55
The resulting equivalent circuit referred to the primary at 50 Hz is shown below:
+
-
V
S
V
P
I
S
+
-
I
P
R
C
jX
M
R
EQ
jX
EQ
1.31 MΩ j473 kΩ
260 Ω j917 Ω

2-20. Prove that the three-phase system of voltages on the secondary of the Y-
∆
transformer shown in Figure 2-
38b lags the three-phase system of voltages on the primary of the transformer by 30°.
S

+
+
++
+
+
V
A
'
V
B
'
V
C
'
V
A
V
B
V
C56
Assume that the phase voltages on the primary side are given by

°∠= 0
PA
V
φ
V °−∠= 120

V
φ
V

where
aVV
PS
/
φφ
= . Since this is a Y-
∆
transformer bank, the line voltage
ab
V on the primary side is

°∠=°−∠−°∠=−= 3031200
PPPBAab
VVV
φφφ
VVV
and the voltage
°∠=
′
=
′′
0
SAba
V
φ
VV


Assume that the phase voltages on the primary side are given by

°∠= 0
PA
V
φ
V °−∠= 120
PB
V
φ
V °∠= 120
PC
V
φ
V

57
Then the phase voltages on the secondary side are given by

°∠=
′
0
SA
V
φ
V

°−∠=
′

φ
V
. The line voltage on the secondary side is given by

°−∠=°∠−°∠=−=
′′
3031200
PPPCAba
VVV
φφφ
VVV
Note that the line voltage on the secondary side lags the line voltage on the primary side by 30
°
.
2-22. A single-phase 10-kVA 480/120-V transformer is to be used as an autotransformer tying a 600-V
distribution line to a 480-V load. When it is tested as a conventional transformer, the following values are
measured on the primary (480-V) side of the transformer:
Open-circuit test Short-circuit test
V
OC
= 480 V V
SC
= 10.0 V
I
OC
= 0.41 A I
SC
= 10.6 A
V
OC


,
,
0.41 A
0.000854 S
480 V
OC
EX
OC
I
Y
V
φ
φ
== =()( )
11
38 W
cos cos 78.87
480 V 0.41 A
OC
OC OC
P
VI
θ
−−
Ω

1193
51.8 pu
23.04
M
X
Ω
==
Ω

The short circuit test yields the values for the series impedances (referred to the primary side):

10.0 V
0.943
10.6 A
SC
EQ
SC
V
Z
I
== = Ω58

()()
11
26 W

Ω

0.915
0.0397 pu
23.04
EQ
X
Ω
==
Ω

The per-unit equivalent circuit is
+
-
V
S
V
P
I
S
+
-
I
P
R
C
jX
M
R
EQ


The copper losses (in resistor
EQ
R ) would be

()( )
2
2
CU EQ
1.0 0.010 0.010 puPIR== =

The output power of the transformer would be

OUT OUT CU core
1.0 0.010 0.0038 0.986PPPP=−−=− − =

and the transformer efficiency would be

OUT
IN
0.986
100% 100% 98.6%
1.0
P
P
η
=× = × =

The output voltage of this transformer is


+
-
-

(c) When used as an autotransformer, the kVA rating of this transformer becomes:

()
SE
IO
SE
41
10 kVA 50 kVA
1
C
W
NN
SS
N
++
== =

(d) As an autotransformer, the per-unit series impedance
EQ
Z
is decreased by the reciprocal of the power
advantage, so the series impedance becomes

0.010
0.002 pu
5

P
R
== =

The copper losses (in resistor
EQ
R ) would be

()( )
2
2
CU EQ
1.0 0.002 0.002 puPIR== =

The output power of the transformer would be

OUT OUT CU core
1.0 0.002 0.0038 0.994PPPP=−−=− − =

and the transformer efficiency would be

OUT
IN
0.994
100% 100% 99.4%
1.0
P
P
η
=× = × =

= 1000 kVA
base2
S
= 1000 kVA
base3
S
= 1000 kVA
base2,L
V = 480 V
base2,L
V = 14,400 V
base3,L
V = 480 V
S
OLUTION
This problem can best be solved using the per-unit system of measurements. The power system
can be divided into three regions by the two transformers. If the per-unit base quantities in Region 1 are
chosen to be
base1
S = 1000 kVA and
base1,L
V = 480 V, then the base quantities in Regions 2 and 3 will be as
shown above. The base impedances of each region will be:

(
)
2
2
1
base1

)
2
2
3
base3
base3
3
3277 V
0.238
1000 kVA
V
Z
S
φ
== =Ω

(a) To get the per-unit, per-phase equivalent circuit, we must convert each impedance in the system to
per-unit on the base of the region in which it is located. The impedance of transformer
1
T
is already in per-
unit to the proper base, so we don’t have to do anything to it:

010.0
pu,1
=R040.0
pu,1

8314 V 500 kVA
R ==()( )
()( )
2
2,pu
2
8314 V 1000 kVA
0.085 0.170
8314 V 500 kVA
X ==61
The per-unit impedance of the transmission line is

line
line,pu
base2
1.5 10
0.00723 0.0482
207.4
Zj
Zj
Z
+Ω
== = +
Ω


The resulting per-unit, per-phase equivalent circuit is shown below:
+
-
1∠0°
T
1
T
2
Line
L
1
L
2
0.010 j0.040 0.00723 j0.0482 0.040 j0.170
1.513
j1.134
-j3.36

(b) With the switch opened, the equivalent impedance of this circuit is

EQ
0.010 0.040 0.00723 0.0482 0.040 0.170 1.513 1.134Zj j j j=+ + + ++ ++EQ
1.5702 1.3922 2.099 41.6Zj=+ =∠°

The resulting current is


0.344 1000 kVA 344 kWPPS== =

The power supplied by the generator is

()( )
,pu
cos 1 0.4765 cos41.6 0.356
G
PVI
θ
== °=()( )
,pu
sin 1 0.4765 sin 41.6 0.316
G
QVI
θ
== °=()( )
,pu
1 0.4765 0.4765
G
SVI== =()( )

1.513 1.134 3.36
0.010 0.040 0.00723 0.0482 0.040 0.170
1.513 1.134 3.36
jj
Zj j j
jj
+−
=+ + + ++ +
++−EQ
0.010 0.040 0.00788 0.0525 0.040 0.170 (2.358 0.109)Zj j j j=+++++++EQ
2.415 0.367 2.443 8.65Zj=+ =∠°

The resulting current is

10
0.409 8.65
2.443 8.65
∠°
==∠−°
∠°
I

The load voltage under these conditions would be


θ
== °=()( )
,pu
sin 1 0.409 sin 6.0 0.0428
G
QVI
θ
== °=()( )
,pu
1 0.409 0.409
G
SVI== =()( )
,pu base
0.407 1000 kVA 407 kW
GG
PPS== =()( )
,pu base
0.0428 1000 kVA 42.8 kVAR


()( )
2
2
line,pu line
0.409 0.00723 0.00121PIR== =()( )
line l ,pu base
0.00121 1000 kVA 1.21 kW
ine
PPS== =

Load 2 improved the power factor of the system, increasing the load voltage and the total power supplied to
the loads, while simultaneously decreasing the current in the transmission line and the transmission line
losses. This problem is a good example of the advantages of power factor correction in power systems. 63
Chapter 3:
Introduction to Power Electronics
3-1. Calculate the ripple factor of a three-phase half-wave rectifier circuit, both analytically and using
MATLAB.
S
OLUTION
A three-phase half-wave rectifier and its output voltage are shown below
π
/6
5

If we find the average and rms values over the interval from
π
/6 to 5
π
/6 (one period), these
values will be the same as the average and rms values of the entire waveform, and they can be used to
calculate the ripple factor. The average voltage is

()
5/6
/6
13
() sin
2
DC M
VvtdtVtdt
T
π
π
ωω
π
==
5
6
6
333333
cos 0.8270

π
ωω
π
==
5/6
2
rms
/
6
311
sin 2
22 4
M
V
Vtt
π
π
ωω
π

=−

64 =−−=−−− 22
rms
31333 3
0.8407
23422 234
MM
M
VV
VV
ππ
ππ=−−−=+=


is just the largest voltage of
()
tv
A
,
()
tv
B
, and
()
tv
C
at any particular time. The function is shown below:

function volts = halfwave3(wt)
% Function to simulate the output of a three-phase
% half-wave rectifier.
% wt = Phase in radians (=omega x time)

% Convert input to the range 0 <= wt < 2*pi
while wt >= 2*pi
wt = wt - 2*pi;
end
while wt < 0
wt = wt + 2*pi;
end

% Simulate the output of the rectifier.
a = sin(wt);
b = sin(wt - 2*pi/3);


% M-file:
test_halfwave3.m

% M-file to calculate the ripple on the output of a
% three phase half-wave rectifier.

% First, generate the output of a three-phase half-wave
% rectifier
waveform = zeros(1,128);
for ii = 1:128
waveform(ii) = halfwave3(ii* pi/64);
end

% Now calculate the ripple factor
r = ripple(waveform);

% Print out the result
string = ['The ripple is ' num2str(r) '%.'];
disp(string);
When this program is executed, the results are

»
test_halfwave3

The ripple is 18.2759%.
This answer agrees with the analytical solution above.
3-2. Calculate the ripple factor of a three-phase full-wave rectifier circuit, both analytically and using
MATLAB.
S

=+

S
OLUTION
By symmetry, the rms voltage over the interval from 0 to T/12 will be the same as the rms
voltage over the whole interval. Over that interval, the output voltage is:

() () ()
22
sin sin
33
CB M M
vt vt vt V t V t
ππ
ωω

=−= +− −

()
22 22
sin cos cos sin sin cos cos sin
33 33
MM
vt V t t V t t
ππ ππ
ωω ωω


/
6
0
0
16 63
() 3 cos sin
DC M M
Vvtdt VtdtVt
T
πω
πω
ω
ωω
ππ
== =
33
1.6540
DC M M
VV V
π
==

The rms voltage is

/6
222
rms


=+

rms
18 1 3 9 3
sin 1.6554
12 4 3 2 4
MMM
VV V V
ωπ π
πωω π

=+=+=
The resulting ripple factor is

2
2
rms
DC
1.6554
1 100% 1 100% 4.2%
1.6540
M
M

% full-wave rectifier.
% wt = Phase in radians (=omega x time)

% Convert input to the range 0 <= wt < 2*pi
while wt >= 2*pi
wt = wt - 2*pi;
end
while wt < 0
wt = wt + 2*pi;
end

% Simulate the output of the rectifier.
a = sin(wt);
b = sin(wt - 2*pi/3);
c = sin(wt + 2*pi/3);

volts = max( [ a b c ] ) - min( [ a b c ] );
The test driver program is shown below.

% M-file: test_fullwave3.m
% M-file to calculate the ripple on the output of a
% three phase full-wave rectifier.

% First, generate the output of a three-phase full-wave
% rectifier
waveform = zeros(1,128);
for ii = 1:128
waveform(ii) = fullwave3(ii*pi/64);
end


for the SCR. Therefore, no current flows to the
load and v
LOAD
= 0.
(2) Voltage v
1
is applied to the control circuit, charging capacitor C
1
with time constant RC
1
.
(3) When v
C
> V
BO
for the DIAC, it conducts, supplying a gate current to the SCR.
(4) The gate current in the SCR lowers its breakover voltage, and the SCR fires. When the SCR fires,
current flows through the SCR and the load.
(5) The current flow continues until i
D
falls below I
H
for the SCR (at the end of the half cycle). The
process starts over in the next half cycle.

69

If switch S
1
is shut, the charging time constant is increased, and the DIAC fires later in each half cycle.

, the period of a half cycle. For a firing angle of 0°, the average voltage will be

/
/
ave
0
00
11
() sin cos
T
MM
VvtdtVtdtVt
T
πω
πω
ω
ωω
ππ
== =−
[]
()( )
ave
12
1 1 0.637 169.5 V 108 V
MM
VV V
ππ

22
MM
VV V
ππ

+
=− − − = = =
(c) For a firing angle of 90°, the average voltage will be

/
/
ave
/
2
/2 /2
11
() sin cos
T
MM
VvtdtVtdtVt
T
πω
πω
π
ππ
ω
ωω

a differential equation with a forcing function. This problem should only be
assigned if the class has the mathematical sophistication to handle it.
S
OLUTION
At the beginning of each half cycle, the voltages across the DIAC and the SCR will both be
smaller then their respective breakover voltages, so no current will flow to the load (except for the very tiny
current charging capacitor C), and v
load
(t) will be 0 volts. However, capacitor C charges up through
resistor R, and when the voltage v
C
(t) builds up to the breakover voltage of D
1
, the DIAC will start to
conduct. This current flows through the gate of SCR
1
, turning the SCR ON. When it turns ON, the
voltage across the SCR will drop to 0, and the full source voltage v
S
(t) will be applied to the load,
producing a current flow through the load. The SCR continues to conduct until the current through it falls
below I
H
, which happens at the very end of the half cycle.
Note that after D
1
turns on, capacitor C discharges through it and the gate of the SCR. At the end of the
half cycle, the voltage on the capacitor is again essentially 0 volts, and the whole process is ready to start
over again at the beginning of the next half cycle.
To determine when the DIAC and the SCR fire in this circuit, we must determine when v

C
(t). 0
21
=+ ii
(since the DIAC is an open circuit at this time)

0
1
=+
−
C
C
v
dt
d
C
R
vv71

1
11
v
RC
v

d
vv
dt RC
+=

The solution to the natural response differential equation is

()

,
e
t
RC
Cn
vtA
−
=
where the constant A must be determined from the initial conditions in the system. The forced response is
the steady-state solution to the equation

1
sin
M
CC
dV
vv t
dt RC RC
ω
+=


++ +=()()
12 12
1
cos in sin cos sin
M
V
BtBst B tB t t
RC RC
ωωω ω ω ω ω
−+ +=cosine equation:

12
1
0
BB
RC
ω
+=
⇒

21
BRCB
ω
=−

V
RC B
RC RC
ω

+=

222
1
1
M
RC V
B
RC RC
ω

+
=
Finally,

72

1
222

MM
Cf
VRCV
vt t t
RC RC
ω
ωω
ωω
=−
++

and the total solution is

(
)
(
)
(
)
,,CCnCf
vt v t v t=+

()

222 222

sin cos
11
t
MM

−
=+ − =
++222

0
1
M
RC V
A
RC
ω
ω
−=
+222

1
M
RC V
A
RC
ω
ω
=
+

)
100
42 11.14 sin 42 cos
t
C
vt e t t
ωω
−
=+ −
This equation is plotted below:

It reaches a voltage of 30 V at a time of 3.50 ms. Since the frequency of the waveform is 60 Hz, the
waveform there are 360
°
in 1/60 s, and the firing angle
α
is

73

()
360
3.50 ms 75.6
1/60 s
α
°

==°
Vtt
πω
α
ωω
π

=−

()()
2
rms
11
sin 2 sin 2
24
M
V
V
πα π α
π

=−−−

rms
0.3284 0.573 97.1 V

R74
3-7. Explain the operation of the circuit shown in Figure P3-2, and sketch the output voltage from the circuit.

S
OLUTION
This circuit is a single-phase voltage source inverter.
(1) Initially, suppose that both SCRs are OFF. Then the voltage on the transformer T
3
will be 0, and
voltage V
DC
will be dropped across SCR
1
and SCR
2
as shown.
(2) Now, apply a pulse to transformer T
1
that turns on SCR
1
. When that happens, the circuit looks like:

Since the top of the transformer is now grounded, a voltage V
DC
appears across the upper winding as
shown. This voltage induces a corresponding voltage on the lower half of the winding, charging capacitor
C

filters it
somewhat.