INDUCTORS
Inductance is ubiquitous:
Ampere's Law: ∇ × H = J + ∂D ∂t
i(t)
Let ∂/∂t ≅ 0 Example—printed circuit: x σ=∞ da W c z y c i(t) d << W
îc H • ds = ∫A J • da ∫
= 0 around both wires = i(t) around the top wire
Magnetic Field H:
Outside: H ≅ 0 Inside: ˆ H ≅ − yi W (since d << W we ignore fringing fields)
L11-1
INDUCTANCE
Quasistatic Behavior (∂/∂t ≅ 0): ∂∂
+ v(t) Side view: y 1 d z=D Ey(t,z) 2 z
H(t)
i(t)
Maxwell’s Equations:
Faraday’s Law:
0
∇ × E = −∂B ∂t E y ( t,z ) d = − (µzd W ) di(t) dt [recall H = i/W] ⇒ ∫c E • ds = − d ∫A µH • da dt
Therefore when z = D:
v ( t ) = ∫1 E • ds = −E y d = (µDd W ) (di ( t ) dt )
2
Note: Dd = cross-sectional area A v(t) = L di(t)/dt where L = µA/W Henries Note: Kirchoff’s voltage law not obeyed here; Ey = f(z)
L11-2
SOLENOIDAL INDUCTORS
N-Turn Solenoidal Inductor:
v ( t ) = ∫1 E • ds = −E y d = (µDd W ) (di ( t ) dt )
2
y W
Wj
v j ( t ) = (µDd W j ) (di ( t ) dt ) for jth turn But: W = NW j v ( t ) = Nv j ( t ) Therefore: v ( t ) = N2µDd W (di ( t ) dt ) ; Dd = A D
(
)
i(t) d
Inductance of N-Turn Solenoid: Magnetic Energy Storage:
Wm = µ H ( t )
2
L = N2µA/W Henries
2
J m-3
2 2 2 = µAW (Ni W ) 2 [J]
w m = µDdW H ( t ) Therefore:
[recall H = Ni W ]
L11-3
w m = Li2 ( t ) 2
[J]
RESISTIVE INDUCTORS
Single-Turn Inductor:
Conductance of slab of cross-sectional area S: σS [Siemens m]
Resistance of slab of length D: D/σS [ohms]
Resistance of a single-turn inductor: 2(D + d)/σS [ohms] (S = δW)
Resistance of an N-turn inductor: 2N(D + d)/σ(S/N) = 2N2(D + d)/σS
L/R Time Constant of Solenoidal Inductor:
τ = L/R seconds = N2µA W
(e.g.
i(t) = io e −τ τ
)
R
L W(N turns)
(
) (2N2 (D + d) σS ) ≅ µdδσ 2N
where D >> d , S = δW, A = Dd For finite size and mass, τ is limited ! Want d → D , δ → d 3 , N → 1 , d → W D s δ i(t) d
L11-4
TRANSFORMERS
Air-Wound Solenoidal Transformers:
E • ds = − d ∫A µH • da îc ∫ dt Say A1 = A2, W1 = W2 here Ni = number of turns in coil i Therefore: + v2(t) i1(t) W1
+ v1(t)
W2
A
Step-up and Step-down Transformers:
The voltage induced in one turn of coil 2 is the same as induced in one turn of coil 1, And the total voltage induced in coil 2 is N2/N1 times the total voltage induced in coil 1, regardless of whether it is generated by i1 or i2. N2/N1 is called the transformer turns ratio N2 N1 N N
1
2
Step-up or step-down the output voltage, correspondingly. The flux coupling between the two coils may be imperfect and the output voltage is correspondingly reduced [flux Λ = µHA, and linked flux = NµHA].
L11-5
IRON-CORE TRANSFORMERS (1)
Boundary Conditions:
H// and B⊥ are continuous across the boundary
(∇ × H = J + ∂D ∂t; ∇ • B = 0 )
µo B2
µ >> µo
H // B and B = µH
µ µo can be as large as 106 . Since µ >> µo , B1 is essentially parallel to the interface, and trapped within the high permeability medium. The magnetic flux B is "trapped" inside.
H2//
B1
L11-6
IRON-CORE TRANSFORMERS (2)
Flux trapping inside high permeability materials:
The magnetic flux density B is trapped inside high-permeability materials, e.g.: i(t) Flux Λ = ∫A B • da ≅ constant Cross-sectional area A B
Transformer Output:
v2(t) = (N2/N1)v1(t) The flux is highly linked with little leakage v1(t) N1 turns + B
+ v2(t) N2
e.g. iron
L11-7
INDUCTANCE OF IRON TOROID (1)
Inductance L of N turns around toroid (N2 = 0):
Recall: wm = Li2 (t) 2 = ∫V Wm dv [J] = ∫V µ H where Wm = µ H(t)
2
(
2
2 dv
)
2
−3 Jm i(t) N turns 2R Area A Volume V
Example: Constant Area Toroid
Since: L = µ ∫V H(t) dv i2 (t)
(
2
)
∇ × H = J + ∂D ∂t ⇒ îc H • ds ≅ Ni(t) ∫ ≅0 Therefore: 2πR H ≅ Ni (R varies slightly over A)
2 2 L ≅ µ ∫V (N 2πR ) dv ≅ µ (N 2πR ) V ≅ µN2 A 2πR [Henries]
where V ≅ 2πRA 3 m
L11-8
INDUCTANCE OF IRON TOROID (2)
Inductance L of a toroid with a gap:
] w m = Li2 (t) 2 = ∫V Wm dv [J
Recall: 2 −3 Wm = µ H(t) 2 Jm where 2 Therefore: L = µ ∫V H(t) dv i2 (t), as before
i(t) 2R gap d[m]
(
)
Finding H(t) :
Since:
îc H • ds ≅ Ni(t) ∫
∇ • B = 0, so µoHo = µH where we assume µ >> µo ! (we require d > 2πR (µo µ )2 )
Therefore: Hµ (2πR − d ) + Hµo d ≅ Ni(t) But Therefore: Magnetic energy density in gap is (µ/µo)2 greater than inside the torus, and dominates unless (µ/µo)2 << 2πR/d
L11-9
INDUCTANCE OF IRON TOROID (3)
Inductance L of a toroid with a gap:
Recall: L= µ(t)∫V H dv i2 Hµ ( 2πR − d ) + Hµo d ≅ Ni(t) Where: Hµo d ≅ Ni(t) for a small gap with little fringing (d << A0.5) and we neglect the energy storage inside the torus L ≅ µAd (N d) or, for a small-gap torus:
2
(
2
)
i(t)
2R gap d[m]
Therefore:
L ≅ µo AN2 d
Henries
2
! provided that d > 2πR (µo µ )
L11-10