Download Solutions for Assignment 3 - Power Electronics | ECE 464 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity!
Solutions for Assign 3
Sep 15 2008
Problem 3.
We can simply replace the switches and voltage source with an ideal voltage source.
We are already given the output voltage, so there is no need to solve for it as we will need
to in future problems.
T
3T/20 7T/
13T/20 17T/
0
25 V
-25 V
0V
v
out
i
out
5 ohm
20 mH
Problem 3.2)
For the pulse waveform, the coefficients are
n := 1 .. 65 V 0
:= 25 R
load
L
load
T
f 1
T
:= f 1
Calculate the angle for the dead spot (0 volt) in the waveform:
δ 1
:= ⋅ 2 π δ 1
π
⋅ = 108 In degrees...
The coefficients are available from the look up table inthe back of your textbook.
a n
sin n( ⋅ π⋅0.5)
n
cos
n δ 1
π
⋅ V
:= ⋅ b n
a not
v out
( )t
n
a n
cos 2 ⋅ π⋅ nf 1
⋅ t
T
⋅ b n
sin 2 ⋅ π⋅n f 1
⋅ t
T
a not
0 5 10
− 4 × 1 10
− 3 × 1.5 10
− 3 × 2 10
− 3 ×
− 20
0
20
v out
( )t
t
j := − 1
Vout
n
a n
j b n
Iout
n
Vout n
R
load
j 2⋅ ⋅π ⋅ nf 1
⋅ L
load
a out1not
a not
R
load
aout n
Re Iout n ( )
:= bout n
Im Iout n ( )
i out
( )t
n
aout n
cos 2 ⋅ π⋅ nf 1
⋅ t
T
⋅ bout n
sin 2 ⋅ π⋅ nf 1
⋅ t
T
∑
a out1not
0 0.05 0.1 0.15 0.
− 20
0
20
v out
( )t
t
j := − 1
Vout
n
a n
j b n
Iout
n
Vout n
R
load
j 2⋅ ⋅π ⋅ nf 2
⋅ L
load
a out2not
a not
R
load
aout n
Re Iout n ( )
:= bout n
Im (Iout2 ) n
i out
( )t
n
aout n
cos 2 ⋅ π⋅ nf 2
⋅ t
T
⋅ bout n
sin 2 ⋅ π⋅ nf 2
⋅ t
T
∑
a out2not
0 0.05 0.1 0.15 0.
− 10
− 5
0
5
10
i out
( )t
t
Notice that the second curent waveform follows the exponential shapes - this is since
the period of on/off times of the pulses is much longer than the time constant L/R
which is 4 ms.
The first current waveform has the opposite case, and the transitions are lines since
the iR drops are small compared to the voltage and the inductance dominates.
Problem 3.3)
The assumption is that di/dt = delta i / delta t, meaning that the iR drop is negligible.
With this assumption, the current will change with delta i = delta t * V/L, where V
is constant for delta t.
The voltage is 25, -25, or 0. So the delta i is either
4 T
4 T
So the delta i is 0.5 A or 50 A, based on the linear assumption. In either case,
the line-segment approximation would look as follows:
P
out
T
0
T 2
v t out
( ) it out
⋅ ( )t
:= ⋅ d P out
Another way is to sum up the harmonics of the power waveform:
P
out
n
a n
aout n
⋅ b n
bout n
∑(^ )
:= ⋅ P
out
P
out
n
a n
aout n
⋅ b n
bout n
∑(^ )
:= ⋅ P
out
(Note in the last equations the an and bn are the same in both cases, also, note that
bn is zero in both cases, so it doesn't actually contribute any average power)
The power factors are straightforward from here
pf 1
P
out
V
rms
I
rms
:= pf 1
pf 2
P
out
V
rms
I
rms
:= pf 2
The second case says that the power factor is much higher, which is because the load
appears very resistive at this low frequency.
Problem 3.5)
using the THD formulas, it is easiest to apply the rms values calculated the previous
problems.
THDV
V
rms
2
Vout 1
2
Vout 1
2
:= THDV =0.
THDI
I
rms
2
Iout 1
2
Iout 1
2
:= THDI
THDI
I
rms
2
Iout 1
2
Iout 1
2
:= THDI
These numbers make since since the voltage is not sinusoidal at all (high THD), the
current is the high frequency case is not sinusoidal but of the three, is the most
sinsoidal (lowest THD).
The current for the low-frequency case more closely resembles the voltage waveform, and
thus has high THD as well.
