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SPICE Modeling of Magnetic Components
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better to stay with the physical model and implement it using the ideal components that are available in PSpice. There may be another problem with the coupled inductor model. In a typical transformer, the magnetizing inductance (L12 ) might be 5 mH. The leakage inductances may be only 0.5 H. The value of k must be speci ed with enough accuracy to recreate this difference accurately; that is, a difference of 104 . For n = 1, k12 = 0.99990 for the preceding values. Inversion of Eq. (2.5) illustrates the problem: L11 = L1 k12 n L1 L2 (2.6)
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L22 = L2 nk12 L1 L2 L12 = k12 n L1 L2
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L11 and L22 are the small difference between two large numbers. In general, you should compute ki j to four decimal places. Reluctance and Physical Models The basic problem when simulating a magnetic component is to translate the physical structure of the device into an equivalent electric circuit. PSpice will use the equivalent circuit to simulate the device. Reluctance modeling, combined with a duality transformation, provides a means to accomplish this task. Reluctance modeling creates a magnetic circuit model that can then be converted into an electric circuit model. Table 2.1 shows a number of analogous quantities between electric and magnetic circuits. By comparing the form of the equations in each column, the following analogous quantities can be identi ed: EMF (V ) and MMF (F ) Electric eld (E) and magnetic eld (H ) intensities Current density (J ) and ux density (B) Current (I ) and ux ( ) Resistance (R) and reluctance (R ) Conductivity ( ) and permeability ( ) Reluctance is computed in the same manner as resistance, that is, from the dimensions of the magnetic path and the magnetic conductivity ( ).
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TABLE 2.1
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Electric and Magnetic Circuit Analogous Quantities Magnetic F NI = magnetic circuit voltage (magnetomotive force) H magnetic eld intensity F = H dlm = Hlm F NI H= = lm lm B magnetic ux density B = H = permeability 0 = 4 10 7 H/m magnetic ux = s B d s = BAm R = reluctance lm N2 R = F = = Am L P = 1/R = permeance
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Electric V electric circuit voltage (Electromotive force) E electric eld intensity V = E dlc = Elc V E= lc J current density J= E = conductivity I electric current I = s J d s = JAc R = resistance V lc R= = I Ac G = 1/R = conductance
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For a constant cross-sectional area (Am ), the reluctance is R = lm Am (2.7)
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where = o r r = relative permeability The inductance of a magnetic circuit is directly related to R and N (the number of winding turns): L= and M12 = N1 N2 = N1 N2 P12 N12 N2 = N 2P R (2.8)
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where P = permeance = 1/R . The example in Fig. 2.8 illustrates the development of the reluctance model for a simple inductor with an air gap in the core. The model develops as follows: Divide the core, including the air gaps, into sections and assign a reluctance to each one (as shown in Fig. 2.8B). Compute the reluctance for each section. Assign a magnetic voltage source to the winding with F = NI. Draw the equivalent network as shown in Fig. 2.9.
SPICE Modeling of Magnetic Components
i N turns Air Gaps c b
Material permeability of both cores = m
e lg
Mean Path Lengths R2 R2 Rg Rg R2 R1
(b-c)
R1 R2
(e 12 c )
Figure 2.8 The development of the reluctance model for a simple inductor with an air gap.
Figure 2.9 is the reluctance model that represents the magnetic structure at the top of Fig. 2.8. Now we need to convert this reluctance model to an equivalent electric circuit model, but before we can do that, it will help to brie y review the duality transformation. We can then proceed to convert the reluctance model. An example of a duality transformation is given in Fig. 2.10. A node is placed within each mesh, including the outer mesh. Branches, which
R1 =
(b c )