Chapter 10
Q. 10.3
Consider a Micrometals toroidal iron powder core with 52/μ75 material. The initial permeability versus magnetic field intensity curve is given in Figure 10.5.
Step-by-Step
Verified Solution
An inductor was constructed using a toroid with core length l_{c} = 4.23 cm, core cross-sectional area A_{c} = 0.179 cm^{2} and with N = 72 turns. Calculate the inductance L and the effective inductance L_{\mathrm{eff}} as a function of current for the range 1 to 4 A.
For H = 30 Oe, μ_{r} is read from Figure 10.5:
i = \frac{Hl_{c}}{N}= \frac{(2387)(4.23\times10^{-2})}{72} = 1.403 A
L(i) = \mu_{r}(i)\times \mu_{0}\times N^{2}\times \frac{A_{c}}{l_{c}} = (56.1)(4\pi\times10^{-7})(72)^{2}\left(\frac{0.179\times 10^{-4}}{4.23\times 10^{-2}} \right) = 154.7 \muH
L_{\mathrm{eff}}=L(i)+ i\frac{\Delta L}{\Delta i} = (154.7\times10^{-6})+ (1.403)\frac{(154.7-176.7)\times10^{-6}}{(1.403-0.935)} = 88.7 \mu H
The full set of calculations are summarized in Table 10.1, and L and L_{\mathrm{eff}} are plotted in Figure 10.6.
| Table 10.1 Calculations for inductance and effective inductance | |||||
| H (Oe) | H (A/m) | \mu_{r} | i (A) | L (\muH) | L_{\mathrm{eff}}(\mu H) |
| 1 | 80 | 75 | 0.047 | 206.8 | 206.8 |
| 1.4 | 111 | 75 | 0.065 | 206.8 | 206.8 |
| 2 | 159 | 74.9 | 0.094 | 206.5 | 205.6 |
| 3 | 239 | 74 | 0.14 | 204.0 | 196.6 |
| 5 | 398 | 73.5 | 0.234 | 202.6 | 199.2 |
| 7 | 557 | 72.5 | 0.327 | 199.9 | 190.2 |
| 10 | 796 | 71.1 | 0.468 | 196.0 | 183.1 |
| 14 | 1114 | 68.4 | 0.655 | 188.6 | 162.5 |
| 20 | 1592 | 64.1 | 0.935 | 176.7 | 137.2 |
| 30 | 2387 | 56.1 | 1.403 | 154.7 | 88.7 |
| 50 | 3979 | 44 | 2.338 | 121.3 | 37.9 |
| 70 | 5571 | 35 | 3.273 | 96.5 | 9.6 |
| 100 | 7958 | 27 | 4.675 | 74.4 | 0.9 |
