EEE572 Advanced Power Electronics
Solutions to Homework 1
Problem 1 (30 points)
+
+
+
+
+
(a)
(a)
(b)
a
b
c
d
(b)
(c)
(d)
(e)
In the above circuits all the components are assumed ideal. The value of inductance is 25
 H and the capacitance is 10  F . The switching frequency is 100 kHz and the duty ratio
is 0.5. Which of the above circuit(s) result in valid steady-state operation? Justify your
answers.
If the inductor is assumed to have a series resistance of 50  (all other components still
ideal), which of the above circuits will result in valid steady-state operation?
With RL = 0
With RL = 50
When the switch is turned off, there is no path for the Inductor
current
is
still
inductor current.
Inductor current can’t be discontinuous, hence, circuit is not
discontinuous; hence the circuit is not valid. If the valid.
circuit is constructed practically, a very large voltage
will result at the turn-off instant, destroying the
switch.
Turning on the switch results in short circuiting the Series resistance is not in the short
dc voltage source, through the diode and switch. circuit path; hence, the circuit is still
Hence, not a valid circuit. Also, any time a diode is invalid.
in parallel (either polarity) with an inductor, voltsecond balance cannot be achieved.
The diode and switch provide a continuous path for With the introduction of the series
the inductor current. However, the diode across the resistance, negative voltage can be
inductor results in unipolar voltage being applied applied across the inductor during the
across the inductor. Hence, volt-second balance is switch OFF interval. The negative
violated and the circuit will not reach steady state. voltage is the product of resistance
The current will keep rising indefinitely.
and the inductor current. Voltsecond balance can be maintained
leading to steady state operation.
Notice that both during on and off
intervals, the circuit is a series R-L
circuit, with time constant less than
1/10th of ON or OFF interval.
The series combination of diode and capacitor can The capacitor current is still
provide current continuity as well as volt-second unidirectional and the circuit
balance for the inductor. But, the diode in series with remains invalid.
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e
the capacitor makes the capacitor current
unidirectional. Therefore, ampere-second balance is
violated for the capacitor, and the circuit is invalid.
The diode in series with the capacitor makes the The capacitor current is still
capacitor current unidirectional. Ampere-second unidirectional and the circuit
balance is violated and the circuit is invalid.
remains invalid.
The circuit is actually a boost converter, but with no
load. The capacitor keeps charging, and with no load
to discharge it, the voltage will exceed its max rating.
Problem 2 (30 points)
For each of the circuits shown below, determine if steady state operation is possible. If yes,
calculate Vo in steady-state (for each). If not, explain why steady state operation is not possible.
(a)
+
5A
D = 0.4
f =100kHz
+
100m F
10W
Vo
_
Steady state operation is possible, since average current through the capacitor can be zero (due to
resistor across the capacitor) and the current source is never open circuited (note that it is
allowed to short circuit a current source by the switch).
average diode current, iD = 5(1- D) = 5´0.6 = 3 A
In steady state, ic = 0  iR = iD = 3 A
\ Vo = R iR = 30V
(b)
D
Vin1 > Vin 2
+
+
Vin1
Vin 2
vA
+
Vo
_
When switch is ON, vA =Vin1 and when switch is OFF, vA = Vin 2 . Note that this is true only when
Vin1 >Vin 2
.
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\ Vo = vA = DVin1 + (1 - D)Vin 2 or Vin 2 + D (Vin1 - Vin 2 )
(c)
iin

+

D  0.4
 24V
10
5
Valid
network
Vo

iin
6 A
4 A
0
4
10
t s
14
Steady state operation is possible, since average current through the capacitor can be zero (due to
resistor across the capacitor) and the voltage source is never short circuited. It is assumed that
there is no power loss in the ‘valid network’ part.
iin  5  0.4  2 A
Since Vin is a constant,
pin  Vin iin  24  2  48W
po  pin  48W
Vo 
po R  21.91V
(d)
10 F
24V
D  0.4
+
10 F

5 Vo

The above circuit does not result in steady-state, since the upper capacitor has a diode in series
and hence can have only uni-directional current. Average current through this capacitor is nonzero, and hence, current-sec balance for a capacitor is violated.
(e)
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A switch is directly in parallel with the output capacitor and hence the circuit is invalid. It will
result in large currents through the switch discharging the output capacitor and the output voltage
will also be zero.
(f)
This is a valid circuit since the output capacitor is no longer short circuited by the switch (compared
to part (e)). Inductor voltage can be positive and negative, and the capacitor current can be positive
and negative satisfying volt-sec balance and current-sec balance respectively.
vL  12V when switch is on
= 12  Vo when switch is off (and diode conducts)
Applying volt-sec balance across the inductor,
12  0.6  (12  Vo ) 0.4  0
Vo  30 V
Optional bonus point problem (20 points)
The PLECS file for the required simulation is attached.
The waveforms in the time frame of 0 to 1 ms is shown below. This corresponds to a transient
condition. The average capacitor current is non-zero and this leads to the charging of the capacitor
from zero to its steady state value of 30 V. The diode current switches between the current source
magnitude (when main switch is off) and zero (when main switch is on) as expected and its CCA
value is always 5x(1-D) = 3A. The output capacitor charges at the time constant of RC = 1 ms.
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The waveforms corresponding to the time frame 9.98 ms to 10 ms is shown below. The circuit is
now in steady state with CCA value of capacitor current at 0 A and output at 30 V as expected.
The instantaneous value of capacitor current switches between load current of 3A (discharging)
and iD-iR = 2A (charging). This pulsating current results in small voltage ripple in the capacitor.
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The PLECS schematic used is shown below.
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Problem 3 (30 points)
A buck converter has the following parameters: L = 100 H, C = 470 F, ESR = 10 m.
Corresponding to an operating condition of Vin = 100 V, Vo = 40 V, load resistance of 5  and
switching frequency of 200 kHz
(a) Draw the schematic of the converter with the given parameters and operating conditions
(b) Draw the waveforms of switch voltage and current over one switching period
(c) Draw the waveforms of inductor voltage and current over one switching period
(d) Draw the waveforms of output capacitor voltage (assume voltage ripple is only due to ESR)
and current over one switching period.
In each of the waveforms above give all possible details including CCA values.
Schematic of the buck converter
Duty ratio, D 
Ts 
Vo
40

 0.4
Vin 100
1
 5  s; DTs  2  s
200, 000
Vo
 8A
R
Vo 1  D  Ts
iL  I o 
I L 
L

40  (1  0.4)  5
 1.2 A
100 
Vc  ESR  I L  10  103  1.2  12 mV (since only the ripple due to ESR is considered)
All the waveforms corresponding to given operating conditions are shown in the figure in next
page.
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