An inductor constructed with an EI assembly EI42 (corresponding to IEC YEI 1-14), with M530-50A material as described in Example 10.1, was constructed with G = 1.0 mm, g = 0.2 mm, lc = 8.4 cm, Ag = 2.027 cm², d = 1.4394 cm and N = 365 turns. Calculate the inductance L and the effective inductance

Question

An inductor constructed with an EI assembly EI42 (corresponding to IEC YEI 1-14), with M530-50A material as described in Example 10.1, was constructed with G = 1.0 mm, g = 0.2 mm, [latex]l_{c} = 8.4 cm, A_{g} = 2.027 cm²[/latex], d = 1.4394 cm and N = 365 turns. Calculate the inductance L and the effective inductance [latex]L_{eff}[/latex] as a function of current for the range 0.25 to 4 A.

Accepted Answer

For low values of current before the onset of saturation, the inductance is a function of the gap and is given by Equation 10.24:

[latex]L_d=\frac{\mu_0 \cdot N^2 \cdot A_g}{G-g} \ln \left(\frac{G}{g} ight)=\frac{\left(4 \pi imes 10^{-7} ight)(365)^2\left(2.072 imes 10^{-4} ight)}{(1.0-0.2) imes 10^{-3}} \ln \left(\frac{1.0}{0.2} ight) imes 10^3=69.8 mH[/latex]

We may divide the core into ten discrete elements, so that Δx = 0.1 d.

The MATLAB program to perform these calculations is included at the end of this chapter.

The plot of L and Leff is shown in Figure 10.11.

The shape of the graph, in this case, is controlled by the sloped gap, which means that we can achieve the desired characteristic by controlling the geometry of the gap.

Question 10.5: An inductor constructed with an EI assembly EI42 (correspond...

Related Answered Questions

Design the filter inductor for a 25 W load at 25 V rms input at 50 Hz that will maintain conduction down to 10% of rated load. ...

Verified Answer:

Verified Answer:

The voltage and current at the maximum power point in a 200 W solar panel are listed in Table 10.5, along with the maximum power for various level of solar insolation. Calculate the internal resistance of the panel for each level of solar power, the duty cycle of the buck converter for impedance ...

Verified Answer:

The rectifier in Figure 10.12 has the specifications given in Table 10.4. Determine the harmonics when the SAG inductor of Example 10.5 is used at the input. For simulation purposes, the circuit is represented in Figure 10.19; the input represents the rectifier output. ...

Verified Answer:

Verified Answer:

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. ...

Verified Answer:

An inductor constructed with an EI assembly EI42 (corresponding to IEC YEI 1-14) with M530-50A material, as described in Example 10.1, was constructed with g = 0.5 mm, core length lc = 8.4 cm, core cross-sectional area Ag = 2.072 cm² and with N = 365 turns. Calculate the inductance L and the ...

Verified Answer:

Establish the constants H0 and Hm for M530-50A material whose permeability versus magnetic field intensity is shown in Figure 10.2. ...

Verified Answer: