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Page 1 of 24
SIMULATIONS WITH THE BOOST TOPOLOGY
EE562: POWER ELECTRONICS I
COLORADO STATE UNIVERSITY
Modified February 2006
Page 2 of 24
PURPOSE: The purpose of this lab is to simulate the Boost converter using ORCAD
to better familiarize the student with some of its operating characteristics. This lab will
explore some of the following aspects of the boost converter:
Discontinuous Conduction Mode
Inductor sizing
Differential voltage across the inductor
Time it takes for the converter to reach steady state
Output Ripple voltage and selection of the capacitor.
Ripple current through the capacitor
Equivalent Series Resistance (ESR) of the output capacitor.
Effects of changing and removing load resistance
Effects of the ON resistance of the switch
Efficiency
Effects of changing frequency
NOTE: The simulations that follow are intended to be completed with ORCAD Capture
CIS. It is assumed that the student has a fundamental understanding of the operation of
ORCAD. ORCAD provides tutorials for users that are not experienced with its
functions.
PROCEDURE: Build the schematic shown in Figure 1.
V1 is a DC voltage source (VDC) from the source library. It needs to be set for 24 volts.
L1 is an ideal inductor from the Analog Library.
Set for 10µH.
R1 is an ideal resistor from the Analog Library.
Set for 1kΩ.
D1 is an ideal diode (Dbreak) and can be found in the Breakout library.
C1 is an ideal capacitor from the Analog library.
Change the value to 100µ F.
Page 3 of 24
S2 is a voltage controlled switch and can be found in the Analog library.
Change RON from 1Ω to 1mΩ.
V2 is a pulsed voltage source and is intended to act as the output of a pulse width
modulator. V2 needs the following parameters set:
DC=0, AC = 10, V1=0, V2= 10, TR=10ns, TF=10ns, PW = 20µ, PER = 40µ.
This results in a switching frequency of 25 kHz.
R2 Set to 1000 kΩ. The purpose of R1 is to prevent any floating nodes.
Two voltage markers need to be placed as shown in the schematic of figure 1.
Figure 1
Boost schematic ORCAD.
Once the above schematic is built simulations can be ran. First, the type of simulation
will need to be specified. Most of these simulations are Transient simulations. The
Transient simulation can be set by selecting PSpice on the menu then New Simulation
Profile. The Run Time will need to be set to 1000µsec.
Page 4 of 24
Figure 2
Transient Analysis setup. Running the simulation will result in the following output.
Figure 3
Page 5 of 24
Remove the voltage markers, and use a current marker to measure the inductor L1
current. Place the marker in series next to L1.
Figure 4
QUESTION 1: What is the peak operating current, and what is the operating mode of
the converter? Verify mathematically the mode and the peak current.
Hint:RTs
L2K = , 2)D1(DKcrit −= ,
DTs
Ipk
L
Vin =
Figure 5
Page 6 of 24
QUESTION 2: What is the output voltage of the converter at steady state? Verify your
results mathematically.
Hint: 2
K
1D411
Vin
Vout
2⋅++=
Figure 6
Page 7 of 24
Now change L1 from 10µH to 1mH and rerun the simulation. Remember you can vary
the “Final Time” in “Transient Analysis”. Keep the “Print Step” at 0. (Hint: Start this
analysis with a “Final Time” = 1msec)
QUESTION 3: What is the peak operating current now? What is the operating mode of
the converter (remember that you can observe this by zooming in)? Also, verify the
mode mathematically.
Figure 7
Page 8 of 24
QUESTION 4: How long does it take for the converter to reach steady state? What is
the peak inductor current during steady state? Verify peak current result mathematically.
Hint: DTs
Ipk
L
Vin =
Figure 8
QUESTION 5: Calculate the size of the inductor required to put this converter in CCM.
Hint: 2
)D1(DTsRL
2−⋅⋅≥
Page 9 of 24
Remove the current marker and add a differential voltage marker across L1. Change
the Run to time to 1800u and Start saving data after 1500u.
Figure 9
QUESTION 6: What can be said about the differential voltage measurement across L1?
Figure 10
Page 10 of 24
Now change the “Final Time” to 2500µ and remove any “No Print Delay” from the
Transient Analysis setup. Remove the differential voltage markers across L1 and add a
voltage marker to the top of C1. From this simulation we can see the output voltage stair
step up to its final value.
Figure 11
QUESTION 7: How long does it take for the output voltage to reach its peak?
Figure 12
Page 11 of 24
QUESTION 8: How long does it take for the output voltage to reach its final value?
What is the output voltage? Prove your simulation results mathematically (Vout).
Hint: 2
K
1D411
Vin
Vout
2⋅++=
Figure 13
Now run the simulation for 100 µsec at a time greater than 2000 µsec.
QUESTION 9: What is the peak-to-peak ripple voltage?
Figure 14
Page 12 of 24
QUESTION 10: With everything else left as is, what is the minimum output capacitance
be to limit the output voltage ripple to 2 volts peak to peak?
Figure 15
Now, place a current marker on one of the pins of the capacitor. Run the simulation for
200 µsec at a time greater than 10 msec.
QUESTION 11: What can be said about the current through the capacitor?
Figure 16
Page 13 of 24
QUESTION 12: If the ESR of the capacitor is modeled by a 10Ω resistor in series with
the capacitor. What happens to the output voltage ripple and the capacitor current?
Figure 17
Figure 18
Page 15 of 24
Change the load resistance from 1kΩ to 100Ω with C=1µFand L=1mH.
QUESTION 13: What happens to the inductor ripple current, capacitor ripple voltage,
and capacitor current with respect to the original values?
Figure 20
Figure 21
Page 16 of 24
QUESTIONS 14: What operating mode is the converter in now?
QUESTION 15: What happens if the load resistance is removed?
Figure 22
Figure 23
Replace the load resistance.
QUESTION 16: What observations can be made from increasing the on resistance of the
switch?
Page 17 of 24
Figure 24
Figure 25
*Remember that your on resistance value of the switch will provide you a complete
different output value from your classmates.
QUESTION 17: What can be said about the efficiency of the converter?
(Comment on the different configurations of the circuit used throughout this lab.)
PinPout ÷=η
)1(*)( DVgVo −÷=η
Page 18 of 24
QUESTION 18: What can be observed by increasing the switching frequency to
100KHz?
Hints: With everything else left as it is, change your “PW” and “PER” on PULSED
voltage to 5u and 10u. Also change your “Run to time” in the simulation profile to
finaltime 251
Switchfrequency⋅:=
.
Figure 26
Page 19 of 24
Vg 24:= V L 10 106−⋅:= H C 100 10
6−⋅:= F R 1 103⋅:= Ω
Duty cycle
Switchfrequency 25 103⋅:= Hz
T1
Switchfrequency:=
T 4 105−×=
PW 20 106−⋅:=
PWD
f
Duty cycle
D SwitchfrequencyPW⋅:= D 0.5=
The output voltage
Vo
Vg
1
1 D−
1−1− D+
Vg⋅
Vo1−
1− D+Vg⋅:=
Vo 48=
Inductor current
Vin
L
∆iL
D T⋅
∆iL1
LVg⋅ D⋅ T⋅:= ∆iL 48=
IoVo
R:=
Io 0.048=
Page 20 of 24
ILIo
1 D−:=
IL 0.096=
ILmin IL∆iL
2−:= ILmin 23.904−=
ILmax IL∆iL
2+:=
ILmax 24.096=
t1
0
D T⋅
D T⋅
T
:= IL1
ILmin
ILmax
ILmax
ILmin
:=
VL
Vg
Vg
Vg
Vg Vo−
Vg Vo−
Vg Vo−
Vg Vo−
Vg
:= t
0
D T⋅
D T⋅
1 D−( ) T⋅
1 D−( ) T⋅
1 D−( ) T⋅
T
T
:=
0 2.105
4 .105
40
20
0
20
40Inductor voltage
duty cycle
indu
ctor
Vol
tage
V
VL
t
0 2.105
4 .105
40
20
0
20
40inductor current
duty cycle
indu
ctor
cur
rent
A
IL1
t1
is
ILmin
ILmax
0
0
:= iD
0
0
ILmax
ILmin
:= ic
Io−
Io−
ILmax Io−
ILmax Io−
:=
Page 21 of 24
0 2.105
4 .105
10
0
10
20
30Capacitor current
duty cycle
capa
cito
r cu
rren
t (A
)
ic
t1
Icmax ILmax Io−:= Icmax 24.048=
0 2.105
4 .105
40
20
0
20
40switch current
duty cycle
curr
ent A
is
t1
0 2.105
4 .105
40
20
0
20
40diode current Vs. duty cycle
duty cycle
diod
e cu
rren
t (A
)
iD
t1
Is D IL⋅:= Is 0.048=
ID 1 D−( ) IL⋅:= ID 0.048=
K2 L⋅R T⋅
:= K 5 104−×=
Kcrit D 1 D−( )2⋅:=
Kcrit 0.125=
D1 0.1 0.2, 1..:=
DC gain for K>Kcrit
V
Vg
1
1 D−( )
M D1( )1
1 D1−( ):=
M1 t( )
1 14 t
2⋅K
++
2:=
Page 22 of 24
0.1 11
10DC gain
Duty cycle
V/V
g M D1( )
D1
∆Vo1
CIo⋅ D⋅ T⋅:=
∆Vo 9.6 103−×=
0 0.5 10
20
40
60DC gain vs. duty cycle
duty cycle
DC
gai
nM1 D1( )
D1
for K<Kcrit
Page 23 of 24
Input and output Power
Po Io Vo⋅:= Po 2.304= W
Ig IL:=
Pg Ig Vg⋅:= Pg 2.304= W
Efficiency
ηPout
pin
ηVo
Vg1 D−( )⋅:=
η 1=
DCM/CCM boundary
Case 1: When inductor, switching frequency and other circuit parameters are constant, but R is varied
iLmin 0:= iLmax ∆iL:=
iL1∆iL
2:=
iL1 24= IL1
iLmin
iLmax
iLmax
iLmin
:=
Io1 1 D−( ) iL1⋅:= Io1 12=
RVo
Io1:= R 4=
Case 2: when R , switching frequency and other circuit parameters are constant, but L is varied
iLmin 0:= iLmax 2 IL⋅:= iLmax 0.192=
∆iL iLmax:=
0 2.105
4 .105
0
20
40
60inductor current Vs. duty cycle
duty cycle
indu
ctor
cur
rent
IL1
t1