Download Sample Solutions for Assignment #7 - Power Electronics | ECE 464 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity!
Solutions for Assign 7 Nov 16 2008
f sw
:=25kHz
D
b
I
out
:=3A
losstan :=0.
Δv := 0.05V Assume p-p as in previous problems
The total ripple is due to the ESR and teh capacitor discharge.
delta-v = Ioutdelta-t/C + ESRdelta-i_cap
The delta_icap is the change in capacitor current when the diode is on vs. off.
The cap current goes from IL-Iout to -Iout, so the change is just IL.
The loss tangent is tand = wESRC
For an ideal converter, IL = Iout/(1-D). Solving for C,
C
b
I
out
D
b
f sw
I
out
1 D
b
losstan
2 ⋅ πf sw
Δv
:= C
b
3 = × ⋅μF
nH 10
− 9 := H
R
w
:= 0.05Ω ESL := 5nH C s
:= 0.01μF R leak
:=100kΩ
Z s( ) R w
+s ESL⋅
s C s
R
leak
s C s
R
leak
j := − 1
100 1 10
4 × 1 10
6 × 1 10
8 × 1 10
10 ×
1
10
100
1 10
3 ×
1 10
4 ×
1 10
5 ×
1 10
6 ×
Z j( ⋅ω)
ω
l c
:= 2cm w c
:= 0.1cm t c
:=5cm
A
c
w c
t c
:= ⋅ A
c
0.5 cm
2 = ⋅
R
core
l c
75 μ 0
A
c
:= R
core
6 ×
H
B
sat
:=1T
N
t
:= 50 L
c
N
t
2
R
core
:= L
c
=0.589 mH⋅
We know that Li = NBA for a linear core with assumptions of uniformity, etc.
So, I max = NBsat*A/L
w core/
l win
l win
wcore
t core
g
The area of the window is dictated by the current density and fill factor.
ff := 0.35 A win
I
dc
ff 300⋅
A
cm
2
:= A
win
0.038 cm
2 = ⋅
here we have assumed the current density is 300 A/cm^2 and conservative ff = 0.
If we assume a square window, then the length of one side of the window is:
l win
A
win
:= l win
=0.195 cm⋅
We will assume that an average turn is roughly circular and has an average radius of
sqrt(Ac)/2 + lwin/2 - since the center leg of the core will be square (by design) and the turns
will be roughly even throughout the window.
So a the total length of all turns is Npi(sqrt(Ac) + lwin).
The area of one strand of wire is then Awire = Awin*ff/N.
So the total resistance is Rdc = Npi(sqrt(Ac) + lwin) / (sigma * Awire)=
Rdc = N^2pi(sqrt(Ac) + lwin)/(sigmaAwinff)
where sigma is the conductivity of copper
For a ferrite core (according to text) we need to figure about 0.3 T for saturation.
B
sat
:=0.3T
NA
core
L
I
dc
B
sat
:= NA
core
66.667 cm
2 = ⋅
Peak flux density ripple is
mT :=0.001T
ΔB
p
L
Δi
NA
core
:= ΔB
p
=2.5 mT⋅
The total volume of hte core is given by
Vcore = Acorelwin + 6lwinAcore/2 + 4sqrt(Acore)Acore/4 + 2sqrt(Acore)*Acore
We can sub NAcore / N for Acore in this equation.
the total loss is then Idc^2 * Rdc +VocreP0f^2*delta-B^b
a := 1.6 b := 2.07 P 0
− 5 ⋅
kW
m
3
We can also see the gap length is much smaller than the other dimensions, so most our
assumptions should be fine.
And (^) A core
NA
core
N
t
:= A
core
11.111 cm
2 = ⋅
w core
A
core
:= w core
=33.333 mm⋅
t core
w core
:= t core
=33.333 mm⋅
