文章目录
- 1：高导磁+气隙（铁氧体）
- 1.1设计原理
- 1.2 设计步骤
- 2 铁粉芯
- 2.1：设计原理
- 2.2：设计步骤
TI电感设计
学习视频原链接
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1：高导磁+气隙（铁氧体）
1.1设计原理
I d c ⋅ L = B ⋅ A e ⋅ B d c I d c ⋅ N = J ⋅ A w ⋅ k u } → A w ⋅ A e = I d c 2 ⋅ L k u ⋅ J ⋅ B B : 磁通密度 J 电流密度 K u 窗口系数 \left.\begin{matrix} I_{dc}\cdot L=B \cdot A_{e}\cdot B_{dc}\\ I_{dc}\cdot N=J \cdot A_{w}\cdot k_{u} \end{matrix}\right\} \to A_{w}\cdot A_{e}=\frac{I_{dc}^{2}\cdot L}{ k_{u}\cdot J\cdot B} \\B:磁通密度\\J 电流密度\\K_{u}窗口系数 Idc⋅L=B⋅Ae⋅BdcIdc⋅N=J⋅Aw⋅ku}→Aw⋅Ae=ku⋅J⋅BIdc2⋅LB:磁通密度J电流密度Ku窗口系数
1.2 设计步骤
| 1 | 根据磁芯材料确定磁密B（0.2到0.4）最好0.28, PC40为0.3 根据散热条件，确定电流密度J，最好2.5A/MM |
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| 2 | 根据指标（ I I I和 L L L）,以及给定的 B B B和 J J J,就可以确定磁芯的面积积（AP法） I d c ⋅ L = B ⋅ A e ⋅ B d c I r m s ⋅ N = J ⋅ A w ⋅ k u < b r / > } → A w ⋅ A e = I p k ⋅ I r m s ⋅ L k u ⋅ J ⋅ B \left.\begin{matrix}I_{dc}\cdot L=B \cdot A_{e}\cdot B_{dc}\\I_{rms}\cdot N=J \cdot A_{w}\cdot k_{u}<br/>\end{matrix}\right\} \to A_{w}\cdot A_{e}=\frac{I_{pk}\cdot I_{rms} \cdot L}{ k_{u}\cdot J\cdot B} Idc⋅L=B⋅Ae⋅BdcIrms⋅N=J⋅Aw⋅ku<br/>}→Aw⋅Ae=ku⋅J⋅BIpk⋅Irms⋅L  I r m s = I d c 2 + △ I 2 ⋅ 1 3 ( ( 1 − D ) + D ) 2 I_{rms}=\sqrt{I_{dc}^{2}+\bigtriangleup I^{2}\cdot \frac{1}{3} \left ( \sqrt{\left ( 1-D \right ) } + \sqrt{D} \right )^{2} } Irms=Idc2+△I2⋅31((1−D)+D)2 I p k = I d c + △ I 2 I_{pk}=I_{dc}+\frac{\bigtriangleup I }{2} Ipk=Idc+2△I |
| 3 | 根据指标 L L L，确定气隙 I a I_{a} Ia ，对应公式 I a = μ N 2 A e L I_{a}=\frac{\mu N^{2} A_{e}}{L} Ia=LμN2Ae， 其中 μ = μ 0 = ( 4 ⋅ π ⋅ 1 0 − 7 ) T ⋅ m A \mu= \mu_{0}=\left ( 4\cdot \pi \cdot 10^{-7} \right ) \frac{T\cdot m}{A} μ=μ0=(4⋅π⋅10−7)AT⋅m:真空磁导率 |
| 4 | 计算磁芯损耗 P c P_{c} Pc和绕组损耗 P w P_{w} Pw |
| 5 | 计算温升 △ T = ( P t S ) 0.833 \bigtriangleup T= \left ( \frac{P_{t}}{S} \right )^{0.833} △T=(SPt)0.833 |
2 铁粉芯
2.1：设计原理
电感是储能 E 储能 = 1 2 L ⋅ I 2 = 1 2 B ⋅ H ⋅ V e ∝ μ ⋅ H 2 → ( μ H 2 ) = B H E_{储能}=\frac{1}{2} L\cdot I^{2}=\frac{1}{2} B\cdot H\cdot V_{e}\propto \mu \cdot H^{2}\to \left ( \mu H^{2} \right )= BH E储能=21L⋅I2=21B⋅H⋅Ve∝μ⋅H2→(μH2)=BH，磁材料储能因子
2.2：设计步骤
| 1 | 根据磁粉芯材料特性，选择具有最大磁能积的工作点的磁通密度 B d B_{d} Bd |
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| 2 | 根据 I I I和 L L L，以及选定的 B B B和 J J J，根据AP法选择磁芯尺寸 A w ⋅ A e = I d c 2 ⋅ L k u ⋅ J ⋅ B d A_{w} \cdot A_{e}= \frac{I_{dc}^{2}\cdot L}{ k_{u}\cdot J\cdot B_{d}} Aw⋅Ae=ku⋅J⋅BdIdc2⋅L |
| 3 | 确定匝数 N N N，并验证电感量 表达式1： L = μ N 2 A e l e L=\frac{\mu N^{2} A_{e}}{l_{e}} L=leμN2Ae 表达式2： I ⋅ L = B d ⋅ N ⋅ A e I\cdot L=B_{d}\cdot N\cdot A_{e} I⋅L=Bd⋅N⋅Ae |
| 4 | … |
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