Application Notes Magnetics Power Inductors Improve Reliability in High Temperature Designs

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Application Notes
Magnetics
Power Inductors Improve Reliability in High Temperature Designs
First calculate the needed inductance value:
V = L * dI/dt where:
V = Vin - Vout (voltage across the inductor)
dT = On time of drive = Vout/Vin/frequency
∆I = Chosen above to be 20%
Calculate the required inductance:
L = V * dt / ∆I = (12-5)*(12/5/600k)/(0.2*4.5)
L= 4.8 µH
Choose 4.7 µH, the nearest standard value
Recalculate ripple current at 23% using 4.7 µH
Second determine peak to peak flux density, Bp-p:
Bp-p = K * L * ∆I where:
The Cooper Coiltronics® brand of High Current FP3™
power inductors from Cooper Bussmann is designed for
high density, medium current applications using a high
temperature iron powder core material. These inductors
do not exhibit the thermal aging issue frequently
associated with iron powder core inductors. In fact the
FP3 core is rated for 200°C without thermal degradation.
The FP3 family is rated for 155°C operation. The
calculations below will allow users to take advantage of
this high temperature capability.
5V
4.5A
12V
Input
K: K-factor from the adjacent table
L: Inductance µH
∆I: Peak to peak ripple current (Amps)
Bp-p = 105*4.7*0.23*4.5 = 510 Gauss
Part Number
FP3-R10-R
FP3-R20-R
FP3-R47-R
FP3-R68-R
FP3-1R0-R
FP3-1R5-R
FP3-2R0-R
FP3-3R3-R
FP3-4R7-R
FP3-8R2-R
FP3-150-R
K-factor
803
482
344
268
219
185
161
127
105
78
59
PWM
In this example, a buck regulator will be used to convert
a 12V input to a 5V output with a load current of 4.5A.
The operating frequency was chosen to be 600 kHz to
reduce the size of the filter components, while still
maintaining good efficiency. The converter is designed to
have 20% ripple current, so a relatively low ESR output
filter capacitor will be used, as is typical in switching
power supplies.
Next determine the total losses in the inductor:
Total losses = DC loss + AC loss
DC loss = I2 *DCR = 4.52 * 0.040 = 0.81 W
(DCR from FP3 datasheet)
AC loss from table at Bp-p of 510 = 0.15 W
Total Loss = DC loss + AC loss = 0.96W
Finally determine the temperature rise.
Total loss = 0.96W, using the table,
Temperature rise is 80°C
Assuming an ambient temperature of 70°C,
The temperature of the inductor is
T = 70 +75 = 150°C
Application Notes
Magnetics
Note the data assumes no cooling airflow. Cooling will reduce the temperature of the inductor.
The FP3 is rated for 155°C operation.
FP3 AC Loss at Frequency, kHz
CoreLoss (W)
CoreLoss vs. Flux Density
1.00
0.90
0.80
0.70
0.60
0.50
0.40
0.30
0.20
0.10
0.00
250
200
300
400
500
600
700
800
900
1000
500
750
1000 1250 1500 1750 2000 2250
B p-p (Gauss)
Temperature Rise (C)
Temperature Rise vs. Watt Loss
140
130
120
110
100
90
80
70
60
50
40
30
20
10
0
0.12
0.24
0.36
0.48
0.61
0.73
0.85
0.97
1.09
1.21
1.33
1.45
1.57
Total Loss (W)
FP3 App Note 3/07
© Cooper Electronic
Technologies 2007
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