ELEC4240/9240
Power Electronics
The University of New South Wales
School of Electrical Engineering & Telecommunications
Solution for Tutorial 1
1. (a)
I
time
/2
/2
3 /2
t=0
i( t ) =
4I d
cos n t
n =1,3,5... n
where
an =
/2
I d cos( n t )d t +
/2
( I d ) cos( n t )d t =
4I d
n
sin
n
2
n = 1,2,3,4.....
a0 = 0
bn = 0
i1 =
4I d
cos t
I0 = 0
I1 =
4I d
= 0.900I d , I 3 =
2
4I d
= 0.300I d , I 5 =
2 (3 )
4I d
= 0.180I d
2 (5 )
The spectrum of the current is like this:
In
n
1
Solution to Tutorial 1
ST-1
M. F. Rahman/28 March, 2003
1(b)
a0 = 0
bn = 0
/ 3
time
/3
2
/ 2
an =
t=0
/3
I d cos( n t )d t +
4I d
sin( n / 3 ) ;
n
i( t ) =
2 / 3
( I d ) cos( n t )d t
n = 1,3,5......
4I d
sin( n / 3 )cos n t
n =1,3,5... n
I0 = 0
4I d
I1 =
sin( / 3 ) = 0.7797 I d ; I 3 =
2
4I d
I5 =
sin( 5 / 3 ) = 0.1559I d
2( 5 )
4I d
sin( 3 / 3 ) = 0
2( 3 )
1(c )
/2
time
/2
/2
t=0
a0 = 0
an = 0
4
bn =
/2
Id
4
t sin( n t )d t +
/2
/2
/2
I d sin( n t )d t
ELEC4240/9240
Power Electronics
8Id
sin( n / 2) ;
n2
n = 1,3,5....
8I d
sin( / 2 ) = 0.8958I d , = 20;
2
8I d
8I d
I3 =
sin( 3 / 2 ) = 0.28658I d ;
I5 =
sin( 5 / 2 ) = 0.1581I d
9 2
25 2
I1 =
1(d)
Id
time
/3
2
t=0
a0 = 0
an = 0
bn =
/ 3+
[ / 32
2
Id
( t
/ 3
2
)sin( n t )d t + 2/ 3+ I d sin( n t )d t ]
2
8I
n
n
bn = 2 d [sin(
) cos(
)] ;
n = 1,3,5......
n
2
6
8I d
8I d
3
I1 =
sin( / 2 ) cos( ) = 0.7757 I d ; I 3 =
sin( 3 / 2 ) cos(
)=0
6
6
2
9 2
8I d
5
I5 =
sin( 5 / 2 ) cos(
) = 0.1369I d
6
25 2
1(e)
I max
/2
time
a0 = 0
an = 0
Solution to Tutorial 1
ST-3
M. F. Rahman/28 March, 2003
bn =
=
I max
t sin( n t )d t
/2
8I max
n 2 2
n = 1,3,5......
8I max
8I max
8I max
= 0.573I max ; I 3 =
= 0.0637 I max ; I 5 =
= 0.0229I max
2
2
2
9 2
25 2 2
I1 =
1.(f)
Vmax
t=0
Vdc = a0 / 2 =
=
an =
/2
/2
time
Vmax s in( t )d t,
Vmax cos( t )d t =
2Vmax
Vmax sin( t ) cos( n t ) d t =
4Vmax
for n = 2,4,6...... ;
( n 2 1)
Vmax sin( t )cos( n t ) d t
an = 0 for n = 1,3,5......
bn = 0
4Vmax
4V
= 0.300Vmax ; V4 = 2 max
= 0.0600Vmax ;
( 2 1) 2
( 4 1) 2
4V
V6 = 2 max
= 0.0257Vmax ......
( 6 1) 2
V2 =
1(g).
V dc
t=0
DT
Vd = a0 / 2 = DVdc ,
an =
2Vdc
sinDn
n
; n = 1,2,3......
time
ELEC4240/9240
Power Electronics
1(h).
V dc
2 /3
t=0
time
1
3 V
Vd =
max cos td( t )
2 / 3
3
3 3Vmax
2
Notice the period of the waveform, we can describe it as:
t
f ( t ) = Vmax cos( )
3
bn = 0
[ Vmax cos(
)cos( n t )d t ]
3
V
sin( n + 1 / 3 ) sin( n 1 / 3 )
]
= max [
+
n +1/ 3
n 1/ 3
an =
n = 1,2,3,4,...
V1 = 0.2923Vmax ; V2 = 0.0668Vmax ......
1(I).
V dc
/3
Vd =
=
time
t=0
1 /6
Vmax cos( t )d t
/ 3 / 6
3Vmax
Notice the period of the waveform, we can describe it as:
f ( t ) = Vmax cos(
t
6
bn == 0
an =
[ Vmax cos(
Solution to Tutorial 1
)cos( n t )d t ] =
Vmax
ST-5
sin( n + 1 / 6 ) sin( n 1 / 6 )
] n = 1,2,3...
+
n +1/ 6
n 1/ 6
M. F. Rahman/28 March, 2003
V1 = 0.0386Vmax ; V2 = 0.00945Vmax ......
2 calculate the THD for waveforms from (a) to (e),
/2
(a) I rms =
I d2 d = I d
/2
4I d
According to the answer of 1(a), I 1 =
I d2 (
THD =
4I d
2
)2
=
4I d
1 0.9 2
= 0.48 = 48%
0.9
2
1
I1 =
/3
(b) I rms =
/3
4I d
2
2
I d = 0.816 I d
3
I d2 d =
sin( / 3 ) = 0.7797 I d
( 0.816 I d )2 (
THD =
4I d
2
4I d
2
sin( / 3 ))2
0.816 2 0.7797 2
= 0.31 = 31%
0.7797
sin( / 3 )
(c)
I rms =
/2
[(
2I d
Because of = 20
2I d
)2
/2
/2
I d d ] =
/2
2I d
)2 2 d + I d (
2
( / 2 )3
2
5
2
] = Id
[ ( )]
+ Id (
3
2
2 12
I rms = 0.9625I d
According to 1(c ), I 1 =
THD =
)2 2 d +
8I d
2
sin( / 2 ) = 0.8957 I d
( 0.9625I d )2 ( 0.8957 I d )2
0.9625 2 0.8957 2
=
= 0.39 = 39%
0.8957 I d
0.8957
ELEC4240/9240
Power Electronics
(d)
2
I rms =
=
/2
Id
( I d )2 (
)2 2 d +
/ 3+
I d d ] =
2
[(
Id
)2
3
3
+ Id (
)]
) = 0.861I d
I1 =
According to 1(d),
THD =
8I d
2
3
= 0.776 I d
2
sin( / 2 )
( 0.861I d )2 ( 0.776 I d )2
0.776 I d
0.8612 0.776 2
= 0.48 = 48%
0.776
(e)
I rms =
/2
2I d
According to 1(e) result: I 1 =
)2 2 d ] =
8I max
2 2
[(
2I d
)2
( / 2 )3
1
] = ( I d )2 ( ) = 0.577 I d
3
3
= 0.573I d
( 0.577 I d )2 ( 0.573I d )2
0.577 2 0.5732
THD =
=
= 0.12 = 12%
0.573I d
0.573
Calculate the total ripple component (f-i) as a ratio of the dc value i.e. the Ripple Factor
(f)
Vrms =
[ (Vmax sin )2 d =
According to 1(f),
Vdc = a0 / 2 =
Vmax (
)=
Vmax
2
= 0.707Vmax
Vmax S in( t )d t =
Vmax = 0.6366Vmax
( 0.707Vmax )2 ( 0.637Vmax )2
0.707 2 0.637 2
=
= 0.48 = 48%
0.637Vmax
0.637
RF =
(g)
Vrms =
1 DT
1 2
[
(Vdc )2 dt =
Vdc D = Vdc D
T 0
T
According to 1(g),
RF =
Vd = a0 / 2 = DVdc ,
DVdc D 2Vdc
=
DVdc
D D2
=
D
1
1
D
(h)
Solution to Tutorial 1
ST-7
M. F. Rahman/28 March, 2003
/3
3
3
3
2
[
(Vmax cos )2 d =
Vmax ( +
) = 0.841Vmax
/
3
2
2
3 2
Vrms =
According to 1(h),
Vdc =
3
2
/3
/3
Vmax cos( t )d t =
3 3
Vmax = 0.827Vmax
2
2
( 0.841Vmax )2 ( 0.827Vmax )2
0.841Vmax 0.827Vmax
RF =
=
= 0.18 = 18%
0.827Vmax
0.827Vmax
(i)
Vrms =
/6
/ 6
According to 1(h),
RF =
(Vmax cos )2 d =
Vdc =
/6
/6
Vmax (
Vmax cos( t )d t =
( 0.9557Vmax )2 ( 0.9549Vmax )2
0.9549Vmax
3
) = 0.9558Vmax
4
Vmax = 0.9549Vmax
0.9557 2 0.9549 2
= 0.04 = 4%
0.9549
4.
10
= 2A
5
V 1 / 2 ( 20 25 ) / 2 ( 20 25 ) / 2 ( 20 25 ) / 2
=
=
=
I1 =
R + j L 5 + j314* 0.015
5 + j4.71
6.87543.3
= 2.057 68.3 A( rms )
I0 =
I2 =
V2 / 2
( 30 20 ) / 2 ( 30 20 ) / 2 ( 3020 ) / 2
=
=
=
= 1.987 42 A( rms )
R + j L 5 + j628* 0.015
5 + j9.42
10.66 62
DCPower = 10 2 = 20W
Power from Source 1:
V1 I 1 cos1 = 21.19W
where
1 = 43.3
Power from Source 2:
V1 I 1 cos2 = 19.746W
ELEC4240/9240
Power Electronics
where 2 = 62
PTotal = 20 + 21.19 + 19.746 = 60.94W
5.
340 15
cos30 + 0 + 0
2
2
= 2209W
P =0+
VSrms = 340 / 2 = 240V
I Srms = 8 2 + ( 15 / 2 )2 + ( 6 / 2 )2 + ( 2 / 2 )2 = 14.0 A
I 1 = 15 / 2 = 10.61A
IDF = I 1 / I s = 10.61 / 14.0 = 0.76
2
2
I Srms
I 1rms
14.0 2 10.612
=
I Srms
14.0
THD =
= 0.86
PF = IDF Cos = 0.76 Cos30 = 0.658
6.
Id
30
v = 340sin t;
an =
V = 340 / 2 = 240.45 Volt
4I d
sin ( n / 3)
n
i( t ) =
I1
time
30
4I d
sin( n / 3 )cos n t
n =1,3,5... n
4I d
2
sin( / 3 ) = 0.45I d = 0.78I d = 7.8 A
IDF = Input Displacement Factor = cos
= cos30 = 0.866
Solution to Tutorial 1
ST-9
M. F. Rahman/28 March, 2003
Input Power = P = VIcos = 240.45 7.8 0.866 = 1463 Watts
Note! Other harmonics do not draw any net power.
5 / 6
I srms =
/6
I 2 d d
= Id
Input Distortion Factor =
Input Power Factor =
THD =
120
= 0.816 I d = 8.165 A
180
I1
7.8
=
= 0.956
I s 8.16
I1
IDF = 0.956 0.866 = 0.828
Is
I 2 S I 21
8.265 2 7.8 2
=
= 30.95%
I1
7 .8
7.
V(t)
B( t )
time
10 sec
V pn = N pri Ac
BPeak =
0.35 =
dB
dt
1
N pn Ac
T /4
V pn dt ,
1
10
50 10 6
N pn 1.2 Ac
4
N pn = 5.62
We must choose N pn = 6 turns to keep the B below 0.35 T.
ELEC4240/9240
Power Electronics
8.
1
Ps = Vd I O f S ( tCON + tCOFF )
2
tCON = tri + t fv = 150n sec
tCOFF = trv + t fi = 300n sec
Ps =
1
300 4 f S ( 150 + 300 ) 10 9 10 3
2
where f S is in kHz.
Ps = 0.27 f SW = 6.75W @ 25kHz
and Ps = 27W @100kHz
The switching power loss increases linearly with frequency.
9.
27
Ps(W)
6.75
25
fs in kHz
100
tON = tri + t fv = 150n sec
tOFF = trv + t fi = 300n sec
For 0 < t < tON
I
iT = d t
tON
VT = Vd (1
t
)
tON
P(t )ON = Vd I d
W
O N
Solution to Tutorial 1
tO
0
t
t
)
(1
tON
tON
P ( t )O N d t =
Id
tO N
ST-11
M. F. Rahman/28 March, 2003
Similarly,
Vt=Vd
Id
IT
tON
WOFF =
time
Vd I d
tOFF
6
Ws = WON + WOFF =
Vd I d
(tON + tOFF ) = 9W at f s = 100kHz .
6
(This is 1/3 of the power loss in problem 8).
