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Material Type: Assignment; Class: Power Electronics; Subject: Electrical and Computer Engr; University: University of Illinois - Urbana-Champaign; Term: Unknown 1989;
Typology: Assignments
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Solutions for Assign 6 Nov 4 2008
f sw
:= 100Hz ω sw
2 ⋅ πf sw
max
f sw
f
:=10mH
load
f
2 ⋅ π ⋅ 10 Hz
2 1
f
f
4 = × μF
n :=1 2, .. 35
out n
π
sin
n ⋅π
n
v out
( )t
n
out n
cos n ω sw
⋅ ⋅t
0 0.01 0.
− 300
− 200
− 100
0
100
200
300
v out
( )t
t
j := − 1
Z s( ) R load
s L f
s C f
out n
out n
Z j n⋅ ω sw
i out
( )t
n
out n
cos n ω sw
⋅ ⋅ t arg I out n
0 0.01 0.
− 60
− 40
− 20
0
20
40
60
i out
( )t
t
dc
We can solve for the output voltage as Vdcdmcos(w*t) if the duty cycle command has the
form of d1(t) = 0.5 + 0.5dmcos(wt) for one leg, and d2(t) = 0.5 - 0.5dmcos(wt) for the
other.
So the rms becomes: dm*Vdc/sqrt(2), and therefore
d m
dc
:= d m
d1(t) = 0.5 + 0.5dmcos(2pi45*t)
for the three phase inverter, we have exactly the same approach except that we have a
da, db, and dc that are spaced in phase by 120 degrees each. This gives us a sqrt(3)/
factor for the rms: dm*Vdc/sqrt(2) * sqrt(3)/
d m
dc
:= ⋅ d m
