
100W SMT PLANAR
TRANSFORMER AND INDUCTOR
For use with Linear Technology’s LT1725
Notes from Tables
P569.B (12/02)
For More Information :
Performance warranty of products offered on this data sheet is limited to the parameters specified. Data is subject to change without notice. Other brand and product
names mentioned herein may be trademarks or registered trademarks of their respective owners.
Printed on recycled paper. 2002, Pulse Engineering, Inc.
1. The
PA0423
transformer and
PA0465/PA04680
inductor were
designed for use with Linear Technology's LT1725
and
LTC1693
IC's and form the foundation of a low cost, discrete
component alternative to telecom power modules. The
PA0423
transformer and
PA0465/PA04680
inductor were designed for
(but not limited to) the following applications:
Topology:
Single switch Forward
Frequency:
230kHz
Pri./Sec. Isolation:
Basic Insulation (1500Vdc)
Input Voltage:
36-75v telecom input
Output Voltage:
12v / 11.7A output
For PA0423: Basic Insulated Planar Transformer:
2. To determine if the transformer is suitable for your application, it
is necessary to ensure that the temperature rise of the compo-
nent (ambieint plus temp. rise) does not exceed its operating
temperature. To determine the temperature rise of the compo-
nent it is necessary to calculate the total power losses (core
and copper) in the application.
Total Copper Losses (Pcu total(W)):
Pcu total(W) = sum of the losses in each winding
The losses in each winding can be calculated by:
Pcu(W) = .001* DCR(m
) * (Irms
2
)
Core Losses (Pcore(W))
To calculate core loss, use the following formula:
CoreLoss (W) = 1.92 * 10
-13
(
B)
2.5
* (Freq kHz)
1.8
where:
B = 22653.1 * Vin min * Dutycycle max / Freq kHz
Total Losses:
P total = Pcu total + CoreLoss
Temperature Rise:
The approximate temperature rise can be found by looking
up the calculated total losses in the temperature rise vs.
power dissipation curve.
For P0465 and P0480: Planar Inductor
3. The rated current as listed is either 85% of the saturation cur-
rent or the heating current depending on which value is lower.
4. The saturation current is the current which causes the induc-
tance to drop by 15% at the stated ambient temperatures
(25°C, 100°C). This current is determined by placing the com-
ponent in the specified ambient environment and applying a
short duration pulse current (to eliminate self-heating effects) to
the component.
5. The heating current is the dc current which causes the temper-
ature of the part to increase by approximately 45°C. This cur-
rent is determined by mounting the component on a PCB with a
.25" wide, 2oz. Equivalent copper traces, and applying the cur-
rent to the device for 30 minutes with no force air cooling.
6. In high volt*time applications additional heating in the compo-
nent can occur due to core losses in the inductor which may
neccessitate derating the current in order to limit the tempera-
ture rise of the component. In order to determine the approxi-
mate total losses (or temperature rise) for a given application
both copper an core losses should be taken into account
Total Copper Losses (Pcu_total(W)):
Pcu(W)= .001* DCR(m
) * (Irms
2
)
where:
Irms = (Idc
2
+ (
1/2)
2
)
.5
1 = ripple current through inductor
Core Losses (Pcore(W)):
Use the Inductor Voltage versus CoreLoss table to
determine the approximate core losses
Total Losses:
P total = Pcu total + CoreLoss
Temperature Rise:
The approximate temperature rise can be found by looking
up the calculated total losses in the temperature rise vs.
power dissipation curve.
7. LT1725
and LTC1693
are registered trademarks of Linear
Technology Corporation.
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