Transformer

Chapter 15 Transformer Design
Some more advanced design issues, not considered in previous
chapter:
• Inclusion of core loss
• Selection of operating flux
density to optimize total loss
• Multiple winding design: as in
the coupled-inductor case,
allocate the available window
area among several windings
• A transformer design
procedure
• How switching frequency
affects transformer size
n1 : n2
+
v1(t)
–
+
v2(t)
–
i1(t) i2(t)
R1 R2
: nk
Rk
+
vk(t)
–
ik (t)
Fundamentals of Power Electronics 1 Chapter 15: Transformer design

Chapter 15 Transformer Design
15.1 Transformer design: Basic constraints
15.2 A step-by-step transformer design procedure
15.3 Examples
15.4 AC inductor design
15.5 Summary

15.1 Transformer Design:
Basic Constraints
Core loss
Pfe = K fe(ΔB)β Ac lm
Typical value of for ferrite materials: 2.6 or 2.7
B is the peak value of the ac component of B(t), i.e., the peak ac flux
density
So increasing B causes core loss to increase rapidly
This is the first constraint

Flux density
Constraint #2
Flux density B(t) is related to the
applied winding voltage according
to Faraday’s Law. Denote the volt-seconds
applied to the primary
winding during the positive portion
of v1(t) as 1:
t2
λ1 = v1(t)dt
t1
v1(t)
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
area λ1
t1 t2 t
To attain a given flux density,
the primary turns should be
chosen according to
ΔB = λ1
2n1Ac
n1 = λ1
2ΔBAc

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 =
ρ(MLT)n1 2
2
I tot
WAKu
with
Itot =
nj
n1
k
I Σ j j = 1
Eliminate n1, using result of previous slide:
Pcu = ρ λ1 2
2
I tot
4Ku
(MLT)
2
WAAc
1
ΔB
2
Note that copper loss decreases rapidly as B is increased

Total power loss
4. Ptot = Pcu + Pfe
There is a value of B
that minimizes the total
power loss
Ptot = Pfe + Pcu
Pfe = K fe(ΔB)β Ac lm
Pcu = ρ λ1 2
2
I tot
4Ku
(MLT)
2
WAAc
Power
loss
1
ΔB
2
ΔB
Ptot
Copper loss Pcu
Core loss Pfe
Optimum ΔB

5. Find optimum flux density B
Given that
Ptot = Pfe + Pcu
Then, at the B that minimizes Ptot, we can write
dPtot
d(ΔB)
=
dPfe
d(ΔB)
+
dPcu
d(ΔB)
= 0
Note: optimum does not necessarily occur where Pfe = Pcu. Rather, it
occurs where
dPfe
d(ΔB)
= –
dPcu
d(ΔB)

Take derivatives of core and copper loss
Pcu = ρ λ1 2
2
I tot
2 Pfe = K fe(ΔB)β Ac lm
4Ku
(MLT)
2
WAAc
dPfe
d(ΔB)
= βK fe(ΔB) β – 1 Aclm dPcu
d(ΔB)
= – 2 ρλ1 2
2
I tot
4Ku
(MLT)
WAAc
dPfe
d(ΔB)
= –
dPcu
d(ΔB)
Now, substitute into and solve for B:
1
ΔB
2 (ΔB)– 3
Optimum B for a
given core and
application
ΔB = ρλ1 2
2
I tot
2Ku
(MLT)
WAAc
3lm
1
βK fe
1
β + 2

Total loss
Substitute optimum B into expressions for Pcu and Pfe. The total loss is:
Ptot = AclmK fe
2
β + 2 ρλ1 2
Rearrange as follows:
2
I tot
4Ku
WA Ac
2(β – 1)/β
2/β
(MLT)lm
β2
– β
β + 2
+ β2
Left side: terms depend on core
geometry
β
β + 2 β2
– β
β + 2
+ β2
2
β + 2
– β + 2
β
=
ρλ1 2
2/β
2 K fe
I tot
4Ku Ptot
β + 2 /β
Right side: terms depend on
specifications of the application
(MLT)
2
WAAc
2
β + 2

The core geometrical constant Kgfe
Define
Kgfe =
WA Ac
2(β – 1)/β
2/β
(MLT)lm
β2
– β
β + 2
Design procedure: select a core that satisfies
+ β2
2
β + 2
– β + 2
Kgfe ≥
ρλ1 2 I tot
2/β
2 K fe
4Ku Ptot
β + 2 /β
Appendix D lists the values of Kgfe for common ferrite cores
Kgfe is similar to the Kg geometrical constant used in Chapter 14:
β
• Kg is used when Bmax is specified
• Kgfe is used when B is to be chosen to minimize total loss

15.2 Step-by-step
transformer design procedure
The following quantities are specified, using the units noted:
Wire effective resistivity (-cm)
Total rms winding current, ref to pri Itot (A)
Desired turns ratios n2/n1, n3/n1, etc.
Applied pri volt-sec 1 (V-sec)
Allowed total power dissipation Ptot (W)
Winding fill factor Ku
Core loss exponent
Core loss coefficient Kfe (W/cm3T)
Other quantities and their dimensions:
Core cross-sectional area Ac (cm2)
Core window area WA (cm2)
Mean length per turn MLT (cm)
Magnetic path length l
e (cm)
Wire areas Aw1, … (cm2)
Peak ac flux density B (T)

Procedure
1. Determine core size
Kgfe ≥
ρλ1 2
2/β
2 K fe
I tot
4Ku Ptot
β + 2 /β
108
Select a core from Appendix D 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.

2. Evaluate peak ac flux density
ΔB= 108 ρλ1 2
2
I tot
2Ku
(MLT)
WAAc
3lm
1
βK fe
1
β + 2
At this point, one should check whether the saturation flux density is
exceeded. If the core operates with a flux dc bias Bdc, then B + Bdc
should be less than the saturation flux density Bsat.
If the core will saturate, then there are two choices:
• Specify B using the Kg method of Chapter 14, or
• Choose a core material having greater core loss, then repeat
steps 1 and 2

3. and 4. Evaluate turns
Primary turns:
n1 = λ1
2ΔBAc
104
Choose secondary turns according to
desired turns ratios:
n2 = n1
n2
n1
n3 = n1
n3
n1

5. and 6. Choose wire sizes
Fraction of window area
assigned to each winding:
α1 =
n1I1
n1Itot
α2 =
n2I2
n1Itot
αk =
nkIk
n1Itot
Choose wire sizes according
to:
Aw1 ≤
α1KuWA
n1
Aw2 ≤
α2KuWA
n2

Check: computed transformer model
Predicted magnetizing
inductance, referred to primary:
LM =
μn1 2
Ac
lm
Peak magnetizing current:
iM, pk = λ1
2LM
Predicted winding resistances:
R1 = ρn1(MLT)
Aw1
R2 = ρn2(MLT)
Aw2
n1 : n2
iM(t)
i1(t) i2(t)
LM
R1 R2
: nk
Rk
ik(t)

15.4.1 Example 1: Single-output isolated
Cuk converter
+–
Vg
25 V
– vC2+ v (t) + C1(t) –
i1(t) n : 1
i2(t)
Ig
4 A
+
v2(t)
–
–
v1(t)
+
100 W fs = 200 kHz
D = 0.5 n = 5
I
20 A
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 = 2.6
+
V
5 V
–

Waveforms
v1(t)
i1(t)
i2(t)
V Area λ1 C1
DTs
D'Ts
– nVC2
I/n
– Ig
I
– nIg
Applied primary volt-seconds:
λ1 = DTsVc1 = (0.5) (5 μsec ) (25 V)
= 62.5 V–μsec
Applied primary rms
current:
2
I1 = D In
+ D' Ig
2 = 4 A
Applied secondary rms
current:
I2 = nI1 = 20 A
Total rms winding
current:
Itot = I1 + 1
n I2 = 8 A

Choose core size
Kgfe ≥
(1.724⋅10– 6)(62.5⋅10– 6)2(8)2(24.7) 2/2.6
4 (0.5) (0.25) 4.6/2.6
108
= 0.00295
Pot core data of Appendix D lists 2213 pot core with
Kgfe = 0.0049
Next smaller pot core is not large enough.

Evaluate peak ac flux density
ΔB= 108 (1.724⋅10– 6)(62.5⋅10– 6)2(8)2
2 (0.5)
(4.42)
(0.297)(0.635)3(3.15)
1
(2.6)(24.7)
= 0.0858 Tesla
This is much less than the saturation flux density of approximately
0.35 T. Values of B in the vicinity of 0.1 T are typical for ferrite
designs that operate at frequencies in the vicinity of 100 kHz.
1/4.6

Evaluate turns
n1 = 104 (62.5⋅10– 6)
2(0.0858)(0.635)
= 5.74 turns
n2 =
n1
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.

Determine wire sizes
Fraction of window area allocated to each winding:
α1 =
4 A
8 A
= 0.5
α2 =
15
20 A
8 A
= 0.5
(Since, in this example, the ratio of
winding rms currents is equal to the
turns ratio, equal areas are
allocated to each winding)
Wire areas:
Aw1 =
(0.5)(0.5)(0.297)
(5)
= 14.8⋅10– 3 cm2
Aw2 =
(0.5)(0.5)(0.297)
(1)
= 74.2⋅10– 3 cm2
From wire table,
Appendix D:
AWG #16
AWG #9

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

Effect of switching frequency on transformer size
for this P-material Cuk converter example
0.1
0.08
0.06
0.04
0.02
0
2213
1811 1811
25 kHz 50 kHz 100 kHz 200 kHz 250 kHz 400 kHz 500 kHz 1000 kHz
Switching frequency
Bmax , Tesla
Pot core size
4226
3622
2616
2213
2616
• As switching frequency is
increased from 25 kHz to
250 kHz, core size is
dramatically reduced
• As switching frequency is
increased from 400 kHz to
1 MHz, core size
increases

15.3.2 Example 2
Multiple-Output Full-Bridge Buck Converter
+–
D1
Q1
DQ 2 2
D3
Q3
i1(t)
DQ 4 4
Vg
160 V
Switching frequency 150 kHz
Transformer frequency 75 kHz
Turns ratio 110:5:15
Optimize transformer at D = 0.75
: n2
+
v1(t)
–
+
5 V
–
D5
D6
I5V
i 100 A 2a(t)
+
15 V
–
D7
D8
i2b(t)
i3a(t)
n1 :
: n2
: n3
: n3
i2b(t)
I15V
15 A
T1

Other transformer design details
Use Magnetics, Inc. ferrite P material. Loss parameters at 75 kHz:
Kfe = 7.6 W/Tcm3
= 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
= 1.724·10–6 -cm

Applied transformer waveforms
t
v1(t)
i1(t)
i2a(t)
Area λ1
= VgDTs
0 0
– Vg
0
i3a(t)
0
Vg
n2
n1
I5V +
n3
n1
I 15V
–
n2
n1
I5V +
n3
n1
I 15V
I5V
0.5I5V
I15V
0.5I15V
0
0 DTs Ts 2Ts Ts+DTs
: n2
+
v1(t)
–
D3
i1(t)
D4
D5
D6
i2a(t)
D7
D8
i2b(t)
i3a(t)
n1 :
: n2
: n3
: n3
i2b(t)
T1

Applied primary volt-seconds
v1(t)
Area λ1
= VgDTs
0 0
Vg
– Vg
λ1 = DTsVg = (0.75) (6.67 μsec ) (160 V) = 800 V–μsec

Applied primary rms current
i1(t)
0
n2
n1
I5V +
n3
n1
I 15V
–
n2
n1
I5V +
n3
n1
I 15V
I1 =
n2
n1
I5V +
n3
n1
I15V D = 5.7 A

Applied rms current, secondary windings
t
i2a(t)
0
i3a(t)
I5V
0.5I5V
I15V
0.5I15V
0
0 DTs Ts 2Ts Ts+DTs
I2 = 12
I3 = 12
I5V 1 + D = 66.1 A
I15V 1 + D = 9.9 A

Itot
RMS currents, summed over all windings and referred to primary
Itot =
nj
n1
I Σ j all 5
windings
= I1 + 2
n2
n1
I2 + 2
n3
n1
I3
= 5.7 A + 5
110
66.1 A + 15
110
9.9 A
= 14.4 A

Select core size
Kgfe ≥
(1.724⋅10– 6)(800⋅10– 6)2(14.4)2(7.6) 2/2.6
4 (0.25) (4) 4.6/2.6
108
= 0.00937
A
From Appendix D

Evaluate ac flux density B
2I tot
Bmax= 108 ρλ1
2
2Ku
(MLT)
WAAc
3lm
1
βKfe
1
β + 2
Eq. (15.20):
Plug in values:
ΔB= 108 (1.724⋅10– 6)(800⋅10– 6)2(14.4)2
2(0.25)
(8.5)
(1.1)(1.27)3(7.7)
1
(2.6)(7.6)
= 0.23 Tesla
This is less than the saturation flux density of approximately 0.35 T
1/4.6

Evaluate turns
Choose n1 according to Eq. (15.21):
n1 = λ1
2ΔBAc
104
n1 = 104 (800⋅10– 6)
2(0.23)(1.27)
= 13.7 turns
Choose secondary turns
according to desired turns ratios:
n2 =
5
110
n1 = 0.62 turns
n3 =
15
110
n1 = 1.87 turns
Rounding the number of turns
To obtain desired turns ratio
of
110:5:15
we might round the actual
turns to
22:1:3
Increased n1 would lead to
• Less core loss
• More copper loss
• Increased total loss

Loss calculation
with rounded turns
With n1 = 22, the flux density will be reduced to
ΔB =
(800⋅10– 6)
2(22)(1.27)
104 = 0.143 Tesla
The resulting losses will be
Pfe = (7.6)(0.143)2.6(1.27)(7.7) = 0.47W
Pcu =
(1.724⋅10– 6)(800⋅10– 6)2(14.4)2
4 (0.25)
(8.5)
(1.1)(1.27)2
1
(0.143)2 108
= 5.4W
Ptot = Pfe + Pcu = 5.9W
Which exceeds design goal of 4 W by 50%. So use next larger core
size: EE50.

Calculations with EE50
Repeat previous calculations for EE50 core size. Results:
B = 0.14 T, n1 = 12, Ptot = 2.3 W
Again round n1 to 22. Then
B = 0.08 T, Pcu = 3.89 W, Pfe = 0.23 W, Ptot = 4.12 W
Which is close enough to 4 W.

Wire sizes for EE50 design
Window allocations Wire gauges
Aw1 =
α1KuWA
n1
=
(0.396)(0.25)(1.78)
(22)
= 8.0⋅10– 3 cm2
⇒ AWG #19
Aw2 =
α2KuWA
n2
=
(0.209)(0.25)(1.78)
(1)
= 93.0⋅10– 3 cm2
⇒ AWG #8
Aw3 =
α3KuWA
n3
=
(0.094)(0.25)(1.78)
(3)
= 13.9⋅10– 3 cm2
⇒ AWG #16
α1 =
I1
Itot
= 5.7
14.4
= 0.396
α2 =
n2I2
n1Itot
= 5
110
66.1
14.4
= 0.209
α3 =
n3I3
n1Itot
= 15
110
9.9
14.4
= 0.094
Might actually use foil or Litz wire for secondary windings

Discussion: Transformer design
• Process is iterative because of round-off of physical number of
turns and, to a lesser extent, other quantities
• Effect of proximity loss
– Not included in design process yet
– Requires additional iterations
• Can modify procedure as follows:
– After a design has been calculated, determine number of layers in
each winding and then compute proximity loss
– Alter effective resistivity of wire to compensate: define
eff = Pcu/Pdc where Pcu is the total copper loss (including proximity
effects) and Pdc is the copper loss predicted by the dc resistance.
– Apply transformer design procedure using this effective wire
resistivity, and compute proximity loss in the resulting design.
Further iterations may be necessary if the specifications are not
met.

15.4 AC Inductor Design
Window area WA
Core mean length
per turn (MLT)
i(t) +
Core
v(t)
–
L
n
turns
Wire resistivity ρ
Fill factor Ku
Core area
Ac
Air gap
lg
Area λ
v(t)
t1 t2 t
i(t)
Design a single-winding inductor, having
an air gap, accounting for core loss
(note that the previous design procedure of
this chapter did not employ an air gap, and
inductance was not a specification)

Outline of key equations
Obtain specified inductance:
L = μ0Acn2
lg
Relationship between
applied volt-seconds and
peak ac flux density:
ΔB = λ
2nAc
Copper loss (using dc
resistance):
Pcu = ρn2(MLT)
KuWA
I 2
Total loss is minimized when
ΔB = ρλ2I 2
2Ku
(MLT)
WAAc
3lm
1
βK fe
1
β + 2
Must select core that satisfies
Kgfe ≥
2/β
ρλ2I 2K fe
2Ku Ptot
β + 2 /β
See Section 15.4.2 for step-by-step
design equations

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