Chapter 14. Transformer Design
Some more advanced design issues, not considered in previous
chapter:
•
•
•
•
•
n1 : n2
Inclusion of core loss
i1(t)
Selection of operating flux
density to optimize total loss
Multiple winding design: how
to allocate the available
window area among several
windings
+
+
v1(t)
v2(t)
–
–
R1
R2
+
A transformer design
procedure
ik(t)
vk(t)
–
How switching frequency
affects transformer size
Fundamentals of Power Electronics
i2(t)
: nk
1
Rk
Chapter 14: Transformer design
Chapter 14. Transformer Design
14.1. Winding area optimization
14.2. Transformer design: Basic constraints
14.3. A step-by-step transformer design procedure
14.4. Examples
14.5. Ac inductor design
14.6. Summary
Fundamentals of Power Electronics
2
Chapter 14: Transformer design
14.1. Winding area optimization
Given: application with k windings
having known rms currents and
desired turns ratios
v1(t) v2(t)
n1 = n2 =
n1 : n2
rms current
rms current
I1
I2
v (t)
= nk
k
Core
Window area WA
rms current
Ik
Core mean length
per turn (MLT)
Wire resistivity ρ
Fill factor Ku
Fundamentals of Power Electronics
3
: nk
Q: how should the window
area WA be allocated among
the windings?
Chapter 14: Transformer design
Allocation of winding area
Winding 1 allocation
α1WA
Winding 2 allocation
α2WA
{
{
Total window
area WA
etc.
0 < αj < 1
α1 + α2 +
Fundamentals of Power Electronics
4
+ αk = 1
Chapter 14: Transformer design
Copper loss in winding j
Copper loss (not accounting for proximity loss) is
Pcu, j = I 2j R j
Resistance of winding j is
lj
Rj = ρ
A W, j
with
Hence
l j = n j (MLT)
length of wire, winding j
WAK uα j
A W, j =
nj
wire area, winding j
n 2j i 2j ρ (MLT)
Pcu, j =
WAK uα j
Fundamentals of Power Electronics
5
Chapter 14: Transformer design
Total copper loss of transformer
Sum previous expression over all windings:
Pcu,tot = Pcu,1 + Pcu,2 +
ρ (MLT)
+ Pcu,k =
WAK u
Σ
k
j=1
n 2j I 2j
αj
Need to select values for α1, α2, …, αk such that the total copper loss
is minimized
Fundamentals of Power Electronics
6
Chapter 14: Transformer design
Variation of copper losses with α1
P
cu,k
Copper
loss
For α1 = 0: wire of
winding 1 has zero area.
Pcu,1 tends to infinity
+..
.+
cu,
3
Pcu,tot
+
P
P cu,1
For α1 = 1: wires of
remaining windings have
zero area. Their copper
losses tend to infinity
u,
2
There is a choice of α1
that minimizes the total
copper loss
Pc
0
Fundamentals of Power Electronics
1
7
α1
Chapter 14: Transformer design
Method of Lagrange multipliers
to minimize total copper loss
Minimize the function
Pcu,tot = Pcu,1 + Pcu,2 +
ρ (MLT)
+ Pcu,k =
WAK u
Σ
k
j=1
n 2j I 2j
αj
subject to the constraint
α1 + α2 +
+ αk = 1
Define the function
f (α 1, α 2,
, α k, ξ) = Pcu,tot(α 1, α 2,
where
g(α 1, α 2,
, α k) = 1 –
Σα
, α k) + ξ g(α 1, α 2,
, α k)
k
j=1
j
is the constraint that must equal zero
and ξ is the Lagrange multiplier
Fundamentals of Power Electronics
8
Chapter 14: Transformer design
Lagrange multipliers
continued
Optimum point is solution of
the system of equations
Result:
ρ (MLT)
ξ=
WAK u
∂ f (α 1, α 2, , α k,ξ)
=0
∂α 1
∂ f (α 1, α 2, , α k,ξ)
=0
∂α 2
αm =
j j
= Pcu,tot
∞
Σ nI
j j
An alternate form:
αm =
V mI m
∞
Σ VI
n=1
Fundamentals of Power Electronics
j=1
2
n mI m
n=1
∂ f (α 1, α 2, , α k,ξ)
=0
∂α k
∂ f (α 1, α 2, , α k,ξ)
=0
∂ξ
ΣnI
k
9
j j
Chapter 14: Transformer design
Interpretation of result
αm =
V mI m
∞
Σ VI
n=1
j j
Apparent power in winding j is
V j Ij
where
Vj is the rms or peak applied voltage
Ij is the rms current
Window area should be allocated according to the apparent powers of
the windings
Fundamentals of Power Electronics
10
Chapter 14: Transformer design
Example
PWM full-bridge transformer
i1(t)
n1 turns
i2(t)
{
}
}
I
n2 turns
n2 turns
n2
I
n1
i1(t)
i3(t)
• Note that waveshapes
(and hence rms values)
of the primary and
secondary currents are
different
0
–
i2(t)
n2
I
n1
I
0.5I
0.5I
0
i3(t)
I
• Treat as a threewinding transformer
0.5I
0.5I
0
0
Fundamentals of Power Electronics
0
11
DTs
Ts
Ts+DTs
2Ts
Chapter 14: Transformer design
t
Expressions for RMS winding currents
I1 =
I2 = I3 =
1
2Ts
2T s
i 21(t)dt =
0
1
2Ts
2T s
0
n2
I
n1
i1(t)
n2
I D
n1
0
0
–
i 22(t)dt = 12 I 1 + D
i2(t)
n2
I
n1
I
0.5I
0.5I
see Appendix 1
0
i3(t)
I
0.5I
0.5I
0
0
Fundamentals of Power Electronics
12
DTs
Ts
Ts+DTs
2Ts
Chapter 14: Transformer design
t
Allocation of window area:
αm =
V mI m
∞
Σ VI
n=1
j j
Plug in rms current expressions. Result:
α1 =
1+
1
1+D
D
1
α 2 = α 3 = 12
1+
Fundamentals of Power Electronics
Fraction of window area
allocated to primary
winding
D
1+D
13
Fraction of window area
allocated to each
secondary winding
Chapter 14: Transformer design
Numerical example
Suppose that we decide to optimize the transformer design at the
worst-case operating point D = 0.75. Then we obtain
α 1 = 0.396
α 2 = 0.302
α 3 = 0.302
The total copper loss is then given by
ρ(MLT) 3
Pcu,tot =
n jI j
WAK u jΣ
=1
ρ(MLT)n 22 I 2
=
1 + 2D + 2 D(1 + D)
WAK u
2
Fundamentals of Power Electronics
14
Chapter 14: Transformer design
14.2 Transformer design:
Basic constraints
Core loss
Pfe = K feB βmax A cl m
Typical value of β for ferrite materials: 2.6 or 2.7
Bmax is the peak value of the ac component of B(t)
So increasing Bmax causes core loss to increase rapidly
This is the first constraint
Fundamentals of Power Electronics
15
Chapter 14: Transformer design
Flux density
Constraint #2
v1(t)
Flux density B(t) is related to the
applied winding voltage according
to Faraday’s Law. Denote the voltseconds applied to the primary
winding during the positive portion
of v1(t) as λ1:
λ1 =
area λ1
t1
t2
t2
v1(t)dt
t1
This causes the flux to change from
its negative peak to its positive peak.
From Faraday’s law, the peak value
of the ac component of flux density is
λ1
Bmax =
2n 1A c
Fundamentals of Power Electronics
To attain a given flux density,
the primary turns should be
chosen according to
λ1
n1 =
2Bmax A c
16
Chapter 14: Transformer design
t
Copper loss
Constraint #3
• Allocate window area between windings in optimum manner, as
described in previous section
• Total copper loss is then equal to
Pcu =
with
ρ(MLT)n I
WAK u
2 2
1 tot
Σ
k
I tot =
j=1
nj
n1 I j
Eliminate n1, using result of previous slide:
ρ λ 21 I 2tot
Pcu =
Ku
(MLT)
WAA 2c
1
B 2max
Note that copper loss decreases rapidly as Bmax is increased
Fundamentals of Power Electronics
17
Chapter 14: Transformer design
Total power loss
4. Ptot = Pcu + Pfe
Power
loss
Co
Ptot = Pfe + Pcu
fe
ss P
Co
ss P c
r lo
ppe
Ptot
re l
o
There is a value of Bmax
that minimizes the total
power loss
u
Pfe = K feB βmax A cl m
ρ λ 21 I 2tot
Pcu =
Ku
Fundamentals of Power Electronics
Optimum Bmax
(MLT)
WAA 2c
Bmax
1
B 2max
18
Chapter 14: Transformer design
5. Find optimum flux density Bmax
Given that
Ptot = Pfe + Pcu
Then, at the Bmax that minimizes Ptot, we can write
dPfe
dPtot
dPcu
=
+
=0
d Bmax d Bmax d Bmax
Note: optimum does not necessarily occur where Pfe = Pcu. Rather, it
occurs where
dPfe
dPcu
=–
dBmax
dBmax
Fundamentals of Power Electronics
19
Chapter 14: Transformer design
Take derivatives of core and copper loss
Pfe = K feB
β
max
ρ λ 21 I 2tot
Pcu =
Ku
A cl m
dPfe
β–1
= βK feB max
A cl m
dBmax
Now, substitute into
ρλ I
Bmax =
2K u
2 2
1 tot
Fundamentals of Power Electronics
(MLT)
WAA 2c
ρλ 21 I 2tot
dPcu
=–2
4K u
d Bmax
dPfe
dPcu
=–
dBmax
dBmax
(MLT) 1
WAA 3c l m βK fe
20
1
B 2max
(MLT) – 3
B max
2
WAA c
and solve for Bmax:
1
β+2
Optimum Bmax for a
given core and
application
Chapter 14: Transformer design
Total loss
Substitute optimum Bmax into expressions for Pcu and Pfe. The total loss is:
Ptot = A cl mK fe
ρλ I
4K u
2 2
1 tot
2
β+2
β
β+2
(MLT)
WAA 2c
β
2
–
β
β+2
β
+
2
2
β+2
Rearrange as follows:
WA A c
2(β – 1)/β
(MLT) l
2/β
m
β
2
–
β
β+2
–
+
β
2
Left side: terms depend on core
geometry
Fundamentals of Power Electronics
2
β+2
β+2
β
ρλ 21I 2totK fe
2/β
=
4K u Ptot
β + 2 /β
Right side: terms depend on
specifications of the application
21
Chapter 14: Transformer design
The core geometrical constant Kgfe
Define
K gfe =
WA A c
2(β – 1)/β
(MLT) l m2/β
β
2
–
β
β+2
–
+
β
2
2
β+2
β+2
β
Design procedure: select a core that satisfies
K gfe ≥
ρλ I K fe
2 2
1 tot
4K u Ptot
2/β
β + 2 /β
Appendix 2 lists the values of Kgfe for common ferrite cores
Kgfe is similar to the Kg geometrical constant used in Chapter 13:
• Kg is used when Bmax is specified
• Kgfe is used when Bmax is to be chosen to minimize total loss
Fundamentals of Power Electronics
22
Chapter 14: Transformer design
14.3 Step-by-step
transformer design procedure
The following quantities are specified, using the units noted:
Wire effective resistivity
ρ
(Ω-cm)
Total rms winding current, ref to pri
Itot
(A)
Desired turns ratios
n2/n1, n3/n1, etc.
Applied pri volt-sec
λ1
(V-sec)
Allowed total power dissipation
Ptot
(W)
Winding fill factor
Ku
Core loss exponent
β
Core loss coefficient
Kfe
(W/cm3Tβ)
Other quantities and their dimensions:
Core cross-sectional area
Ac
Core window area
WA
Mean length per turn
MLT
Magnetic path length
le
Wire areas
Aw1, …
Peak ac flux density
Bmax
Fundamentals of Power Electronics
23
(cm2)
(cm2)
(cm)
(cm)
(cm2)
(T)
Chapter 14: Transformer design
Procedure
1.
Determine core size
ρλ 21I 2totK fe
2/β
K gfe ≥
4K u Ptot
β + 2 /β
10 8
Select a core from Appendix 2 that satisfies this inequality.
It may be possible to reduce the core size by choosing a core material
that has lower loss, i.e., lower Kfe.
Fundamentals of Power Electronics
24
Chapter 14: Transformer design
2.
Evaluate peak ac flux density
1
β+2
2 2
ρλ
8
1I tot (MLT)
1
Bmax = 10
2K u WAA 3c l m βK fe
At this point, one should check whether the saturation flux densityis
exceeded. If the core operates with a flux dc bias Bdc, then Bmax + Bdc
should be less than the saturation flux density.
If the core will saturate, then there are two choices:
• Specify Bmax using the Kg method of Chapter 13, or
• Choose a core material having greater core loss, then repeat
steps 1 and 2
Fundamentals of Power Electronics
25
Chapter 14: Transformer design
3. and 4.
Primary turns:
n1 =
Evaluate turns
λ1
10 4
2Bmax A c
Choose secondary turns according to
desired turns ratios:
Fundamentals of Power Electronics
n2 = n1
n2
n1
n3 = n1
n3
n1
26
Chapter 14: Transformer design
5. and 6.
Choose wire sizes
Fraction of window area
assigned to each winding:
Choose wire sizes according
to:
n 1I 1
α1 =
n 1I tot
n 2I 2
α2 =
n 1I tot
α 1K uWA
n1
α 2K uWA
A w2 ≤
n2
A w1 ≤
n kI k
αk =
n 1I tot
Fundamentals of Power Electronics
27
Chapter 14: Transformer design
Check: computed transformer model
Predicted magnetizing
inductance, referred to primary:
n1 : n2
i1(t)
µn A c
LM =
lm
2
1
iM(t)
i2(t)
LM
Peak magnetizing current:
R1
λ1
i M, pk =
2L M
R2
ik(t)
Predicted winding resistances:
ρn 1(MLT)
A w1
ρn (MLT)
R2 = 2
A w2
R1 =
Fundamentals of Power Electronics
: nk
28
Rk
Chapter 14: Transformer design
14.4.1
Example 1: Single-output isolated
Cuk converter
Ig
4A
+ vC1(t) –
– vC2(t) +
25 V
+
–
20 A
+
–
Vg
I
+
v1(t)
v2(t)
+
V
5V
–
–
i1(t)
n:1
i2(t)
100 W
fs = 200 kHz
D = 0.5
n=5
Ku = 0.5
Allow Ptot = 0.25 W
Use a ferrite pot core, with Magnetics Inc. P material. Loss
parameters at 200 kHz are
Kfe = 24.7
Fundamentals of Power Electronics
β = 2.6
29
Chapter 14: Transformer design
Waveforms
v1(t)
VC1
Area λ1
Applied primary voltseconds:
λ 1 = DTsVc1 = (0.5) (5 µsec ) (25 V)
= 62.5 V–µsec
D'Ts
DTs
i1(t)
– nVC2
Applied primary rms
current:
I/n
I1 =
– Ig
i2(t)
2
+ D' I g
2
=4A
Applied secondary rms
current:
I 2 = nI 1 = 20 A
I
Total rms winding
current:
I tot = I 1 + 1n I 2 = 8 A
– nIg
Fundamentals of Power Electronics
D nI
30
Chapter 14: Transformer design
Choose core size
(1.724⋅10 – 6)(62.5⋅10 – 6) 2(8) 2(24.7) 2/2.6
8
K gfe ≥
10
4 (0.5) (0.25) 4.6/2.6
= 0.00295
Pot core data of Appendix 2 lists 2213 pot core with
Kgfe = 0.0049
Next smaller pot core is not large enough.
Fundamentals of Power Electronics
31
Chapter 14: Transformer design
Evaluate peak ac flux density
1/4.6
(1.724⋅10 – 6)(62.5⋅10 – 6) 2(8) 2
(4.42)
1
Bmax = 10
3
2 (0.5)
(0.297)(0.635) (3.15) (2.6)(24.7)
8
= 0.0858 Tesla
This is much less than the saturation flux density of approximately
0.35 T. Values of Bmax in the vicinity of 0.1 T are typical for ferrite
designs that operate at frequencies in the vicinity of 100 kHz.
Fundamentals of Power Electronics
32
Chapter 14: Transformer design
Evaluate turns
–6
(62.5⋅10
)
n 1 = 10 4
2(0.0858)(0.635)
= 5.74 turns
n1
n 2 = n = 1.15 turns
In practice, we might select
n1 = 5
and
n2 = 1
This would lead to a slightly higher flux density and slightly higher
loss.
Fundamentals of Power Electronics
33
Chapter 14: Transformer design
Determine wire sizes
Fraction of window area allocated to each winding:
α1 =
α2 =
4A
8A
1
5
(Since, in this example, the ratio of
winding rms currents is equal to the
turns ratio, equal areas are
allocated to each winding)
= 0.5
20 A
8A
= 0.5
From wire table,
Appendix 2:
Wire areas:
(0.5)(0.5)(0.297)
= 14.8⋅10 – 3 cm 2
(5)
(0.5)(0.5)(0.297)
A w2 =
= 74.2⋅10 – 3 cm 2
(1)
A w1 =
Fundamentals of Power Electronics
34
AWG #16
AWG #9
Chapter 14: Transformer design
Wire sizes: discussion
Primary
5 turns #16 AWG
Secondary
1 turn #9 AWG
•
Very large conductors!
•
One turn of #9 AWG is not a practical solution
Some alternatives
•
Use foil windings
•
Use Litz wire or parallel strands of wire
Fundamentals of Power Electronics
35
Chapter 14: Transformer design
Effect of switching frequency on transformer size
for this P-material Cuk converter example
4226
0.1
2616
2616
2213
2213
1811
0.08
0.06
1811
0.04
Bmax , Tesla
Pot core size
3622
0.02
0
25 kHz
50 kHz
100 kHz
200 kHz
250 kHz
400 kHz
500 kHz
1000 kHz
Switching frequency
• As switching frequency is
increased from 25 kHz to
250 kHz, core size is
dramatically reduced
Fundamentals of Power Electronics
• As switching frequency is
increased from 400 kHz to
1 MHz, core size
increases
36
Chapter 14: Transformer design
14.4.2
Example 2
Multiple-Output Full-Bridge Buck Converter
Q1
D1
Q3
T1
D3
n1 :
I5V
: n2
160 V
+
–
100 A
+
D5
+
Vg
i2a(t)
i1(t) v1(t)
5V
–
Q2
D2
Q4
Switching frequency
D6
i2b(t)
–
: n2
D4
I15V
: n3
i3a(t)
+
D7
150 kHz
15 A
15 V
Transformer frequency
75 kHz
D8
Turns ratio
110:5:15
Optimize transformer at
D = 0.75
Fundamentals of Power Electronics
i2b(t)
–
: n3
37
Chapter 14: Transformer design
Other transformer design details
Use Magnetics, Inc. ferrite P material. Loss parameters at 75 kHz:
Kfe = 7.6 W/Tβcm3
β = 2.6
Use E-E core shape
Assume fill factor of
Ku = 0.25
(reduced fill factor accounts for added insulation required
in multiple-output off-line application)
Allow transformer total power loss of
Ptot = 4 W
(approximately 0.5% of total output power)
Use copper wire, with
ρ = 1.724·10-6 Ω-cm
Fundamentals of Power Electronics
38
Chapter 14: Transformer design
Applied transformer waveforms
v1(t)
T1
D3
n1 :
: n2
i2a(t)
0
0
D5
+
– Vg
i1(t) v1(t)
n
n2
I 5V + 3 I 15V
n1
n1
i1(t)
–
D4
Area λ1
= Vg DTs
Vg
D6
i2b(t)
0
: n2
: n3
i3a(t)
D7
–
i2a(t)
n2
n
I 5V + 3 I 15V
n1
n1
I5V
0.5I5V
0
D8
i2b(t)
i3a(t)
I15V
0.5I15V
: n3
0
0
Fundamentals of Power Electronics
DTs
39
Ts
Ts+DTs 2Ts
t
Chapter 14: Transformer design
Applied primary volt-seconds
v1(t)
Vg
Area λ1
= Vg DTs
0
0
– Vg
λ 1 = DTsVg = (0.75) (6.67 µsec ) (160 V) = 800 V–µsec
Fundamentals of Power Electronics
40
Chapter 14: Transformer design
Applied primary rms current
i1(t)
n
n2
I 5V + 3 I 15V
n1
n1
0
–
n2
n
I 5V + 3 I 15V
n1
n1
n2
n3
I 1 = n I 5V + n I 15V
1
1
Fundamentals of Power Electronics
41
D = 5.7 A
Chapter 14: Transformer design
Applied rms current, secondary windings
i2a(t)
I5V
0.5I5V
0
i3a(t)
I15V
0.5I15V
0
0
DTs
Ts
Ts+DTs 2Ts
t
I 2 = 12 I 5V 1 + D = 66.1 A
I 3 = 12 I 15V 1 + D = 9.9 A
Fundamentals of Power Electronics
42
Chapter 14: Transformer design
Itot
RMS currents, summed over all windings and referred to primary
I tot =
Σ
all 5
windings
nj
n2
n3
n1 I j = I 1 + 2 n1 I 2 + 2 n1 I 3
= 5.7 A + 5 66.1 A + 15 9.9 A
110
110
= 14.4 A
Fundamentals of Power Electronics
43
Chapter 14: Transformer design
Select core size
(1.724⋅10 – 6)(800⋅10 – 6) 2(14.4) 2(7.6) 2/2.6
8
K gfe ≥
10
4 (0.25) (4) 4.6/2.6
= 0.00937
A2.2
EE core data
From Appendix 2
A
Core
type
Geometrical
constant
Geometrical
constant
(A)
(mm)
Kg
cm5
Kgfe
cmx
Crosssectional
area
Ac
(cm2)
Bobbin
winding
area
WA
(cm2)
Mean
length
per turn
MLT
(cm)
Magnetic
path
length
lm
(cm)
Core
weight
(g)
EE22
8.26·10-3
1.8·10-3
0.41
0.196
3.99
3.96
8.81
EE30
-3
85.7·10
6.7·10
-3
1.09
0.476
6.60
5.77
32.4
EE40
0.209
11.8·10-3
1.27
1.10
8.50
7.70
50.3
EE50
0.909
28.4·10-3
2.26
1.78
10.0
9.58
116
Fundamentals of Power Electronics
44
Chapter 14: Transformer design
Evaluate ac flux density Bmax
Eq. (14.41):
2 2
ρλ
8
1I tot (MLT)
1
Bmax = 10
2K u WAA 3c l m βK fe
1
β+2
Plug in values:
1/4.6
(1.724⋅10 – 6)(800⋅10 – 6) 2(14.4) 2
(8.5)
1
Bmax = 10
2 (0.25)
(1.1)(1.27) 3(7.7) (2.6)(7.6)
8
= 0.23 Tesla
This is less than the saturation flux density of approximately 0.35 T
Fundamentals of Power Electronics
45
Chapter 14: Transformer design
Evaluate turns
Choose n1 according to Eq. (14.42):
n1 =
λ1
10 4
2Bmax A c
To obtain desired turns ratio
of
(800⋅10 – 6)
n 1 = 10
2(0.23)(1.27)
= 13.7 turns
4
110:5:15
we might round the actual
turns to
Choose secondary turns
according to desired turns ratios:
22:1:3
Increased n1 would lead to
5
n2 =
n = 0.62 turns
110 1
• Less core loss
• More copper loss
15
n = 1.87 turns
n3 =
110 1
Fundamentals of Power Electronics
Rounding the number of turns
• Increased total loss
46
Chapter 14: Transformer design
Loss calculation
with rounded turns
With n1 = 22, the flux density will be reduced to
(800⋅10 – 6)
Bmax =
10 4 = 0.143 Tesla
2 (22) (1.27)
The resulting losses will be
Pfe = (7.6)(0.143) 2.6(1.27)(7.7) = 0.47 W
(8.5)
(1.724⋅10 – 6)(800⋅10 – 6) 2(14.4) 2
8
1
Pcu =
10
4 (0.25)
(1.1)(1.27) 2 (0.143) 2
= 5.4 W
Ptot = Pfe + Pcu = 5.9 W
Which exceeds design goal of 4 W by 50%. So use next larger core
size: EE50.
Fundamentals of Power Electronics
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Chapter 14: Transformer design
Calculations with EE50
Repeat previous calculations for EE50 core size. Results:
Bmax = 0.14 T, n1 = 12, Ptot = 2.3 W
Again round n1 to 22. Then
Bmax = 0.08 T, Pcu = 3.89 W, Pfe = 0.23 W, Ptot = 4.12 W
Which is close enough to 4 W.
Fundamentals of Power Electronics
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Chapter 14: Transformer design
Wire sizes for EE50 design
Window allocations
α1 =
I1
= 5.7 = 0.396
I tot 14.4
α2 =
n 2I 2
= 5 66.1 = 0.209
n 1I tot 110 14.4
α3 =
n 3I 3
= 15 9.9 = 0.094
n 1I tot 110 14.4
Wire gauges
α 1K uWA (0.396)(0.25)(1.78)
=
= 8.0⋅10 – 3 cm 2
(22)
n1
⇒ AWG #19
αKW
(0.209)(0.25)(1.78)
= 93.0⋅10 – 3 cm 2
A w2 = 2 u A =
(1)
n2
A w1 =
⇒ AWG #8
αKW
(0.094)(0.25)(1.78)
A w3 = 3 u A =
= 13.9⋅10 – 3 cm 2
(3)
n3
⇒ AWG #16
Might actually use foil or Litz wire for secondary windings
Fundamentals of Power Electronics
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Chapter 14: Transformer design