参数资料
型号: FP1007R3-R23-R
厂商: Cooper Bussmann
文件页数: 49/56页
文件大小: 0K
描述: INDUCTOR LO PROFLE 230NH 61A SMD
特色产品: FP/HCP Series Inductors
标准包装: 650
系列: FLAT-PAC™, FP1007
电感: 230nH
电流: 61A
电流 - 饱和: 48A
类型: 铁氧体芯体
容差: ±10%
屏蔽: 无屏蔽
DC 电阻(DCR): 0.29 毫欧
材料 - 芯体: 铁氧体
封装/外壳: 0.402" L x 0.307" W x 0.287" H(10.20mm x 7.80mm x 7.30mm)
安装类型: 表面贴装
包装: 带卷 (TR)
工作温度: -40°C ~ 125°C
频率 - 测试: 100kHz
Coiltronics ? High Frequency Inductor Catalog
Power Inductors Improve Reliability in High Temperature Designs
Cooper Bussmann Coiltronics ? high current FP3? power inductors are
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
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.
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
Δ I = Chosen above to be 20%
Calculate the required inductance:
L = V * dt / Δ 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 * Δ I where:
K: K-factor from the adjacent table
L: Inductance μH
Δ I: Peak to peak ripple current (Amps)
Bp-p = 105*4.7*0.23*4.5 = 510 Gauss
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 data sheet)
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
Part Number
FP3-R10-R
FP3-R20-R
FP3-R47-R
FP3-R68-R
FP3-1R0-R
FP3-1R5-R
K-factor
803
482
344
268
219
185
Part Number
FP3-2R0-R
FP3-3R3-R
FP3-4R7-R
FP3-8R2-R
FP3-150-R
K-factor
161
127
105
78
59
For product information and data sheets, visit www.cooperbussmann.com/datasheets/elx
49
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