**Amplifiers/Converters**

# Designing a half-bridge converter using a CoreMaster E2000Q core

**Keywords:power
**

/ARTICLES/2001SEP/2001SEP07_AMD_POW_AN4.PDF |

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AN108
Designing a Half Bridge Converter
Using a CoreMaster E2000Q Core
By
Colonel Wm. T. McLyman
The half bridge converter is shown in Figure 1.
Figure 1. Half bridge converter
The dynamic BH loop for a half bridge push-pull converter is shown in Figure 2.
Figure 2 The dynamic BH loop of a push-pull converter.
Half Bridge Converter Transformer Design Specification
1. Input voltage nominal Vnom = 28 V
2. Input voltage minimum Vmin = 24 V
3. Input voltage maximum Vmax = 32 V
4. Output voltage VO = 5 V
5. Output current IO = 10 A
6. Frequency f=100 kHz
7. Efficiency = 98 %
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8. Maximum duty ratio Dmax = 0.5
9. Regulation = 0.5 %
10. Operating flux density BAC = 0.2 T
11. Diode voltage drop Vd = 1 V
12. Window utilization KU =0.4*
13. Waveform factor KF =4.0
14. Temperature rise Tr=300C
* Note - window utilization will be re-calculated.
Figure 3. Typical half bridge converter waveforms.
The waveforms shown in Figure 3, are typical waveforms of the half bridge converter. The
collector current Ic is shown in Figure 3-A. The collector voltage, Vc is shown in figure 3-B. The
inductor L1 current, IL, made up from the rectifier CR2 and CR4 are shown in Figure 3-C.
Select a wire so that the relationship between the AC resistance and the DC resistance is 1:
1AC
DC
R
R
=
The skin depth in cm is:
6.62
f
=
6.62
0.0209 [cm]
100,000
= =
Then, the wire diameter is:
Wire diameter = 2
Wire diameter = 2 0.0209= 0.0418 [cm]
Then, the bare wire area AW is:
2
4
W
D
A
=
2
23.1416 0.0418
0.00137[cm ]
4
WA
= =
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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.
Wire AWG Bare Area Area Ins. Bare/Ins. 5/cm
#26 0.00128 0.001603 0.798 1345
#27 0.001021 0.001313 0.778 1687
#28 0.000804 0.000105 0.765 2142
When operating at high frequencies, the engineer has to review the window utilization factor, Ku.
Operating at 100kHz and having to use a #26 wire, because of the skin effect, the ratio of the bare
copper area to the total area is 0.79. The window utilization factor was developed using a #20
wire, with a bare copper area to the total area ratio of 0.86. Therefore, the overall window
utilization Ku is reduced. To return the design back to the norm, the core geometry Kg is to be
multiplied by 1.1, and the current density J is calculated, using a window utilization factor of
0.367.
Step No. 1 Calculate the total period, T.
1
T
f
=
61
10 10 [s]
100,000
T -
= =
Step No. 2 Calculate the maximum transistor on time, ton.
on MAXt TD=
6
10 10 0.5 5 [ s]ont 5-
= =
Step No. 3 Calculate the secondary output power, PO.
( )O O O dP I V V= +
( )10 5 1 60 [W]OP = + =
Step No. 4 Calculate the total input power, Pin
O
in
P
P
=
60
61.2 [W]
0.98
inP = =
Step No. 5 Calculate the apparent power, Pt.
1
2t OP P
= +
1
60 1.41 146 [W]
0.98
tP
= + =
Step No. 6 Calculate the electrical conditions, Ke
2 2 2 4
0.145 10e f ACK K f B -
=
2 2 2 4
0.145 4 100,000 0.2 10 92800eK -
= =
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Step No. 7 Calculate the core geometry, Kg .
t
2
g
e
P
K
K
=
5146
0.00157 [cm ]
2 92800 0.5
gK = =
The core geometry Kg in the data table is calculated using a window utilization of 0.4. Operating
at 100kHz and using a #26 AWG the window utilization Ku has to be multiplied 1.1 to bring it
back to the norm.
1.1g gK K=
5
1.1 0.00157 0.00173 [cm ]gK = =
Step No. 8 Select from the data sheet a E 2000Q core comparable in core geometry, Kg.
Core number TEA0111Q
Manufacturer CMI
Magnetic material E 2000Q
Magnetic path length, MPL 4.06 cm
Core weight, Wtfe 4.6 g
Copper weight, Wtcu 5.6 g
Mean length turn, MLT 2.7 cm
Iron area, Ac 0.14 cm2
Window area, Wa 0.541 cm2
Area product, Ap 0.0757 cm4
Core geometry, Kg 0.00158 cm5
Surface area, At 15.9 cm2
Step No. 9 Calculate the low line input current, Iin.
IN
IN
INMIN
P
I
V
=
61.2
2.55 [A]
24
INI = =
Step No. 10 Calculate the primary rms current, Iprms. When using a half bridge the input current is
multiplied by 2 to calculate the primary current.
Pr
2
2
IN
ms
MAX
I
I
D
=
2 2.55
5.1 [A]
1
PI
= =
Step No. 11 Calculate the number of primary turns, Np. Because this is a half bridge the input
voltage is divided by 2.
(min
2
in
p
V
N =
24
12[ ]
2
pN V= =
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4
( ) 10P MIN
p
c AC f
V
N
fA B K
=
4
12 10
10.7 use 10, [turns]
100,000 0.14 0.2 4.0
PN
= =
Step No. 12 Calculate the current density J using a window utilization Ku = 0.367.
4
10t
P AC u f
P
J
f A B K K
=
4
2146 10
657 [A/cm ]
100,000 0.0757 0.2 0.367 4.0
J
= =
Step No. 13 Calculate the required primary bare wire area, Awp.
Prms
wp
I
A
J
=
25.1
0.00776 [cm ]
657
wpA = =
Step No. 14 Calculate the required number of strands NSp. Using the area of a #26 wire.
( )
#26
wp B
P
A
NS =
0.00776
6.06 use 6
0.00128
PNS = =
Step No. 15 Calculate the primary new 5/cm from the number 26AWG.
/
/
p
cm
new cm
NS
5
5
=
1345
/ 224
6
new cm5 = =
Step No. 16 Calculate the primary winding resistance, Rp.
6
10P PR MLT N
cm
5 -
=
6
2.7 10 224 10 0.00605 [ ]PR -
= =
Step No. 17 Calculate the primary copper loss, PP.
2
PrP ms PP I R=
2
3.607 0.026 0.338 [W]PP = =
Step No. 18 Calculate the transformer secondary voltage, Vs.
S O dV V V= +
5 1 6 [V]SV = + =
Step No. 19 Calculate the number of secondary turns, NS.
1
100
P S
S
PMIN
N V
N
V
= +
10 6 1.0
1 5.05 use 5 [turns]
12 100
SN
= + =
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Step No. 20 Calculate the secondary rms current, ISrms.
Srms S MAXI I D=
10 0.707 7.07 [A]SrmsI = =
Step No. 21 Calculate the secondary bare wire area, Aws.
( )
S rms
ws B
I
A
J
=
2
( )
7.07
0.0108 [cm ]
657
ws BA = =
Step No. 22 Calculate the required number of secondary strands, NSs.
( )
#26
ws B
S
A
NS =
0.0108
8.4 use 8
0.00128
SNS = =
Step No. 23 Calculate the secondary new mW per centimeter using number 26 AWG.
/
( ) /
S
cm
new cm
NS
5
5
=
1345
( ) / 168
8
new cm5 = =
Step No. 24 Calculate the winding resistance, Rs.
6
10S SR MLT N
cm
5 -
=
6
2.7 5 168 10 0.0227 [ ]SR -
= =
Step No. 25 Calculate the secondary copper loss, PS.
2
S Srms SP I R=
2
10 0.00227 0.227 [W]SP = =
Step No. 26 Calculate the total copper loss, Pcu.
CU P SP P P= +
0.157 0.227 0.384 [W]CUP = + =
Step No. 27 Calculate the regulation, .
100%CU
O
P
P
=
0.384
100 0.64%
60
= =
Step No. 28 Calculate the window utilization KU.
2P P S SN N NS N NS= +
10 6 2 5 8 140N = + =
(#26)W
U
NA
K
W
=
140 0.00128
0.331
0.541
UK
= =
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Step No. 29 Calculate the true flux density, BAC.
4
min 10P
AC
C P f
V
B
f A N K
=
4
12 10
0.214 [Tl]
100,000 0.14 10 4
ACB
= =
Step No. 30 Calculate mW/g.
7 1.834 2.1122
/ 8.64 10 ACmW g f B-
=
7 1.834 2.1122
/ 8.64 10 100,000 0.214 49.2mW g -
= =
Step No. 31 Calculate the core loss, PFe.
( ) 3
/ 10Fe tfeP mW g W -
=
3
49.2 4.6 10 0.226 [W]FeP -
= =
Step No. 32 Calculate the total loss, P.
Cu FeP P P = +
0.384 0.226 0.61 [W]P = + =
Step No. 33 Calculate the watt density, .
t
P
A
=
20.61
0.0384 [W/cm ]
15.9
= =
Step No. 34 Calculate the temperature rise, Tr.
0.826
450rT =
0.826
450 0.0384 30.5 [ C]rT = = o
Step No. 35 Calculate the transformer efficiency,
O
O
P
P P
=
+
60
100 99 [%]
60 0.61
= =
+
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: [email protected]

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