Inductance Defintion- Inductance is the result of charge movement(current). Inductance is unique defined for a closed loop of circuit. Inductance is part of the lumped-circuit model. It is only valid when the physical dimension of the circuit is electrically small lamda/10, in other words, much less than the wavelength of interest. With introduction of mutual inductance, we can
L=\frac{\Phi}{I}=\frac{\int\int_{S}\mathbf{B}\cdot d\mathbf{s}}{I}
=\frac{\int\int_{S}(\bigtriangledown \times \mathbf{A})\cdot d\mathbf{s}}{I}
=\frac{\oint_{C}\mathbf{A}\cdot d\mathbf{l}}{I}
where A is a static magnetic vector potential, and given by
A=\int_{V}\frac{\mu \mathbf{J}d\mathbf{v}}{4\pi r}
The above definition is a only valid for a loop with in finitesimal small cross section. However, for a conductor with finite cross section. Divide the cross section into small section. Using the average for A,
A=\frac{1}{a}\int_{a}\mathbf{A}da}
where a is the conductor cross section perpendicular to the current flow.
Flux coupling and Mutual inductance
Given two loops with uniform cross section and constant current distribution, i and j, the average flux on loop i due to current on loop j(Ij) is
\Phi_{IJ}=<\Phi_{ij}>=<\oint_{Ci}\mathbf{A}\cdot d\mathbf{l}>
=\oint_{Ci}<\mathbf{A}>\cdot d\mathbf{l}=\oint_{Ci}\frac{1}{a_i}\int_{a_i}\mathbf{A}da_i\cdot d\mathbf{l}
=\frac{1}{a_i}\oint_{C_i}\int_{a_i}\mathbf{A_{ij}}\cdot d\mathbf{l_i}da_i
=\frac{\mu}{4\pi a_i}\oint_{C_i}\int_{a_i}\int_{V_j}\frac{
1}{r_{ij}}\mathbf{J_j}dv\cdot d\mathbf{l_i}da_i
where
\int_{V_j}\mathbf{J_j}dv=\oint_{C_j}\int_{a_j}\frac{\mathbf{I_j}da_jdl_j}{a_j}
=\oint_{C_j}\int_{a_j}\frac{I_jda_jd\mathbf{l_j}}{a_j}Thus,
<\Phi_{ij}>=\frac{\mu I_j}{4\pi a_ia_j}\oint_{C_i}\oint_{C_j}\int_{a_i}\int_{a_j}\frac{d\mathbf{l_i}\cdot d \mathbf{l_j} }{r_{ij}}da_i da_j
The mutual inductance between loop i and loop j is,
L_{ij}=\frac{<\Phi_{ij}>}{I_j}=\frac{\mu}{4\pi a_ia_j}\oint_{C_i}\oint_{C_j}\int_{a_i}\int_{a_j}\frac{d\mathbf{l_i}\cdot d \mathbf{l_j} }{r_{ij}}da_i da_j
For thin filamentary wire i and j, the above formula reduce to Newman formula
L_f_{ij}=\frac{\mu }{4\pi}\oint_{C_i}\oint_{C_j}\frac{d\mathbf{l_i}\cdot d \mathbf{l_j} }{r_{ij}}
Rewrite Lij in terms of Lfij
L_{ij}=\frac{1}{a_i a_j}\int_{a_i}\int_{a_j}L_f_{ij}da_ida_j
Partial Inductance
Divide the loops Ci and Cj into K and M segments respectively.
L_{ij}\approx \sum_{k=1}^K \sum_{m=1}^M\frac{\mu}{4\pi a_ka_m}\int_{a_k}\int_{a_m}\int_{b_k}^{c_k}\int_{b_m}^{c_m}\frac{d\mathbf{l_k}\cdot d \mathbf{l_m} }{r_{km}}da_k da_m
Here comes the partial inductance between segment m and segment k,
L_{pkm}=\frac{\mu}{4\pi a_k a_m}\int_{a_k}\int_{a_m}\int_{b_k}^{c_k}\int_{b_m}^{c_m}\frac{|d\mathbf{l_k}\cdot d \mathbf{l_m}| }{r_{km}}da_k da_m
if
\inline \theta_{km} is the angle between \inline d\mathbf{l}_{k}} and \inline d\mathbf{l}_{m}} , then
L_{ij}=\sum_{k=1}^K\sum_{m=1}^MS_{km}L_{pkm}
where
S_{km} =
\begin{cases}
1, & \mbox{if } \theta_{km} < \frac{\pi}{2} \\ -1, & \mbox{if } \theta_{km} > \frac{\pi}{2} \\
0, & \mbox{if } \theta_{km} = \frac{\pi}{2}
\end{cases}
For thin wire conductors i and j, partial inductance between segment k and m will be reduced to
L_{pkm}=\frac{\mu}{4\pi}\int_{b_k}^{c_k}\int_{b_m}^{c_m}\frac{|d\mathbf{l_k}\cdot d \mathbf{l_m}| }{r_{km}}
L_{bk-ck-inf|m}=\frac{1}{I_m}\int_{a_k'}\mathbf{B}\cdot d\mathbf{a_k'}
=\frac{1}{I_m}\oint_{l_k}\mathbf{A_{km}}\cdot d\mathbf{l_k}
=\frac{\mu}{4\pi}\oint_{l_k}\int_{b_m}^{c_m}\frac{d\mathbf{l_m}\cdot d\mathbf{l_k} }{r_{km}}
=\frac{\mu}{4\pi}\int_{b_k}^{c_k}\int_{b_m}^{c_m}\frac{d\mathbf{l_m}\cdot d\mathbf{l_k} }{r_{km}}
=S_{km}L_p_{km}where Akm=0 for path perpendicular to bm-cm as well as at infinity.
\mathbf{A}_{km}=\frac{\mu I_m}{4\pi}\int_{b_m}^{c_m}\frac{d\mathbf{l_m}}{r_{km}}
From these, we can see partial inductance Lpkm is associated with a loop inductance due to Im. This loop is made up segment bk-ck and infinity on the ends and two sides normal to bm-cm and passing both bk and ck respectively
Loop Inductance
Divide the loop C into N segments
L=\frac{\Phi}{I}=\frac{\oint_{C}\mathbf{A}\cdot d\mathbf{l}}{I}=\sum_{i=1}^N{\frac{\int_{C_{i}}\mathbf{A}\cdot d\mathbf{l}}{I}=L_{1}+L_{2}+..+L_{N}
where
L_i=\frac{1}{I}\int_{c_i}\mathbf{A}\cdot d\mathbf{l}}
=\sum_{j=1}^{N}\frac{1}{I}\int_{c_i}\mathbf{A_{ij}}\cdot d\mathbf{l_i}}
=\sum_{j=1}^{N}\frac{\mu}{4\pi}\int_{c_i}\int_{c_j}\frac{d\mathbf{l_j}\cdot d\mathbf{l_i}}{r_{ij}}
=\sum_{j=1}^{N}S_{ij}L_p_{ij}
Where Li is a net inductance of segment i. The net inductance of segment i,Li, determines the contribution of the segment current to the overall magnetic flux through the loop.
Li can be further divided into self and mutual partial inductance, Lpii and Lpij, respectively.
Example 1. 4-side thin wire loop with rectangle shape.
L_O=\sum_{i=1}^4 L_i=\sum_{i=1}^4\sum_{j=1}^4 L_{pij}
Due to rectangle shape, side 1 normal to 2,hence Lp12/Lp21=0, same thing for other sides, as such
L_O=(L_{11}-L_{31})+(L_{22}-L_{42})+(L_{33}-L_{13})+(L_{44}-L_{24})=L_{11}+L_{22}+L_{33}+L_{44}-2*L_{13}-2*L_{24}
The flux associated with L11-L31 can be seen on the top of the figure.
Example 2. A current flowing on the wire and return ground that form a current loop.
With partial inductance modeling, wire=Lsig, ground=Lret, the coupling=M
With net inductance modeling, wire=Lnet=Lsig-M, ground=Lret-M.
ex. H due to an infinity long thin wire with current I at DC
\oint_c \mathbf{H}\cdot d\mathbf{l}=\int_0^{2\pi}H_0Rd\theta=2\pi RH_0=I
=>H_0=\frac{I}{2\pi R}
ex. Inductance of a rectangular loop(ignoring fringing fields at the 4 corners)
with l and w on the sides and radius of rw.
L&=\frac{\phi}{I}=2\int_0^w \int_{l=rw}^l\frac{u_0I}{2\pi l}dldw+2\int_{w=rw}^w \int_0^l\frac{u_0I}{2\pi w}dldw \\
=&\frac{u_0l}{\pi}ln(\frac{w}{r_w})+\frac{u_0w}{\pi}ln(\frac{l}{r_w})
ex. two parallel thin wire of radii r with length h and spacing s. The self inductance of this
loop formed by two wires are: L=2*(L11-M). L11=self partial inductance. M= mutual partial inductance. Using flux linkage concept, we can calculate the self inductance L=PHI/I
with s>>r.
L&=\frac{\phi}{I}=2\int_{h=0}^h \int_{r=r}^s\frac{u_0I}{2\pi r}dsdh\\
=&\frac{u_0h}{\pi}ln(\frac{s}{r})\\
where..\frac{u_0}{2\pi}=2x10^{-7}H/m=5.08nH/inch
ex. More accurate self and mutual partial inductance of wires
L_{pii}&=\frac{\mu l}{2\pi}\{ln[\frac{l}{r}+\sqrt{(\frac{l}{r})^2+1}]+\frac{r}{l}-\sqrt{(\frac{r}{l})^2+1}\}\\
L_{pij}&=\frac{\mu l}{2\pi}\{ln[\frac{l}{d}+\sqrt{(\frac{l}{d})^2+1}]+\frac{d}{l}-\sqrt{(\frac{d}{l})^2+1}\}
For l>>r and l>>d, r=radius of wire, d=spacing between wire, l=wire length
L_{pii}&=\frac{\mu l}{2\pi}[ln\frac{2l}{r}-1]\\
L_{pij}&=\frac{\mu l}{2\pi}[ln\frac{2l}{d}-1+\frac{d}{l}]
The above Lpii is for f->infinity. For DC, the constant 1 need to be changed to 3/4
For loop inductance of two wires Lloop. Assume current distribution are uniform and of opposite
direction on two wires.
L_{lp}=2(L_{p11}-L_p{12})=\frac{2\mu l}{2\pi}[ln\frac{2l}{r}-1-ln\frac{2l}{d}+1-\frac{d}{l}]\\
=\frac{2\mu l}{2\pi}[ln\frac{d}{r}-\frac{d}{l}]
=\frac{\mu l}{\pi}ln\frac{d}{r}
That's same result as the one derived previously using flux linkage concept. However, if we using DC constant 3/4. Lloop will be
L_{lp}=\frac{\mu l}{\pi}[ln\frac{d}{r}+\frac{1}{4}]
L=Lext+Lint
The first term corresponding to Lext,external inductance of a wire, and the 2nd term equals to Lint, internal inductance within the wire. As frequency get higher, skin effect forces the current to flow on the surface, internal inductance will tend to reduce to a small value comparing to external inductance.
e.x. how to calculate internal inductance of a wire. First, calculate the H within wire
\oint H dl=i \\
H2\pi r=\frac{I}{\pi a^2}\pi r^2\\
H=\frac{Ir}{2\pi a^2}
W=\int \frac{1}{2}uH^2da=\frac{1}{2}\int_{r=0}^a\int_{\phi=0}^{2\pi} \mu H^2 rdrd\phi\\
=\frac{1}{2}\int_{r=0}^a\int_{\phi=0}^{2\pi}\mu (\frac{Ir}{2\pi a^2})^2rdrd\phi
=\frac{1}{2}\int_{r=0}^a\int_{\phi=0}^{2\pi} \mu \frac{Ir}{2\pi a^2}rdrd\phi
=\frac{1}{2}\frac{u}{8\pi}I_0^2
Thus, per unit length internal inductance of a wire equals to
\inline L=\frac{2W}{I^2}=\frac{u}{8\pi} Lint=1.27nH/inch per unit length.
* Skin depth of a wire
R_{lf}=R_{dc}=\frac{1}{\sigma \pi r_w^2} for r_w<<\delta \\
where..\sigma=conductivity,r_w= radius of wire\\
As frequency goes higher, current tends to c
R_{hf}=R_{dc}=\frac{1}{\sigma \pi (r_w^2-(r_w-\delta)^2}, for r_w>>\delta \\
\approx \frac{1}{\sigma 2\pi r_w\delta}\\
=\frac{r_w}{2\delta}r_{dc}\\
=\frac{1}{2r_w}\sqrt{\frac{\mu_0 f}{\pi \sigma}}\\
where.. \delta=\frac{1}{\sqrt{\pi f \mu_0 \sigma}}
Ex. wire inductance due to skin effect.
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