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AN109
Push-Pull Converter Design Using a
CoreMaster E2000Q Core
By
Colonel Wm. T. McLyman
The single output push-pull converter is shown in Fig.1.
Figure 1 Single output push-pull converter.
Push-Pull Converter Transformer Design Specification
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
Input voltage nominal
Input voltage minimum
Input voltage maximum
Output voltage
Output current
Output circuitry
Frequency
Regulation
Efficiency
Operating flux density
Window utilization
Diode voltage drop
Duty ratio
Temperature rise
Vnom = 28 V
Vmin = 24 V
Vmax = 32 V
VO = 5 V
IO = 10 A
Center tapped
f=100 kHz
α = 0.5 %
η = 98 %
Bm = 0.1T
KU =0.4
Vd = 1 V
Dmax =0.5
Tr=25°C
At this point, select a wire so that the relationship between the ac resistance and the dc resistance
is 1.
RAC
=1
RDC
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The skin depth in cm is:
δ =
δ =
6.62
f
6.62
100,000
= 0.0209 [cm]
Then, the wire diameter is:
Wire diameter = 2δ
Wire diameter = 2 ⋅ 0.0209= 0.0418 [cm]
Then, the bare wire area AW is:
π D2
4
3.1416 ⋅ 0.04182
AW =
= 0.00137[cm2 ]
4
From the Wire Table, number 26 has a bare wire area of 0.001280 centimeters. This will be the
minimum wire size used in this design. If the design requires more wire area to meet the
specification, then, the design will use a multifilar of #26. Listed Below are #27 and #28, just in
case #26 requires too much rounding off.
AW =
Wire AWG
#26
#27
#28
Bare Area
0.00128
0.001021
0.000804
Area Ins.
0.001603
0.001313
0.000105
µΩcm
/
Bare/Ins.
0.798
0.778
0.765
1345
1687
2142
Step No. 1. Calculate the transformer output power, PO.
PO = I O (VO + Vd )
PO = 10 ⋅ (5 + 1) = 60 [W]
Step No. 2. Calculate the total apparent power, Pt.
 2

Pt = PO 
+ 2
 η

 1.41

Pt = 60 
+ 1.41 = 171 [W]
0.98


Step No. 3. Calculate the electrical conditions, Ke
Ke = 0.145 ⋅ K 2f ⋅ f ⋅2 Bm2 ⋅ 10 −4
K f = 4 for square wave
Ke = 0.145 ⋅ 4 2 ⋅ 100,0002 ⋅ 0.12 ⋅ 10−4 = 23200
Step No. 4. Calculate the core geometry, Kg.
Pt
Kg =
2 ⋅α ⋅ K e
171
Kg =
= 0.00369 [cm5 ]
2 ⋅ 23,200 ⋅ 1
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Step No. 5. Select from the data sheet a E2000Q core comparable in core geometry, Kg
Core number
Manufacturer
Magnetic material
Magnetic path length, MPL
Core weight, Wtfe
Mean length turn, MLT
Iron area, Ac
Window area, Wa
Area product, Ap
Core geometry, Kg
Surface area, At
TEA0112Q
CMI
E 2000Q
5.11 cm
9.5 g
3.4 cm
0.24 cm2
0.866 cm2
0.208 cm4
0.00594 cm5
24.9 cm2
Step No. 6. Calculate the number of primary turns NP using Faradays Law.
VP ( MIN ) ⋅ 104
Np =
f ⋅ Ac ⋅ B AC ⋅ K f
12 ⋅ 104
= 25 [turns]
100,000 ⋅ 0.24 ⋅ 0.1 ⋅ 4.0
Step No. 7. Calculate the current density J using a window utilization, Ku = 0.4.
Pt ⋅ 104
J=
f ⋅ AP ⋅ BAC ⋅ Ku ⋅ K f
NP =
171 ⋅104
= 514 [A/cm2 ]
100,000 ⋅ 0.208 ⋅ 0.1 ⋅ 0.4 ⋅ 4.0
Step No. 8. Calculate the input current, Iin.
P
I IN = O
VIN ⋅η
60
I IN =
= 2.55 [A]
24 ⋅ 0.98
Step No. 9. Calculate the primary bare wire area, Awp(B)
I ⋅ DMAX
Awp ( B ) = IN
J
2.55 ⋅ 0.707
Awp ( B ) =
= 0.00351 [cm2 ]
514
Step No. 10. Calculate the required number of primary Snp. Using the area of a #26 wire.
A
Snp = wp ( B )
# 26
0.00351
Snp =
= 2.74 use 3
0.00128
Step No. 11. Calculate the primary new µΩ /cm from the number 26AWG.
µΩ / cm
newµΩ / cm =
Snp
J=
newµΩ / cm =
1345
= 448
3
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Step No. 12. Calculate the primary winding resistance, Rp.
 µΩ  −6
RP = MLT ⋅ N P 
 ⋅ 10
 cm 
RP = 3.4 ⋅ 25 ⋅ 448 ⋅ 10 −6 = 0.0381 [Ω]
Step No. 13. Calculate the primary copper loss, PP.
PP = I Pr ms 2 RP
PP = 2.552 ⋅ 0.0381 = 0.247 [W]
Step No. 14. Calculate the number of secondary turns, NS.
N ⋅V 
α 
N S = P S 1 +

VPMIN  100 
VS = VO + Vd
VS = 5 + 1 = 6 [V]
25 ⋅ 6  1.0 
1 +
 = 6.31 use 6 [turns]
24  100 
Step No. 15. Calculate the secondary bare wire area, Aws.
I ⋅ DMAX
Aws = O
J
10 ⋅ 0.707
Aws ( B ) =
= 0.0138 [cm 2 ]
514
Step No. 16. Calculate the required number of secondary strands, Sns.
Aws ( B )
Sns =
# 26
0.0138
Sns =
= 10.78 use 10
0.00128
Step No. 17. Calculate the secondary new mW per centimeter using number 26 AWG.
µΩ / cm
( new) µΩ / cm =
NS S
1345
( new) µΩ / cm =
= 134.5
10
Step No. 18. Calculate the secondary resistance, Rs.
 µΩ  −6
RS = MLT ⋅ N S 
 ⋅ 10
 cm 
Rs = 3.4 ⋅ 6 ⋅ 134.5 ⋅ 10 −6 = 0.0027 [Ω]
Step No. 19. Calculate the secondary copper loss, PS.
PS = I S 2 RS
NS =
PP = 102 ⋅ 0.00274 = 0.274 [W]
Step No. 20. Calculate the total copper loss, Pcu.
PCU = PP + PS
PCU = 0.247 + 0.274 = 0.521 [W]
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Step No. 21. Calculate the transformer regulation, α.
P
α = CU ⋅ 100%
PO
0.521
α=
⋅ 100 = 0.868%
60
Step No. 22. Calculate the milliwatts per gram, mW/g.
2.1122
mW / g = 8.64 ⋅ 10−7 ⋅ f 1.834 ⋅ BAC
mW / g = 8.64 ⋅ 10−7 ⋅ 100,0001.834 ⋅ 0.12.1122 = 9.875
Step No. 23. Calculate the core loss, PFe.
PFe = ( mW / g ) ⋅ Wtfe ⋅10−3
PFe = 9.875 ⋅ 9.4 ⋅ 10−3 = 0.0928 [W]
Step No. 24. Calculate the total loss, PΣ.
PΣ = PCu + PFe
PΣ = 0.521 + 0.0928 = 0.614 [W]
Step No. 25. Calculate the watt density, Ψ.
P
Ψ= Σ
At
0.614
Ψ=
= 0.0247 [W/cm2 ]
24.9
Step No. 26. Calculate the temperature rise, Tr.
Tr = 450 ⋅ Ψ 0.826
Tr = 450 ⋅ 0.02470.826 = 21 [o C]
Step No. 27. Calculate the total window utilization KU.
KU = KuP + KuS
2 ⋅ N S ⋅ S N ⋅ Aws
KuS =
Wα
2 ⋅ 6 ⋅ 10 ⋅ 0.00128
= 0.154
0.866
2 ⋅ N P ⋅ SnP ⋅ AwP
KuP =
Wα
2 ⋅ 25 ⋅ 3 ⋅ 0.00128
KuP =
= 0.222
0.866
Ku = 0.222 + 0.154 = 0.376
Kus =
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BIBLIOGRAPHY
Colonel William T. McLyman, Transformer and Inductor Design Handbook, Second Edition,
Marcel Dekker Inc., New York, 1988.
Colonel William T. McLyman, Magnetic Core Selection for Transformers and Inductors, Second
Edition, Marcel Dekker Inc., 1997
Colonel William T. McLyman, Designing Magnetic Components for High Frequency, dc-dc
Converters, Kg Magnetics, Inc., 1993.
For information regarding the above Books and Companion
Software for Windows 95', 98' and NT, contact:
Kg Magnetics, Inc.
38 West Sierra Madre Blvd, Suite J
Sierra Madre, Ca. 91024
Phone: (626) 836-7233, FAX: (626) 836-7263
Web Page: www.kgmagnetics.com
Email: sheasso@pacbell.net
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