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Solutions for Assign 2
Sep 8 2008
Problem 2.
We need a 2x3 switch matrix as shown:
this matrix includes an inductor connected to the source so that it behaves like a
current source, or else there is no way to connect the input to output without violating
KVL or having zero power flow. You could also connect inductors in series with each
RC combination.
Numbering the switches in the usual way, KCL on the input side requires that
q_11 + q_12 + q_13 >=1 and q_21 + q_22 + q_23 >= 1
Considering the capacitors as voltage sources, KVL requires:
q_11 + q_12 + q_13 <=1 and q_21 + q_22 + q_23 <= 1
So: q_11 + q_12 + q_13 =1 and q_21 + q_22 + q_23 = 1
The general requirements in terms of blocking and conducting depend on several factors
We are only given that the source is dc, but we are not told anything about the way the
switches are controlled, so we don't know if the load has ac or dc voltages. Most 3-phase
systems have AC loads, so if we assume AC voltages on the output, then the switches must
block both ways.
It is possible to generate AC power from a DC source, so we can for the sake of argument
assume that the current in the inductor is unidirectional (DC + maybe some small AC ripple).
In this case, the switches must conduct forward only at a minimum.
The switch we would need would be FCBB, which can be made from an IGBT or MOSFET
in series with a diode, or a reverse-blocking IGBT (an uncommon device).
Since the load is wye-connected, it is worth pointing out hat the load currents sum to zero,
so naturally when one or two are positive, the other(s) are negative.
Since little information is given on the types of voltage and current waveforms, it would
be fair to assume BCBB switches to be sure all cases are covered, but for most practical
converters this is overkill.
Problem 2.2)
For the pulse waveform, the coefficients are
n := 1 .. 35 D := 0.25 V 0
f 1
a n
sin n( โ
ฯโ
D)
n
ฯ
โ
V
:= โ
b n
a not
V
:= โ
D
v in
( )t
n
a n
cos 2 โ
ฯ โ
nf 1
โ
โ
t
โ
b n
sin 2 โ
ฯ โ
nf 1
โ
โ
t
a not
R
load
:= 10 L
f
R
load
2 โ
ฯ f 1
L
f
โ 4 = ร
0 1 10
โ 4 ร 2 10
โ 4 ร
โ 100
0
100
200
300
v in
( )t
v out
( )t
t
0 1 10
5 ร 2 10
5 ร 3 10
5 ร 4 10
5 ร
0
10
20
30
40
50
Vout1n
fplot n
Note that this plot excludes the DC component, which is 50 V.
C
load
:= 0.000047 L
f
C
load
2 โ
ฯf 1
โ 2
:= L
f
โ 5 = ร
Vout n
R load
1 j 2โ
โ
ฯโ
nf 1
โ
R load
โ
C load
R
load
1 j 2โ
โ
ฯโ
nf 1
โ
R
load
โ
C
load
j 2โ
โ
ฯ โ
nf 1
โ
L
f
Vph n
:= โ
a out2not
a not
aout n
Re Vout
( n)
:= bout n
Im Vout
( n)
v out
( )t
n
aout n
cos 2 โ
ฯ โ
nf 1
โ
โ
t
โ
bout n
sin 2 โ
ฯ โ
n f 1
โ
โ
t
โ(^ )
a out2not
0 1 10
โ 4 ร 2 10
โ 4 ร
โ 100
0
100
200
300
v in
( )t
v out
( )t
t
For the second case, the 3wt term is irrelevant since there is no voltage at that frequency.
The wt terms are both in phase, so their contribution is maximized, therefore the power is
P
:= + P
V
rms
2
:= + I
rms
2
2
:= + pf
P
V
rms
I
rms
pf2 =0.
note that hte power factor is affected by the harmonic content at 3wt.
In the last case, the voltage has an anverage value of 8 V * (0.4), or 3.2 V. It has
infinite harmonics, but the current has only DC, so the harmonics don't add to the power
P
:= 8 0.4โ
โ
3 P
V
rms
:= 0.4 8โ
I
rms
:= 3 pf
P
V
rms
I
rms
:= pf3 =0.
the Vrms3 is derived using the integral formula for RMS, or just by looking in the
appendix. Note that the power factor is substantially reduced by the harmonic content.
Problem 2.4)
This circuit has a KCL problem with the inductor in series with the switch. therefore,
it is not generally legal. however, if the switch is set up so that it only turns off when
the current is zero, then it can work. This can be achieved by using a diode or SCR,
for example, or any other device that is controlled with feedback so that it is turned
off when the current is exactly zero (or very close in practical cases).
"Soft switching" converters normally are set up to trigger switches on such zero
crossings.
Problem 2.5)
A) It will increase since a positive voltage is connected across the inductor in its positive
orientation (defined by the current arrow).
B) The current must be zero in this configuration since the left end of the inductor is open.
The current must "stay the same" since it is zero indefinitely until another switch is on.
If the current is not zero, then it is a KCL violation.
C) The voltage source is connected to the inductor in a negative configuration (against
the arrow), btu the source now is < 0, so current is decreasing in a negative direction,
meaning the current is actually increasing.
D) In this case, zero volts is applied to the inductor, so the current stays the same.
E) The voltage is applied in the positive orientation, but the voltage is positive or negative
depending on the time instant. Therefore, we cannot say if the current is increasing or
decreasing unless we know the sign of the voltage.
F) This case has a KVL violation, so it is irrelevant, but the inductor is open circuited,
so its current must be zero - i.e., stays the same.
