π
2
= 0.435V
(11.31)
(V)
hence the load voltage ripple factor is
RFv
VRI / Vo = FFv2 − 1
RFv = 1 − ( 2
2
) /
2
π
2 2
π
(11.32)
= 0.483
which is significantly less (better) than the half-wave rectified value of 1.211 from equation (11.17).
The rms value of the load circuit voltage v0 is
Vrms =
232
(V)
(11.29)
The output voltages and currents (rms and average) can be derived from the voltage Fourier expansion
form for a half-sine wave:
2 2V
2 2V ∞
2
+
(11.33)
vo (ωt ) =
cos nω t
2
π
π n∑
= 2,4,6 n − 1
The first term is the average output voltage, as given by equation (11.28). Note the harmonic
2
q =2 r =1 s =1
p=qxrxs
p=2
2
2
2
magnitudes decrease rapidly with increased order, namely 3 : 15 : 35 : 63 : ... The output voltage is
therefore dominated by the dc component and the harmonic at 2ω.
The output current can be derived by dividing each voltage component by the appropriate load
impedance at that frequency. That is
V
2 2V
Io = o =
R
πR
2
(11.34)
Vn 2 2 V
n2 − 1
In =
n = 2, 4, 6..
=
×
for
2
2
Zn
π
R + ( nω L )
The load rms current whence load power, critical load inductance, and power factor, are given by
I rms = I o2 +
∞
∑ ½× I
n = 2,4
2
n
2
PL = I rms
R
(11.35)
pf =
I R
PL
= rms
V I rms
V
Lcritical =
R
3× ω
(see equation 11.38)
Each diode rms current is I rms / 2 . For the circuit in figure 11.4a, the transformer secondary winding rms
current is Irms, while for the centre-tapped transformer, for the same load voltage, each winding has an
rms current rating of Irms / √2. The primary current rating is the same for both transformers and is related
to the secondary rms current rating by the turns ratio.
11.1.3i - Single-phase full-wave bridge rectifier circuit with an output L-C filter
A – with an output L-C filter and continuous load current
Table 11.1 shows three typical single-phase, full-wave rectifier output stages, where part c is a typical
output filtering stage used to obtain a near constant dc output voltage.
If it is assumed that the load inductance is large and the load resistance small such that continuous load
current flows, then the bridge average output voltage Vo is the same as the average voltage across the
load resistor since the average voltage across the filter inductor is zero. From equation (11.33), the
dominant load voltage harmonic is due to the second harmonic therefore the ac current is predominately
the second harmonic current, I o , ac ≈ I o , 2 . By neglecting the higher order harmonics, the various circuit
currents and voltages can be readily obtained as shown in table 11.1. From equation (11.33) the output
voltage is given by
Figure 11.4. Single-phase full-wave rectifier bridge: (a) circuit with four rectifying diodes;
(b) circuit with two rectifying diodes; and (c) circuit waveforms.
Power Electronics
Naturally commutating converters
Vo +
Vo , 2 cos 2ωt
2 2V 2 2V
2
cos nωt for n = 2
=
+
× 2
π
π
n −1
(11.36)
2 2V 2 2V 2
=
+
× cos 2ωt
π
π
3
= 0.90V + 0.60V × cos 2ωt
With the filter capacitor across the load resistor, the average inductor current is equal to the average
resistor current, since the average capacitor current is zero.
With continuous inductor current, the inductor current is
io (ωt ) = I o + I o , 2 cos 2ωt
Vo
2 Vo
0.90V
0.60V
(11.37)
cos 2ωt =
=
+
+
× cos 2ωt
2
R
3 Z2
R
R 2 + ( 2ω L )
Since the load resistor must be small enough to ensure continuous inductor current, then 2ω L > R such
that Z 2 = R 2 + ( 2ω L )2 ≈ 2ω L . Equation (11.38) therefore gives the following load identity for continuous
inductor current
1
2 1
1
>
=
that is L > 1
(11.39)
3ω
R
3 Z 2 3ω L
R
The load and supply (peak) ac currents are I o , ac = I s , ac = I o , 2 . The output and supply rms currents are
233
vo (ωt ) =
From equation (11.37) for continuous inductor current, the average current must be larger than the peak
second harmonic current magnitude, that is
I o > I o, 2
(11.38)
2 Vo
Vo
>
3 Z2
R
Table 11.1. Single-phase full-wave uncontrolled rectifier circuits – continuous inductor current
Full-wave rectifier circuit
Load
circuit
(a)
2nd harmonic
current
Io,2
average output
current
Io
(A)
(A)
Vs
Vo
R
R 2 + ( 2ω L )
I o2, rms R
2
=
2
2
I + ½ Io, 2
(11.40)
o
(11.41)
B – with an output L-C filter and discontinuous load current
If the inductor current reduces to zero, at angle β, all the load current is provided by the capacitor. Its
voltage falls to Vo (<√2 V) and inductor current recommences when
v L = 2V sin ωt −Vo
at an angle
α = sin−1
Vo
2V
By integrating v = L di/dt for i, the inductor current is of the form
1
2V ( cos α − cos ωt ) −Vo (ωt − α )
i L (ωt ) =
ωL
(11.42)
(
(11.43)
)
(11.44)
where α ≤ ωt ≤ β. The voltage Vo is found from equation (11.44) by iterative techniques.
Full-wave uncontrolled rectifier with an L-C filter and continuous load current
A single-phase, full-wave, diode rectifier is supplied from a 230V ac, 50Hz voltage source and uses an
L-C output filter with a resistor load, as shown in the last circuit in Table 11.1. The average inductor
current is 10A with a 4A rms ripple current dominated by the 100Hz component. Ignoring diode voltage
drops and initially assuming the output voltage is ripple free, determine
i. the dc output voltage, hence load resistance and power
ii. the dc filter inductance and its average voltage, whilst neglecting any capacitor voltage ripple
iii. the dc filter capacitance if its peak-to-peak ripple voltage is 5% the average voltage
iv. diode average, rms, and peak current
v. the supply power factor
R
Vo ,2
o
and the power delivered to resistance R in the load is
PR = I o2, rms R
Example 11.3:
(W)
Vo
Is
2
I + ½ I o , ac =
PR+PE
L
Io
R-L
see
section
11.1.3
and
11.3.3
α=0
output power
2
I o , rms = I s , rms =
234
2 2V
πR
Solution
(b)
Io
L
R-L-E
Vo − E
R
R
Vs
Vo ,2
Is
Vo
see
section
11.3.4
α=0
E+
R 2 + ( 2ω L )

1  2 2V
= 
−E

R  π

Io,dc
Vo
Io,ac
L-R//C
Vs
see
section
11.1.3i
L
Is
R
Vo ,2
Vo
C
R
2
I o2, rms R = I R
2ω L
o
=
2 2V
πR
2 × 4A < 10A , the output current is continuous.
PR = Vo × I o = 207V × 10A = 2070W
ii.
Io
(c)
)
2 I o rms , 2 < I o
The dc output voltage is Vo = 0.9×230V = 207V. Assuming the 207V is ripple free, that is, Vrms =
Vdc, then the load resistance and power dissipated are
V
207V
R= o =
= 20.7Ω
10A
Io
I o2, rms R + I o E
2
(
Since
i.
The 100Hz voltage component in the output voltage is given by equation (11.36), that is
2 2V
2
× 2
Vo , 2 =
cos nωt
π
n −1
2 2V 2
=
π × 3 cos 2ωt
= 0.60 × 230V × cos 2ωt = 138 × cos 2ωt
which has an rms value of 138/√2 = 97.6V. The 100Hz rms current I o , 2 / 2 produced by this
voltage is 4A thus
Power Electronics
235
from
I o, 2
2
=
Naturally commutating converters
Vo , 2
2ω L
Vo , 2
97.6V
=
= 38.8mH
2ω I o , 2 2 × 2π 50Hz × 4A
The average inductor voltage is zero.
L=
iii.
From part i, the dc output voltage is 207V. The peak-to-peak ripple voltage is 5% of 207V, that is
10.35V. This gives an rms value of 10.35V /2√2 = 3.66V. From
I o, 2
Vo , 2
I
2
= o, 2
× X c , 100 Hz =
2
2
2ωC
I o, 2
4A
⇒C =
=
= 1.7mF
2ω × Vo , 2 2 × 2π 50Hz × 3.66V
iv.
The diode currents are
2
I D , rms = I o , rms / 2 =
I o + ½I
2
o, 2
/ 2 = 10A + 4A / 2 = 10.8A / 2 = 7.64A
2
2
I D = I o / 2 = 10A/2 = 5A
236
As the ac supply voltage rises to its extremes each half cycle, as shown in figure 11.5b, a pair of rectifier
diodes D1-D2 or D3-D4, alternately become forward biased at time ωt = α. The ac supply provides load
resistor current and simultaneously charges the capacitor, its voltage having drooped whilst providing
the load current during the previous diode non-conduction period. The capacitor charging current period
θc around the ac supply extremes is short, giving a high peak to rms ratio of diode and supply current.
When all the rectifier diodes are reverse biased at ωt = β because the capacitor voltage is greater than
the instantaneous supply ac voltage, the capacitor supplies the load current and its voltage decreases
with an R-C time constant until ωt = π+α. The output voltage and diode voltages, plus load current vo /R,
and capacitor current C dvo /dt are defined in Table 11.2.
The start of diode conduction, α, the diode current extinction angle, β, hence diode conduction period,
θc, are specified by the following equations.
From ic + iR = 0 at ωt = β :
2V
2V
cos β +
sin β = 0
(11.49)
X
R
β = tan −1 ( −ω RC ) = π − tan −1 (ω RC )
½π ≤ β ≤ π
By equating the two expression for output voltage at the boundary ωt = π+α gives
− π +α − β ) / tan β
2V sin (π + α ) = 2V sin β × e (
(11.50)
and a transcendental expression for α results:
− π +α − β ) / ω RC
sin α − sin β × e (
=0
(11.51)
I D = I o + I o , 2 = 10A + 2 × 4A = 15.7A
v.
The input and output rms current is
I s = I o , rms =
D4
iD
D1
2
2
2
2
I o + ½ I o , 2 = 10A + 4A = 10.8A
ic
iR
is
vo
Vs
Assuming the input power equals the output power, then from part i, Po = Pi = 2070W. The
supply power factor is
P
P
2070W
= 0.83
pf = i = i =
S Vs I s 230V × 10.8A
♣
C
D2
D3
(a)
11.1.3ii - Single-phase full-wave bridge circuit with highly inductive load – constant load current
With a highly inductive load, which is the usual practical case, virtually constant load current flows, as
shown dashed in figure 11.4c. The bridge diode currents are then square wave 180º blocks of current of
magnitude I o . The diode current ratings can now be specified and depend on the pulse number p. For
this full-wave single-phase application each input cycle comprises two 180º output current pulses, hence
p = 2.
The mean current in each diode is
I D = 1p Io = ½ Io
(A)
(11.45)
R
vo
D1
D3
D2
D4
θc
θc
∨
− vo
D1
D3
D2
D4
∆vo
vo
conducting
diodes
√2 V
and the rms current in each diode is
ID = 1
p
Io = Io / 2
(A)
o
whence the diode current form factor is
RFID = I D / I D =
π
½π
(11.46)
α
β
2π
ωt
π+α
Vs
p= 2
(11.47)
Since the load current is approximately constant, power delivered to the load is
Po ≈ Vo I o =
8
π2
× V 2R
(W)
(11.48)
ICap
The supply power factor is pf =Vo /V = 2√2/π = 0.90, since I o = I rms .
iD
11.1.3iii - Single-phase full-wave bridge rectifier circuit with a C-filter and resistive load
The capacitor smoothed single-phase diode rectifier circuit shown in figure 11.5a is a common power
rectifier circuit used to obtain unregulated dc voltages. The circuit is simple and cheap but the input
current has high peak and rms values, high harmonics, and a poor power factor.
The capacitor reduces the ripple voltage, so large voltage-polarised capacitance is used to produce an
almost constant dc output voltage. Isolation and voltage matching (step-up or step down) are obtained
by using a transformer before the diode rectification stage as shown in figures 11.4a and b. The resistor
R across the filter capacitor represents a resistive dissipative load.
α
β
π+α
iR
ωt
(b)
Figure 11.5. Single-phase full-wave rectifier bridge:
(a) circuit with C-filter capacitor and (b) circuit waveforms.
Power Electronics
237
Naturally commutating converters
Table 11.2. Single-phase, full-wave rectifier voltages and currents
Diodes conducting
vs(ωt) = √2Vsinωt
Diodes non-conducting
α ≤ ωt ≤ β
Output voltage
Diode voltage
0 and - 2 V sin ωt
vD(ωt)
Capacitor current
ic(ωt)
Resistor current
iR(ωt)
Diode bridge current
β ≤ ωt ≤ π + α
2V sin ωt
vo(ωt)
2V sin β × e
− 2 V sin β × e
2V
cos ωt
X
2V
sin ωt
R
ID(ωt)=
ic(ωt)+ iR(ωt)
2V
× sin (ωt + φ )
R cos φ
−(ωt − β ) / tan β
−(ωt − β ) / tan β
+ 2 V sin ωt
2V
× e−(ωt −β ) / tan β
Z
∆Vo = Vo − Vo = 2 V − 2 V sin α = 2 V (1 − sin α )
By assuming α ≈ ½π, β ≈ ½π, and a series expansion for the exponent
2 Vπ
V
∆Vo ≈
=
ω RC
2 f RC
The ac source current is the sum of the diode currents, that is
is = iD1,2 − iD 3,4 = iR + ic
2V
− ωt − β / tan β
×e ( )
= −ic (ωt )
Z
0
The supply voltage is vs = √2×230 sin2π 50t, which has a peak value of Vs = 325.3V.
ω RC = 2π 50Hz × 100Ω × 1000µF = 31.416 rad
Thus X = 1/ωC = 3.1831Ω and Z = 100.0507Ω.
From equation (11.49) the diode current extinction angle β is
β = π − tan −1 (ω RC ) = π − tan −1 ( 31.416rad ) = 1.6026 rad = 91.8°
The diode current turn-on angle α is solve iteratively from equation (11.51), that is
sin α − sin β × e
i.
A single-phase, full-wave, diode rectifier is supplied from a 230V ac, 50Hz voltage source and uses a
capacitor output filter, 1000µF, with a resistor 100Ω load, as shown in Figure 11.5a. Ignoring diode
voltage drops, determine
2V sin ωt
= 325.27V × sin ωt
66.5° ≤ ωt ≤ 91.8°
vo (ωt ) = 2 × 230V × sin1.6026rad × e
(
)
− ωt −1.6026rad / 31.416rad
= 325.13 × e
(
)
− ωt −1.6026rad / 31.416rad
91.8° ≤ ωt ≤ 246.5°
The output voltage ripple ∆vo is given by equation (11.54), that is
∆Vo = 2 V (1 − sin α ) = 2 230V × (1 − sin1.16026 ) = 26.94V p-p
From equation (11.55)
V
230V
∆Vo ≈
=
= 32.5V
2 f RC
2 × 50Hz × 100Ω × 1000µF
The approximation predicts a higher ripple: a +21% over-estimate.
(11.56)
Single-phase full-wave bridge rectifier circuit with C-filter and resistive load
From table 11.2, the output voltage, which is the capacitor voltage, is given by
vo (ωt ) =
(11.55)
Similar expressions can be derived for the half-wave rectifier case. For the non-conduction period,
β=2π+α. The output ripple voltage is about twice that given by equation (11.55) and the average resistor
voltage in equation (11.53) (after modification), is reduced.
=0
−(π +α −1.603) / 31.416
ii.
I c = 2 V ωC cos α
(11.57)
From table 11.2, the peak diode current occurs at the same time as the peak capacitor current, ωt = α:
I D = ic (π + α ) + iR (π + α )
(11.58)
2V
2V
2V
2V Z
sin α =
cos α +
sin α =
sin (α + φ )
= 2 V ωC cos α +
R
X
R
R X
−(π +α − β ) / ω RC
sin α − sin1.603 × e
=0
gives α = 1.16095 rad or 66.5°. The diode conduction period is θc = β - α =1.6026 -1.16095 = 0.44167rad
or 25.3°.
(11.54)
when α < ωt < β . Otherwise is = 0.
Since the capacitor voltage is in steady-state, the average capacitor current is zero, thus for full-wave
rectification, the average diode current is half the average load current.
The peak capacitor current occurs at ωt = α, when the diodes first conduct. From the capacitor current
equation in table 11.2:
Example 11.4:
i. expressions for the output voltage
ii. output voltage ripple ∆vo and the % error in using the approximation equation (11.55)
iii. expressions for the capacitor current
iv. diode peak current
v. the average load voltage and current
Assuming the output ripple voltage is triangular, estimate
vi. the average output voltage and rms output ripple voltage
vii. capacitance C for ∆vo = 2% of the maximum output voltage
Solution
The diode current conduction period θc is given by
θc = β − α
(11.52)
When the diodes conduct, R and C are in parallel and tan φ = ω C R .
When the diodes are not conducting, the output circuit current flows in a series R-C circuit with a
fundamental impedance of:
Z = R 2 + X 2 and X = 1
ωC
The resistor average voltage and current are
2V (1 − cos θ c )
VR =
= IR R
(11.53)
− cos β
π
The maximum output voltage occurs at ωt = ½π when vo = Vo = Vs =√2V, while the ∨minimum output
voltage occurs at the end of the capacitor discharge period when ωt = α and vo = Vo =√2Vsinα. The
output peak-to-peak ripple voltage is therefore the difference:
∨
238
.
iii.
.
From table 11.2, the capacitor current is
ic (ωt ) = 2V ωC cos ωt = 2 230V × 2π 50Hz × 1000µF × cos ω t = 102.2 × cos ω t
66.5° ≤ ωt ≤ 91.8°
ic (ωt ) =
iv.
2V sin β −(ωt −β ) / ω RC
×e
=
R
2 230V × sin1.16 −(ωt −1.16) / 31.4
−(ωt −1.16 ) / 31.4
×e
= 3.0 × e
100Ω
91.8° ≤ ωt ≤ 246.5°
The peak diode current is given by equation (11.58):
2V
I D = 2V ωC cos α +
sin α
R
= 2 230V × 2π 50Hz × 1000µF × cos1.16026 +
2 230V
× sin1.16026
100Ω
= 40.7A + 3A = 43.7A
The peak diode current is dominated by the capacitor initial charging current of 40.7A
Power Electronics
239
v.
11.2
The average load voltage and current are given by equation (11.53)
2V (1 − cos θ c )
VR =
− cos β
π
=
IR =
vi.
Naturally commutating converters
2 230V
π
×
(1 − cos 0.4417 ) = 312.3V
− cos1.603
V R 312.3V
=
= 3.12A
R
100Ω
If the ripple voltage is assumed triangular then
(a) The average output voltage is the peak output voltage minus half the ripple voltage, that is
Vs - ½∆vo = √2×230V - ½×26.9V = 311.8V
which is less than that given by the accurate equation (11.54), 312.3V.
(b) If the 26.9V p-p ripple voltage is assumed triangular then its rms value is ½×26.9/√3 =7.8V rms.
vii.
Re-arrangement of equation (11.55), which under-estimates the capacitance requirement for
2% ripple, gives
Vs
1
V
C=
=
=
2 f R × 2%
2 f R × ∆Vo 2 f R × 2% of Vs
1
=
= 5, 000µF
2 × 50Hz × 100Ω × 0.02
♣
240
Single-phase full-wave half-controlled converter
When a converter contains both diodes and thyristors, for example as shown in figure 11.7 parts a to d,
the converter is termed half-controlled (or semi-controlled). These four circuits produce identical load
and supply waveforms, neglecting any differences in the number and type of semiconductor voltage
drops. The power to the load is varied by controlling the angle α, shown in figure 11.7e, at which the
bridge thyristors are triggered (after first becoming forward biased). The circuit diodes prevent the load
voltage from going negative, extend the conduction period, and reduce the output ac ripple.
The particular application will determine which one of the four circuits should be employed. For
example, circuit figure 11.7a contains five devices of which four are thyristors, whereas the other circuits
contain fewer devices, of which only two are thyristors. The circuit in figure 11.7b uses the fewest
semiconductors, but requires a transformer which introduces extra cost, weight, and size. Also the
thyristors experience twice the voltage of the thyristors in the other circuits, 2√2 V rather than √2 V. The
transformer does provide isolation and voltage matching.
T1
q =2 r =1 s =1
p=qxrxs
p=2
T2
.
(a)
11.1.3iv - Other single-phase bridge rectifier circuit configurations
Df
Figure 11.6a shows a transformer used to create a two-phase supply (each phase is 180° apart), which
+
upon rectification produce equal split-rail dc output voltages, V and V . The electrical characteristics
can be analysed as in the case of the single-phase full-wave bridge rectifier circuit with a capacitive Cfilter and resistive load, in section 11.1.3iii. In the split rail case, the rectifiers conduct every 180°,
alternately feeding each output voltage rail capacitor. Thus the diode average and rms currents are
increased by 2 and √2 respectively, above those of a conventional single phase rectifier.
The voltage doubler in figure 11.6b can be used in equipment that must be able to operate from both
115Vac and 230V ac voltage supplies, without the aid of a voltage-matching transformer. With the
switch in the 115V position, the output is twice the peak of the input ac supply. The capacitor C1 charges
through diode D1, and when the supply reverses, capacitor C2 charges through D2. Since C1 and C2 are
in series, the output voltage is the sum VC1+VC2, where each capacitor is alternately charged (half-wave
rectified) from the ac source Vs. The other, unused, two diodes remain reverse biased, and are only
necessary if the dual input voltage function is required.
With the switch in the 230V ac position (open circuit), standard rectification occurs, with the two series
capacitors charging simultaneously every half cycle. In dual frequency applications (110V ac, 60Hz and
230V ac, 50Hz), the capacitance requirements are based on the supply with the lower frequency, 50Hz.
Np:1:1
C +
V
Vs
+
V
D1
vo
(c)
(d)
C1 +
Vo
115V
C2 +
D2
(a)
vo
230V
Vs
0V
C +
115V/230V
ac
input
D1
(b)
(b)
Circuit
a
T1 and T4
D1
b
T1
D1
T2
D1
c
T1 and D2
T1 and D1
T2 and D1
T2 and D2
d
T1 and D2
D2 and D1
T2 and D1
D1 and D2
T2 and T3
D1
Figure 11.6. Bridge rectifiers: (a) split rail dc supplies and (b) voltage doubler.
(e)
Figure 11.7. Full-wave half-controlled converters with freewheel diodes:
(a), (b), (c), and (d) different circuit configurations producing the same output; and (e) circuit
voltage and current waveforms and device conduction table.
Power Electronics
Naturally commutating converters
The thyristor triggering requirements of the circuits in figures 11.7b and c are simple since both
thyristors have a common cathode connection. Figure 11.7c may suffer from prolonged shut-down times
with highly inductive loads. The diode in the freewheeling path will hold on the freewheeling thyristor,
allowing conduction during that thyristors next positive cycle without any gate drive present. The extra
diode Df in figure 17.11c bypasses the bridge thyristors allowing them to drop out of conduction. This is
achieved at the expense of an extra device, but the freewheel path conduction losses are decreased
since that series circuit now involves only one semiconductor voltage drop. This continued conduction
problem does not occur in circuits 11.7a and d since freewheeling does not occur through the circuit
thyristors, hence they will drop out of conduction at converter shut-down. The table in figure 11.7e
shows which semiconductors are active in each circuit during the various periods of the load cycle.
The various semiconductor average current ratings can be determined from the average half-cycle
freewheeling current, I ½ F , and the average half-cycle supply current, I ½ s . For discontinuous load current
2V
α −π / tan φ
(11.67)
I ½F = ½
sin φ sin φ − sin (α − φ ) e( )
πR
241
Circuit waveforms are shown in figure 11.7e. Since the load is a passive L-R circuit, independent of
whether the load current is continuous or discontinuous, the mean output voltage and current
(neglecting diode voltage drops) are
Vo = I o R =
I o = Vo R =
1
π
∫
2V
πR
π
α
2V
sin(ωt ) d ωt =
(1 + cos α )
2V
π
(1 + cos α )
(V)
(11.59)
(A)
where α is the delay angle from the point at which the associated thyristor first becomes forward-biased
and is therefore able to be turned on and conduct current. The maximum mean output voltage,
Vo = 2 2 V / π (also predicted by equation (11.28)), occurs at α = 0.
The normalised mean output voltage Vn is
Vn = Vo / Vo = ½(1 + cos α )
(11.60)
The Fourier coefficients of the 2-pulse output voltage are given by equation (11.176). For the singlephase, full-wave, half-controlled case, p = 2, thus the output voltage harmonics occur at n = 2, 4, 6, …
Equation (11.59) shows that the load voltage is independent of the passive load (because the diodes
clamp the load to zero volts thereby preventing the load voltage from going negative), and is a function
only of the phase delay angle for a given supply voltage.
The rms value of the load circuit voltage vo is
2
1 π
π − α + ½ sin 2α
Vrms =
(11.61)
(V)
∫ ( 2 V sin ωt ) dωt = V
π
π
α
From the load voltage definitions in section 11.10, the load voltage form factor is
FFv =
π (π − α + ½ sin 2α )
Vrms
=
Vo
2 (1 + cos α )
(11.62)
The ripple voltage is
2
Vrms
− Vo2
VRi
(11.63)
hence the voltage ripple factor is
Kv
VRi / Vo = FF − 1
(11.64)
2
v
The load and supply waveforms and equations, for continuous and discontinuous load current, are the
same for all the circuits in figure 11.7. The circuits differ in the device conduction paths as shown in the
table in figure 11.7e. After deriving the general load current equations, the current equations applicable
to the different circuit devices can be decoded.
11.2i - Discontinuous load current, with α < π and β – α < π, the load current (and supply current) is
based on equation (11.1) namely
i (ωt ) = is (ωt ) =
2V
Z
(sin(ωt − φ ) − sin (α − φ ) e
− ωt +α
tanφ
)
(A)
(11.65)
α ≤ ωt ≤ π
where Z = R 2 + (ω L ) and φ = tan −1 ω L
2
R
After ωt = π the load current decreases exponentially to zero through the freewheel diode according to
i (ω t ) = iDf (ω t ) = I 01π e −ωt / tan φ
(A)
0 ≤ ωt ≤ α
(11.66)
where for ωt = π in equation (11.65)
I o1π = 2 V
Z
sin(φ − α )(1 − e −π / tanφ )
(
I ½s = ½I o - I ½F = ½
2V
πR
242
)
( cos φ + cos α + sin φ sin (α − φ ) e
2
(α −π ) / tan φ
)
(11.68)
11.2ii - Continuous load current, with α < φ and β − α ≥ π , the load current is given by equations similar
to equations (11.20) and (11.21), specifically

sin φ e −α / tan φ − sin(α − φ ) −ωt +α tan φ 
i (ω t ) = is (ω t ) = 2 V  sin(ωt − φ ) +
e

Z
(11.69)
1 − e −π / tan φ


(A)
α ≤ ωt ≤ π
while the load current when the freewheel diode conducts is
i (ω t ) = iDf (ω t ) = I 01π e −ωt / tan φ
(A)
(11.70)
0 ≤ ωt ≤ α
where, for ωt = π in equation (11.69)
sin φ − sin(α − φ )e −π +α / tanφ
(A)
I 01π = 2 V
Z
1 − e −π / tanφ
The various semiconductor average current ratings can be determined from the average half cycle
freewheeling current, I ½ F , and the average half cycle supply current, I ½ s . For continuous load current
sin φ − sin (α − φ ) e −π +α / tan φ
2V
(11.71)
I ½F = ½
sin φ
(1 − e−α / tan φ )
1 − e −π / tan φ
πR
I ½s = ½I o - I ½F
=½
2V
πR
(
)
− π +α / tan φ
−α


1− e ( )
cos φ  tan φ
e tanφ sin φ − sin (α − φ ) + cos φ + cos (α − φ ) 
−π / tan φ
e
1
−


(11.72)
Table 11.3. Semiconductor average current ratings
Bridge circuit
Number
Average device current
figure 11.7
of devices
Thyristor
Diode
a
4T+1D
1× I ½ s
2× I ½ F
b
2T+1D
1× I ½ s
c
2T+2D
½× I o
2× I ½ F
½× I o
d
2T+2D
1× I ½ s
1× I ½ s + 2× I ½ F
The device conduction table in figure 11.7e can be used to specify average devices currents, for both
continuous and discontinuous load current for each of the circuits in figure 11.7, parts a to d.
For a highly inductive load, constant load current, the supply power factor is pf = 2 π √2cosα.
Critical load inductance
The critical load inductance, to prevent the current falling to zero, is given by
ω Lcrit
α + sin α + π cos θ
(11.73)
= θ − α − ½π +
1 + cos α
R
for α ≤ θ where
V
1 + cos α
θ = sin −1 o = sin −1
(11.74)
π
2V
The minimum current occurs at the angle θ, where the mean output voltage Vo equals the instantaneous
load voltage, vo. When the phase delay angle α is greater than the critical angle θ, θ = α in equation
(11.74) yields (see figure 11.21)
ω Lcrit
α + sin α + π cos α
= −½π +
(11.75)
1 + cos α
R
It is important to note that converter circuits employing diodes cannot be used when inversion is
required. Since the converter diodes prevent the output voltage from being negative, (and the current is
unidirectional), regeneration from the load into the supply is not achievable.
Figure 11.7a is a fully controlled converter with an R-L load and freewheel diode. In single-phase
circuits, this converter essentially behaves as a half-controlled converter.
Power Electronics
243
11.3
Naturally commutating converters
Single-phase controlled thyristor converter circuits
11.3.1 Half-wave controlled circuit with an R-L load
The rectifying diode in the circuit of figure 11.1 can be replaced by a thyristor as shown in figure 11.8a to
form a half-wave controlled rectifier circuit with an R-L load. The output voltage is now controlled by the
thyristor trigger angle, α. The output voltage ripple is at the supply frequency. Circuit waveforms are
shown in figure 11.8b, where the load inductor voltage equal areas are shaded.
The output current, hence output voltage, for the series circuit are given by
di
L + Ri = 2V sin ωt
(V)
(11.76)
dt
α ≤ ωt ≤ β
(rad)
where phase delay angle α and current extinction angle β are shown in the waveform in figure 11.8b and
are the zero load (and supply) current points.
Solving equation (11.76) yields the load and supply current
2V
{sin(ωt - φ ) - sin(α - φ )e(α - ωt ) / tan φ } (A)
i (ωt ) =
Z
where Z =
R + (ω L)
2
2
α ≤ ωt ≤ β
(ohms)
(11.77)
tan φ = ω L / R
and zero elsewhere.
244
The current extinction angle β is dependent on the load impedance and thyristor trigger angle α, and can
be determined by solving equation (11.77) with ωt = β when i(β) = 0, that is
sin( β - φ ) = sin(α - φ ) e(α -β )/ tanφ
(11.78)
This is a transcendental equation. A family of curves of current conduction angle versus delay angle,
that is β - α versus α, is shown in figure 11.9a. The straight line plot for φ = ½π is for a purely inductive
load, whereas ø = 0 is a straight line for a purely resistive load.
The mean load voltage, whence the mean load current, is given by
Vo =
1
2π
∫
β
2 V sin ωt d ωt
α
(11.79)
2V
(cos α − cos β )
2π
Vo = Io R =
(V)
where the angle β can be extracted from figure 11.9a.
The rms load voltage is
Vrms =  12π ∫
α

β
(
2V
) sin ωt dωt 
2
½
2
= V  12π {( β − α ) − ½(sin 2 β − sin 2α )}
(11.80)
½
The rms current involves integration of equation (11.77), squared, giving
½
sin ( β − α ) cos(α + φ + β )  
(11.81)
  (β − α ) −
 
Z  2π 
cos φ
 

Iterative solutions to equation (11.78) are shown in figure 11.9a, where it is seen that two straight-line
relationships exist that relate α and β-α. Exact solutions to equation (11.78) exist for these two cases.
That is, exact tractable solutions exist for the purely resistive load, Φ = 0, and the purely inductive load,
Φ = ½π.
I rms =
q =1 r =1 s =1
p=qxrxs
p=1
V  1 
(a)
(a)
straight line
β=π-α
straight line β = π
π
⅓π
½π
vo
(b)
α
π
2π-α
vL
Eqn (11.89)
√2V × Io
Z
iL
vL
0
⅔π
ωt
vT
v
(c)
Figure 11.8. Single-phase half-wave controlled converter:
(a) circuit diagram; (b) circuit waveforms for an R-L load; and (c) purely inductive load.
1/π
Eqn (11.83)
π
⅓π
½π
π
Delay angle α (degrees)
Figure 11.9. Half-wave, controlled converter thyristor trigger delay angle α versus:
(a) thyristor conduction angle, β-α, and (b) normalised mean load current.
Power Electronics
Naturally commutating converters
11.3.1i - Case 1: Purely resistive load. From equation (11.77), Z = R, φ = 0 , and the current is given
by
2V
sin(ωt )
(A)
i (ωt ) =
(11.82)
R
α ≤ ωt ≤ π and β = π ∀α
The average load voltage, hence average load current, are
11.3.1iii - Case 3: Back emf E and R-L load. With a load back emf, current begins to flow when the
supply instantaneous voltage exceeds the back emf magnitude E, that is when
245
Vo =
1
2π
∫
π
2V sin ωt d ωt
α
(11.83)
2V
(1 + cos α )
2π
Vo = I o R =
(V)
where the maximum output voltage is 0.45V for zero delay angle.
π
(
2V
) sin ωt dωt 
2
½
(11.84)
= V  12π {(π − α ) + ½ sin 2α )}
Since the load is purely resistive, I rms = Vrms / R and the voltage and current factors (form and ripple) are
2
equal. The power delivered to the load is Po = Irms
R.
The supply power factor, for a resistive load, is Pout /Vrms Irms, that is
½
2V
ωL
( sin(ωt − ½π )
- sin (α -½π ) )
(A)
2V
α ≤ ωt ≤ β and β = 2π − α
( cos α - cos ωt )
ωL
The average load voltage, based on the equal area criterion, is zero
1
2π
∫
2π-α
α
(11.87)
2V sin ωt d ωt = 0
(11.88)
The average output current is
I o = 12π
∫
2π −α
2 V
ωL
α
{cos α - cos ωt} d ωt
(11.89)
 (π − α ) cos α + sin α 
πω L 
The rms output current is derived from
½
2π −α
2
2 V 1
I rms =
( cos α - cos ωt ) dωt 
2π α
ω L 
=
2 V
3
V 1 

=
 (π − α )( 2 + cos 2α ) + sin 2α  
2
X  π 

The rms output voltage is
∫ ( 2 V ) sin ωt dωt 
= V 1π {(π − α ) + ½ sin 2α )}
Vrms =  12π

2
(11.93)
which yields
i (ωt ) =
 2V
−ωt
2V
E
E
sin (ωt − φ ) − − e tanφ 
sin (α − φ ) −
 Z
Z
R
R

 α tanφ
 e

(11.94)
2
PL = I rms
R + I oR
(11.95)
while the supply power factor is given by
pf =
I 2 R + I oR
PL
= rms
V I rms
V I rms
(11.96)
∨
The solution for the uncontrolled converter (a half-wave rectifier) is found by setting α = α , eqn (11.92).

2
α
The ac supply of the half-wave controlled single-phase converter in figure 11.8a is v = √2 240 sinωt. For
the following loads
Load #1: R = 10Ω, ωL = 0 Ω
Load #2: R = 0 Ω, ωL = 10Ω
Load #3: R = 7.1Ω, ωL = 7.1Ω
Determine in each load case, for a firing delay angle α = π/6
• the conduction angle γ=β - α, hence the current extinction angle β
• the dc output voltage and the average output current
• the rms load current and voltage, load current and voltage ripple factor, and power
dissipated in the load
• the supply power factor
Solution
Load #1: Z = R = 10Ω, ωL = 0 Ω
From equation (11.77), Z = 10Ω and φ = 0° .
From equation (11.82), β = π for all α, thus for α = π/6, γ = β - α = 5π/6.
From equation (11.83)
Vo = I o R =
=
∫
2π −α
di
+E
dt
Example 11.5: Half-wave controlled rectifier
=
Vo =
2V sin ωt = Ri + L
(11.85)
11.3.1ii - Case 2: Purely inductive load. Circuit waveforms showing equal inductor voltage areas are
shown in figure 11.8c. From equation (11.77), Z = ωL, φ = ½π , and the output voltage and current ares
given by
 2V sin ωt α ≤ ωt ≤ 2π − α
(11.86)
v o (ωt ) = 
elsewhere
0
i (ωt ) =
(11.92)
2V
When current flows, Kirchhoff’s voltage law gives
α ≤ ωt ≤ β and α ≥ α
The extinction angle β is found from the boundary condition i(ωt) = i(β) = 0, for β > π.
The load power is given by
2
α sin 2α
+
pf = ½ 2π
4π
E
∨
The rms output voltage is
Vrms =  12π ∫
α

∨
α = sin−1
246
½
(11.90)
2V
(1 + cos α )
2π
2V
(1 + cos π / 6) = 100.9V
2π
The average load current is
I o = Vo / R =
2V
(1 + cos α ) = 100.9V/10Ω =10.1A.
2π R
The rms load voltage is given by equation (11.84), that is
Vrms = V  12π {(π − α ) + ½ sin 2α )}
½
(11.91)
½
Since the load is purely inductive, Po = 0 and the load voltage ripple factor is undefined since Vo = 0.
By setting α = 0, the equations (11.82) to (11.91) are valid for the uncontrolled rectifier considered in
section 11.1.1, for a purely resistive and purely inductive load, respectively.
½
= 240V ×  12π {(π − π / 6 ) + ½ sin π / 3} = 167.2V
½
Since the load is purely resistive, the power delivered to the load is
2
Po = Irms
R = Vrms2 / R = 167.2V 2 /10Ω = 2797.0W
I rms = Vrms / R = 167.9V /10Ω = 16.8A
Power Electronics
247
For a purely resistive load, the voltage and current factors are equal:
167.2V 16.8A
FFi = FFv =
=
= 1.68
100.9V 10.1A
RFi = RFv = FF 2 − 1 = 1.32
The power factor is
2797W
pf =
= 0.70
240V × 16.7A
Alternatively, use of equation (11.85) gives
π /6 sin π /6
pf = ½ +
= 0.70
2π
4π
=
2 V
  (β − α ) −
Z  2π 

× ( 5π / 6 ) cos π / 6 + sin π / 6  = 14.9A
I rms =
240V  1
10Ω  π
= 240V  12π

3
2 sin
π
}
½

= 37.9A
3 
Since the load is purely inductive, the power delivered to the load is zero, as is the power factor,
and the output voltage ripple factor is undefined. The output current ripple factor is
I
37.9A
FFi = rms =
= 2.54 whence RFi = 2.542 − 1 = 2.34
14.9A
Io
Load #3: R = 7.1Ω, ωL = 7.1Ω
From equation (11.77), Z = 10Ω and φ = ¼π .
From figure 11.9a, for φ = ¼π and α = π / 6 , γ = β – α =195º whence β = 225º.
Iteration of equation (11.78) gives β = 225.5º = 3.936 rad
From equation (11.79)
=
2V
(cos α − cos β )
2π
2 × 240
(cos 30° − cos 225°) = 85.0V
2π
The average load current is
Io = Vo / R
= 85.0V/7.1Ω = 12.0A
Alternatively, the average current can be extracted from figure 11.9b, which for φ = ¼π and
α = π / 6 gives the normalised current as 0.35, thus
Io = 2V
× 0.35
Z
×0.35 = 11.9A
= 2×240V
10Ω
From equation (11.81), the rms current is
½
)
½
½
{( 3.94 − 16 π ) − ½ × (sin ( 2 × 3.94) − sin ( 2 × 16 π ))}
½
= 175.1V
The load current and voltage ripple factors are
18.18A
FFi =
RFi = FFi 2 − 1 = 1.138
= 1.515
12.0A
175.1V
FFv =
RFv = FFv2 − 1 = 1.8
= 2.06
85V
The supply power factor is
2346W
= 0.54
240V × 18.18A
♣
pf =
}
{(π − π 6 ) ( 2 + cos 2α ) +
Vo = Io R =
(
Vrms = V  12π {( β − α ) − ½(sin 2 β − sin 2α )}
Using equations (11.90) and (11.91), the load rms voltage and current are
½
1

Vrms = 240V  π − π + ½ sin π  = 236.5V
6
3
π

{
sin ( β − α ) cos(α + φ + β )  
 
cos φ

248

sin 3.93 − π cos(π + ¼π + 3.93)  
6
6
 (3.93 − π ) −
  = 18.18A


6
cos¼π




The power delivered to the load resistor is
2
Po = Irms
R = 18.18A 2 × 7.1Ω = 2346W
The load rms voltage, from equation (11.80), is
(π − α ) cos α + sin α 
2 240
V  1 

1
=
× 
10Ω
2π

π ωL 
π × 10
I rms =
240V
Load #2: R = 0 Ω, Z = X = ωL = 10Ω
From equation (11.77), Z = X =10Ω and φ = ½π .
From equation (11.87), which is based on the equal area criterion, β = 2π - α, thus for α = π/6, β
= 11π/6 whence the conduction period is γ = β – α = 5π/3.
From equation (11.88) the average output voltage is
Vo = 0V
The average load current is
Io =
Naturally commutating converters
11.3.2 Half-wave half-controlled
The half-wave controlled converter waveform in figure 11.8b shows that when α < ωt < π, during the
positive half of the supply cycle, energy is delivered to the load. But when π < ωt < 2π, the supply
reverses and some energy in the load circuit is returned to the supply. More energy can be retained by
the load if the load voltage is prevented from reversing. A load freewheel diode facilitates this objective.
The single-phase half-wave converter can be controlled when a load commutating diode is incorporated
as shown in figure 11.10a. The diode will prevent the instantaneous load voltage v0 from going negative,
as with the single-phase half-controlled converters shown in figure 11.7.
The load current is defined by equation (11.18) for α ≤ ωt ≤ π and equation (11.19) for π ≤ ωt ≤ 2π + α,
namely:
di
(A)
α ≤ ωt ≤ π
L + Ri = 2 V sin ωt
dt
(11.97)
di
π ≤ ωt ≤ 2π + α
(A)
L + Ri = 0
dt
At ωt = π the thyristor is line commutated and the load current, and hence freewheel diode current, is of
the form of equation (11.21). As shown in figure 11.10b, depending on the delay angle α and R-L load
time constant (L/R), the load current may fall to zero, producing discontinuous load current.
The mean load voltage (hence mean output current) for all conduction cases, with a passive L-R load, is
Vo =
1
2π
∫
π
α
Vo = Io R =
2 V sin ω t d ω t
2V
(1 + cos α )
2π
(11.98)
(V)
which is half the mean voltage for a single-phase
half-controlled converter, given by equation (11.59).
∧
The maximum mean output voltage, V o = 2V / π (equation (11.14)), occurs at α = 0. The normalised
mean output voltage Vn is
∧
Vn = Vo / V o = ½ (1 + cos α )
(11.99)
Power Electronics
249
is
Naturally commutating converters
i
250
11.3.2ii - For continuous conduction the load current is defined by

sin φ e −α / tan φ − sin(α − φ ) −ωt +α tanφ 
i (ω t ) = is (ω t ) = 2 V ×  sin(ωt − φ ) + (
)e

Z
1 − e −2π / tan φ


q =1 r =1 s =1
p=qxrxs
p=1
α ≤ ωt ≤ π
(A)
i (ω t ) = iDf (ω t ) = I 01π e−ωt / tan φ
(11.104)
sin φ − sin(α − φ ) e −π −α / tan φ

= 2V ×
1 − e −π / tan φ
 Z
(A)
π ≤ ωt ≤ 2π + α
The advantages of incorporating a load freewheel diode are
• the input power factor is improved and
• the load waveform is improved (less ripple) giving a better load performance
vo = 0 = vL+vR
vo = v = vL+vR
 −ωt +π / tan φ
e

11.3.3 Full-wave controlled rectifier circuit with an R-L load
I o 1π R
Full-wave voltage control is possible with the circuits shown in figures 11.11a and b. The circuit in figure
11.11a uses a centre-tapped transformer and two thyristors which experience a reverse bias of twice the
supply. At high powers where a transformer may not be applicable, a four-thyristor configuration as in
figure 11.11b is suitable. The voltage ratings of the thyristors in figure 11.11b are half those of the
devices in figure 11.11a, for a given converter input voltage.
Load voltage and current waveforms are shown in figure 11.11 parts c, d, and e for three different phase
control angle conditions.
The load current waveform becomes continuous when the phase control angle α is given by
α = tan −1 ω L / R = φ
(rad)
(11.105)
at which angle the output current is a rectified sine wave. For α > ø, discontinuous load current flows as
shown in figure 11.11c. At α = ø the load current becomes continuous as shown in figure 11.11d,
whence β = α + π. Further decrease in α, that is α < ø, results in continuous load current that is always
greater than zero (no zero current periods), as shown in figure 11.11e.
vR
vL = vR v
L
for ωt > π
Figure 11.10. Half-wave half-controlled converter:
(a) circuit diagram and (b) circuit waveforms for an inductive load.
The Fourier coefficients of the 1-pulse output voltage are given by equation (11.176). For the singlephase, half-wave, half-controlled case, p = 1, thus the output voltage harmonics occur at n = 1, 2, 3, …
The rms output voltage for both continuous and discontinuous load current is

∫ ( 2 V ) sin ωt dωt 
= V  12π {(π − α ) + ½ sin 2α )}
Vrms =  12π

π
2
2
(11.100)
½
(
[sin(ωt - φ ) - sin(α - φ ) e
)
2 V × sin(ωt − φ ) − sin α − φ e −ωt +α tanφ
(
)
Z
α ≤ ωt ≤ π
(A)
i (ω t ) = iDf (ω t ) = I 01π e −ωt / tan φ
−π / tan φ 

)  e −ωt +π / tan φ
=  2 V × sin(φ − α ) (1 − e

 Z

π ≤ ωt ≤ 2π + α
The average thyristor current is
V
IT =
× ( cos 2 φ + cos α + sin φ × sin (α − φ ) × eα −π / tan φ )
2π R
while the average freewheel diode current is
V sin φ
× ( sin φ − × sin (α − φ ) × eα −π / tan φ )
I Df = I o − I T =
2π R
{(α
- ωt ) / tan φ }
]
(A)
(11.106)
(rad)
The mean output voltage for this full-wave circuit will be twice that of the half-wave case in section
11.3.1, given by equation (11.79). That is
Z
Vo = Io R =
11.3.2i - For discontinuous conduction the load current is defined by equation (11.77) during thyristor
conduction
i (ω t ) = is (ω t ) =
2V
i (ωt ) =
α ≤ ωt < β
½
α
11.3.3i - α > φ , β - α < π , discontinuous load current
The load current waveform is the same as for the half-wave situation considered in section 11.3.1, given
by equation (11.77). That is
(11.101)
(A)
=
2V
π
1
π
β
α
d ωt
(11.107)
(V)
where β can be extracted from figure 11.9. For a purely resistive load, β = π. The average output current
is given by I o = Vo / R and the average and rms thyristor currents are ½ I o and I rms / 2 , respectively.
The rms load voltage is
β
Vrms =  π1 ∫ 2 V 2 sin 2 ωt d ωt  ½
 α

(11.108)
½
1

( β − α ) − ½(sin 2β − sin 2α )}
=V
 π {

The rms load current is
V 1 
  (β − α ) −
Z  π 

2
The load power is therefore P = I rms
R.
(11.103)
2 V sin ωt
(cos α - cos β )
I rms =
(11.102)
∫
sin ( β − α ) cos(α + φ + β )  
 
cos φ
 
½
(11.109)
Power Electronics
251
Naturally commutating converters
252
11.3.3iii - α < φ , β - π = α, continuous load current (and also a purely inductive load)
Vo
Under a continuous load current conduction condition, a thyristor is still conducting when another is
forward-biased and is turned on. The first device is instantaneously reverse-biased by the second device
which has been turned on. The first device is commutated and load current is instantaneously
transferred to the oncoming device.
The load current is given by
2V
i (ωt ) =
q =2 r =1 s =1
p=qxrxs
p=2
Z
2 sin(α - φ )
[sin(ωt - φ ) -
1− e
− π / tan φ
e
{(α
- ωt ) / tan φ }
(11.112)
]
This equation reduces to equation (11.110) for α = φ and equation (11.26) for α = 0.
The mean output voltage, whence mean output current, are defined by equation (11.111)
Vo = Io R =
2 2V
π
cos α
(V)
∧
which is uniquely defined by α. The maximum mean output voltage, Vo = 2 2 V / π (equation (11.28)),
occurs at α = 0. Generally, for α > ½π, the average output voltage is negative, resulting in a net energy
transfer from the load to the supply.
The normalised mean output voltage Vn is
∧
Vn = Vo / Vo = cos α
The rms output voltage is equal to the rms input supply voltage and is given by
Vrms =
1
π
∫ (
π +α
α
2V
) sin ωt dωt
2
2
(11.113)
= V
(11.114)
The ac in the output voltage is
2
V ac = V rms
−Vo2 = V 1 +
8
π
cos 2 α
(11.115)
The ac component harmonic magnitudes in the load are given by
2V  1
1
2 cos 2α 
×
+
−
Vn =

 ( n − 1)2 ( n + 1)2 ( n − 1)( n + 1) 
2π


for n even, namely n = 2, 4, 6…
The load voltage form factor, (thence ripple factor), is
FFv =
(11.116)
π
(11.117)
2 2 cos α
The current harmonics are obtained by division of the voltage harmonic by its load impedance at that
frequency, that is
V
Vn
(11.118)
In = n =
n = 0, 2, 4, 6,..
2
Zn
R 2 + ( nω L )
Integration of equation (11.112), squared, yields the load rms current (or equation (11.27) for α = 0)
½
2


 2sin (α − φ ) 
 2sin (α − φ ) 
V  1 
−2π / tan φ
− π / tan φ  
− 4
(11.119)
 tan φ (1 − e
 sin α sin φ (1 − e
)
)
π + 

− π / tan φ
− π / tan φ

Z π 

 1− e

 1− e

 
Thyristor average current is ½ I o , while thyristor rms current rating is I rms / 2 . The same thyristor current
rating expressions are valid for both continuous and discontinuous load current conditions.
I rms =
Figure 11.11. Full-wave controlled converter:
(a) and (b) circuit diagrams; (c) discontinuous load current; (d) verge of continuous load
current, when α = ø; and (e) continuous load current.
11.3.3ii - α = φ , β - α = π , verge of continuous load current
When α = φ = tan ω L / R , the load current given by equation (11.106) reduces to
-1
i (ωt ) =
2V
sin(ωt − φ )
(A)
(11.110)
Z
(rad)
for φ ≤ ωt ≤ φ + π
and the mean output voltage, on reducing equation (11.107) using β = α+π, is given by
2 2V
cos α
(V)
Vo =
(11.111)
π
which is dependent on the load such that α = φ = tan -1 ω L / R . From equation (11.108), with β − α = π ,
the rms output voltage is V, Irms = V/Z, and power = VI rms cos φ .
For a highly inductive load, constant load current, the supply power factor is pf = 2 π (1 + cos α ) π π
The harmonic factor or voltage ripple factor for the output voltage is
RFv =
2
V rms
−Vo2  π 2

=
− cos 2 α 
Vo
 8

−α .
½
which is a minimum of 0.483 for α=0 and a maximum of 1.11 when α=½π.
Critical load inductance (see figure 11.21)
The critical load inductance, to prevent the load current falling to zero, is given by
ω Lcrit
π 

2
2
=
 cos θ + π sin α − π cos α (½π + α + θ ) 
R
2 cos α 

for α ≤ θ where
V
2 cos α
θ = sin −1 o = sin −1
π
2V
(11.120)
(11.121)
(11.122)
Power Electronics
253
Naturally commutating converters
The minimum current occurs at the angle θ, where the mean output voltage Vo equals the instantaneous
load voltage, vo. When the phase delay angle α is greater than the critical angle θ, substituting α = θ in
equation (11.121) gives
ω Lcrit
= − tan α
(11.123)
R
For a purely resistive load
2V
Vo =
(11.124)
(1 + cos α )
.
π
Example 11.6: Controlled full-wave converter – continuous and discontinuous conduction
The fully controlled full-wave, single-phase converter in figure 11.11a has a source of 240V rms, 50Hz,
and a 10Ω 50mH series load. If the delay angle is 45°, determine
i. the average output voltage and current, hence thyristor mean current
ii. the rms load voltage and current, hence thyristor rms current and load ripple factors
iii. the power absorbed by the load and the supply power factor
If the delay angle is increased to 75° determine
iv. the load current in the time domain
v. numerically solve the load current equation for β, the current extinction angle
vi. the load average current and voltage
vii. the load rms voltage and current hence load ripple factors and power dissipated
viii. the supply power factor
Solution
The load natural power factor angle is given by
φ = tan -1 ω L / R = tan -1 ( 2π 50 × 50mH /10Ω ) = 57.5°=1 rad
Continuous conduction
Since α < φ ( 45° < 57.5° ) , continuous load current flows, which is given by equation (11.112).
2 × 240V
2 × sin(1.31 -1)
i (ωt ) =
e
[sin(ωt - 1) 18.62Ω
1 − e −π / 1.56
= 18.2 × [sin(ωt - 1) - 1.62 × e - ωt / 1.56 ]
{(1.31 - ωt ) / 1.56}
]
The dc output voltage component is given by equation (11.111).
From the calculations in the table, the rms load current is
I rms = I o2 + ½
Vn
Zn
Vn
0
(152.79)
10.00
15.28
(233.44)
2
55.65
32.97
1.69
1.42
4
8.16
63.62
0.13
0.01
6
3.03
94.78
½ I n2
0.07
0.00
∑
234.4
½ I n2 =
2
of 15.3A = 10.8A
iii. The power absorbed by the load is
2
PL = I rms
R = 15.3A 2 × 10Ω = 2344W
The supply power factor is
PL
2344W
pf =
=
= 0.64
Vrms I rms 240V × 15.3A
Discontinuous conduction
iv. When the delay angle is increased to 75° (1.31 rad), discontinuous load current flows since the
natural power factor angle φ = tan -1 ω L / R = tan -1 ( 2π 50 × 50mH /10Ω ) = 57.5° ≡ 1rad is exceeded. The load
current is given by equation (11.106)
2 × 240V
(1.31 - ωt ) / 1.56}
i (ωt ) =
[sin(ωt - 1) - sin(1.31 -1) e {
]
18.62Ω
- ωt / 1.56
= 18.2 × [sin(ωt - 1) - 0.71× e
]
v. Solving the equation in part iv for ωt = β and zero current, that is
0 = sin( β - 1) - 0.71× e - β / 1.56
gives β = 4.09 rad or 234.3°.
Vo
90.8V
=
= 9.08A
R
10Ω
vii. The rms load voltage is given by equation (11.108)
harmonic
n
I o2 +
1
π
ii. The rms load current is determined by harmonic analysis. The voltage harmonics (peak magnitude)
are given by equation (11.116)
2V  1
1
2 cos 2α 
×
+
−
for n = 2, 4, 6,..
Vn =

 ( n − 1)2 ( n + 1)2 ( n − 1)( n + 1) 
2π


and the corresponding current is given from equation (11.118)
V
Vn
In = n =
2
Zn
R 2 + ( nω L )
In =
= 234.4 = 15.3A
The rms load voltage is given by equation (11.114), that is 240V.
I
15.3A
FFi = rms =
RFi = FFi 2 − 1 = 1.002 − 1 = 0.0
= 1.0
I o 15.3A
V
240V
FFv = rms =
RFv = FFv2 − 1 = 1.57 2 − 1 = 1.21
= 1.57
V o 152.8V
Io =
Io = Vo / R = 152.8V /10Ω = 15.3A
Each thyristor conducts for 180°, hence thyristor mean current is ½ of 15.3A = 7.65A.
2
2
n
Since each thyristor conducts for 180°, the thyristor rms current is
π
Z n = R 2 + ( nω L )
∑I
vi. The average load voltage from equation (11.107) is
2 240V
Vo =
(cos 75° - cos 234.5°) = 90.8V
i. The average output current and voltage are given by equation (11.111)
2 2V
2 2V
Vo = Io R =
cos α =
cos 45° = 152.8V
π
254
½
Vrms = 240V ×  1 {(4.09 − 1.31) − ½(sin 8.18 − sin 2.62)} = 216.46V
 π

The rms current from equation (11.109) is
½
1 
sin ( 4.09 − 1.31) × cos(1.31 + 1 + 4.09)  
×   ( 4.09 − 1.31) −
= 13.55A
 
18.62Ω  π 
cos1


The load voltage and current form and ripple factors are
I
13.55A
FFi = rms =
RFi = FFi 2 − 1 = 1.492 − 1 = 1.11
= 1.49
9.08A
Io
V
216.46V
FFv = rms =
RFv = FFv2 − 1 = 2.382 − 1 = 2.16
= 2.38
90.8V
Vo
The power dissipated in the 10Ω load resistor is
2
P = I rms
R = 13.552 × 10Ω = 1836W
I rms =
240V
viii. The supply power factor is
PL
1836W
pf =
=
= 0.56
Vrms I rms 240V × 13.55A
♣
Power Electronics
255
Naturally commutating converters
11.3.4 Full-wave, fully-controlled circuit with R-L and emf load, E
An emf source and R-L load can be encountered in dc machine modelling. The emf represents the
machine speed back emf, defined by E = kφω . DC machines can be controlled by a fully controlled
converter configuration as shown in figure 11.12a, where T1-T4 and T2-T3 are triggered alternately.
If in each half sine period the thyristor firing delay angle occurs after the rectified sine supply has fallen
below the emf level E, then no load current flows since the bridge thyristors will always be reversebiased. Thus the zero current firing angle α is:
(
α = sin −1 E / 2V
)
(rad)
for
∧
½π < α < π
q =2 r =1 s =1
p=qxrxs
p=2
T1
T2
T3
T4
(11.125)
where it has been assumed the emf has the polarity shown in figure 11.12a. With discontinuous output
current, load current cannot flow until the supply voltage exceeds the back emf E. That is
(
∨
α = sin −1 E / 2V
)
for
(rad)
256
∨
0< α < ½π
∨
α
(11.126)
Load current can always flow with a firing angle defined by
∨
α ≤α ≤α
(11.127)
(rad)
The load circuit current can be evaluated by solving
di
2 V sin ωt = L + Ri + E
(V)
dt
The load voltage and current ripple are both at twice the supply frequency.
α
(11.128)
11.3.4i - Discontinuous load current
The load current is given by
i (ωt ) =
2V
R
[cos φ sin(ωt - φ ) - E +
2V
{
E
2V
}
- cos φ sin(α - φ ) e
{(α
- ωt ) / tan φ }
]
(11.129)
∨
α ≤ ωt ≤ β < π + α
(rad)
For discontinuous load current conduction, the current extinction angle β, shown on figure 11.12b, is
solved by iterative techniques for i(ωt=β) = 0 in equation (11.129).
cos φ sin( β - φ ) - E +
2V
{
E
2V
}
- cos φ sin(α - φ ) e
{(α
- β ) / tan φ }
=0
(11.130)
The mean output voltage can be obtained from equation (11.107), which is valid for E = 0.
For any E, including E = 0
∫ (
)
β
Vo = 1π
2 V sin ω t + E d ω t
α

E 
Vo =
(V)
 cos α − cos β + (π + α − β )

π
2V 

0 < β −α < π
(rad)
The current extinction angle β is load-dependent, being a function of Z and E, as well as α.
2V
(11.131)
α
Since Vo = E + I o R , the mean load current is given by
Io =
V −E
o
R
=
2V
πR
E


(β − α ) 
 cos α − cos β −
2V


0 < β −α < π
(A)
(11.132)
(rad)
Figure 11.12. A full-wave fully controlled converter with an inductive load which includes an emf
source: (a) circuit diagram; (b) voltage waveforms with discontinuous load current; (c) verge of
continuous load current; and (d) continuous load current.
The rms output voltage is given by
½
2
 2 β −α

β −α
V
2
 V π + E (1 − π ) − 2π (sin 2 β − sin 2α ) 


The rms voltage across the R-L part of the load is given by
Vrms =
2
VRLrms = Vrms
− E2
(V)
(11.133)
(11.134)
The total power delivered to the R-L-E load is
2
Po = I rms
R + IoE
(11.135)
where the rms load current is found by integrating the current in equation (11.129), squared, etc.
11.3.4ii - Continuous load current
With continuous load current conduction, the load rms voltage is V.
The load current is given by
i (ωt ) =
2V
Z
[sin(ωt - φ ) -
E
2V
cos φ
+2
sin(α - φ )
e
e −π / tan φ − 1
{(α
- ωt ) / tan φ }
]
α ≤ ωt ≤ π + α
(11.136)
(rad)
The periodic minimum current is given by
∨
 ½π  E
2V
e −π / tan φ + 1 E
2V
(11.137)
I=
sin (α − φ ) −π / tan φ
sin (α − φ ) tanh 
−
=
−
−1 R
Z
e
Z
 tan φ  R
For continuous load current conditions, as shown in figures 11.12c and 11.12d, the mean output voltage
is given by equation (11.131) with β = π − α
Power Electronics
257
Vo = π1
=
∫
π +α
α
2 2V
π
2 V sin ωt d ωt
cosα
Naturally commutating converters
258
(= E + I R)
o
(11.138)
(V)


p
π
 = 2 V π sin p cos α 


The average output voltage is dependent only on the phase delay angle α (independent of E), unlike the
mean load current, which is given by
V −E
2V  2
E 
Io =
=
(A)
(11.139)
 cosα −

R
R
2V 
π
The power absorbed by the emf source in the load is P = I o E , while the total power delivered to the R-L2
E load is Po = I rms
R + IoE .
The output current and voltage ripple is at multiples of twice the supply frequency. The output voltage
harmonic magnitudes for continuous conduction, given by equation (11.116), are
2V  1
1
2 cos 2α 
Vn =
×
+
−
for n = 2, 4, 6, ..
(11.140)

2
2
2π  ( n − 1) ( n + 1) ( n − 1)( n + 1) 
The dc component across the R-L (and just the resistor) part of the load is
Vo R−L = Vo − E
(11.141)
2 2V
=
× cos α − E
T1
T2
T3
T4
q =2 r =1 s =1
p=qxrxs
p=2
o
α
.
π
The ac component of the output voltage is
 2 2 cos α
2
2
Vac = Vrms
−V o = 1− 

π

and the output voltage form factor is
FFv =
π



2
(11.142)
α
(11.143)
2 2 cos α
Thyristor average current is ½ I o , while thyristor rms current rating is I rms / 2 . These same two thyristor
expressions are valid for both continuous and discontinuous load current conditions.
Critical load inductance
From equation (11.137) set to zero (or i = 0 in equation (11.136)), the boundary between continuous and
discontinuous inductor current must satisfy
 ½π 
R
E
(11.144)
sin (α − φ ) tanh 
 >
Z
2V
 tan φ 
Inversion
If the polarity of the back emf E is reversed as shown in figure 11.13a, waveforms as in parts b and c of
figure 11.13 result. The emf supply can provide a forward bias across the bridge thyristors even after the
supply polarity has gone negative. The zero current angle α now satisfies π < α < 3π/2, as given by
equation (11.125). Thus load and supply current can flow, even for α > π.
The relationship between the mean output voltage and current is now given by
p
π
Vo = − E + I o R = 2 V
sin cos α with p = 2
(11.145)
π
p
That is, the emf term E in equations (11.125) to (11.144) is appropriately changed to - E.
The load current flows from the emf source and if α > ½π, the average load voltage is negative. Power is
being delivered to the ac supply from the emf source in the load, which is an energy transfer process
called power inversion. In general
0 < α < 90° → Vo > 0 Po > 0 io > 0 rectification
90° < α < 180° → Vo < 0 Po < 0 io > 0 inversion
Figure 11.13. A full-wave controlled converter with an inductive load and negative emf source:
(a) circuit diagram; (b) voltage waveforms for discontinuous load current; and (c) continuous load
current.
Example 11.7: Controlled converter – continuous conduction and back emf
The fully controlled full-wave converter in figure 11.11a has a source of 240V rms, 50Hz, and a 10Ω,
50mH, 50V emf opposing series load. The delay angle is 45°.
Determine
i. the average output voltage and current
ii. the rms load voltage and the rms voltage across the R-L part of the load
iii. the power absorbed by the 50V load back emf
iv. the rms load current hence power dissipated in the resistive part of the load
v. the load efficiency, that is percentage of energy into the back emf and power factor
vi. the load voltage and current form and ripple factors
Solution
From example 11.5, continuous conduction is possible since α < φ
i. The average output voltage is given by equation (11.138)
2 2V
cosα
Vo =
π
=
2
2 × 240
π
× cos45° = 152.8V
( 45° < 57.5° ) .
Power Electronics
259
Naturally commutating converters
Example 11.8: Controlled converter – constant load current, back emf, and overlap
The average current, from equation (11.139) is
V − E 152.8V − 50V
=
= 10.28A
Io =
R
10Ω
o
ii. From equation (11.114) the rms load voltage is 240V. The rms voltage across the R-L part of the load
is
2
VRLrms = Vrms
− E2
= 240V 2 − 50V 2 = 234.7V
The fully controlled single-phase full-wave converter in figure 11.11a has a source of 230V rms, 50Hz,
and a series load composed of ½Ω, infinite inductance, 150V emf non-opposing. If the average load
current is to be 200A, calculate the delay angle assuming the converter is operating in the inversion
mode, taking into account 1mH of commutation inductance.
Solution
The mean load current is
Io =
iii. The power absorbed by the 50V back emf load is
P = I o E = 10.28A × 50V = 514W
iv. The R-L load voltage harmonics (which are even) are given by equations (11.140) and (11.141):
2 2V
Vo R−L =
× cos α − E
.
π
 1
1
2 cos 2α 
×
+
−
for

 ( n − 1)2 ( n + 1)2 ( n − 1)( n + 1) 


The harmonic currents and voltages are shown in the table to follow.
Vn =
harmonic
2V
2π
Z n = R 2 + ( nω L )
(Ω )
Vn
n
0
102.79
2
60.02
4
8.16
6
3.26
2
In =
n = 2, 4, 6,..
Vn
Zn
(A)
10.00
10.28
105.66
32.97
1.82
1.66
63.62
0.13
0.01
94.78
0.04
0.00
∑½I
2
n
=
107.33
From the table the rms load current is given by
∑I
2
n
V o (α ) − E
R
Vo (α ) − −150V
200A =
½Ω
which implies a load voltage Vo(α) = -50V.
The output voltage is given by equation (11.111) Vo = 2 π2 V cos α . Commutation of current from one
rectifier to the other takes a finite time. The effect of commutation inductance is to reduce the output
voltage, thus according to equation (11.225), the output voltage becomes
Vo =
½ I n2
I o2 +
I rms = I o2 + ½
260
2V
sin π n cos α − n ω Lc I o / 2π
π /n
where n = 2
2 × 230V
× cos α − 2 × 50Hz × 1mH × 200A
π /2
= 207V × cos α − 20V
which yields α = 98.3º. The commutation overlap causes the output voltage to reduce to zero volts and
the overlap period γ is given by equation (11.226)
−50V =
Io =
2V
2π f Lc
( cos α − cos (γ + α ) )
2 230V
( cos 93.8° − cos (γ + 93.8° ) )
2π 50Hz × 1mH
This gives an overlap angle of γ = 11.2º.
♣
200A =
= 107.33 = 10.36A
The power absorbed by the 10Ω load resistor is
2
PL = I rms
R = 10.36A 2 × 10Ω = 1073.3W
v. The load efficiency, that is, percentage energy into the back emf E is
514W
η=
× 100 o o = 32.4 o o
514W + 1073.3W
The power factor is
PL
514W+1073.3W
pf =
=
= 0.64
Vrms I rms
240V × 10.36A
vi. The output performance factors are
I
10.36A
= 1.011
FFi = rms =
RFi = FFi 2 − 1 = 1.0232 − 1 = 0.125
I o 10.28A
V
240V
= 1.57
FFv = rms =
RFv = FFv2 − 1 = 1.57 2 − 1 = 1.211
V o 152.8V
Note that the voltage form factor (hence voltage ripple factor) agrees with that obtained by substitution
into equation (11.143), 1.57.
♣
11.4
Three-phase uncontrolled converter circuits
Single-phase supply circuits are adequate below a few kilowatts. At higher power levels, restrictions on
unbalanced loading, line harmonics, current surge voltage dips, and filtering require the use of threephase (or higher - polyphase) converter circuits. Generally it will be assumed that the output current is
both continuous and smooth. This assumption is based on the dc load being highly inductive.
11.4.1 Half-wave rectifier circuit with an inductive R-L load
Figure 11.14 shows a half-wave, three-phase diode rectifier circuit along with various circuit voltage and
current waveforms. A transformer having a star connected secondary is required for neutral access, N.
The diode with the highest potential with respect to the neutral conducts a rectangular current pulse. As
the potential of another diode becomes the highest, load current is transferred to that device, and the
previously conducting device is reverse-biased and naturally (line) commutated. Note that the load
voltage, hence current never reaches zero, when the load is passive (no opposing back emf).
In general terms, for an n-phase p-pulse system, the mean output voltage is given by
2V π/p
cos ωt d ωt
(V)
Vo =
2π / p ∫ −π / p
(11.146)
sin(π / p )
= 2V
(V)
π/p
Power Electronics
261
Naturally commutating converters
mmf = N ( i a − i c + i R )
mmf = N ( i b − i a + iY
q =3 r =1 s =1
p=qxrxs
p=3
iR
)
R-L
iY
mmf = N ( i c − i b + i B )
iB
1:N:N
i R + iY + i B = 0 = mmf
c
2π
q =3 r =1 s =1
p=qxrxs
p=3
ωt
2π
3N
2π
3
262
a
3
b
π
Figure 11.15. Three-phase zig-zag interconnected star winding, with three windings per limb, 1:N:N:
(a) transformer connection showing zero dc mmf in each limb (phase) and
(b) phasor diagram of transformer voltages.
π
π
The diode conduction angle is 2π/n, namely ⅔π. The peak diode reverse voltage is given by the
maximum voltage between any two phases, √3√2 V = √6 V.
From equations (11.45), (11.46), and (11.47), for a constant output current, I o = I o rms , the mean diode
current is
I D = 1 n I o = 13 I o
(A)
(11.151)
and the rms diode current is
ID =
1
n Io
rms
≈
1
n Io
=
1
3
Io
(A)
The diode current form factor is
FFID = I D / I D = 3
Figure 11.14. Three-phase half-wave rectifier:
(a) circuit diagram and (b) circuit voltage and current waveforms.
For a three-phase, half-wave circuit (p = 3) the mean output voltage, (thence average current) is
5π / 6
Vo = I o R = 1 2π ∫
2 V sin ωt d ωt
3 π /6
(11.147)
½ 3
= 2V
= 1.17 × V (V)
π /3
The rms load voltage is
2
 3 π
5π / 6
3 
Vrms = 1 2π
2 V sin 2 ωt d ωt = 2 V 
(11.148)
 +
  = 1.19 × V
3 π /6
4  
 2π  3
∫ (
)
The load form factor is
VFF =
Vrms
= 1.19V /1.17V = 1.01
V
ac voltage across the load
RF = ripple factor =
=
dc voltage across the load
(11.149)
Vrms
V
− 1 = 0.185
(11.150)
(11.152)
(11.153)
If neutral is available, a transformer is not necessary. The full load current is returned via the neutral
supply. This neutral supply current is generally not acceptable other than at low power levels. The
simple delta-star connection of the supply in figure 11.14a is not appropriate since the unidirectional
current in each phase is transferred from the supply to the transformer. This may result in increased
magnetising current and iron losses if dc magnetisation occurs. This problem is avoided in most cases
by the special interconnected star winding, called zig-zag, shown in figure 11.15a. Each transformer limb
has two equal voltage secondaries which are connected such that the magnetising forces balance. The
resultant phasor diagram is shown in figure 11.15b. 15% more turns are needed than with a star
connection.
As the number of phases increases, the windings become less utilised per cycle since the diode
conduction angle decreases, from π for a single-phase circuit, to ⅔π for the three-phase case.
11.4.2 Full-wave rectifier circuit with an inductive R-L load
Figure 11.16a shows a three-phase full-wave rectifier circuit where no neutral is necessary and it will be
seen that two series diodes are always conducting. One diode (one of D1, D3, or D5, at the highest
potential) can be considered as being in the feed circuit, while the other (one of D2, D4, or D6, at the
lowest potential) is in the return circuit. As such, the line-to-line voltage is impressed across the load.
Power Electronics
263
Naturally commutating converters
Given no two series connected diodes conduct simultaneously, there are six possible diode pair
combinations. The rectifier circuit waveforms in figure 11.16b show that the load ripple frequency is six
times the supply. Each diode conducts for ⅔π and experiences a reverse voltage of the peak line
voltage, √2 VL.
o
= 2V
π
3
3
π /3
∫
L
π /3
=
3
π
n
.
The critical load inductance (see figure 11.21) for continuous load current, is Lcritical = R
3
2
ω × p ( p 2 − 1)
.
The output harmonics of a p-pulse voltage output are
The mean load voltage is given by twice equation (11.147), that is
2π / 3
V =I R= 1
2 V sin ωt d ωt
(V)
o
Generally the peak-to-peak ripple voltage for n-phases is 2V − 2 cos π
264
(11.154)
V an = −1
2 VL = 1.35VL = 2.34V
= −1
where VL is the line-to-line rms voltage (VL =√3V).
n
n
p
p
×
2V
π
sin π
p
2
p 2
n − 1
(11.155)
2
×Vo ×
2
n − 1
where n = mp and m = 1, 2, 3, … and Vo is the mean output voltage given by equation (11.146).
The output voltage harmonics for p = 6 are given by
a
b
c
Vo n =
6 VL
(11.156)
π ( n 2 − 1)
for n = 6, 12, 18, ..
The rms output voltage is given by
 1
Vrms = 
 2π / 6
q =3 r =1 s =2
p=qxrxs
p=6
∫
2π / 3
π /3
½

2VL sin 2 ωt d ωt 

(11.157)
3 3
= 1.352 VL
2π
Generally, for a p-pulse rectifier output, the rms output voltage is
= VL 1 +
source
voltages
.
V rms = V L 1 +
ab
ac
bc
ba
ca
ca
6,1
1,2
2,3
3,4
4,5
5,6
p
2π
sin 2π
p
(11.158)
The load voltage form factor = 1.352/1.35 = 1.001 and the ripple factor = √form factor -1 = 0.06.
11.4.2i Three-phase full-wave bridge rectifier circuit with continuous load current
output
voltage
If it is assumed that the load inductance is large, then (even with a load back emf), continuous load
current flows and the dominate load current harmonic is due to the sixth harmonic current, that is
let I o,ac = I o,6 . By neglecting the higher order harmonics, the various circuit currents and voltages can be
readily obtained as shown in table 11.3. From equations (11.154) and (11.156) the output voltage is
given by
i a = iD1 - i D4
i b = iD3 - i D6
i c = iD5 - i D2
vo (ωt ) = V o +
Vo , 6 cos 6ωt
3
3
2
2 VL +
2 VL 2
=
cos nωt for n = 6
π
π
( n − 1)
(11.159)
3
3
2
= π 2 VL + π 2 VL × cos 2ωt
35
= 1.35VL + 0.077VL cos 2ωt
The fundamental voltage, hence current, Vo /R, is therefore much larger than the sixth harmonic current,
Vo,6 / Z6, that is I o > I o , 6 . The load and supply ac currents are I o , ac = I s , ac = I o , 6 . The output and supply
rms currents are
2
2
I o , rms = I s , rms = I o + I o2, ac = I o + I o2, 6
and the power delivered to resistance R in the load is
PR = I o2, rms R
Figure 11.16. Three-phase full-wave bridge rectifier:
(a) circuit connection and (b) voltage and current waveforms.
(11.160)
(11.161)
Power Electronics
265
Naturally commutating converters
A phase voltage and current are given by
Table 11.3. Three-phase full-wave uncontrolled rectifier circuits
Full-wave rectifier circuit
load
average output
current
Io
(A)
(A)
(W)
Vs
Vo
Is
Vo
R
I o2, rms R
output power
PR+PE
L
Io
(a)
R-L
6th harmonic
current
Io, 6
circuit
Vo , 6
R
R 2 + ( 6ω L )
see
section
11.4.2i
2
266
v a = 2V sin ωt
(11.168)
S = 3V L I s rms
(11.171)

sin ( n − 1) ωt sin ( n + 1) ωt 
ia =
I o sin ωt +
n = 6, 12, 18, ..
(11.169)
+

n −1
n +1
π


with phases b and c shifted by ⅔π. That is substitute ωt in equations (11.168) and (11.169) with ωt±⅔π.
Each load current harmonic n produces harmonics n+1 and n-1 on the input current.
The total load instantaneous power is given by
cos n ωt 

(11.170)
p (ωt ) = 3 × 2V I o ×  ½ − 2
n − 1 

The apparent power is
2 3
Example 11.9: Three-phase full-wave rectifier
Io
L
R
(b)
Vs
Vo , 6
Is
Vo
E+
R 2 + ( 6ω L )
R-L-E
2
Vo − E
R
I o2, rms R + Io E
The full-wave three-phase rectifier in figure 11.16a has a three-phase 415V 50Hz source (240V phase),
and a 10Ω, 50mH, series load. During the problem solution, verify that the only harmonic that need be
considered is the sixth.
Determine
i. the average output voltage and current
ii. the rms load voltage and the ac output voltage
iii. the rms load current hence power dissipated and power factor
iv. the load power percentage error in assuming a constant load current
v. the diode average and rms current requirements
Solution
Io
L
i. From equation (11.154) the average output voltage and current are
Vo = I o R = 1.35VL = 1.35 × 415V = 560.45V
Io,dc
Io =
Io,ac
(c)
Vs
Is
Vo , 6
Vo
C
R-L-C
6ω L
R
Vo
R
2
I o2, rms R = I o R
Vo 560.45V
=
= 56.045A
R
10Ω
ii. The rms load voltage is given by equation (11.157)
Vrms = 1.352 VL = 1.352 × 415V = 560.94V
The ac component across the load is
2
Vac = Vrms
− Vo2
= 560.94V 2 − 560.447V 2 = 23.52V
11.4.2ii Three-phase full-wave bridge circuit with highly inductive load – constant load current
For a highly inductive load, that is a constant load current:
the mean diode current is
I D = 1 n I o = 13 I o
(A)
and the rms diode current is
I D rms = 1 n I o rms ≈
1
n Io
=
1
3
Io
and the power factor for a constant load current is
3
pf = = 0.955
π
(A)
(11.162)
(11.163)
(11.164)
The rms input line currents are
I L rms =
2
I o rms
3
The diode current form factor is
FFID = I D rms / I D = 3
(11.165)
harmonic
n
Vn =
6 VL
(11.166)
(11.167)
Z n = R 2 + ( nω L )
π ( n 2 − 1)
2
In =
Vn
Zn
½ I n2
0
(560.45)
10.00
56.04
6
32.03
94.78
0.34
0.06
12
7.84
188.76
0.04
0.00
I o2 + ∑ ½ I n2 =
3141.07
Note the 12th harmonic current is not significant
The rms load current is
I rms = I o2 +
The diode current ripple factor is
RFID = FF ID2 − 1 = 2
iii. The rms load current is calculated from the harmonic currents, which are calculated from the
harmonic voltages given by equation (11.156).
∑½I
2
n
= 3141.07 = 56.05A
The power absorbed by the 10Ω load resistor is
2
PL = I rms
R = 56.05A 2 × 10Ω = 31410.7W
The power factor is
.
(3141.01)
Power Electronics
267
pf =
PL
=
Vrms I rms
PL
Naturally commutating converters
31410.7W
= 0.955
2
× 56.05A
3
This power factor 0.955 is as predicted by equation (11.164), 3 π , for a constant current load.
3 VL I L
268
=
3 × 415V ×
R-L
iv. The percentage output power error in assuming the load current is constant is given by
I 2R
P
56.045A 2 × 10Ω
31410.1W
1 − L = 1 − 2o = 1 −
= 1−
0 oo
PL
I rms R
56.05A 2 × 10Ω
31410.7W
v. The diode average and rms currents are given by equations (11.162) and (11.163)
I D = 13 I o = 13 × 56.045 = 18.7A
I D rms =
1
3 I o rms =
1
q =3 r =1 s =2
p=qxrxs
p=6
3 × 56.05 = 23.4A
♣
11.5
Three-phase half-controlled converter
Figure 11.17a illustrates a half-controlled (semi-controlled) converter where half the devices are
thyristors, the remainder being diodes. As in the single-phase case, a freewheeling diode can be added
across the load so as to allow the bridge thyristors to commutate and decrease freewheeling losses. The
output voltage expression consists of √2V 3√3/2π due to the uncontrolled half of the bridge and √2V 3√3
× cos α /2π due to the controlled half which is phase-controlled. The half-controlled bridge mean output
is given by the sum, that is
3 3
3
Vo = 2 V
(1 + cos α ) = 2 VL
(1 + cos α )
2π
2π
(V)
= 2.34 V (1 + cos α )
(11.172)
0≤α ≤π
(rad)
Vo = I o R
At α = 0, V o = √2 V 3√3/π = 1.35 VL, as in equation (11.46). The normalised mean output voltage Vn is
Vn = Vo / V o = ½(1 + cos α )
(11.173)
The diodes prevent any negative output, hence inversion cannot occur. Typical output voltage and
current waveforms for a highly inductive load (constant current) are shown in figure 11.17b.
11.5i - α ≤ ⅓π
When the delay angle is less than ⅓π the output waveform contains six pulses per cycle, of alternating
controlled and uncontrolled phases, as shown in figure 11.17b. The output current is always continuous
(even for a resistive load) since no output voltage zeros occur. The rms output voltage is given by
2
2
α + 2π / 3
3 2π / 3
Vrms =
2VL sin 2 ωt d ωt + ∫
2VL sin 2 ωt d ωt
π /3
2π ∫α +π / 3
{
(
)
(
 3 3

= VL 1 +
(1 + cos 2α ) 

4π


)
}
½
(11.174)
for α ≤ π / 3
11.5ii - α ≥ ⅓π
For delay angles greater than ⅓π the output voltage waveform is made up of three controlled pulses per
cycle, as shown in figure 11.17c. Although output voltage zeros result, continuous load current can flow
through a diode and the conducting thyristor, or through the commutating diode if employed. The rms
output voltage is given by
2
3 π
Vrms =
2 VL sin 2 ω t d ωt
2π ∫α
(
)
½
 3
= VL 
(π − α + ½ sin 2α ) 
 2π

(11.175)
for α ≥ π / 3
Figure 11.17. Three-phase half-controlled bridge converter: (a) circuit connection; (b) voltage and
current waveforms for a small firing delay angle α; and (c) waveforms for α large.
The Fourier coefficients of the p-pulse output voltage are given by
cos ( n + 1) α cos ( n − 1) α 
2V  −2
an =
−
+


2π  n 2 − 1
n +1
n −1


p
bn =
2V  sin ( n + 1) α sin ( n − 1) α 
−


2π 
n +1
n −1

p
(11.176)
Power Electronics
Naturally commutating converters
where n = mp and m = 1, 2, 3, .. For the three-phase, full-wave, half-controlled case, p = 6, thus the
output voltage harmonics occur at n = 6, 12, …
The mean output voltage for an n-phase half-wave controlled converter is given by (see example 11.8)
2 V α +π / n
cos ωt d ωt
Vo =
2π / n ∫ α −π / n
(11.177)
sin(π / n )
= 2V
cos α
(V)
π /n
which for the three-phase circuit considered with continuous or discontinuous (R) load current gives
3 3
2 V cos α = 1.17 V cos α
Vo = Io R =
(11.178)
0 ≤α ≤π /6
2π
269
11.6
Three-phase, fully controlled thyristor converter circuits
11.6.1 Half-wave, fully-controlled circuit with an inductive load
When the diodes in the circuit of figure 11.14 are replaced by thyristors, as in figure 11.18a, a threephase fully controlled half-wave converter results. The output voltage is controlled by the delay angle α.
This angle is specified from the thyristor commutation angle, which is the earliest point the associated
thyristor becomes forward-biased, as shown in parts b, c, and d of figure 11.18. (The reference is not the
phase zero voltage cross-over point). The thyristor with the highest instantaneous anode potential will
conduct when fired and in turning on will reverse bias and turn off any previously conducting thyristor.
The output voltage ripple is three times the supply frequency and the supply currents contain dc components. Each phase progressively conducts for periods of π, displaced by α, as shown in figure 11.18b.
q =3 r =1 s =1
p=qxrxs
p=3
270
For discontinuous conduction, and a resitive load, the mean output voltage is
3
Vo = Io R =
2 V (1 + cos (α + π / 6 ) )
π / 6 ≤ α ≤ 5π / 6
(11.179)
2π
The mean output voltage is zero for α = ½π. For 0 < α < π, the instantaneous output voltage is always
greater than zero. Negative average output voltage occurs when α > ½π as shown in figure 11.18d.
Since the load current direction is unchanged, for α > ½π, power reversal occurs, with energy feeding
from the load into the ac supply. Power inversion assumes a load with an emf to assist the current flow,
as in figure 11.13. If α > π no reverse bias exists for natural commutation and continuous load current
will freewheel.
The maximum mean output voltage V o = √2V 3√3 /2π occurs at α = 0. The normalised mean output
voltage Vn is
Vn = Vo / Vˆo = cos α
(11.180)
With an R-L load, at Vo = 0, the load current falls to zero. Thus for α > ½π, continuous load current does
not flow for an R-L load.
The rms output voltage is given by
2
3 α +π / 3
2
Vrms =
2 V sin (ωt ) d ωt
2π ∫ α −π / 3
(11.181)
½
1
3
= 3 2 V  + sin 2α 
 6 8π

From equations (11.178) and (11.181), the ac in the output voltage is
(
)
1
2
−Vo2 = 3 2 V  +
V ac = V rms
6

3
8π
sin 2α −
3
2π
½
cos2 α 

(11.182)
The output voltage distortion ripple factor is
RFv =
11.6.2
2π 2
27
+
3π
18
sin 2α − cos2 α
(min ( at α =0 ) = 0.173; max ( at α =½π ) = 0.66)
(11.183)
Half-wave converter with freewheel diode
Figure 11.19 shows a three-phase, half-wave controlled rectifier converter circuit with a load freewheel
diode, Df. This diode prevents the load voltage from going negative, thus inversion is not possible.
q =3 r =1 s =1
p=qxrxs
p=3
Figure 11.18. Three-phase half-wave controlled converter:
(a) circuit connection; (b) voltage and current waveforms for a small firing delay angle α; (c) and (d)
load voltage waveforms for progressively larger delay angles.
Df
Figure 11.19. A half-wave fully controlled three-phase converter with a load freewheel diode.
Power Electronics
271
Naturally commutating converters
11.6.2i - α < π/6. The output is as in figure 11.18b, with no voltage zeros occurring. The mean output
voltage (and current) is given by equation (11.178), that is
3 3
Vo = Io R =
(11.184)
2V cos α = 1.17V cos α
(V)
0 ≤α ≤π /6
(rad)
2π
The maximum mean output Vo = √2V 3√3/2π occurs at α = 0. The normalised mean output voltage, Vn is
given by
Vn = Vo / V o = cos α
(11.185)
The Fourier coefficients of the 3-pulse output voltage are given by (11.176). For the three-phase, halfwave, half-controlled case, p = 3, thus the output voltage harmonics occur at n = 3, 6, 9, …
11.6.2ii - α > π/6. Because of the freewheel diode, voltage zeros occur and the negative portions in the
waveforms in parts c and d of figure 11.18 do not occur. The mean output voltage is given by
2V π
sin ωt d ωt
Vo = Io R =
2π / 3 ∫ α −π / 6
2V
=
(V)
(11.186)
(1 + cos (α + π / 6 ) )
2π / 3
π / 6 ≤ α ≤ 5π / 6
The normalised mean output voltage Vn is
Vn = Vo / V o = [1 + cos(α + π / 6)]/ 3
The average load current (with an emf E in the load) is given by
V −E
Io = o
R
These equations assume continuous load current.
(11.187)
(11.188)
11.6.2iii - α > 5π/6. A delay angle of greater than 5π/6 would imply a negative output voltage, clearly
not possible with a freewheel load diode.
272
di c
di c
= Xc
dt
d ωt

2V 
π

ic =
 cos  α +  − cos ωt 
Xc 
6


Solving for when the current rises to the load current I oγ gives
2V 
π
π



I oγ =
 cos  α +  − cos  α + + γ  
Xc 
6
6



2V sin ωt = Lc
but
I oγ =
Voγ
2V
=
(1 + cos (α + π / 6 + γ ) )
R R 2π / 3
Xc
π

cos  α +  −
π
6  R 2π / 3



= cos  α + + γ 
Xc
6


+1
R 2π / 3
Xc 

π

 cos  α +  −
π
R
π / 3  
6
2


− α +  = 0.68°
γ = cos 
Xc

 
6
+
1


R 2π / 3


The load current and voltage are therefore
2V 
π
π
2 240V



I oγ =
( cos ( 90° ) − cos ( 90.68 ) ) = 16.11A
 cos  α +  − cos  α + + γ   =
6
6
Xc 
¼Ω



−1
Voγ = I oγ R = 16.11A × 10Ω = 161.1V
♣
11.6.3 Full-wave, fully-controlled circuit with an inductive load
Example 11.10:
Three-phase half-wave rectifier with freewheel diode
The half-wave three-phase rectifier in figure 11.19 has a three-phase 415V 50Hz source (240V phase),
and a 10Ω resistor and infinite series inductance as a load. If the delay angle is 60º determine the load
current and output voltage if:
i. the phase commutation inductance is zero
ii. the phase commutation reactance is¼Ω
A three-phase bridge is fully controlled when all six bridge devices are thyristors, as shown in figure
11.20a. The frequency of the output ripple voltage is six times the supply frequency and each thyristor
always conducts for ⅔π. Circuit waveforms are shown in figure 11.20b. The output voltage is
continuous, and the mean output voltage for both inductive and resistive loads is given by
3 α +½ π
Vo = ∫
3 2 V sin (ω t + π / 6 ) d ω t
π
=
Solution
i.
The output voltage, without any line commutation inductance and a 60º phase delay angle, is
given by equation (11.186)
Vo = I o R =
2V
2π / 3
(1 + cos (α + π / 6 ) )
2 240 V
(1 + cos ( 60° + π / 6 ) ) = 162V
2π / 3
The constant load current is therefore
V
162V
= 16.2A
Io = o =
10Ω
R
=
ii.
When the current changes paths, any inductance will control the rate at which the commutation
from one path to the next occurs. The voltage drops across the commutating inductors modifies the
output voltage. Since the voltage across the freewheel diode is not associated with commutation
inductance, the output voltage is not effected when the current swaps from a phase to the freewheel
diode. But when the current transfers from the diode to a phase, while the commutation inductance
current in the phase is building up to the constant load current level, the output remains clamped at the
diode voltage level, viz. zero. The average voltage across the load during this overlap period is therefore
reduced. The commutation current is defined by
α +π / 6
3 3
π
2 V cos α = 2.34 V cos α
(V)
(11.189)
0 ≤ α ≤ 2π / 3
which is twice the voltage given by equation (11.178) for the half-wave circuit, but for a purely resistive
load the output voltage is discontinuous and equation (11.189) becomes
3 3
Vo =
(V)
2 V 1 + cos (α + π / 6 ) 
(11.190)
π
π / 3 ≤ α ≤ 2π / 3
The average output current is given by I o = Vo / R in each case. If a load back emf exists the average
current becomes
Vo − E
Io =
(11.191)
R
The maximum mean output voltage V o = √2V 3√3/π occurs at α = 0. The normalised mean output Vn is
Vn = Vo / V o = cos α
(11.192)
For delay angles up to ⅓π, the output voltage is at all instances non-zero, hence the load current is
continuous for any passive load (both resistive and inductive). Beyond ⅓π the load current may be
discontinuous (always discontinuous for a resistive load). For α > ½π the current is always
discontinuous for passive loads (no back emf, E) and the average output voltage is less than zero.
Power Electronics
273
Naturally commutating converters
274
The rms value of the output voltage for a purely resistive load is given by
q =3 r =1 s =2
p=qxrxs
p=6
3
Vrms = 
π
∫
α +π / 2
α −π / 6
3
(
= 3 2 V 1+
½
2V
) sin (ωt ) dωt 
3 3
2π
cos 2α
2
2
0 ≤α ≤π /3
(11.197)
π / 3 ≤ α ≤ 2π / 3
(11.198)
and
Vrms = 3 2 V 1 −
3α
2π
−
3
4π
sin ( 2α − π / 3)
The output voltage ripple factor (with continuous current) is
π2
RFv =
18
3π
+
12
(min ( at α =0 ) = 0.025; max ( at α =½π ) = 0.3)
cos 2α − cos 2 α
(11.199)
The normalise voltage harmonic peak magnitudes in the output voltage, with continuous load current,
are
1
VL n = 2 V
.
2
3 3 1
1
2 cos 2α 
+
−


π  ( n − 1)2 ( n + 1)2 ( n − 1)( n + 1) 
(11.200)
.
for n = 6, 12, 18…
The harmonics occur at multiples of six times the fundamental frequency.
For discontinuous load current, at high delay angles, when the output current becomes discontinuous
with an inductive load, the output current is given by
i (ωt ) =
(
3 2V 
sin ωt + π

Z
6
)
(
− φ − sin α + π
3
)
−φ e
− ωt − π 6 + α
tan φ
 − E 1 − e −ωt −π 6 +α tan φ 
 R 

α ≤ ωt ≤ α + θc
(11.201)
where θc is the conduction period, which is found by solving the transcendental equation formed when in
equation (11.201), i(ωt = α+π+θc) = 0. The average output voltage can then be found from
(
3 3 2V 
cos α + π
Vo =
π

20°
30°
3
) − cos (α + π 3 + θ ) − 3πE π 3 − θ 
c
(11.202)
c
fully-controlled converter
delay angle α
0
¼π
60°
70°
80°
½π°
10
Figure 11.20. A three-phase fully controlled converter:
(a) circuit connection and (b) load voltage waveform for four delay angles.
Single-phase
2 pulse
5
sing
3 2V
i ( ωt ) =
Z
(
sin ωt + π
6
)
−φ −
E
R
+
3 2V
Z
sin (α − φ )
− ωt + π 6 + α
e
e
− π 3 tan φ
tan φ
−1
(11.193)
The maximum and minimum ripple current magnitudes are
I =
3 2V
Z
(
sin α + π
2
)
−φ −
E
R
+
3 2V
Z
e
sin (α − φ )
e
−
π 6
π 3
(11.194)
−1
I =
3 2V
Z
(
sin α + π
3
)
−φ −
R
+
3 2V
Z
1
sin (α − φ )
e
−
π 3
tan φ
fully
-con
tr oll
ed
Semi-controlled
0.5
three-phase
1
0.1
0.8
0.05
E
as e
single-phase
1
tan φ
tan φ
at ωt = α +nπ for n = 0, 6, 12, ..
∨
ω Lcrit / R
For continuous load current, the load current is given by
le -ph
thre e
-pha
se fu
0.6
lly-c
ontr
(11.195)
0.4
0.2
←
olle d
0
per unit dc output voltage Vo
Three-phase
6 pulse
−1
at ωt = α - π +n π for n = 0, 6, 12, ..
∨
With a load back emf the critical inductance for continuous load current must satisfy I = 0 in equation
(11.195), that is
sin (α − φ ) 
R 
E
× sin (α − φ + 13 π ) + −π / 3 tan φ
(11.196)
≥
−1
Z 
e
3 2V
where tan φ = ω L / R .
0
30°
¼π
60°
75°
½π
105°
120°
¾π
150°
π
delay angle α
semi-controlled converter
Figure 11.21. Critical load inductance (reactance) of single-phase (two pulse) and three-phase (six
pulse), semi-controlled and fully-controlled converters, as a function phase delay angle α whence dc
output voltage Vo. For rectifier, α = 0.
Power Electronics
275
11.6.3i - Resistive load
For a resistive load, the load voltage harmonics for p pulses per cycle, are given by
cos ( n + 1) α cos ( n − 1) α 
2V  −2
an =
−
−


2π  n 2 + 1
n +1
n −1


p
bn =
2V  sin ( n + 1) α sin ( n − 1) α 
−


2π 
n +1
n −1

p
Naturally commutating converters
pf =
3 Vrms ×
(11.203)
for n = pm and m = 1, 2, 3, .. The harmonics occur at multiples of six times the fundamental frequency,
for a 6 - pulse (p = 6) per cycle output voltage.
As with a continuous load current, with a constant load current the input current comprises ⅔π
alternating polarity blocks of current, with each phase displaced relative to the others by ⅔π,
independent of the thyristor triggering delay angle. At maximum voltage hence maximum power output,
the delay angle is zero and the phase voltage and current fundamental are in phase. As the phase
angle is increased, the inverter output voltage, hence power output is decreased, and the line current
block of current (fundamental) shifts by α with respect to the line voltage. Reactive input power
increases as the real power decreases. At α = ½π, the output voltage reduces to zero, the output power
is zero, and the ⅔π current blocks in the ac input are shifted ½π with respect to the line voltage,
producing only VAr’s from the ac input. When the delay angle is increased above ½π, the inverter dc
output reverses polarity and energy transfers back into the ac supply (inversion), with maximum inverted
power reached at α = π, where the reactive VAr is reduced to zero, from a maximum at α = ½π.
For a highly inductive load, that is a constant load current:
the mean diode current is
I Th = 1n I o = 13 I o
(A)
(11.204)
1
=
n Io
1
3
Io
(A)
The diode current form factor is
FFITh = ITh rms / I Th = 3
i.
ii.
(11.207)
(11.208)
v a = 2 V sin ωt
(11.209)
iii.
A phase voltage is given by
π n
2
π
Io
iv.
(11.210)
v.
(11.211)
Io =
1
3
× 100A = 57.7A
The rms and fundamental line currents are
2
2
I L rms =
I o rms =
× 100A = 81.6A
3
3
2
2
I1rms = 3
Io = 3
× 100A = 78.0A
π
The apparent power is
The supply power factor, from equation (11.213), is
PL
25kW
 3

pf =
=
= 0.426
 = π cos α 
3 Vrms I rms
3 × 415V × 81.6A


The output voltage is
Vo = power = 25kW
= 250V dc
100A
Io
The load resistance is
The supply fundamental apparent power, S1, active power P and reactive power Q, are given by
RL =
S 1 = 3VI 1rms = P 2 + Q 2
P = P1 = S 1 cos α
3
S 1 = 3VI 1rms = 3 × 415V × 78A = 56.1kVA
The rms fundamental input current is
I 1rms = 3
1
π
2
I o rms
3
with phases b and c shifted by ⅔π. That is substitute ωt with ωt±⅔π.
From equation (14.34), the line current harmonics are
4 1
cos ½ ( 1 3 π ) n
for n odd
ii = I o
From equations (11.204) and (11.205) the thyristor average and rms currents are
I Th = 1 3 I o = 1 3 × 100A = 33 1 3 A
I Th rms =
The rms input line currents are
I L rms =
Three-phase full-wave controlled rectifier with constant output current
Solution
The diode current ripple factor is
RF ITh = FF ITh2 − 1 = 2
(11.213)
The full-wave three-phase controlled rectifier in figure 11.20a has a three-phase 415V 50Hz source
(240V phase), and provides a 100A constant current load.
Determine:
i.
the average and rms thyristor current
ii.
the rms and fundamental line current
iii.
the apparent fundamental power S1
If 25kW is delivered to the dc load, calculate:
iv.
the supply power factor
iv.
the dc output voltage, load resistance, hence the converter phase delay angle
v.
the real active and reactive Q1 ac supply power
vi.
the delay angle range if the ac supply varies by ±5% (with 25kW and 100A dc).
(11.205)
(11.206)
2
Io
3
2
I o cos α
3
π
= cos α
π
2
3 Vrms ×
Io
3
3 Vrms 3
=
Converter shut down is best achieved regeneratively by increasing (and controlling) the delay angle to
greater than ½π such that the output voltage goes negative, which results in controlled power inversion
back into the ac supply. Undesirably, if triggering pulses to all the thyristors are removed, the dc current
decays slowly and uncontrolled to zero through the last pair of thyristors that were triggered.
Example 11.11:
11.6.3ii - Highly inductive load – constant load current
and the rms diode current is
ITh rms = 1 n I o rms ≈
3 Vrms I1rms cos α
276
Vo 250V
=
= 2.5Ω
I o 100A
Thyristor delay angle is given by equation (11.189), that is
Vo = 2.34 V cos α
(11.212)
Q = Q1 = S 1 sin α
250Vdc = 2.34 × 415V
× cos α
3
which yields a delay angle of α = 1.11rad = 63.5°
The supply apparent power is constant for a given constant load current, independent of the thyristor
turn-on delay angle.
The power factor for a constant load current is
vi.
For a constant output power at 100A dc, the output voltage must be maintained at 250V
dc independent of the ac input voltage magnitude, thus for equation (11.189)
Power Electronics
277
α = cos −1
Naturally commutating converters
278
In each case the average output current is given by I o = Vo / R , which can be modified to include any load
back emf, that is, I o = (Vo − E ) / R .
250Vdc
2.34 × ( 415 ± 5% )
3
∨
α = cos
−1
250Vdc
2.34 × ( 415 − 5% )
= 1.08 rad = 61.9°
3
α = cos −1
250Vdc
2.34 × ( 415 + 5% )
= 1.13 rad = 64.9°
3
♣
− πp
q =3 r =1 s =2
p=qxrxs
p=6
+ πp
0
2π
(a)
p
2V 2 V s i n n π
sin p π
p π nπ
+
rectify
½
π
π
invert
Figure 11.22. A full-wave three-phase controlled converter
with a load freewheeling diode (half-controlled).
−
11.6.4 Full-wave converter with freewheel diode
Both half-controlled and fully controlled converters can employ a discrete load freewheel diode. These
circuits have the voltage output characteristic that the output voltage can never go negative, hence
power inversion is not possible. Figure 11.22 shows a fully controlled three-phase converter with a
freewheel diode D.
• The freewheel diode is active for α > ⅓π. The output is as in figure 11.20b for α < ⅓π. The
mean output voltage is
Vo = I o R =
3 3
π
2 V cos α = 2.34V cos α
0 ≤α ≤π /3
(V)
(11.214)
(rad)
The maximum mean output voltage V o = √2V 3√3/π occurs at α = 0.
The normalised mean output voltage Vn is given by
Vn = Vo / V o = cos α
•
3 3
π
(
2 V 1 + cos (α + π / 3)
Example 11.12:
π / 3 ≤ α ≤ 2π / 3
(V)
Solution
Figure 11.23 defines the general output voltage waveform where p is the output pulse number per cycle
of the ac supply. From the output voltage waveform
π / n+α
1
Vo =
2 V cos ω t d ω t
2π / p ∫ −π / n+α
•
2π / 3 ≤ α
(V)
(rad)
2V
( sin(α + π / p) − sin(α − π / p) )
2π / p
=
2V
2sin(π / p ) cos α
2π / p
Vo =
(11.217)
while
Vo = 0
=
(11.216)
(rad)
The normalised mean output, Vn, is
Vn = Vo / V o = 1 + cos (α + π / 3)
Converter average load voltage
Derive a general expression for the average load voltage of an p-pulse controlled converter.
(11.215)
)
(11.218)
(b)
Figure 11.23. A half-wave n-phase controlled converter:
(a) output voltage and current waveform and (b) transfer function of voltage versus delay angle α.
while
Vo = I o R =
2V2 V
sins ipn πn π
−
πn π
p
2V
π/p
sin(π / p ) cos α
= Vo cos α
(V)
where
for p = 2 for the single-phase (n = 1) full-wave controlled converter in figure 11.11.
for p = 3 for the three-phase (n = 3) half-wave controlled converter in figure 11.18.
for p = 6 for the three-phase (n = 3) full-wave controlled converter in figure 11.20.
♣
Power Electronics
279
11.7
Naturally commutating converters
280
0º
Overlap
In the previous sections of this chapter, impedance of the ac source has been neglected, such that
current transfers or commutates instantly from one switch to the other with higher anode potential, when
triggered. However, in practice the source has inductive reactance Xc and current takes a finite time to
fall in the device turning off and rise in the device turning on.
Consider the three-phase half-wave controlled rectifying converter in figure 11.18a, where it is assumed
that a continuous dc load current, Io, flows. When thyristor T1 is conducting and T2 (which is forward
biased) is turned on after delay α, the equivalent circuit is shown in figure 11.24a. The source
reactances X1 and X2 limit the rate of change of current in T1 as i1 decreases from Io to 0 and in T2 as i2
increases from 0 to Io. These current transitions in T1 and T2 are shown in the waveforms of figure
11.24d. A circulating current, i, flows between the two thyristors. If the line reactances are identical, the
output voltage during commutation, vγ, is mid-way between the conducting phase voltages v1 and v2, as
shown in figure 11.24b. That is vγ = ½(v1 + v2), creating a series of notches in the output voltage
waveform as shown in figure 11.24c. This interval during which both T1 and T2 conduct (i ≠ 0) is termed
the overlap period and is defined by the overlap angle γ. Ignoring thyristor voltage drops, the overlap
angle is calculated as follows:
α α+γ
Vγ = ½ (v1 + v2)
(b)
√2 VLL
VT3
With reference t = 0 when T2 is triggered
v2 − v1 = vL = 3 v phase = 3 2 V sin (ωt + α )
v3-1
α+γ
where V is the line to neutral rms voltage.
Equating these two equations
2 L di / dt = 3 2 V sin (ωt + α )
0º
α
(e)
π
Rearranging and integrating gives
3 2V
i (ωt ) =
2ω L
( cos α − cos (ωt + α ) )
(11.219)
commutation
voltage
δ
recovery angle = ωtq
Commutation from T1 to T2 is complete when i = Io, at ωt = γ, that is
(11.220)
(A)
( cos α − cos (γ + α ) ) = 23πωVLo ( cos α − cos (γ + α ) )
2ω L
Figure 11.24b shows that the load voltage comprises the phase voltage v2 when no source inductance
exists minus the voltage due to circulating current vγ (= ½(v1 + v2)) during commutation.
Io =
3 2V
(c)
The mean output voltage Voγ is therefore
Voγ = Vo − v γ
=
1 
2π / 3 
α + 5π / 6
∫v
2
d ωt −
α +π / 6
∫


γ +α +π / 6
v d ωt
γ
α +π / 6
where vγ = ½(v1 + v2)
Voγ =

3 
2π 

∫
α +5π / 6
α +π / 6



2 V sin (ω t + 2π 3 ) + sin ωt d ω t 

commutation
voltage
2 V sin (ω t + α ) d ω t
−
∫
γ +α +π / 6
α +π / 6
{
}
v2 − v1 = 2 L di / dt
Voγ =
3
2π
3 2 V cos α −
3 3
2 V ( cos α − cos (α − γ ) )
2π 2
(11.221)
(d)
3 3
Voγ =
(11.222)
2 V  cos α + cos (α + γ )  = ½Vo  cos α + cos (α + γ ) 
4π
which reduces to equation (11.178) when γ = 0. Substituting cos α - cos (α + γ) from equation (11.220)
into equation (11.221) yields
3 3
3
3
3 3
Voγ =
ω LI o = Vo −
ω LI o where Vo =
(11.223)
2 V cos α −
2V
2π
2π
2π
2π
The mean output voltage Vo is reduced or regulated by the commutation reactance Xc = ωL and this
regulation varies with load current magnitude Io. Converter semiconductor voltage drops also regulate
(decrease) the output voltage.
The component 3ωL/2π is called the equivalent internal resistance. Being an inductive phenomenon, it
does not represent a power loss component.
Figure 11.24. Overlap: (a) equivalent circuit during overlap; (b) angle relationships; (c) load voltage for
different delay angles α (hatched areas equal to IoL; last overlap shows commutation failure); (d) thyristor
currents showing eventual failure; and (e) voltage across a thyristor in the inversion mode, α >90°.
Power Electronics
281
Naturally commutating converters
282
v2
vo
As shown in figure 11.25, the overlap occurs immediately after the delay α. The commutation voltage,
v2 - v1, is √3 √2 V sin α. The commutation time is inversely proportional to the commutation voltage v2 v1.
For rectification, as α increases from zero to ½π, the commutation voltage increases from a minimum of
zero
volts to a maximum of √3
√2 V at ½π, whence the overlap angle γ decreases from a maximum of
∧
∨
γ at α = 0 to a minimum of γ at ½π.
∨
∧
[For inversion, the overlap angle γ decreases from a minimum of γ at ½π to a maximum of γ at π, as
the commutation voltage reduces from a maximum, back to zero volts.]
From equation (11.220), with α = π
½(v1+v2 )
v1
∨
The general expressions for the mean load voltage Voγ of an n-pulse, fully-controlled converter, with
underlap, are given by
2V
(11.224)
Voγ =
sin π n cos α + cos (α + γ ) 
2π / n
and
2V
Voγ =
sin π n cos α ∓ nX c I o / 2π
(11.225)
π /n
where V is the line voltage for a full-wave converter and the phase voltage for a half-wave converter and
the plus sign in equation (11.225) accounts for inversion operation.
Effectively, as shown in figure 11.26, during rectification, overlap reduces the mean output voltage by
nfLIo or as if α were increased. The supply voltage is effectively distorted and the harmonic content of
the output is increased. Equating equations (11.224) and (11.225) gives the mean output current
2V
Io =
sin π n ( cos α − cos (γ + α ) )
(A)
(11.226)
Xc
which reduces to equation (11.220) when n = 3.
Harmonic input current magnitudes are decreased by a factor sin (½ nγ ) / ½ nγ .
In the three-phase case, for a constant dc link current Io, without commutation effects, the rms
phase current and the magnitude of the nth current harmonic are
2I o
2 3
(11.227)
I rms =
Ihn = Io
nπ
3
When accounting for commutation reactance effects the fundamental current is
I h1
2
δ
2 3V
ωL
i1
γ
Io
i2
Io
α
= 135°
rectifying
Vo > 0
Inverting
Vo < 0
i1
Io
∨
γ
Io
i2
2 3V
ωL
α
= 90°
i1
γ
Io
Io
equation
(11.219)
i2
α
= 60°
i1
γ
Io
Io
i2
ωt
½
α = 0°
(11.228)
The single-phase, full-wave, converter voltage drop is 2ωLIo /π and the overlap output voltage is zero.
The general effects of line inductance, which causes current overlap are:
• the average output voltage is reduced
• the input voltage is distorted
• the inversion safety angle to allow for thyristor commutation, is increased
• the output voltage spectrum component frequencies are unchanged but there
magnitudes are decreased slightly.
11.8
vγ
i(ωt)
γ = arc sin(2ω LI o / 2 3 V )
2
2



 
2 3  cos 2α - cos 2 (α + γ )  + 2γ + sin 2α − sin 2 (α + γ ) 
= Io
π
4  cos α − cos (α + γ ) 
v2-v1
π
½π
Figure 11.25. Overlap γ for current commutation from thyristor 1 to thyristor 2, at delay angle α.
nX/2π
Voγ
Io
Io=0, γ
rectification
Overlap - inversion
γ
A fully controlled converter operates in the inversion mode when α > 90° and the mean output voltage is
negative and less than the load back emf shown in figure 11.24a. Since the direction of the load current
Io is from the supply and the output voltage is negative, energy is being returned, regenerated into the
supply from the load. Figure 11.27 shows the power flow differences between rectification and inversion.
As α increases, the returned energy magnitude increases. If α plus the necessary overlap γ exceeds ωt
= π, commutation failure occurs. The output goes positive and the load current builds up uncontrolled.
The last commutation with α ≈ π in figures 11.24c and d results in a commutation failure of thyristor T1.
Before the circulating inductor current i has reduced to zero, the incoming thyristor T2 experiences an
anode potential which is less positive than that of the thyristor to be commutated T1, v1 - v2 < 0. The
incoming device T2 fails to stay on and conduction continues through T1, impressing positive supply
cycles across the load. This positive converter voltage aids the load back emf and the load current
builds up uncontrolled.
=0
from equation 11.180
Voγ = n 2½ V/π×sinπ/n×cosα
γ
Vo =0, γ
Vo
n 2 V/π×sinπ/n×cosα
Mean
output
voltage
=π
from equation 11.184
½
Io = 2×2 V/X×sinπ/n×cosα
½
slope =
-nX/2π
Io
0
Load current
(a)
inversion
(b)
Figure 11.26. Overlap regulation model: (a) equivalent circuit and (b) load plot of overlap model.
Power Electronics
283
Naturally commutating converters
Equations (11.224) and (11.225) are valid provided a commutation failure does not occur. The
controllable delay angle range is curtailed to
0 ≤ α ≤ π −γ
∧
∧
The maximum allowable delay angle α occurs when α + γ = π and from equations (11.224) and (11.225)
with α + γ = π gives
XI o


− 1 < π
α = cos−1 
(rad)
(11.229)
 2V sin π / n 
In practice commutation must be complete δ rad before ωt = π, in order to allow the outgoing thyristor to
regain a forward blocking state. That is α + γ + δ ≤ π . δ is known as the recovery or extinction angle, and
is shown in figure 12.23e. The thyristor recovery period increases with increased anode current and
temperature, and decreases with increased voltage.
where the mean output voltage without commutation inductance effects is 560.4V.
The power output for 100A is 560.4V×100A = 56.04kW and the load resistance is 560.4V/100A =
5.6Ω.
From equation (11.224)
2V
Voγ =
sin π / n  cos α + cos (α + γ ) 
2π / n
2 × 415
× sin π / 6 × [1 + cos γ ]
2π / 6
that is γ = 8.4°
557.44 =
(ii)
α = 60º
2V
sin π n cos α − nX c I o / 2π
2π / n
Voγ =
i
rectification
vs
i
vs
+
i>0
vs > 0
power
in
0 < α < ½π
2 × 415
sin π 6 × cos 60° − 3V = 280.22V - 3V = 277.22V
2π / 6
where the mean output voltage without commutation inductance effects is 280.2V.
The power output for 100A is 280.2V×100A = 28.02kW and the load resistance is 280.2V/100A =
2.8Ω.
=
i>0
vs < 0
½π < α < π
+
power
out
power
in
power
out
+
inversion
284
2V
sin π / n cos α + cos (α + γ ) 
2π / n
Voγ =
+
2 × 415
× ½ × cos 60° + cos ( 60° + γ ) 
2π / 6
that is γ = 0.71°
277.22 =
(a)
(b)
Figure 11.27. Controlled converter model showing: (a) rectification and (b) inversion.
Equation (11.229) gives the maximum allowable delay angle as


X Io
− 1
 2 V sin π / n 
α = cos −1 
The input power is equal to the dc power
P = 3VI cos φ = Vo I o
The input power factor is therefore
cos φ =
Example 11.13:
Vo I o
3VI
≈ ½ cos α + cos (α + γ ) 
(11.230)
 2π50×10-4 ×102 
= cos-1 
-1
 2 ×415×½

γ
= 171.56° and Vo = −557.41V
♣
(11.231)
Converter overlap
A three-phase full-wave converter is supplied from the 415 V ac, 50 Hz mains with phase source
inductance of 0.1 mH. If the average load current is 100 A continuous, for phase delay angles of (i) 0º
and (ii) 60º determine
i.
the supply reactance voltage drop,
ii.
mean output voltage (with and without commutation overlap), load resistance, and output
power, and
iii.
the overlap angle
Ignoring thyristor forward blocking recovery time requirements, determine the maximum allowable delay
angle.
Solution
Using equations (11.224) and (11.225) with n = 6 and V = 415 V ac, the mean supply reactance voltage
n
6
vγ =
2π f LI o =
× 2π 50 × 10−4 × 102
2π
2π
= 3V
(i)
α = 0° - as for uncontrolled rectifiers. From equation (11.225), the maximum output
voltage is
2V
Voγ =
sin π n cos α − nX c I o / 2π
2π / n
2 × 415 π
=
sin 6 × cos 0 − 3V = 560.44V - 3V = 557.44V
2π / 6
11.9
DC MMFs in converter transformers
Half-wave rectification – whether controlled, semi-controlled or uncontrolled, is notorious for producing a
dc mmf in transformers and triplen harmonics in the ac supply neutral of three-phase circuits.
Generally, a transformer based solution can minimise the problem. In order to simplify the underlying
concepts, a constant dc load current Io is assumed, that is, the load inductance is assumed infinite. The
transformer is assumed linear, no load excitation is ignored, and the ac supply is assumed sinusoidal.
Independent of the transformer and its winding connection, the average output voltage from a rectifier,
when the rectifier bridge input rms voltage is VB and there are q pulses in the output, is given by
∧
Vo =
VB
2π / q
π /q
∫
∧
cos ωt d ωt = V B
−π / q
sin π / q
π /q
(11.232)
The rectifier bridge rms voltage output is dominated y the dc component and is given by
Vo rms
q
=
2π
+
π
q
∫ 2V
2
B
cos 2 (ωt ) d ωt = V B 1 +
π
−
q
q
2π
sin
2π
q
(11.233)
The Fourier expression for the output voltage, which is also dominated by the dc component, is
∞
v o (ωt ) = Vo + Vo ∑
k =1
2 ( −1)
k +1
k n −1
2
2
cos kn ωt
(11.234)
Table 11.4 summarizes the various rectifier characteristics that are independent of the transformer
winding configuration.
Power Electronics
285
Naturally commutating converters
Table 11.4: Rectifier characteristics
q phases
Parallel connected secondary windings
Series connected secondary windings
Star, thus neutral always exists
Polygon, hence no neutral
v 1 = 2V sin ωt 

v 2 = 2V sin ωt −

.
.
i.
The E-I three-limb transformer (shell)
The key feature of the three-limb shell is that the three windings are on the centre limb, as shown in
figure 11.28a. The area of each outer limb is half that of the central limb. Assuming a constant load
current Io and equal secondary turns, Ns, excitation of only the central limb yields the following mmf
equation
mmf = i p N p + i s 1N s − i s 2N s
(11.235)
ip
2π 
q 
Np
Full-wave
Vo
q
π
2V sin
π
q
Load
harmonics
n=q
q
π
2V sin
π
q
n=q
q even
n=2q
q odd
2V
Vo
∨
Vo
n=q
n=2q
2V
q
π
2V cos
2q
q
even
2 2V
odd
π
2 2V cos
2q
q
q
ID
q diodes
Is
Po = VoIo
S = qVsIs
pf secondary
P
= o
S
Is = Io
1
q
Io
q
Is = Io
even
odd
+is2Ns
2V
π
π
sin
π
q
2 q
π
sin
q odd
2q
-½ipNp
Vp1
D1
I D rms =
2
q
π
q
Ns
½Np
+is1Ns
is2
+½ipNp
ip
mmf2
RL
Ns
Vs2
Ns
+is2Ns
Vo
is1
RL
LL Io
D2
is2
LL
D1
D2
Io
Σ mmf1
mmf
Vo
Vs1
Vp2
½Np
mmf1
Vo
Ns
Vs2
Σ mmf
Vs1
2q diodes
Vo =
Vo
2 2
π
Vs
Σ mmf2
Vs1
Vo
Vo
Io
q
ωt
I s = ½I o
I s = ½I o
q even
q2 −1
q
2 2
2q
ip
Np
Vs1
+is1Ns
q even
π
sin
q
i s2
Vp
Vp
is1
2q diodes
ID =
I D rms
Io
ip
2 sin
No of diodes
Ns
is1
q even
q odd
+ipNp
2V
V DR
Ns
mmf
π
2 2V cos
2q
π
2V cos
q
½Np
Ns
q
2V
π
mmf2
½Np
Ns
2
ip
mmf1
ip

2π 
v q = 2V sin ωt − (q − 1) 
q 

Half-wave
286
2 2
π
q
q2 −1
Vs1
Vs1
q odd
ωt
q even
π
ωt
VD1
π
q odd
2π
π
VD1
VD1
2π
V D = 2 ×V s
is1
Io
D1
11.9.1 Effect of multiple coils on multiple limb transformers
is2
The transformer for a single-phase two-pulse half-wave rectifier has three windings, a primary and two
secondary windings as shown in figure 11.28. Two possible transformer core and winding configurations
are shown, namely shell and core. In each case the winding turns ratios are identical, as is the load
voltage and current, but the physical transformer limb arrangements are different. One transformer,
figure 11.28a, has three limbs (made up from E and I laminations), while the second, figure 11.28b, is
made from a circular core (shown as a square core). The reason for the two possibilities is related to the
fact that the circular core can use a single strip of wound cold-rolled grain-orientated silicon steel as
lamination material. Such steels offer better magnetic properties than the non-oriented steel that must
be used for E core laminations. Single-phase toroidal core transformers are attractive because of the
reduced size and weight but manufacturers do not highlight their inherent limitation and susceptibility to
dc flux biasing, particularly in half-wave type applications. Although the solution is simple, the
advantageous features of the toroidal transformer are lost, as will be shown.
ωt
VD1
ωt
D1
Io
D2
Ns
ip
Np
D2
D1
is2
Io
Ns
ωt
Ns
Np
ip
D2
Np
mmf
ωt
Io
ωt
−
Io
ωt
ωt
D1
D2
Io
−
mmf
ωt
is1
Io
Ns
Np
Io
½NsIo
ωt
(a)
(b)
Figure 11.28. Single-phase transformer core and winding arrangements: (a) E-I core with zero dc
mmf bias and (b) square/circular core with dc mmf bias.
Power Electronics
287
Thus the primary current ip is
ip =
Ns
mmf
(i − i ) +
N p s 2 s1
Np
Naturally commutating converters
I s1 = I s 2 = I s =
(11.236)
V p 1 = V p 2 = ½V p
From the waveforms in figure 11.28a, since is2 – is1 is alternating, an average primary current of zero in
equation (11.236) can only be satisfied by mmf = 0.
The various transformer voltages and currents are
I s1 = I s 2 = I s =
2
Ns
V I
Np p o
Ns
π
=
V
Np 2 2 o


π
Po = 1.11Po 
=
 2 2

S = ½ (S s + S p ) =
(11.238)
Np
1+ 2
π Po = 1.34Po
4 2
Since the transformer primary current is the line current, the supply power factor is
2
Vs I o
P V I
2 2
pf = o = o o = π
=
= 0.9
ii.
Vp I p
Np Ns
V
I
Ns s Np o
(11.239)
π
(11.240)
mmf 2 = +i p ½N p + i s 2N s
(11.242)
mmf 1 = mmf 2 = mmf
mmf = N s ½ ( i s 1 + i s 2 ) = N s ½I o
Ns
(i − i )
N p s1 s 2


π
Po = 1.11Po 
=
 2 2

(11.245)
The average output voltage, hence output power, are
2 2
2 2 Ns
2 2 Ns
Vo =
V =
V
½V p =
π s
π ½N p
π Np p
(11.246)
1+ 2
Po = 1.34Po
(11.247)
4 2
and the supply power factor is pf= Po / S = 0.9.
The two cores give the same rated transformer apparent power and supply power factor, but
importantly, undesirably, the toroidal core suffers an mmf magnetic bias.
In each core case each diode conducts for 180º and
I D = ½I o
I D rms =
Io
2
VD =2 2
Ns
V
Np p
(11.248)
11.9.2 Single-phase toroidal core mmf imbalance cancellation – zig-zag winding
Figure 11.28b shows the windings equally split on each transformer leg. In practice the windings can all
be on one leg and the primary is one coil, but separation as shown allows visual mmf analysis. The load
and diode currents and voltages are the same as for the E-I core arrangement, as seen in the
waveforms in figure 11.28b. The mmf analysis necessary to assess the primary currents and core flux,
is based on analysing each limb.
mmf 1 = −i p ½N p + i s 1N s
ip =
 π

 = 2 Po = 1.57Po 


(11.241)
The two-limb strip core transformer
These equations yield
(11.244)
Ns
1+ 2
V I
Np p o 2
S =π
Thus
S
Ns
Np
Thus
Po = I oVo
S =
=Vp
Po = I oVo
The average output voltage, hence output power, are
N
2 2
2 2 Ns
Vo =
Vs =
V p = 0.9 s V p
π Np
Ns
V I
Np p o
Ns
V I
Np p o
S p = V p 1I p + V p 2 I p =
N
1+ 2
S = ½ (S s + S p ) = p V p I o
2
Ns
π
Ns
½N p
Np π
V
Ns 2 2 o
S s = V s 1I s 1 +V s 2 I s 2 = 2
(11.237)
Therefore the transformer input, output and average VA ratings are
N
 π

S s = V s 1I s 1 + V s 2 I s 2 = 2 s V p I o  = Po = 1.57Po 
Np
 2

S p =Vp I p =
Ns
Np
Therefore the transformer VA ratings are
N
I p = Io s
Np
Vs 1 =Vs 2 =Vs =V p
I p = Io
2
V s 1 = V s 2 = V s = ½V p
Vp =
Io
Io
288
(11.243)
These two equations are used every ac half cycle to obtain the plots in figure 11.28b. It will be noticed
that the core has a magnetic mmf bias of ½NsIo associated with the half-wave rectification process.
The various transformer ratings are
In figure 11.29, each limb of the core has an extra secondary winding, of the same number of turns, Ns.
MMF analysis of each limb in figure 11.29Y yields
limb1: mmf o = -i p N p - i s 2N s + i s 1N s
(11.249)
limb 2: mmf o = i p N p + i s 2N s − i s 1N s
Adding the two mmf equations gives mmfo = 0 and the resulting alternating primary current is given by
ip =
Ns
(i − i )
N p s1 s 2
(11.250)
The transformer apparent and real power are rated by the same equation as for the previous winding
arrangements, namely
 π
π 
S = ½ (S p + S s ) = ½ 
Po + Po  = 1.34Po
2 
2 2
(11.251)
Ns 2 2
where Po = Vo I o and Vo =
Vp
Np π
Since the transformer primary current is the ac line current, the supply power factor is pf= Po / S = 0.9.
The general rule to avoid any core dc mmf is, each core leg must be effectively excited by a net
alternating current.
Power Electronics
289
ip
ip
mmfo
Vo =
mmfo
Np
Np
Ns
Ns
Ns
Ns
4 2
π
Naturally commutating converters
B
Vs
Vs1
Vo
C
N
A
290
b
c
n
Vo
IB
IC
IN
IA
Ib
Ic
In
Io
Vs1
2π
π
n
N
ωt
VD1
V D = 4 ×V s
Vp
Vp1
Np
Vp2
Np
ip
ip
Ns
Ns
+ipNp
is1
D1
Io
-is1Ns
Vs2
D2
Ns
+is1Ns
Vs1
Ns
is1
Σmmfo
D1
ip
RL
Vo
LL
Io
Ns
Np
D2
is2
Σmmfo
VBN
Ns
Np
mmf o
Van
Vbn
VBC
Io
Vbc
IC
ωt
Figure 11.29. Single-phase zig-zag transformer core and winding arrangement using square/circular
core with zero dc mmf bias.
Ic
IA
IB
11.9.3 Single-phase transformer connection, with full-wave rectification
The secondary current are ac with a zero average , thus no core mmf bias occurs.
The average output voltage and peak diode reverse voltage, in terms of the transformer secondary rms
voltage, are
2 2
Vo =
Vs
V Dr = 2V s
(11.252)
π
Po
2 2
Since the line current is the primary current, the supply power factor is
P
2 2
pf = o =
S
π
(11.253)
Ia
Ib
(a)
(b)
Figure 11.30. Three-phase Y-y transformer: (a) winding arrangement and (b) phasor diagrams.
Independent of the three-phase connection of the primary and secondary, for a balance three-phase
load, the apparent power, VA, from the supply to the load is
π
The transformer average VAr rating is
S =
Vab
ωt
−
D2
-Vbn
Vcn
Vca
VAN
ωt
Io
+is2Ns
Vs2
VAB
ωt
D1
is2
Vs1
-VBN
VCN
VCA
Io
mmf
mmf
-is2Ns
Vbn
VBN
i s2
VD1
-ipNp
Ia
ωt
VD1
is1
a
S = 3V line I line = 3V phase I phase
Also the sum of the primary and secondary line voltages is zero, that is
V AB +VBC +VCA = 0
Vab +V bc +Vca = 0
(11.255)
(11.256)
where upper case subscripts refer to the primary and lower case subscripts refer to the secondary.
(11.254)
11.9.4 Three-phase transformer connections
Basic three-phase transformers can have a combination of star (wye) and delta, primary and secondary
winding arrangements.
i. Y-y is avoided due to imbalance and third harmonic problems, but with an extra delta
winding, triplen problems can be minimised. The arrangement is used to interconnect high
voltage networks, 240kV/345kV or when two neutrals are needed for grounding.
ii. Y- δ is commonly used for step-down voltage applications.
iii. ∆-δ is used in 11kV medium voltage applications where neither primary nor neutral
connection is needed.
iv. ∆-y is used as a step-up transformer at the point of generation, before transmission.
Y-y (WYE-wye) connection
Electrically, the Y-y transformer connection shown in figure 11.30, can be summarized as follows.
ηY − y =
N p V AN I a I a V BN I b VCN I c
=
=
=
=
=
=
=
N s V an I A I L1 V bn I B Vcn I C
V AB = V AN + V NB = V AN −VBN = 3VAN e j 30° = 3VAN ∠30°
V BC = V BN + VNC = VBN −VCN VCA = VCN + V NA = VCN −V AN
V ab = V an −V bn = 3Van e j 30° = 3V an ∠30°
V bc = V bn −Vcn Vca = Vcn −Van
(11.257)
(11.258)
Power Electronics
291
B
C
IB
N
IC
Naturally commutating converters
b
A
IN
c
Ib
IA
Ic
IC
c
Ib
IA
a
Ic
ICA
Ia
Ica
Vbc
IBC
Ibc
Iab
IAB
Iab
VBN
VCN
b
A
VBC
Ibc
N
C
IB
Ia
Ica
Vbc
VCA
B
a
292
VCA
Vca
VAB
-VBN
Vca
VAB
Vab
VBC
VAN
Vab
Vbc
Ic
IC
Vbc
VBN
Ica
ICA
Ic
VBC
Iab
IAB
Ica
IC
Iab
IBC
IB
IA
-ICA
Ibc
Ib
IA
(a)
Ibc
Ib
(a)
-Ica
Ia
(b)
Figure 11.31. Three-phase Y- δ transformer: (a) winding arrangement and (b) phasor diagrams.
I N = I A + I B + IC
In = Ia + Ib + Ic
The output current rating is
(11.259)
S
V
3 = S
3V
3
N p V AN I ba V BN I cb VCN I ac
=
=
=
=
=
=
N s V ab
I A V bc
I B Vca
IC
V AB = V AN −VBN = V AN −V AN e
− j 120°
= 3V AN e
S
I∆ =
3 = S
V
3V
η∆−δ =
N p V AB I a I ba VBC I b I cb VCA I c I cb
=
=
=
=
=
=
=
=
=
N s V ab I A I AB V bc I B I BC Vca I C I CA
I A = I AB − I CA = 3 I A B e − j 30° = 3I A B ∠ − 30°
I B = I BC − I AB
I C = I CA − I BC
I A + IB + IC = 0
(11.262)
(11.263)
The output current rating is
S
(11.260)
IY =
j 30°
I a = I ba − I ac = I ba − I ba e − j 240° = 3 I ba e − j 30°
Ia + Ib + Ic = 0
The output current rating is
∆-δ connection
In figure 11.32, the ∆ - δ transformer connection can be summarized as follows.
I a = I ab − I ca = 3 I ab e − j 30° = 3I ab ∠ − 30°
I b = I cb − I ba
I c = I ac − I cb
Ic + Ib + Ic = 0
Y- δ connection
The Y- δ transformer connection in figure 11.31 can be summarized as follows.
ηY −δ =
Ia
Figure 11.32. Three-phase ∆-δ transformer: (a) winding arrangement and (b) phasor diagrams.
IB
IY =
-Ica
(b)
(11.261)
V
3 = S
3V
3
(11.264)
∆-y connection
The ∆-y transformer connection in figure 11.33 can be summarized as follows.
η ∆− y =
*
V AB V AB e − j 30° V AN I a
=
=
= * =
V ab V ab
V
I
3
an
A
I A = I AB − I CA = I AB − I AB e
− j 240°
(
I a*
3 I AB e − j 30°
= 3 Ian e
− j 30°
)
*
(11.265)
Power Electronics
293
Naturally commutating converters
The output current rating is
S
S
IY = 3 =
V
3V
3
B
C
IB
(11.266)
A
IC
b
IA
c
Ib
n
Ic
a
In
Ia
ICA
VBC
IBC
Vcn
Ns  1

2
1
− i + i − i
N p  3 s 1 3 s 2 3 s 3 
ip3 =
Ns
Np
(11.269)
 is1 + is 2 + is 3  1
 = 3 Ns Io
3


Specifically, the core has an mmf dc bias of NsIo.
Waveforms satisfying these equations are show plotted in figure 11.34a. The various transformer
currents and voltages are
I s1 = I s 2 = I s 3 = I s =
Van
VAB
VBC
Io
3
2 Ns
Io
3 Np
I p1 = I p 2 = I p 3 = I p =
Vbn
IC
V p 1 = V p 2 = V p 3 = V p = Vo
ICA
V s 1 = V s 2 = V s 3 = V s = Vo
Ic
3 6
S s = V s 1I s 1 +V s 2 I s 2 + V s 3I s 3 = 3V s I s =
Ia
-ICA
(11.270)
N p 2π
Ns 3 6
Vo
2π
=
1.17
Therefore the various transformer VA ratings are
IAB
IBC
 1

1
2
 − 3 is1 − 3 is 2 + 3 is 3 


mmf o = N s 
Vab
VCA
IA
(a)
S p = V p 1I p + V p 2 I p + V p 3I p = 3V p I p =
Ib
S = ½ ( S s + S p ) = Po
(b)
Figure 11.33. Three-phase ∆-y transformer: (a) winding arrangement and (b) phasor diagrams.
11.9.5 Three-phase transformer, half-wave rectifiers - core mmf imbalance
Note that a delta secondary connection cannot be used for half-wave rectification as no physical neutral
connection exists.
i.
i p2 =
Vbn
Vca
(11.268)
The same mmf equations are obtained if the load is purely resistive.
Any triplens in the primary will add algebraically, while any other harmonics will vectorially cancel to
zero. Therefore the neutral may only conduct primary side triplen currents. Any input current harmonics
are due to the rectifier and the rectifier harmonics of the order h = cp±1 where c = 0,1,2,… and p is the
pulse number, 3. No secondary-side third harmonics can exist hence h ≠ 3k for k = 1, 2, 3, . Therefore no
primary-side triplen harmonic currents exist to flow in the neutral, that is iN = 0. In a balanced load
condition, the neutral connection is redundant. The system equations resolve to
N 

i p1 = s  2 i s 1 − 1 i s 2 − 1 i s 3 
3
3
Np  3

n
IAB
IB
If iN is the neutral current then the equation for the currents is
i p1 + i p 2 + i p 3 = i N
294
star connected primary Y-y
The three-phase half-wave rectifier with a star-star connected transformer in figure 11.34a is prone to
magnetic mmf core bias. With a constant load current Io, each diode conducts for 120º. Each leg is
analysed on an mmf basis, and the current and mmf waveforms in figure 11.34a are derived as follows.
mmf o = N s i s 1 − N p i p 1
mmf o = N s i s 2 − N p i p 2
mmf o = N s i s 3 − N p i p 3
By symmetry and balance, the mmf in each leg must be equal.
(11.267)
2+π 6
3 3
π 2
3
2
3 3
Po = 1.48Po
Po = 1.21Po
(11.271)
= 1.34Po
The average output power is
Po = I oVo
(11.272)
Since with a wye connected transformer primary, the transformer primary phase current is the line
current, the supply power factor is
3
pf =
3
2V s
I
Po
V I
2 o = 0.827
= o o = π
N
3
V
I
S
3 Ns
p p
3 p Vs
I
2 Np o
Ns
(11.273)
Although the neutral connection is redundant for a constant load current, the situation is different if the
load current has ripple at the three times the rectified ac frequency, as with a resistive load. Equations in
(11.269) remain valid for the untapped neutral case. In such a case, when triplens exist in the load
current, how they are reflected into the primary depends on whether or not the neutral is connected:
• No neutral connection – a triplen mmf is superimposed on the mmf dc bias of ⅓NsIo.
• Neutral connected – a dc current (zero sequence) flows in the neutral and the
associated zero sequence line currents in the primary, oppose the generation of any
triplen mmf onto the dc mmf bias of ⅓NsIo.
Power Electronics
295
ii.
Naturally commutating converters
A
delta connected primary ∆-y
mmf o = N s i s 2 − N p i p 2
i L 2 = i p 2 − i p1
Vp1
Vs1
 Ns
3

 Np
=
Ns  1

2
1
− i + i − i
N p  3 s 1 3 s 2 3 s 3 
i L2
i L3
mmfo
Ns
Vs2 Ns
is2
b
N
Np
Np
ip1
ip2
IL3
Vp2
Vp3
Np
ip3
mmfo
mmfo
n
n
mmfo
+isNs
Vs1
Vs3 RL
is1
is3
c
D2
D3
LL
Ns
Ns
Vs2 Ns
is1
is2
b
Vs3 RL
Vo
Vs1
Vs2
Vs3
a
Σ mmfo
Io
Vo =
π
Vs1
3 3
2π
D1
is3
D2
c
LL
D3
Io
Vs
V
Vs1
Vs2
Vs3
Vs1
ωt
2π
π
2π
(11.276)
ωt
Vs2 -Vs1
is2
Vs3 -Vs1
(11.277)
is2
√6Vs
ωt
D1
Io
(11.279)
ip2
⅔Io
Ns
Np
ωt
ωt
Io
D3 ωt
ip1=is1-is2-is3
⅔Io
Ns
Np
ip2
D3 ωt
D3
ip1
D2
D3
ωt
Io
is3
Io
is3
ip1
D2
ωt
D1
D2
Io
D2
(11.278)
Io
D1
D1
N
= s (i p 3 − i p 2 )
Np
is1
VD1
is1
3
Io
2
iN
Vp1
IL2
ωt
Ns
(i − i )
N p p1 p 3
Ns 3
N
I and I L = s
Np 2 o
Np
mmfo
-ipNp
Vp3
Np
ip3
ip2
D1
V
The waveforms for these equations are shown plotted in figure 11.34b, where
Ip =
VL2
IL1
(11.275)
 1

1
2
 − 3 is1 − 3 is 2 + 3 is 3 


These line-side equations are the same as for the star connected primary, hence the same real and
apparent power equations are also applicable to the delta connected primary transformer, viz. equations
(11.271) and (11.272).
The line currents are
N
= s (i p 2 − i p 1 )
Np
N
Vo
N

N
i p3 = is 3 − 1 Io  s = s
3  Np
Np

i L1 =
C
Vp2
Np
Ns
a
The line-side currents have average values of zero and if it is assumed that the core mmf has only a dc
component, that is no alternating component, then based on these assumptions

Np
ip1
iL3 = i p 3 − i p 2
i + i + i 
mmf o = N s  s 1 s 2 s 3  = 1 N s I o
3

 3
The primary currents are then
N 

N

i p1 =  i s 1 − 1 I o  s = s  2 i s 1 − 1 i s 2 − 1 i s 3 
3  Np
3
3
Np  3


B
mmfo
(11.274)
mmf o = N s i s 3 − N p i p 3
i p2 = is 2 − 1 Io 
C
VL1
A
The three-phase half-wave rectifier with a delta-star connected transformer in figure 11.30b is prone to
magnetic mmf core bias. With a constant load current Io each diode conducts for 120º. Each leg is
analysed on an mmf basis, and the current and mmf waveforms in figure 11.34b are derived as follows.
mmf o = N s i s 1 − N p i p 1
i L1 = i p 1 − i p 3
296
B
-⅓Io
ωt
ωt
Ns
Np
ip3
ωt
that is I L = 3I p
S = 1.34Po
The supply power factor is
pf =
-⅓Io
Vo I o
== 0.827
3V p I L
(11.280)
Although with a delta connected primary, the ac supply line currents are not the transformer primary
currents, the supply power factor is the same as a star primary connection since the proportions of the
input harmonics are the same.
Each diode conducts for 120º and
ID =
1
3
Io
I D rms =
ωt
Ns
IL1
ωt
⅓NsIo
(a)
Io
3
VD
N
= 6 s Vp
Np
(11.281)
The primary connection, delta or wye, does not influence any dc mmf generated in the core, although
the primary connection does influence if an ac mmf results.
Io
Ns
Np
ωt
Np
ip3
mmf
i L1 = i p 1 − i p 3
ωt
-Io
mmf
Ns
Np
⅓NsIo
ωt
(b)
Figure 11.34. Three-phase transformer winding arrangement with dc mmf bias:
(a) star connected primary and (b) delta connected primary.
11.9.6 Three-phase transformer with hexa-phase rectification, mmf imbalance
Figure 11.35 shown a tri-hexaphase half-wave rectifier, which can employ a wye or delta primary
configuration, but only a star secondary connection is possible, since a neutral connection is required.
The primary configuration can be shown to dictate core mmf bias conditions.
Power Electronics
297
i.
Naturally commutating converters
Y-y connection
The line currents are
The mmf balance for the wye primary connection in figure 11.35a is
N s i s 1 − N s i s 4 − N p i p1 = 0
Ns is 3 − Nsis 6 − N pi p2 = 0
(11.282)
Ns is5 − Nsis 2 − N pi p3 = 0
i p1 + i p 2 + i p 3 = 0
The primary currents expressed in terms of the secondary current are
Ns 2
( i + 1i − 1i − 2i − 1i + 1i )
N p 3 s1 3 s 2 3 s 3 3 s 4 3 s 5 3 s 6
N
= s ( − 13 i s 1 + 13 i s 2 + 23 i s 3 + 13 i s 4 − 13 i s 5 − 23 i s 6 )
Np
N
= s ( − 13 i s 1 − 23 i s 2 − 13 i s 3 + 13 i s 4 − 23 i s 5 − 13 i s 6 )
Np
i p3
2
Io
3
Note that because of the zero sequence current, triplens, in the delta primary that
iL = 2 i p
not
(11.284)
(11.290)
(11.291)
The transformer power ratings are
 π
I
π
Ss = 6 
Vo  o =
Po
3
3 2  6
 π
Sp = 3
3 2
 Io
Vo 
 3
=
π
6
Po
(11.292)

1 
π
π 
Po +
Po  = ½
1 +
 Po = 1.54Po
6 
3
2
 3
The supply power factor is pf = 3/π = 0.955.
and the diode average and rms currents are
ID =
Io
(11.286)
When the primary is delta connected, as shown in figure 11.35b, the mmf equations are the same as
with a wye primary, namely
N s i s 1 − N s i s 4 − N p i p1 = 0
Ns is 3 − Nsis 6 − N pi p2 = 0
(11.287)
Ns is5 − Nsis 2 − N pi p3 = 0
but Kirchhoff’s electrical current equation becomes of the following form for each phase:
1 2π
mmf =
N s ( i s 1 − i s 4 ) d ωt = 0
(11.288)
2π ∫0
Thus since each limb experiences an alternating current, similar to is1-is4 for each limb, with an average
value of zero, the line currents can be calculated from
Ns
(i − i )
Np s5 s2
I D rms =
V Dr = 2 2V s
)
ip3 =
2
Io
3
iL =
6
The maximum diode reverse voltage is
∆-y connection
Ns
(i − i )
Np s3 s6
Ns
( −i s 2 − i s 3 + i s 5 + i s 6 )
Np
π
(11.285)
The transformer power ratings are
 π
 π
 1
Ss = 6 
Vo  
Io  =
Po
3
 3 2  6 
 π
 2  π
Sp = 3
Vo  
I o  = Po
 3 2   3
 3
 π
π  π
S = ½
Po + Po  =
3 + 1 Po = 1.43Po
3  6
 3
i p2 =
i L3 = i p 3 − i p 2 =
Independent of the primary connection, the average output voltage is
3 2
Vo =
V s = 1.35V s
iL = 3 i p
Ns
(i − i )
N p s1 s 4
Ns
( −i s 1 + i s 3 + i s 4 − i s 6 )
Np
S = ½
iL =
i p1 =
i L2 = i p 2 − i p1 =
 π
The transformer primary currents and the line currents are
2
ip =
Io
3
i.
Ns
(i + i − i − i )
N p s1 s 2 s 4 s 5
(11.283)
1
mmf = N s ( i s 1 − i s 2 + i s 3 − i s 4 + i s 5 − i s 6 )
3
These line side equations are plotted in figure 11.35a. Notice that an alternating mmf exists in the core
related to the pulse frequency, n = 2q = 6.
(
i L1 = i p 1 − i p 3 =
The transformer primary currents and the line currents are
1
ip =
Io
3
i p1 =
i p2
298
(11.289)
Io
6
(11.293)
(11.294)
(11.295)
The line currents are added to the waveforms in figure 11.35a and are also shown in figure 11.35b. The
core mmf bias is zero, without any ac component associated with the 6 pulse rectification process. Zero
sequence, triplen currents, can flow in the delta primary connection. A star connected primary is
therefore not advisable.
Power Electronics
299
Naturally commutating converters
A
A
Vp1
N
RL
Np
Np
ip1
ip2
is4
is6
Vp2
mmfo
Ns
Ns
Ns
D1
D3
D5
Io
is1
is3
is5
Vs1
Vs2
LL
Np
Np
ip2
Vs3
Vs3
Ns
Ns
Vs2
-is4Ns
n
+is1Ns
Vs5
Vs4
RL
Vo
Vs1
LL
Σ mmf
Vs4
Vs5
Vs6
Vo =
V
2
π
Io
Vs1
Vs2
is4
is6
11.9.7 Three-phase transformer mmf imbalance cancellation – zig-zag winding
IL3
Vp2
In figures 11.36a and 11.37a, for balanced input currents and equal turns number Ns in the six windings
N s ( I aa ' + I nc ' ) = N s ( I an − I cn )
Vp3
Np
whence
ip3
mmfo
mmfo
D2
Vs6
C
VL2
IL2
ip1
is2
D6
Vs1
-ip1Np
ip3
D4
Vs4
Vp1
Vp3
Np
mmfo
mmfo
IL1
C
Ns
n
Vo
B
B
VL1
mmfo
D4
D6
Ns
If the same windings were connected in series in a Y configuration the mmf would be 2NIan. Therefore
1.15 times more turns (2/√3) are needed with the zig-zag arrangement in order to produce the same
mmf.
Ns
Ns
D1
D3
D5
is1
is3
is5
Vs3
D2
Vs6
Vs3
Vs4
(11.296)
N s ( I an − I cn ) = 3 N s I an ∠ − 30°
is2
Ns
Vs2
Ns
Ns
Similarly for the output voltage, when compared to the same windings used in series in a Y
configuration:
V na = V na ' + V a ' a
Vs5
= −V a ' n + V a ' a
(11.297)
= 3V a ' a ∠30°
That is, for a given line to neutral voltage, 1.15 times as many turns are needed as when Y connected.
Vs5
Vs6
Vs 1
c
Vc′c=Vb′n
V
Vna′
a′
c′
ωt
ωt
π
300
2π
π
2π
Vna
n
a
Vb′b=Va′n
Va′a=Vc′n
b′
Icn
a′
VD1
b
VD1
Ian
a
c
2√2Vs
Io
is1
D1
is2
Io
D2
is3
D3
Io
is4
ωt
D1
Io
Io
ωt
D2
Io
D1
D6
D3
D4
D4
D5
is6
Io
D6
 Io
D3
D2
D5
D4
D1
D6
ωt
Io
IoNs/Np
o
(a)
ωt
D1
D4
-Vbn
Ian-Icn
b
D4
(a)
D1
(b)
Figure 11.36. Three-phase transformer secondary zig-zag winding arrangement:
(a) secondary windings and (b) current and voltage phasors for the fork case.
D4
ωt
ip2
ωt
 Io
Io
D3
D6
D3
ωt
D6
-Io
Io
ωt
ip3
ωt
IL1
Io
mmf
-Io
ωt
D5
D2
mmf
ωt
D3
D2
Ibn
-Io
Io
D6
iL1
b′
ωt
ip1
is5
ωt
o
D2
-Io
ωt
ωt
o
(b)
Figure 11.35. Three-phase transformer winding arrangement with hexa-phase rectification:
(a) star connected primary with dc mmf bias and (b) delta connected primary.
i.
star connected primary Y-z
In figure 11.37, each limb of the core has an extra secondary winding, of the same number of secondary
turns, Ns.
MMF analysis of each of the three limbs yields
limb1:- mmf o = -i p 1N p + i s 1N s − i s 3N s
limb 2:-
mmf o = −i p 2N p + i s 2N s − i s 1N s
limb 3:-
mmf o = −i p 2N p + i s 3N s − i s 2N s
i p1 + i p 2 + i p 3 = 0
(11.298)
Power Electronics
301
Naturally commutating converters
Vs31
Adding the three mmf equations gives mmfo = 0 and the alternating primary (and line) currents are
i p1
N
= s (i s 1 − i s 3 )
Np
i p2 =
Ns
(i − i )
N p s 2 s1
i p3 =
Ns
(i − i )
Np s3 s2
A
Vp1
(11.299)
B
Np
Np
ip1
ip2
Vp2
Vp3
Np
ip3
Vs2
N
mmfo
Vs1o
Ns
Ns
Vs2o
Vs3o
Ns
n
(11.300)
v s 3 = v s 31 − v s 1o
S p = 3I pV p =
π
(
ip2
Np
Vp3
ip3
N
mmfo
2π
3 3
Ns
a
Ns
Vs21 Ns
Vs31
is2
b
c
LL
D3
Io
is1
D1
Po
)
Ns
Vs2o
Ns
Ns
Vs3o
n
-is3Ns
RL
is3
D2
Ns
Vs11
Vo
Po
3 3
Np
Vs1
Vs11
(11.301)
S = ½ ( S s + S p ) = Po
Np
ip1
Vp2
mmfo
Vs1o
Vs1
The various transformer ratings are
3 3
IL3
IL2
mmfo
-ip1Np
v s 1 = v s 11 − v s 2o
v s 2 = v s 21 − v s 3o
2 2π
Vp1
Vs11
Vs21
VL2
VL1
IL1
302
C
-Vs3o
These equations are plotted in figure 11.37a.
If a 1:1:1 turns ratio is assumed, the power ratings of the transformer (which is independent of the turns
ratio) involves the vectorial addition of the winding voltages.
S s = 3I sV s 0 + 3I sV s 1 = 6I sV so =
B
C
mmfo
mmfo
A
Ns
Vs21 Ns
Vs31
RL
is2
b
c
LL
D3
Io
Vo
+is1Ns
a
is1
D1
Σmmfo
is3
D2
(11.302)
2 + 1 = 1.46Po
V
Vs1
Vs2
Vs3
Vo =
Vs1
3 3
2π
Vs 1
V
Vs1
Vs2
Vs3
Vs1
ωt
ωt
ii.
π
delta connected primary ∆-z
π
2π
2π
Carrying out an mmf balancing exercise, assuming no alternating mmf component, and the mean line
current is zero, yields
i L1 = i p 1 − i p 3 =
Ns
(i + i − 2i s 3 )
N p s1 s 2
i L2 = i p 2 − i p1
N
= s ( i s 2 + i s 3 − 2i s 1 )
Np
i L3 = i p 3 − i p 2
N
= s ( i s 3 + i s 1 − 2i s 2 )
Np
ωt
ID =
3
Io
I D rms =
3
VD
π
Io
is1
√6Vs
D1
Io
is2
2π
D1
D2
Io
Io
is3
Ns
Io N
Ns
Io N
ip1
Io N
ip2
(11.306)
Ns
Io N
iL1
p
ωt
p
Io N
ip3
p
-2Io N
p
ωt
iL2
-Io N
o
p
ωt
ωt
mmf
Io N
ωt
p
Ns
-2Io N
o
(a)
p
p
Ns
p
Ns
mmf
-Io N
Ns
Ns
-Io N
Ns
ωt
p
Ns
Ns
-Io N
Ns
ωt
D3
ωt
p
ωt
D2
ωt
D3
ip1
D2
ωt
D2
ωt
D1
Io
is2
(11.305)
N
= 6 s Vp
Np
VCB
Io
is1
ωt
D1
is3
2
Io
I L = 3I p
3
A zig-zag secondary can be a Y-type fork for a possible neutral connection or alternatively, a ∆-type
polygon when the neutral is not required.
Each diode conducts for 120º and
Io
VAC
Vs3 -Vs1
Ip =
1
VBA
ωt
If a 1:1:1 turns ratio is assumed, the line, primary and load current are related according to
2 2 4 2
I o + I o = 2I o
3
3
VCB
Vs2 -Vs1
(11.303)
The primary and secondary currents are the same whether for a delta or star connected primary,
therefore
S = ½ ( S s + S p ) = 1.46Po
(11.304)
IL =
VL-L
VD1
p
(b)
Figure 11.37. Three-phase transformer winding zig-zag arrangement with no dc mmf bias:
(a) star connected primary and (b) delta connected primary.
ωt
Power Electronics
303
Naturally commutating converters
Such that
11.9.8 Three-phase transformer full-wave rectifiers – zero core mmf
Full-wave rectification is common in single and three phase applications, since, unlike half-wave
rectification, the core mmf bias tends to be zero. In three-phase, it is advisable that either the primary or
secondary be a delta connection. Any non-linearity in the core characteristics, namely hysteresis,
causes triplen fluxes. If a delta connection is used, triplen currents can circulate in the winding, thereby
suppressing the creation of triplen core fluxes. If a Y-y connection is used, a third winding set, delta
connected, is usually added to the transformer in high power applications. The extra winding can be
used for auxiliary type supply applications, and in the limit only one turn per phase need be employed if
the sole function of the tertiary delta winding is to suppress core flux triplens.
The primary current harmonic content is the same for a given output winding configuration, independent
of whether the primary is star or delta connected.
i.
Ip =
IL = 3 I p =
(11.307)
π
iii.
i =1
i =1
(11.308)
i p1 + i p 2 + i p 3 = 0
(11.309)
i L3 = i p 3
Ns
i
Np s2
(11.310)
Ns
i
N p s1
Sp =
2π
3 3
Po = 1.21Po
 2π
Ss =
2π
Po = 1.48Po
3
(11.312)
3 3
π
π
S
iL2 =
Ns
(i − i ) = i p 3 − i p1
N p s 3 s1
The secondary phase currents in figure 11.38b are the same as for the Y-y connection, but the line
currents are composed as follows
i L1 = i p 1 − i p 3
i L 2 = i p 2 − i p1
iL3 = i p 3 − i p 2
(11.315)
Ns
i
Np s3
(11.319)
(11.320)
Thus the transformer currents are related to the supply line currents by
i p1 =
Ns
i = 2i − 2i
N p s 1 3 L1 3 L 2
i p2 =
Ns
i = 2i − 2i
N p s 2 3 L2 3 L3
i p3 =
Ns
2
2
i = i − i
N p s 3 3 L 3 3 L1
(11.321)
i L1 + i L 2 + i L 3 = 0
(11.322)
Ns
N 2 2
I = s
½I o
Np s Np 3
(11.323)
Generally
Ip =
The full-wave, three-phase rectified average output voltage (assuming all winding turns are equal) is
3 3
3
Vo =
V p = VL
(11.324)
π
π
The supply power factor is
pf =
π
delta connected primary ∆-y (Delta-wye)
i p3 =
Ns
(i − i ) = i p 2 − i p 3
Np s2 s3
Po +
Since with a star primary the line currents are the primary currents, the supply power factor is
P
3
(11.314)
pf = o = = 0.955
Ns
i
Np s2
i L2 =
where
2π 
Po  = 1.35Po
3

The full-wave, three-phase rectified average output voltage (assuming all winding turns are equal) is
3 3
3
Vo =
V p = VL
(11.313)
S = ½ 
i p2 =
Ns
(i − i ) = i p 1 − i p 2
N p s1 s 2
(11.311)
whence
(11.318)
i L1 =
Generally
2
Io
3
= 0.955
where
N
= s is 3
Np
N
N
Ip = s Is = s
Np
Np
3
π
and i s 1 + i s 2 + i s 3 = 0
N
= s is1
Np
i L2 = i p 2 =
π
In the Y-δ configuration in figure 11.39a, there are no zero sequence currents hence no mmf bias
arises, mmfo = 0, and both transformer sides have positive and negative sequence currents.
and the secondary currents always sum to zero, then mmfo = 0.
Additionally
i L1 = i p 1
2 Io
star connected primary Y-δ (Wye-delta)
i p1 =
3 × mmf o = N p ∑ i pi − N s ∑ i si
but
(11.316)
Ns
Np
The supply power factor is
mmf o = N s i s 2 − N p i p 2
3
3 Is =
The full-wave, three-phase rectified average output voltage (assuming all winding turns are equal) is
3 3
3
Vo =
V p = VL
(11.317)
Adding the three mmf equations gives
3
2
Io
3
Ns
Np
pf =
The Y-y connection shown in figure 11.38a (with primary and secondary neutral nodes N, n
respectively) is the simplest to analyse since each phase primary current is equal to a corresponding
phase secondary current.
mmf o = N s i s 1 − N p i p 1
ii.
Ns
N
I = s
Np s Np
star connected primary Y-y (Wye-wye)
mmf o = N s i s 2 − N p i p 2
304
3
π
for an output power, Po = Vo Io,
iv.
delta connected primary ∆-δ (Delta-delta)
The phase primary and secondary voltages are in phase.
(11.325)
Power Electronics
305
As shown in figure 11.39b the line currents are composed as follows
i L1 = i p 1 − i p 3
i L 2 = i p 2 − i p1
iL3 = i p 3 − i p 2
The transformer primary and secondary currents are
i p1 =
Ns
i
N p s1
i p2 =
and
Ns
i
Np s2
ip3 =
Ns
i
Np s3
Naturally commutating converters
A
(11.326)
A
Generally
Ip =
Vp1
Np
ip1
ip2
Vs1
Ns
a
(11.329)
pf =
ID =
1
Io
3
Np
Np
ip1
ip2
C
VL2
IL3
Vp2
Vp3
Np
ip3
mmfo
mmfo
Vs1
Vs3
Vs2 Ns
Ns
is1
3 Ns
π Np
c
is3
D4
D1
D3
a
Σ mmf
D5
Io
LL
RL
(11.331)
π
b
is2
D6
D2
3
In summary, when the primary and secondary winding configurations are the same (∆-δ or Y-y) the input
and output line voltages are in phase, otherwise (∆-y or Y-δ) the input and output line voltages are
shifted by 30º relative to one another.
Independent of the transformer primary and secondary connection, for the specified turns ratio, the
following electrical equations hold.
π
N
mmfo
IL2
mmfo
n
Ns
Ns
is1
is2
Vs3
Vs2 Ns
+is1Ns
π
Vp =
ip3
Vp1
-ip1Np
Vp3
Np
mmfo
ab
V
ac
bc
ba
ca
cb
ab
D4
D1
D3
D2
Vo =
ac
Vbn
Vcn
Van
Vs1
Vs2
Vs3
Vs1
3 3
π
Vs 1
ab
V
ac
bc
ba
cb
ab
ac
Vcn
Van
Vs1
Vs2
Vs3
Vs1
ωt
2π
2π
Io
ip1=is1
D1
ωt
D1
D4
π
V DR = 3 2V s
ca
Vbn
π
3
I D rms =
D5
Io
LL
RL
Van
ωt
π
c
Vo
Van
VL
is3
b
D6
Vo
for an output power, Po = Vo Io,
3 3
Vp2
B
VL1
IL1
n
The supply power factor is
Vo =
C
(11.328)
The full-wave, three-phase rectified average output voltage (assuming all winding turns are equal) is
3 3
3
Vo =
V p = VL
(11.330)
pf =
Np
mmfo
Ns
I
Np s
π
B
(11.327)
i p1 + i p 2 + i p 3 = 0
is1 + is 2 + is 3 = 0
i L1 + i L 2 + i L 3 = 0
306
1
3
D4
-Io
Io
VD4
ip2=is2
VD1
√6Vs
Io
Io
Io
ip1=is1
D1
D4
Io
D5 ωt
D2
D2
-Io
2Io
Io
D3
Io
iL1
ωt
D3
D6
ωt
D6
-Io
ip1=is3
-Io
Io
-2Io
D5 ωt
D5
D2
mmf
-Io
D5
D4
-Io
ip2=is2
D6
ip1=is3
ωt
D1
ωt
D3
D6
ωt
o
Io
D3
D2
-Io
(a)
Io
iL2
ωt
o
(b)
mmf
-2Io
o
ωt
ωt
Figure 11.38. Three-phase transformer wye connected secondary winding with full-wave rectification
and no resultant dc mmf bias: (a) star connected primary Y-y and (b) delta connected primary ∆-y.
Power Electronics
307
Naturally commutating converters
A
A
Vp1
B
Np
Np
ip1
ip2
C
Vp2
a
Ns
Ns
is1
is2
Vp3
Np
ip3
mmfo
mmfo
+is1Ns
c
is3
a
D1
D6
D3
D5
Io
LL
IL2
Np
Np
ip1
ip2
VL2
Vp2
Ns
Ns
is1
is2
General expressions for n-phase converter mean output voltage, Vo
Vp3
Np
(i) Half-wave and full-wave, fully-controlled converter
sin(π / n )
Vo = 2 V
cos α
π /n
where V is
the rms line voltage for a full-wave converter or
the rms phase voltage for a half-wave converter.
cos α = cosψ , the supply displacement factor
From L’Hopital’s rule, for n→∞, Vo = √2 V cosα
(ii) Full-wave, half-controlled converter
sin(π / n )
Vo = 2 V
(1 + cos α )
π /n
where V is the rms line voltage.
ip3
mmfo
Vs3
Ns
b
c
is3
D1
D6
D3
D2
D5
Io
LL
RL
308
11.10 Summary
IL3
mmfo
Vo
V
C
D4
Σ mmf
D2
RL
VL1
IL1
mmfo
Vs3
Ns
D4
-ip1Np
N
mmfo
b
Vp1
B
Vo
Van
Vbn
Vcn
Van
Vs1
Vs2
Vs3
Vs1
Vo =
3 3
π
Vs 1
ωt
π
2π
√2Vs
VD1
Io
Io
o
D1
ip2=is2
D4
-Io
ip1=is3
ωt
D2
-Io
ip1=is3
D2
Io
2Io
D2
-Io
D2
-Io
D5 ωt
D5
Io
D5 ωt
D5
D6
ωt
D3
D6
-Io
Io
D3
D6
Io
D3
D6
D4
D3
ωt
D1
D4
ωt
D1
D4
mmf
D1
-Io
Io
ip1=is1
Io
ip1=is1
ωt
Io
iL1
ωt
ωt
o
Figure 11.40. Converter normalised output voltage characteristics as a function of firing delay angle α.
-Io
-2Io
Io
iL2
(a)
(b)
mmf
-2Io
o
ωt
ωt
Figure 11.39. Three-phase transformer with delta connected secondary winding with full-wave
rectification and no resultant dc mmf bias: (a) star connected primary Y- δ and (b) delta connected
primary ∆- δ.
(iii) Half-wave and full-wave controlled converter with load freewheel diode
sin(π / n )
Vo = 2 V
cos α
0 < α < ½π − π / n
π /n
1 + cos (α + ½π − π / n )
Vo = 2 V
½π − π / n < α < ½π + π / n
2π / n
the output rms voltage is given by
cos 2α sin 2π / n
Vrms = V 1 +
α + π / n ≤ ½π
2π / n
Vrms = V ½ +
cos ( 2α − 2π / n )
α
n
−
−
4 2π / n
4π / n
α + π / n > ½π
Power Electronics
309
Naturally commutating converters
The mean converter output voltage Vo can be specified by
where V is
the rms line voltage for a full-wave converter or
the rms phase voltage for a half-wave converter.
n = 0 for single-phase and three-phase half-controlled converters
= π for three-phase half-wave converters
= ⅓π for three-phase fully controlled converters
Vo = s
These voltage output characteristics are shown in figure 11.40 and the main converter circuit
characteristics are shown in table 11.5.
11.11
Definitions
average output voltage
Vo
Vrms rms output voltage
Io
average output current
Irms rms output current
I peak output current
peak output voltage
Vrms
Load voltage form factor = FFv =
Load voltage crest factor = CFv = V
Vo
Vrms
V
Load current form factor = FFi =
Irms
Load current crest factor = CFi = I
Io
Irms
dc load power
Rectification efficiency = η =
ac load power + rectifier losses
Vo I o
=
Vrms I rms + Loss rectifier
Waveform smoothness = Ripple factor = RFv =
=
where
 ∞
VRi = 
 n=1
∑ (v
1
2
2
an

+ vbn2 ) 

effective values of ac V (or I ) VRi
=
average value of V (or I )
Vo
2
Vrms
− Vo2
= FFv2 − 1
Vo2
½
similarly the current ripple factor is RFi =
I Ri
= FFi 2 − 1
Io
RFi = RFv for a resistive load
11.12
310
Output pulse number
Output pulse number p is the number of pulses in the output voltage that occur during one ac input
cycle, of frequency fs. The pulse number p therefore specifies the output harmonics, which occur at p x
fs, and multiples of that frequency, m×p×fs, for m = 1, 2, 3, ...
The pulse number p is specified in terms of
q
the number of elements in the commutation group
r
the number of parallel connected commutation groups
s
the number of series connected (phase displaced) commutating groups
Parallel connected commutation groups, r, are usually associated with (and identified by) intergroup
reactors (to reduce circulating current), with transformers where at least one secondary is effectively star
connected while another is delta connected. The rectified output voltages associated with each
transformer secondary, are connected in parallel.
Series connected commutation groups, s, are usually associated with (and identified by) transformers
where at least one secondary is effectively star while another is delta connected, with the rectified output
associated with each transformer secondary, connected in series.
q =3 r =2 s =2
p=qxrxs
p = 12
q
π
2 Vφ × sin
π
× cos α
q
For a full-wave fully-controlled single-phase converter, r = 1, q = 2, and s = 1, whence p = 2
2 2 Vφ
π
2
Vo = 1 ×
× cos α
2Vφ × sin × cos α =
π
π
2
For a full-wave, fully-controlled, three-phase converter, r = 1, q = 3, and s = 2, whence p = 6
3 2 Vφ
π
3
Vo = 2 ×
× cos α
2 Vφ × sin × cos α =
π
π
3
(11.332)
Power Electronics
311
Naturally commutating converters
312
Table 11.5. Main characteristics of controllable converter circuits
11.13
AC-dc converter generalised equations
Alternating sinusoidal voltages
V1 = 2V sin ωt
(
2π
V 2 = 2V sin ωt −
.
q
)
(
Vq = 2V sin ωt − (q − 1) 2qπ
11.8a
11.11a
11.10b
11.7
11.18a
11.19a
11.17a
)
where q is the number of phases (number of voltage sources)
On the secondary or converter side of any transformer, if the load current is assumed constant I o then
the power factor is determined by the load voltage harmonics.
Voltage form factor
FFv =
V rms
Vo
whence the voltage ripple factor is
½
½
1
2
RFv = V rms
−Vo2  = FFv 2 − 1
V
o
The power factor on the secondary side of any transformer is related to the voltage ripple factor by
P
V I
1
pf = d = o o =
S
qVI rms
RFv 2 + 1
11.10a
11.7
11.19
11.22
11.17a
On the primary side of a transformer the power factor is related to the secondary power factor, but since
the supply is assumed sinusoidal, the power factor is related to the primary current harmonics.
Relationship between current ripple factor and power factor
1 ∞ 2
1
2
RFi =
− I 12
∑ I h = I rms
I1
I1
h =3
I1
pf =
I rms
=
1
1 + RFi 2
The supply power factor is related to the primary power factor and is dependent of the supply
connection, star or delta, etc.
Half-wave diode rectifiers [see figures 11.1, 11.15]
Pulse number p=q. Pulse number is the number of sine crests in the output voltage during one input
voltage cycle. There are q phases and q diodes and each diode conducts for 2π/q, with q crest (pulses)
in the output voltage
Mean voltage
Vo =
q
=
π
q
∫
2π
½π + π q
2V sin ωt d ωt
½π − π q
π
2 V sin
q
RMS voltage
q
 2π
V rms = 
∫
½π + π q
½π − π
q
(
2 V sin ωt

q
2π 
= 2V ½ +
sin
q 
4π

Normalised peak to peak ripple voltage
v p − p = 2V − 2V cos
Vnp − p =
v p −p
=
Vo
π
q
2 V − 2 V cos
q
π
2 V sin
π
q
π
q
)
2


½
=
½
d ωt 
π
q
1 − cos
sin
π
q
π
q
Power Electronics
313
Voltage form factor
FFv =
V rms
Vo

q
2π 
½ + 4π sin q 


=
π
q
sin
π
½
q
whence the voltage ripple factor is
½
½
1
2
RFv = V rms
−Vo2  = FFv 2 − 1
V
Naturally commutating converters
p=q=
Isec rms
2
Io /√2
3
Io /√3
6
Io /√6
Vo
VD
%Vp-p
Ks/c
pfsec
0.90V
2√2 V
0.157
1
0.636
0.90
0.68
1.17V
√6 V
0.604
1.73
0.675
0.827
0.31
1.35V
2√2 V
0.140
6
0.55
0.995
RFv
For three-phase resistive load, with transformer turns ratio 1:N
½
2V 3 3
V
Io =
I o rms =  13 + 4.3π 
R 2π
R
.
o
.
π +
FFi output =  227
2
Diode reverse voltage
V
V
= 2 2V
DR
if q is even
π
= 2 2 V cos
2q
DR
I D = Io
ID =
I p∆ =
if q is odd
For a constant load current Io, diode currents are
Io
q
q
.
For a constant load current Io the output power is
Pd = Vo I o
The apparent power is
S = qVI rms
The power factor on the secondary side of any transformer is
P
V I
1
pf = d = o o =
S
qVI rms
RFv 2 + 1
=
q
π
2 V sin
π
× Io
q
1
qV × I o
=
2q
π
sin
N
1
×
V 1
+
R 3
.3
4π
3
2π 2
−


½
N
π
6 .3


½
I L∆ =
V
N
½
×
1
V 2
+
R 3
q phases and 2q diodes
Mean voltage
q ½π + π q
2V sin ωt d ωt
π ∫½π − π q
π
2q
2V sin
=
π
q
Vo =
q
Pulse number
π
V Io
pf 3φ ,½ =
Vo I o
=
3V I o
3 3
= 0.827
2π
For six-phase half-wave p=q=6
pf 6φ ,½ =
Vo I o
3V I o
==
3
π
= 0.995
(Y conection)
The short circuit ratio (ratio actual s/c current to theoretical s/c current) is
q 2V
K s/c =
ωLc
2 2V
ωLc
Commutation overlap angle
1 − cos µ =
sin
π
q
p=q
p=2q
if q is even
if q is odd
Diode reverse voltage
V DR = 2 2V
if q is even
V DR = 2 2V cos
For three-phase half wave p=q=3
=
q
π
2 sin
q
ωLc I o
π
2V sin
q
The commutation voltage drop
v com = q ωL I where 2Lc = Ls / c
c o
2π
½
Full-wave diode bridge rectifiers - star [see figures 11.14, 11.16]
π
q
The primary side power factor is supply connection and transformer construction dependant.
For two-phase half-wave p=q=2
V I
2 2
pf 1φ ,½ = o o =
= 0.90
.3 
2π 
×  2 + .3 
1 R  9 6π 
Time domain half-wave single phase R-L-E load
E Z
 − ωt −α
E
2V 
i o (ωt ) = − +
 sin (ωt − φ ) + 
− sin (ωt − φ )  e tanφ

R
Z 
 R 2V

k


∞ −2 −1
(
)
v o (ωt ) = Vo 1 + ∑ 2 2
cos ( kq ωt ) 


k =1 k q − 1


I LY =
Io
I D rms =
314
π
if q is odd
2q
For a constant load current Io, diode currents are
I D = Io
ID =
Io
q
I D rms =
Io
q
.
The current and power factor are
I rms = I o
pf =
=
2
q
Pd
V I
= o o
S
qVI rms
π
2q
2V sin × I o
π
q
qV × I o
2
=
2 q
π
q
which is √2 larger than the half-wave case.
For single-phase, full-wave p=q=2
pf 1φ =
Vo I o 2 2
=
= 0.90
V Io
π
sin
π
q




pfprim
Power Electronics
315
For three-phase full-wave p=2 q=6
6
316
The commutation voltage drop
2V × ½I o
Vo I o
= π
3VI o
pf 3φ =
Naturally commutating converters
2
I
3 o
3V ×
=
3
π
v com = q ωL I where 2Lc = Ls / c
π c o
= 0.955
For p=q=2, only
p, q
Isec
RFv
Vo
VD
%Vp-p
Ks/c
pfY prim
pfsec
p=q=2
Io
0.483
1.80V
2√2 V
0.157
2/π
0.90
0.90
p=2 q=6
√⅔ Io
0.31
2.34V
√6 V
0.140
6/π
0.995
0.995
The short circuit ratio (ratio actual s/c current to theoretical s/c current) is
K s/c =
1 − cos µ =
v com = 4 ωL I
π c o
2V
Load characteristics
q
2π sin
2ωLc I o
I o rms
=
Current Form Factor = FF I =
Io
π
q
2
Io
q
=
Io
2
q
which is smaller by a factor π than the half-wave case.
K s/c =
q
for q = 2
π
Full-wave diode bridge rectifiers – delta
Relationship between current ripple factor and supply side power factor on the primary
RFi =
1
1
∞
∑ I h2 =
I 1 h =3
I1
I
1
pf = 1 =
I rms
1 + RFi 2
V
2 sin
1
The mean output voltage is
π 2q
2q
Vo =
2V ∆ sin =
π
2
I rms
− I 12
I1
 1 4
I o2 − 
 2π
1 4
=
2π
1
1 + RFi 2
2

π2 − 8
=
1
1+
= 0.483
8
Io
=
π2 − 8
=
2 2
.
π
= 0.90
π
p2
I rms even =
I rms odd =
2
π
sin
p
−1
Commutation overlap angle
1 − cos µ =
sin
π q
=
q π
2V
=
if q is odd
π
2 sin
2q
I D = Io /q
Io
.
p
π
sin
π
p
ωLc I o
π
q
q
2V I o
Vo I o
2 2
= π
=
qVI rms
qV ½I o
π
.
pf q even =
2
2
I o q − 1
2
q
½
pf q odd =
ωLI o
.
Vo I o
2 2
q
=
qVI rms
π q 2 − 1½


.
v com = q ωL I
c o
2π

1
q
v com =
ωLc I o 1 − q 


2π
2V
.
2V sin
I D rms = I o / q
Commutation angle and voltage
1 − cos µ =
2
1
2V
I D = Io
For p-pulse
1 + RFv 2
π
q
rms current and power factor
I1
I1
=
for k ≥ 1, 2, 3...
h kp ± 1
pf =
2 sin
q
V DR =
.
RFv =
V
2
3V L −N =
p=q
if q is even
p=2q if q is odd
diode reverse voltage and currents
2V
V DR =
if q is even
π
sin
8
The rms of the fundamental component is
1 4
I1 =
Io
2π
The rms of the harmonic components are
Ih =
π
q
.
Pulse number
Io 
.
.
pf =

π
q
For example in three-phase, V is replaced by V/√3, that is, V L −L =
For single phase p=2
RFi =
Same expression as for delta connected supply, except supply voltages V are replaced by
2
I rms
− I 12
ωLI o 
q even
1
q odd
1 − cos µ =
1 − 
q
2V 
The short circuit ratio (ratio actual s/c current to theoretical s/c current) is
q
π
q −1
π
K s/c even = sin
K s/c odd =
sin
.
π
q
π
q
For single-phase resistive load, with transformer turns ratio 1:N
.
3V phase
Power Electronics
317
2V 4
R π
π
Io =
I o rms =
.
FFi output =
2V
L
R
RFv = FF 2 − 1 =
2 2
.
Ip =
N
I sec =
1
N
1
×
2V
1
pf =
R
Naturally commutating converters
RF + 1
2
=
π2
8
−1
2 2
.
π
di
+ Ri = 2V cos ωt
dt
(V)
Continuous current
π
( − +α
2 V
2 V
π
i (ωt ) =
cos(ωt - φ ) + { i o cos(− + α - φ ) e p
Z
where Z = R 2 + (ωL )2
tan φ = ωL / R
318
Z
- ωt ) / tanφ
p
}
(ohms)
where
Half-wave controlled rectifiers – star connected supply [see figures 11.8, 11.18]
q phases and q thyristors, and a phase delay angle of α. The pulse number is p (=q).
Mean output voltage is
q ½ π + π q +α
Vo =
2V sin ωt d ωt
2π ∫½π − π q +α
q
π
2V sin cos α
=
π
q
= Vo (α = 0 ) cos α
The rms output voltage is
q
½π + π q + α
(
)
π
for 0 < α <
q
π
for α >
q
3π π
for α <
+
2 q
3π π
for α >
+
2 q
v = 2V
.
 π

= 2V cos  − + α 
 q

∨
π

v = 2V cos  + α 
q

.
.
= − 2V
.
Normalised peak to peak ripple voltage and amplitude of the output harmonics are
Vnp − p =
Vh =
q
4π
Diode reverse voltage
2V sin
.
2π
q
v p −p π
=
Vo
q
cos α
1 − cos
sin
2
k 2q 2 − 1
π
q

 −
π
π
π
π
+ α ) + tan φ sin( + α ) − cos( − α ) − tan φ sin( − α )  e p tan φ
p
p
p
p


with an average value of
V
π
p 2V
sin cos α
Io = o =
π
R
π
π
+ tanh
p
p tan φ
tan α =
π
π
tan − tan φ tanh
p
p tan φ
tan φ tan
π


π

π
  ( − p +α
cos ωt + tan φ sin ωt +  cos  − α  − tan φ sin  − α   e
p

p



The average output voltage is dependent on the current extinction angle, β

 π

p
Vo =
2V  sin β − sin  − + α  
2π
 p


i (ωt ) =
2V
R sec 2 φ
q phases q thyristors and 1 diode
Overlap angle and inductive voltage
cos α − cos (α + µ ) =
ωLc I o
2V sin
.
v com =
Time domain equations, for an R-L load
q
2π
π
q
ωLc I o



π
π
+ α < ωt < + α
p
p
π
π
for
< ωt < + α
2
p
for −
where the earliest conduction point is
α > ½π -
π
p
Mean rectified output voltage is
Vo =
=
Power factor (is related to the equivalent diode circuit)
2q
π
sin cos α = pf α =0 cos α
π
q
- ωt ) / tanφ
Half-wave controlled rectifiers, with freewheel diode [see figures 11.10, 11.19]
=0
q even
π
q odd
= 2 2V cos
2q
2π
p tan φ
p
Discontinuous current
Boundary condition
π
q
1 + k 2q 2 tan2 α
−
.
R
The thyristor currents are the same as the equivalent diode circuit
pf =
1−e
v (ωt ) = 2V cos ωt
V DR = 2 2V
V DR
cos(
2
2V sin ωt d ωt
2π ∫½π − π q +α
½


2π
q
sin
cos α 
= 2V ½ +
q
4
π


The maximum and minimum voltages in the output are
V rms =
2π
2 V
×
io =
R sec 2 φ
p
2π
∫
p
2π
½π
− π q +α
2V cos ωt d ωt

 π

2V 1 − sin  − + α  
 p


RMS voltage

½

pα
p
2π  
sin  2α −
+
q  
8 4π 8π


π
The maximum ripple occurs at ωt = − + α , with zero volts during diode freewheeling, thus
p
V rms = 2V ¼ −
p
−
Power Electronics
319
π
v p − p = 2V cos 
p
Vnp − p =
v p −p
Vo
=
p
2π
Naturally commutating converters
Overlap angle and inductive voltage

−α −0

π

π

2V cos  − α 
cos  − α 
p
2π
p



=
p

π

π

1 + sin  − α 
2V 1 − sin  − α  
p

p


The freewheel diodes conduct for p periods of duration π/p+α, and the currents are
pα


I Df = I o  ½ +
- ¼p 
2π


½

I rms Df = I o  ½ +
pα

- ¼p 
2π

I rmsTh
v com =
q phases 2q thyristors
Pulse number, p
p=q
p=2q
Mean voltage
q phases q thyristors and q diodes, each of which conduct for 2π/q
Pulse number
p=q
for all q, odd or even
Mean voltage
Vo =
q
π
= Vo'
where Vo' =
π
(1 + cos α )
q
1 + cos α
2 V sin
2
2q
2 V sin
π
Thyristor and diode reverse voltage
V R = 2 2V
V R = 2 2V cos
if q is even
if q is odd
Power factor
2
Is = Io
pf =
q
π
p
Vo = V sin cos α = Vo' cos α
π
p
α

I s = I o 1 − 
π

The rms output voltage is
q
ωL I
π c o
π
q
Half-controlled full bridges – star connected supply [see figures 11.7, 11.157, 11.22]
½
Full-wave fully controlled thyristor converters–star connected supply [see figures 11.11, 11.12, 11.13, 11.20]
2V sin
.

The thyristor conductor for 2π/p without a load freewheel diode and 2π/p-(π/p+α+½π) when the diode is
present. The thyristor rms current is
I 
pα

= o ½ + ¼p 
2π
q 

ωLc I o
cos α − cos (α − µ ) =
320
½
2
q even
π
2q
q odd
π 1 + cos α
q
2
q sin
π
π
q
.
α <π −
½
π
 π 
pf =
sin (1 + cos α ) 

π
q
π −α 
.
2
α >π −
2π
q
2π
q
½


p
2π
sin
cos α 
Vo rms = V ½ +
4π
p


The maximum and minimum voltages in the output are
Full-wave fully-controlled bridges – delta connected supply
for 0 < α <
v =V
 π

= V cos  − + α 
 p

∨
π

v = V cos  + α 
p

for α >
π
p
3π π
+
2
p
3π π
+
for α >
2
p
for α <
= −V
π 1
V =
×Vo'
p sin π
p
π
2q
Vo' =
2V sin
π
q
Thyristor maximum reverse and forward voltage
V R = 2 2V
π
2q
Thyristor maximum forward and reverse voltages
2 2V
VR =
q even
π
q
sin
π
q
2 2 V cos
π
2q
q odd
π
q
sin
π
q
The power factor is the same as for the star case.
Half-controlled full bridges – delta connected supply
q even
q odd
RMS currents and power factor
Io
2
I TH rms = I o
q
q
q
π
pf = sin cos α = pf α =0 cos α
π
q
I TH =
Pulse number in the rectified output is
p=q
for q even
p=2q for q odd
VR =
where
V R = 2 2V cos
π
p
In terms of the semiconductors and rectified voltage star and mesh behave the same.
Mean voltage, for all q is
Vo =
q
π
= Vo'
where Vo' =
2q
π
π
(1 + cos α )
q
1 + cos α
2 V sin
2
2 V sin
π
q
Power Electronics
321
For a 3-phase half-controlled converter, the secondary current is
.
α
1 − 
π
3
11.3.
Derive equations (11.20) and (11.21) for the circuit in figure 11.3.
11.4.
Assuming a constant load current derive an expression for the mean and rms device current
and the device form factor, for the circuits in figures 11.4 and 11.7.
11.5.
Plot load ripple voltage KRI and load voltage ripple factor Kv, against the thyristor phase delay
angle α for the circuit in figure 11.7.
11.6.
Show that the average output voltage of a n-phase half-wave controlled converter with a
freewheel diode is characterised by
sin(π / n )
Vo = 2 V
cos α
(V)
π /n
0 < α < ½ -π / n
1 + cos α + ½π − 1 n π
Vo = 2 V
(V)
2π / n
1
1
½π − n π < α < ½π + n π
11.7.
Show that the average output voltage of a single-phase fully controlled converter is given by
2 2V
Vo =
cos α
For large q, q>6
Is =
Io 
.
pf =
.
½
α  
1 −   
3 
 π  
2
1 + cos α
π
2
½
2

α  
π 1 −   

 π  

Reading list
Dewan, S. B. and Straughen, A., Power Semiconductor Circuits, John Wiley and Sons, New York, 1975.
Sen, P.C., Power Electronics, McGraw-Hill, 5th reprint, 1992.
π
Shepherd, W et al. Power Electronics and motor control, Cambridge University Press, 2nd Edition 1995.
Assume that the output current Io is constant.
Prove that the supply current Fourier coefficients are given by
4I
an = − o sin nα
nπ
4Io
bn =
cos nα
nπ
for n odd.
Hence or otherwise determine (see section 12.6)
i. the displacement factor, cos ψ
ii. the distortion factor, µ
iii. the total supply power factor λ.
Determine the supply harmonic factor, ρ, if
ρ = Ih / I
where Ih is the total harmonic current and I is the fundamental current.
http://www.ipes.ethz.ch/
Problems
11.1.
322
½
Io 
Is =
Naturally commutating converters
For the circuit shown in figure 11.41, if the thyristor is fired at α = ⅓π
i.
derive an expression for the load current, i
ii.
determine the current extinction angle, β
iii.
determine the peak value and the time at which it occurs
iv.
sketch to scale on the same ωt axis the supply voltage, load voltage, thyristor voltage,
and load current.
11.8. Show that the average output voltage of a single-phase half-controlled converter is given by
2V
vo =
(1 + cos α )
π
Assume that the output current Io is constant.
Determine
i.
the displacement factor, cos ψ
ii.
the distortion factor, µ
iii.
the total supply power factor, λ.
Show that the supply harmonic factor, ρ (see problem 11.7), is given by
Figure 11.41. Problem 11.1.
11.2.
For the circuit shown in figure 11.42, if the thyristor is fired at α = ¼π determine
i.
the current extinction angle, β
ii.
the mean and rms values of the output current
iii.
the power delivered to the source E.
iv.
sketch the load current and load voltage vo.
Figure 11.42. Problem 11.2.
 π (π − α )
ρ= 
 4 (1 + cos α )
11.9.

− 1

Draw the load voltage and current waveforms for the circuit in figure 11.10a when a freewheel
diode is connected across the load. Specify the load rms voltage.
Power Electronics
323
11.10. A centre tapped transformer, single-phase, full-wave converter (figure 11.11a) with a load
freewheel diode is supplied from the 240 V ac, 50 Hz supply with source inductance of 0.25 mH.
The continuous load current is 5 A. Find the overlap angles for
i. the transfer of current form a conducting thyristor to the load freewheel diode and
ii. from the freewheel diode to a thyristor when the delay angle α is 30°.
 ω LI o 
γ t −d = cos−1 1 −
 = 2.76°;
2V 



ω LI o 

2V 

γ t −d = cos −1 cos α −
 − α = 0.13°
11.11. The circuit in figure 11.8a, with v = √2 V sin(ωt + α), has a steady-state time response of
2V
i (ωt ) =
{sin(ωt + α − φ ) − sin(α − φ )e− Rt / L }
Z
where α is the trigger phase delay angle after voltage crossover and φ = tan −1 (ω L / R )
Sketch the current waveform for α = ¼π and Z with
i.
R >> ωL
ii.
R = ωL
iii.
R << ωL.
[(√2 V/R) sin(ωt + ¼π); (V/R) sin ωt; (V/ωL) (sin ωt - cos ωt + 1)]
11.12. A three-phase, fully-controlled converter is connected to the 415 V supply, which has a
reactance of 0.25 Ω/phase and resistance of 0.05 Ω/phase. The converter is operating in the
inverter mode with α = 150º and a continuous 50 A load current. Assuming a thyristor voltage
drop of 1.5 V at 50 A, determine the mean output voltage, overlap angle, and available recovery
angle.
[-485.36 V -3 V -5 V -11.94 V = -505.3 V; 6.7°; 23.3°]
11.13. For the converter system in problem 11.12, what is the maximum dc current that can be
accommodated at a phase delay of 165°, allowing for a recovery angle of 5°?
[35.53 A]
11.14. The single-phase half-wave controlled converter in figure 11.8 is operated from the 240 V, 50 Hz
supply and a 10 Ω resistive load. If the mean load voltage is 50 per cent of the maximum mean
voltage, determine the (a) delay angle, α, (b) mean and rms load current, and (c) the input
power factor.
11.15. The converter in figure 11.10a, with a freewheel diode, is operated from the 240 V, 50 Hz
supply. The load consists of, series connected, a 10 Ω resister, a 5 mH inductor and a 40 V
battery. Derive the load voltage expression in the form of a Fourier series. Determine the rms
value of the fundamental of the load current.
11.16. The converter in figure 11.7a is operated from the 240 V, 50 Hz supply with a load consisting of
the series connection of a 10 Ω resistor, a 5 mH inductor, and a 40 V battery. Derive the load
voltage expression in the form of a Fourier series. Determine the rms value of the fundamental
of the load current.
11.17. The converter in figure 11.19 is operated from a Y-connected, 415 V, 50 Hz supply. If the load is
100 A continuous with a phase delay angle of π/6, calculate the (a) harmonic factor of the
supply current, (b) displacement factor cos ψ, and (c) supply power factor, λ.
11.18. The converter in figure 11.19 is operated from the 415 V line-to-line voltage, 50 Hz supply, with
a series load of 10 Ω + 5 mH + 40 V battery. Derive the load voltage expression in terms of a
Fourier series. Determine the rms value of the fundamental of the load current.
11.19. Repeat problem 11.18 for the three-phase, half-controlled converter in figure 11.17.
11.20. Repeat problem 11.18 for the three-phase, fully-controlled converter in figure 11.20.
11.21. The three-phase, half-controlled converter in figure 11.17 is operated from the 415 V, 50 Hz
supply, with a 100 A continuous load current. If the line inductance is 0.5 mH/phase, determine
the overlap angle γ if (a) α = π/6 and (b) α = ⅔π.
Naturally commutating converters
324
11.22. Repeat example 11.1 using a 100Vac 60Hz supply.
11.23. A fully controlled half-wave rectifier has a resistive 30 Ω and a 240V ac 50Hz voltage source.
(a)
If the delay angle α = 60º, determine:
i.
the average voltage across the load resistor
ii.
the power absorbed by the load resistor
iii.
the ac source power factor.
(b)
If the average load current is 5A determine
i.
the average voltage across the load
ii.
the power absorbed by the load
iii.
the supply power factor.
11.24. A fully controlled half-wave rectifier has a 240V ac 50Hz source and a series R-L load of R =
20Ω and L = 50mH. If the delay angle is α = 60º, determine
i.
an expression for the load current
ii.
the average load current
iii.
the power absorbed by the load
iv.
the supply power factor
11.25. A single phase uncontrolled rectifier has a 24Ω resistive load a 240V ac 50Hz supply. Determine
the average, peak and rms current and peak reverse voltage across each rectifier diode for
i.
an isolating transformer with a 1:1 turns ratio
ii.
centre-tapped transformer with turns ratio 1:1:1.
11.26. A single phase bridge rectifier has an R-L of R = 20Ω and L = 50mH and a 240V ac 50Hz
source voltage. Determine:
i.
the average and rms currents of the diodes and load
ii.
rms and average 50Hz source currents
iii.
the power absorbed by the load
iv.
the supply power factor
11.27. An electromagnet is modelled by a series R-L circuit with R=10Ω and L=100mH, and supplied
from a 50Hz 240V ac voltage source.
i.
when supplied from a full-wave uncontrolled rectifier, the average current must be
20A to active the magnetic field. What series resistance must be added to
increase the average current to 20A?
ii.
when supplied from a full-wave fully-controlled converter, what delay angle will
produce the necessary average current of 20A to activate the electro-magnet?
11.28. A single-phase, full-wave uncontrolled rectifier has a back emf Eb in its load. If the supply is
240Vac 50Hz and the series load is R = 20Ω, L = 50mH, and Eb = 120V dc, determine:
i.
the power absorbed by the dc source in the load
ii.
the power absorbed by the load resistor
iii.
the power delivered from the ac source
iv.
the ac source power factor
v.
the peak-to-peak load current variation if only the first ac term of the Fourier
series for the load current is considered.
11.29. Show that the power factor for a fully-controlled single-phase full-wave converter with a purely
resistive load is given by
α sin 2α
pf = 1 − +
2π
π
11.30. A fully controlled single-phase full-wave bridge converter has a 60 Ω resistive load and a 240V
ac 50Hz voltage source. If the firing delay angle is α = 60º, determine:
i.
the average load current
ii.
the rms load current
iii.
the rms source current
iv.
the ac source power factor
325
Power Electronics
11.31. A fully controlled single-phase full-wave bridge converter has an R-L load of 50 Ω and 50mH
and is supplied from a 240V ac 50Hz voltage source. Determine the average load current if:
i.
α = 15º
ii.
α = 75º
11.32. A three-phase uncontrolled rectifier is supplied from a 50Hz 415V ac line-to-line voltage source.
If the rectifier load is a 75 Ω resistor, determine
i.
the average load current
ii.
the rms load current
iii.
the rms source current
iv.
the supply power factor.
11.33. A three-phase uncontrolled rectifier is supplied from a 50Hz 415V ac line-to-line voltage source.
If the rectifier load is a series R-L circuit where R = 10Ω and L = 100mH, determine:
i.
the average and rms load currents
ii.
the average and rms diode currents
iii.
the rms source and power current
iv.
the supply power factor.
11.34. A three-phase, fully-controlled, converter is supplied from a 3.3kV ac 50Hz source. If the load is
a 110Ω resistor determine:
i.
the delay angle which results is an average load current of 20A
ii.
the amplitude of the first voltage harmonic (at 300Hz)
11.35. A three-phase, fully-controlled, converter is supplied from a 3.3kV ac 50Hz source. If the R-L
load is R = 100Ω and L=100mH, and the delay angle is α = 30º, determine:
i.
the average load current
ii.
the amplitude of the first current harmonic (at 300Hz)
iii.
the rms phase current from the ac voltage source.
Naturally commutating converters
Blank
326