Inductance is defined as

$L = N {dφ}/{dI} = N {d(webers)}/{d(amps)}$

Solving for L gives equation (1) and is the start of the derivation for "Inches to the Fifth" for inductors.

(1) $L = {U_o N^2}/{{L_a/{A_a} + {L_m}/{U_m A_c}}$

  (2) $N = {A_p}/{d^2} = {A_p}/{A_w} * {A_w}/{d^2}$

(3) $R = MTL * {A_p}/{A_w} * {A_w}/{d^2}* {p}/{12000}$

(4) $H = {0.4 pi*N I}/{lm}$

With F being the fringing factor, we can define

  (5) $A_a = F A_c$

(6) $U_r = {U_m}/(1 + U_m * {la}/{F * lm}) $

Insert equations (5) and (6) into equation (1) to produce

(7) $ L = {U_o U_r A_c *N^2 }/ {lm} $

(8) $ I = {{H*lm} / (0.4 pi * N)} $  (Equation (4) rearranged)

Note that material permeability, Um (Ur in MPP toroids), is defined at the oersted drive being used which can be greatly different from the initial permeability (Um at low drive levels)! For gapless structures, the oersted level will give Um. For gapped structures, the oersted level in the air gap will be converted to gauss to find Um. It should be noted that in air.

   1 GAUSS = 1 OERSTED

Example: 

   10 gauss = 10 Oersted =  ${{0.4 pi (3 turns) (3.33 amps)} / {1.6 cm (AirGap)}} $

The energy stored in an inductor is

${L*I^2}/2$

Using equations (7) and (8), this is given by:

(9) $ LI^2 = {U_o U_r A_c * N^2 / {lm}} * {({H*lm} / {0.4 pi})^2} $

Reducing and rearranging equation (9) two different ways:

(10) $ LI^2 = (A_c lm) * (U_r H^2) * {U_o / (0.4 pi)^2} $

(11) $ {{LI^2} / {A_c lm}} = {{NI} /{lm}} * U_r * {{U_o H} / {0.4 pi}} $

Equation (10) is useful because it shows that energy storage is proportional to the volume of the core material. Equation (11) is rearranged to the terms used in the Hanna Curves. However, the immediate goal is to relate inductance to "Inches to the Fifth" as used in Lowell Quist's VA equation. Lowell Quist's equation is given below for square wave excitation (use 4.92E-6 for sine waves):[1]

${VA} = 3.999E-6 * {A_p / A_W} * B^2 f^2 * {{A_c^2 A_w} /{MTL}} * {2 /{pd_p^2 + pd_s^2}} * {{\%REG} / (1 + {{\%REG} / 100})^2} * {1 / (1 + ({X_1 / R})^2)^.5} $

$ {{VA} / f^2} = VA * time^2 = Energy * time = F(B,f) * {{A_C^2 A_w} /{MTL}} $

Energy equals power (VA) multiplied by time. Since frequency is the inverse of time by dividing both sides of Lowell Quist's equation by frequency squared (multiplying by time squared), it is shown that energy times time is a function of "Inches to the Fifth".

To convert equation (10) to "Inches to the Fifth," both sides of the energy equation for LI^2 must be multiplied by something with the units of time. The only relationship of interest with the units of time is the time constant of the inductor, L/DCR or equation (7) /equation (3).

(12) $ {L / R} = {{({U_o U_r A_c N^2} /{ lm})} / {MTL (A_p / A_w) (A_w / d^2) (p / 12000)} $

Substitute equation (2) into equation (12):

(13) $ {L / R} = {({U_o U_r A_c (A_p / A_w)^2 (A_w / d^2)^2 }/{lm})} / {MTL (A_p / A_w) (A_w / d^2) (p / 12000)} $

Multiply equation (13) by equation (10) to get energy*time units:

(14) $ {{L^2 I^2} / R} = {{A_c lm U_o U_r H^2} / (0.4 PI)^2} * {{U_o U_r A_c (A_p / A_w)^2 (A_w / d^2)^2 }/ {lm}} / {MTL (A_p / A_w) (A_w / d^2) (p / 12000)} $

which reduces to:

(15) $ {{L^2 I^2} / R} = {(U_o U_r H)^2 / (0.4 PI)^2} * {{A_c^2 A_w} / {MTL}} * {12000 / {pd^2}} * {A_p / A_w} $

Since Ac was originally defined in square centimeters and Ac in square inches is wanted, multiply the above equation by (2.54)^4]. Then the numeric constant becomes:

$ {{12000 U_o^2 (2.54)^4} / (0.4 PI)^2} = {{12000 (0.4 pi*1E-8)^2 (2.54)^4 }/ (0.4 PI)^2} = 49.95 E-12 $

The final equation with Ac^2Aw/MTL in inches and Ur*H in gauss is:

( 16 )  $ {{L^2 I^2} / R} = 49.95 E-12 * (U_r * H)^2 * {{A_c^2 A_w} / {MTL}} * {1 / {pd^2}} * {A_w / A_p} $

2.4 Applying the Final Equation

If Ur*H is in the linear region of the B-H curve, then Ur equals B/H and Ur*oersteds can be substituted with gauss. So for a gapped structure, Ur*H equals the gauss at the knee of the B-H loop. However, for a MPP toroid that has biased at the 50% left oersted level, Ur*H is the initial permeability times H times 50% / 100% or 0.50. For a 300 perm core that is at 50% at 20 oersted:

$ U_r  * H = 300 perm * 20 oe * 0.50.left = 3000 Gauss $

The pd^2 factor in equation (16) is the wire figure of merit. As wires get smaller, the insulation becomes a greater percentage of the total area. Thus, as wires become smaller, pd^2 becomes larger. Since square wire uses more of the available area for copper than round wire, pd^2 will be lower for square wire of edge length d than for round wire with diameter d because round wire occupies the area of its diameter squared.

Listed later in the paper is a table of typical values for pd^2. To calculate pd^2 multiply the Ohms per 1000 feet by the square of the area occupied by the wire when winding it. The factor pd^2 does not include temperature effects on the DC resistance or the AC resistance increase due to eddy current effects. 

The (Ap/Aw) factor is the window utilization factor (fill factor).  For toroids (Ap/Aw) is usually between .2 and .4. For rectangular geometries (Ap/Aw) is usually between .4 to .6 for a transformer and 0.8 to 0.9 for an inductor. Because of the 1/(pd^2) correction factor, we don't have to pad the fill factor as hard as when using other calculation methods.

With a little manipulation of equations (2) through (8), solutions for N and Ur can be obtained. Multiplying equations (7) and (8) to get LI:

$ LI = {{U_o U_r A_c N^2 }/{lm}} * {{lm*H} / {0.4 pi* N}} $

Solving for N:

$ N = {{LI} / {A_c *( {U_o U_r H} / {0.4 PI})} $

Convert Ac to use square inch instead of square cm and reduce:

$ N = {{LI} / {{{{2.54}^2 A_c * 0.4 pi*1E-8*U_r H }/ {0.4 pi}}}} $

Watch the units on this next one. UrH is in gauss, Ac is in square inch and Ac includes the stacking factor, K. Meaning Ac in the equation is (K * physical Ac).

(17) $ N = {15.5E6 {LI} / {(U_r H) A_c}} $

The coated diameter of the wire that is needed is given by

Wire Diameter = square root ( AW * (Ap/Aw)/(N * number of strands))

Solving equation (7) for Ur produces:

 $ U_r = {{L* l_m} / {U_o A_c*N^2}} $

With lm in centimeters and Ac in square centimeters:

(18) $ U_r = 79.6E6 * {{L* l_m} / {A_c*N^2}} $

Converting to inches:

$ U_r = {{L* lm} / {0.4 pi *1E-8* A_c* N^2} }* {2.54 / (2.54)^2} $

With lm in inches and Ac in square inches:

(19) $ U_r = 31.3E6 * {{L* l_m} / {A_c*N^2}} $

If there is no air gap, Ur equals Um of the material at H oersteds bias. If Ur*H was chosen at the 50% point, then pick the initial permeability design point of the material to be Ur/(50%/100%).

For example, if Ur*H was chosen to be 3000 (50% left on MPP) and Ur from equation (19) is 149, then the core permeability to use is:

149 / ( 50% / 100% ) = 149 / 0.50 = 298 or 300u

EI, EE, C-cores, and Pot cores all depend on the air gap to control Ur. The size of the air gap is given by equation (20) below. One thing to keep in mind is that Ur cannot exceed Um (the material perm), and that at high drive levels Um will drop in value. This will become apparent by examining equation (6) (repeated below).

(6) $ U_r = {U_m / {1 + U_m *{{la} /{ F*lm}}}} $

A second approximation for F is to add the air gap to the radius of the physical core area.