Inches to the Fifth

The Other Way to Design Magnetics
2. Derivation 3. Summary of Useful Information (Print these pages) 4. Design Examples

  • 4.1 MPP Toroid Example
  • 4.2 C-Core Design Example

  • The mathematical formulas below are presented using a locally supported version of MathScribe. Mathscribe can be locally hosted which means web pages using it can correctly function as long as HTML code will support the functions it uses.  There isn't a danger of an upgrade to it or the location of the external web site moving and requiring a re-write of the web page. I've had issues with this in the past so this falls under once bitten, twice shy.

    1. Over View

    This design method started out with a need to design a group of inductors that were very small but didn't saturate at a peak current much higher that the part could support thermally. The voltage drop and saturation current had to be controlled separately. Lowell Quist's Inches to to the Fifth for transformers came close, but this equation needed to work for Inductors, not for transformers. With the INCH^5 equation for inductors, once you pick "L","Ipk","DCR", Window fill factor and gauss drive at Ipk, you have defined the core size. It also quickly tells you what the trade offs will be if you want to make the part smaller. That is exactly what I needed for this group of inductors.

    From Quist, Lowell, "A New Converter Transformer Design Technique Relates Complex Performance Parameters," Proceedings of Powercon 8, Section B-2, Power Concepts, Inc., 1981.

    ...

    Rearranging this we get:

    $VA = 3.99*10^{-6}*{B_{NL}}^2f^2*{K_{STK}}^2(A_c^2 A_w)/{MTL}*1/{[[pd_p^2 + pd_s^2]/2]}*{A_p/A_w}*{%REG}/{[1+{%REG}/{100}]}$

    The "Inches to the Fifth" title comes from the factor for the core dimensions:

    $(A_c^2 A_w)/{MTL}$

    (A_c^2 A_w)/MTL having the units of

    ( Inch^2 ) ^2 * Inch ^2 / Inch
    Inch^6 / Inch
    Inch^5

    The Inches to the Fifth Equation equation for inductors with K_STK=1 is:

    $(LI)^2/{DCR} = K_c * (U_r H)^2 * (A_c^2 * A_w)/{MTL} * {1/{{pd}^2} * {A_p}/{A_w}$

    Mr. Schwerman presented a bunch of math about this topic at "PowerCon 11" to provide a derivation of this equations. The PowerCon 11 paper work had an error in it, the fix for it is shown below (From: Schwerman, Paul, “Single Iteration Magnetic Design Based On Winding Resistance,” Proceedings of Powercon 11, pages 222-234, October 1988).

    ...

    2.  Derivation

    What is shown below comes from a copy Mr. Schwerman was handing out before the version he sent to PowerCon 11. He gave me permission to post it here for posterity.

    Magnetic Design Based on (LI)^2/DCR
    Last Revision June 4, 1987

    2.1 Introduction

    There are eight critical specifications in the design of a magnetic device:

      1. Inductance (L)
      2. Current Response
      3. DC resistance (DCR)
      4. AC losses (Q)
      5. Self Resonant Frequency (SRF)
      6. Final Dimensions
      7. Removal of heat loss temperature rise and cooling method)
      8. Corona and insulation breakdown

    Generally, the inductance at an average current and the desired finished dimensions are the only constraints that are specified to a magnetic designer. However, all seven of the above parameters must be considered. The more the inductance is allowed to "swing" (drop with applied current), the higher the DCR, the higher the temperature rise, and the lower the thermal impedance to the heat sink, the smaller a part can be made. Equation (16) shown below relates inductance, current, and DC resistance to the shape of the core and the gauss drive of the core material. This equation allows independent comparison of these variables.

    (16)    ${L^2I^2}/{DCR} = 49.95E-12* (U_r*H)^2 *{K_{STK}}^2 *{{{A_c}^2A_w}/{MTL}* {1/{pd}^2 * {A_P}/{ A_W}$

    Three of the eight design parameters are not explicitly stated in equation (16). These parameters are temperature rise, self resonant frequency, and Q. Temperature rise is dependent not only on the resistive loss and the core loss, but also on the method of getting the heat out of the device. Self resonant frequency is dependent on the insulation thickness, dielectric constant, surface area of the winding, and winding method. Q is dependent on the total AC losses, frequency and the gauss drive (lower gauss drive gives a higher Q). However, these remaining four parameters are indirectly controlled by equation (16) through R, UrH, and Ap/Aw.

    The rest of this paper is divided into four sections. The first section discusses the derivation of equation (16) and some fine points of its use. The second section is a summary of significant equations and constants. The third section contains several examples. The last section is short list of "Inches to the Fifth" for several cores. The most common figure of merit for a transformer core is "Inches to the Fourth," or AwAc. Aw is the winding area in square inches. Ac is the core area in square inches. The product AwAc gives dimensional units of inches raised to the fourth power. Most manufacturers provide AwAc in their catalogs. The core's figure of merit in this paper has dimensional units of inches raised to the fifth power.

    ...

    The conversion from transformer design to inductor design came from a flash of insight that] most inductors are fairly linear:

                  $E    = L {dI}/{dt} + I {dL}/{dt}$

    becomes

                     $E      = L {dI}/{dt}$

    on a linear inductor and critically to this derivation is

                  $E*dt = L*dI$ 

    2.2 Units

    The units used on the page for the derivation of equations are:

    L Inductance in Henries
    lm Magnetic path length in centimeters
    la Effective air gap in centimeters
    Ac Actual iron cross sectional area in square centimeters
    Aa Effective cross sectional area of the air gap
    Aw is the available winding area of the core in square inches
    Ap is the area occupied by the wire in the primary of a transformer or the wire of an inductor in square inches
    d is the insulated wire diameter in inches
    p is the Ohms/1000 feet, milliOhms/foot, or Ohms/12000 inches of the wire
    H oersteds
    I Amps
    Uo 0.4*PI*1E-8 for lm in cm and Ac in square cm
    Ur Relative permeability (dimensionless)
    N Turns
    I Current in Amps
    F Fringing factor
    Kstk Ratio of the actual iron area to physical iron area

               Later, I convert Ac into in^2 to support inches^5.

    ...

    2.3 Derivation of (LI)^2/DCR

    Lowell Quist has shown that a powerful method for choosing a transformer core is by

    ${{A_c}^2 A_w}/{MTL}$

    (dimensional units of "Inches to the Fifth") which is a function of both VA and regulation. [1]

    Since the geometry factor of a regulation-based transformer core is "Inches to the Fifth," it would also be desirable to be able to choose the core of an inductor based on "Inches to the Fifth" -- if for no other reason other than not to have another set of tables to carry around.

    Listed below are several relationships between the available independent variables of inductance, current capacity, and DC resistance. Definitions for N and R are as used in Lowell Quist's equation. [1] First consider the physics of an inductor. An inductor is a device that when ampere-turns are applied, a magnetic field develops.

    $L (Henries) = {Uo Ur N^2 Ac(Core Area)}/{(Lm (path length)}$

    $ 1/{L (Henries)} = {lm} / { Uo Ur N^2 * Ac}$

    In the reluctance model of a magnetic circuit, the magnetic field (webers) flows like a current source through the reluctance (equivalent to a resistance) of the magnetic paths. This flow causes an ampere-turn drop across each reluctance in the same manner as the voltage drop from a current through a resistor. The units for reluctance are 1/Henry or 1 divided by the inductance for one turn for the section of magnetic material under consideration. 

    Hint: This works the same way as resistors


              NI Air            
         NI Iron  
    FIELD ==> 
       ----/|/|/|/----
             +
         -----/|/|/|/---          = NI total
    (weber)

      R air
         R Iron


      la / (UoAa)

         lm / (UoUrAc) 

    Current                   * (Resistance 1        + Resistance 2)          = Voltage

    weber                     * (1/H                  + 1/H )                    = NI (Amp-turn)

    maxwells      * 10E-8 * ( reluctance air      + reluctance core)      = Amp-turn

    gauss * cm^2 * 10E-8 * (cm_airgap/(UoAa) + cm_iron/(UoUrAc)   = Amp-turn

    ...

    When calculating NI:

    1 Amp-Turn is        1.0 turn  at 1 amp or
                              10 turns with each turn at 1/10 A

    and a reminder that:

                     1 weber is           1 H (Henry) * 1 Amp-turn.

    With Aa (area of the airgap) approximately = Ac (area of the core), we can substitute Ac for Aa.

    Using this in an example with air path length la = 1, Ur=10K and iron path length lm = 200:

    $1/{U_o A_c} + {{200}/{(U_o*10,000*A_c)}$

    10 webers * (1/H + .02/H) = 10.2 Amp-turn

    We will have to add a correction factor for fringing at the air gap:

    $Gauss.in.iron = Gauss.in.gap * {Fringing.Factor * Area.Of.Gap}/{Stacking.Factor * Area.Of.Iron}$

    Since the maxwells in the iron and the air gap are equal, the gauss in the iron is increased by the ratio of the effective area of the gap divided by the actual area of the iron. The simplest way to find the field in a gapped structure is to assume that all the ampere-turns are supported by the air gap. This is valid because the reluctance of the air is usually (but not always) much larger than the reluctance of the iron; therefore, the air will support all the ampere-turns. An example is given above with the iron area equal to the gap area, the iron permeability equal to 10,000, and the iron magnetic path length 200 times longer than the air gap length. In this situation, the approximation is correct to about 2%.

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

    N    = number of turns = area of winding / area occupied by 1 wire
    R    = MTL * N * ( Ohms / inch )
    d^2 = Effective area of the wire (i.e. a round peg in square hole.)
    p     = is the Ohms/1000 feet, milliOhms/foot, or Ohms/12000 inches of the wire

    ...

      (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}}}} $

    Solving for the total effective air gap, la:

    $ U_r + la*{ U_m U_r} /{ F*lm} = U_m $

    $ la * {U_m U_r} / {F *lm} = U_m - U_r $

    Sizing the Airgap:

    (20) $ la = F * lm * {U_m - U_r} / {U_m U_r} $

    If lm is in inches, then la is in inches. If lm is in centimeters, then la is in centimeters. Since la is the effective air gap, it will have to be corrected for fringing effects (the F factor).

    First calculate la assuming F = 1, then correct la with the actual F. One iteration is all that is needed. If Um is not known exactly, but is known to be a very high value, then equation (20) can be solved for "infinite" Um:

    $ la = F * lm * {Um}/(Um Ur) $

    For "infinite" Um

    (21)  $ la = {F * {lm} / U_r} $

    ...

    Equation (22) is the correction factor for fringing in a square stack as given in Reuben-Lee^2 modified by Kstack to hold the gauss in the iron at the design target.

    * Lprime/L is the ratio of the inductance with fringing to the inductance without fringing.
    * "S" is the length of the leg with the air gap.
    * Ac is the area of the core without the stacking factor (Kstack).
    * "lg" is the gap length in that leg.
    * If the gap is in both legs of a C-core or all three legs of an EI-core, lg = la/2.

    (22) $ F = {L^{prime} / L} = 1 + {1 / K_{stack}} * {{2 lg} / A_c^0.5} * Ln {{2 S} / {lg}} $

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

    (23) $ F = 1+{(length + 2 lg) * ( width + 2 lg) }/ {K_{stack} * length * width} $

    It is advisable not to iterate on equations (22) and (23). There are several items to remember:


    1. Air gaps are commonly available in 1, 2, and 5 mil increments.
    2. The surfaces of the core will usually have 1 to 2 mils gap. With special handling or on many ferrite cores that have ground gaps, this value can be lower.
    3. Your calculator may be exact; the rest of the world is not.
    4. The coil of wire will have "leakage" inductance that adds to the inductance from the core. If Ur is low, the leakage inductance can cause the measured inductance to read high.
    5. The coil of wire over the air gap affects the amount of fringing that occurs. First order, the fringing flux stops increasing the "gap area" when it hits the winding.  If it does hit the winding, there will be additional AC losses.

    References

    1   Quist, Lowell, "A New Converter Transformer Design Technique Relates Complex Performance Parameters," Proceedings of Powercon 8, Section B-2, Power Concepts, Inc., 1981.

    2   Lee, Reuben, _Electronic Transformers and Circuits_, Second Edition, 1955



    ...

    3.0 Summary of Useful Information

    $ {A_c^2 A_w} /{MTL} = {{pd^2 * {(LI)^2} / {DCR}} /{ K1 * (U_r * H)^2 * {A_p / A_w}} $



    (16)  $ {L^2 I^2} / {DCR} = K1 *(U_r H)^2 * {A_c^2 A_w} / {MTL} * {1 / {pd^2}} * {A_p / A_w} $



    (17)  $ N = K2 * {LI} /{ (U_r H)* A_c} $



    (19)  $ U_r = K3 * {L *lm} / {A_c*N^2} $



    (20)  $ la = F * lm * {U_m - U_r} / {U_m U_r} $



    (22)  $ F = {L^{prime} / L} = 1 + {1 / K_{stack}} * {{2 lg} / A_c^0.5} * ln {{2 S} / {lg}} $

    For UrH in gauss:

         K1      K2     K3    AC    Aw MTL   lm    p   d2
    49.95 E-12 15.5E6 31.33 E6 in4 in2 in in Ohm/ ft in2
    1.200 E-21 100.E6 79.58 E6 cm4 in2 in in Ohm/ ft in2
    942.5 E-21 100.E6 79.58 E6 cm4 cir mil in in Ohm/ ft in2
    472.4 E-15 100.E6 79.58 E6 cm4 cm2 cm in Ohm/ ft in2
    100.0 E-18 100.E6 79.58 E6 cm4 cm2 cm cm Ohm/ cm cm2

    TABLE 1. Units for equations (16), (17) and (19)

    ...

    Pick the wire diameter based on the insulated (coated) diameter of the wire for Aw occupying the complete winding area (after the area for any layer insulation is removed from the winding area), Aw:

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

    Finding the insulated (coated) diameter of the wire for Aw in circular mils, a PI/4 must be added in:

    Wire Diameter = square root { (PI/4) * Aw * (Ap/Aw) / (N * number of strands) }

    The wire diameter equation will give the coated diameter of the wire to use. If Aw is in square inches, the diameter is in inches. If Aw is in square centimeters, the diameter is in centimeters. If Aw is in circular mils, then the wire diameter is in mils, but when working with circular mils, the window area must be corrected because a round wire occupies the space of square circumscribed around it. (The round peg fills the square hole.)
    When K1 = 472.4E-15, pd^2 is the same as given in the table (2) below.

    The most common design mistake is forgetting to square N in the Ur equation 19. The second most common design mistake is forgetting that Ur is the permeability under DC bias, not the ZERO DC initial permeability.


    AWG $ pd^2 $ $ pd^2 $ $ pd^2 $ $ pd^2 $ $ pd^2 $ $ pd^2 $
    # Round Round Square Round Round Round

    Solid Solid Solid Solid Stranded Stranded

    Bare Double Double 81822 22759 81381 Single

    Copper Film Film Teflon Teflon Kapton

    4 .01037
    12 .01037


    .02352 .01979
    16 .01037 .01193

    .02680 .01737
    22 .01037 .01279
    .04225 .03928 .02180

    24 .01037 .01323
    .05188 .04921 .02488
    28 .01037 .01441
    .05410 .07816
    36 .01037 .01646
    40 .01037 .01726

    TABLE 2. Wire Insulation Figure of Merit (p=mOhms/ft d=inches)

    ...

    Material 80%
    UrH
    50%
    UrH
    Notes

    P MagInc 3000 4800 Ferrite - Varies Strongly with Ur
    F MagInc 1800 2500 80% is for 600u, 50% is for 50u
    J MagInc 1300 2000 Ferrites saturate very sharply




    14 MPP 2240 3100 MagInc MPP cores
    26 MPP 2290 3100 very stable with Ur except for 550u.
    60 MPP 2110 3000 80% saturation oersteds are maximum limits. 
    125 MPP 2200 3100 50% points are off a "typical" graph

    147 MPP 2120 3200 For 80% LEFT designs, use UrH = 2100 typ.
    160 MPP 2050 3200 For 50% LEFT designs, use UrH = 2500 typ.
    173 MPP 2080 3200 or increase current by 25% to meet a
    200 MPP 2080 3100 minimum inductance under bias for
    300 MPP 2160 3000 production margin.

    550 MPP 748 1400




    Si Steel 8000 12000 3% Silicon Steel Saturates Slightly faster



    than MPP cores, but not as fast as Ferrites



    The Inductance is hard to hold to tight 



    tolerances. Low Ur can go higher in gauss.

    TABLE 3. Typical UrH (gauss) Drive levels for common materials

    ...

    3.1 The Power in the Equation

    (16) $ {L^2 I^2} / {DCR} = K1 *(U_r H)^2 * {A_c^2 A_w} / {MTL} * {1 / {pd^2}} * {A_p / A_w} $

    Using the following simplification, we'll see some of the power in the inch^2 equation: 

     ${L^2 I^2} / {DCR}$ proportional to $(U_r H)^2 * {A_c^2 A_w}/ {MTL}$

    A. If we want to halve the gauss (UrH) and try to keep all else the same we must change one (or a combination) of these parameters:

    • Half the inductance (L)
    • Half the current (I)
    • Quadruple (4X) the resistance (DCR)
    • Double the core area (Ac) (without changing the MTL). But, doubling Ac usually makes the MTL grow by a factor of about 1.5X so we need closer to 2.5X the core area.
    • Quadruple the window area (Aw)

    B. If we want to double the current (I) and keep all else the same and the core size the same we must accept this:

    • Half the inductance (L)
    • Take 4X the power loss and about 4X higher temperature rise

    We change the inductance by changing the relative permeability. On a gapped EI core, this means we approximately double the airgap size.

    C. If we want to double the current and keep the power loss the same, we need 4X lower DCR. This means we have to accept 4X lower inductance.

    D. If we want to go from 15,000 gauss on steel to 2,400 gauss on ferrite and keep the size, DCR and current rating the same we have to accept 15K/2.5K = 6.25 times lower inductance.

    E. Knowing the AC voltage at the lowest frequency we can calculate the relative impact of flux it generates as "X1" in volt*seconds. "X1" will normally be fixed and independent of the inductor's value. This means we can "optimize" (play with) the L and I dc values that combine to make "X2" volt*seconds to see how it impacts the part. In the end, the part must support X1 + X2 in the design equations.

    If the inductor is reasonably linear, L * I has the units of volt*seconds. If "X2" is much larger than "X1", the dc current flow though the inductance of the device is driving the size of the part. If "X1" is bigger, the AC voltage and low frequency cut off is driving the size of the part. If "X1" is significantly bigger than "X2", we may be able to make the inductance (L) larger, i.e. X2) for little to no penalty in size, weight and/or cost.

    ...

    3.2 Metric Equations for Inch^5 Design (in work)

    The Table of Metric Design Constants is in work and is not ready to present.

    10,000 gauss = 1 tesla

    The earth's magnetic field is 0.22 to 0.65 gauss or 22 to 65 microtesla.

    ...

    3.3 The MPP Toroid Inch^5 List


    ( Data from Late 1980s data book. Current data on the web may be different. )


        OD    ID    HT                       50%     AL    IN^5     IN^5    
    MAX   MIN   MAX   LM     AC     AW    MTL    ----  SINGLE   DOUBLE
      INCH  INCH  INCH   CM    CM^2   INCH   INCH    Ur    STACK    STACK 
      -----------------------------------------------------------------------
      0.150 0.060 0.072  0.817 0.0137 0.0028  0.272  0.207  46.9E-09 123.E-09
      0.165 0.078 0.110  0.942 0.0211 0.0048  0.357  0.280  143.E-09 354.E-09
      0.205 0.076 0.130  1.060 0.0285 0.0045  0.437  0.330  203.E-09 508.E-09
      0.275 0.090 0.135  1.361 0.0470 0.0064  0.511  0.400  661.E-09 1.73E-06
      -----------------------------------------------------------------------
      0.285 0.090 0.125  1.363 0.0476 0.0064  0.500  0.433  692.E-09 1.85E-06
      0.288 0.087 0.218  1.363 0.0920 0.0059  0.690  0.823  1.75E-06 4.29E-06
      0.335 0.135 0.150  1.787 0.0615 0.0143  0.585  0.413  2.22E-06 5.88E-06
      0.405 0.168 0.150  2.180 0.0752 0.0222  0.644  0.427  4.68E-06 12.8E-06
      -----------------------------------------------------------------------
      0.405 0.168 0.180  2.180 0.0945 0.0222  0.704  0.530  6.76E-06 17.9E-06
      0.425 0.180 0.180  2.380 0.1000 0.0254  0.720  0.530  8.50E-06 22.7E-06
      0.468 0.232 0.186  2.690 0.0906 0.0423  0.759  0.423  11.0E-06 29.5E-06
      0.530 0.275 0.217  3.120 0.1140 0.0594  0.869  0.447  21.3E-06 56.9E-06
      -----------------------------------------------------------------------
      0.680 0.375 0.280  4.110 0.1920 0.1104  1.112  0.577  87.9E-06 234.E-06
      0.830 0.475 0.280  5.090 0.2260 0.1772  1.230  0.543  177.E-06 486.E-06
      0.930 0.527 0.330  5.670 0.3310 0.2181  1.412  0.720  407.E-06 1.11E-03
      0.956 0.542 0.382  5.880 0.3880 0.2307  1.537  0.563  543.E-06 1.45E-03
      -----------------------------------------------------------------------
      1.090 0.555 0.472  6.350 0.6540 0.2419  1.841  1.257  1.35E-03 3.57E-03
      1.385 0.888 0.387  8.950 0.4540 0.6193  1.871  0.633  1.64E-03 4.64E-03
      1.332 0.760 0.457  8.150 0.6720 0.4536  1.990  1.017  2.47E-03 6.78E-03
      1.445 0.848 0.444  8.980 0.6780 0.5648  2.050  0.937  3.04E-03 8.49E-03
      -----------------------------------------------------------------------
      1.602 0.918 0.605  9.840 1.0720 0.6619  2.503  1.343  7.30E-03 19.7E-03
      1.875 1.098 0.635 11.630 1.3400 0.9469  2.778  1.423  14.7E-03 40.4E-03
      2.035 1.218 0.565 12.730 1.2500 1.1652  2.761  1.217  15.8E-03 45.0E-03
      1.875 0.918 0.745 10.740 1.9900 0.6619  3.043  2.247  20.7E-03 55.6E-03
      -----------------------------------------------------------------------
      2.285 1.368 0.585 14.300 1.4440 1.4698  3.001  1.247  24.5E-03 70.6E-03
      3.108 1.888 0.550 19.600 1.7700 2.7996  3.584  1.137  58.8E-03 180.E-03


    ...

    For Estimating The Size of Scrapless EI Laminations:

    INCH^5 = 0.1081 * T^5 

    Where T = width of the tongue (center leg) in inches.

    The EI laminations are not a solid piece of iron. The stacking factor for 11 mil laminations runs at 96% with all the "manufacturing burrs" facing the same way.

    $Ac = 0.96 * T^2$  for a square stack
    Where T = the tongue width. 

    For scrapless laminations, the window winding height is 0.5 times the tongue width. The window winding width is 1.5 times the tongue width. This gives Aw = 0.5 T * 1.5 T or

    $Aw = 0.75 T^2$ (tongue width) with out the bobbin. 
    Quite a few EI laminations run a taller window to make room for the bobbin, but for simple estimates, we won't use that assumption. If we assume 30 mil (1/1000 of an inch) bobbin wall thicknesses on a 0.75 inch tongue, the AW drops from 0.422 sq in to 0.367 sq in or 87% of the available area or $Aw = 0.87 * 0.75 *T^2$ with bobbin.

    $Aw = 0.622 *T^2$ with bobbin.

    If we make this assumption, we should base Aw/Ap = 100% for a full bobbin that has no tape between layers.

    The MTL for  a square stack runs 4 * T + 2 pi * (winding height /2)  with height being T/2 =  4T + pi/2 * T

    MTL= 5.571 * T.

    Ac^2 Aw / MTL for a square EI stack with nominal bobbin correction is  {( 0.96 * T^2 ) ^2 * 0.87*0.75 * T^2} / (5.571* T) = 0.108 * T^5.

    INCH^5 = 0.1081 * T^5

    If not correcting for the bobbin, use INCH^5 = Ac^2 Aw / MTL  = 0.1241 T^5.
    If not correcting for both the bobbin and lamination stacking factor, use Ac^2 Aw / MTL  = 0.1346 T^5.

    3.4 The Ferrite Toroid Inch^5 List

    ( Data from Late 1980s data book. Current data on the web may be different. )
      CORE  OD   ID    HT    AW    AC    Lm  50%    IN^5     
        NAME INCH  INCH  INCH  IN^2   CM^2  CM  MTL IN          
    40503 0.155 0.088 0.100 0.0061 0.021 0.921 0.314 2.05E-007
    40601 0.230 0.120 0.060 0.0113 0.021 1.303 0.294 3.85E-007
    40402 0.190 0.090 0.100 0.0064 0.031 1.021 0.347 4.18E-007
    40603 0.230 0.120 0.125 0.0113 0.043 1.303 0.424 1.18E-006
    40705 0.300 0.125 0.188 0.0123 0.100 1.498 0.615 4.80E-006
    41003 0.375 0.187 0.125 0.0275 0.073 2.072 0.536 6.56E-006
    41005 0.375 0.187 0.188 0.0275 0.110 2.072 0.662 1.21E-005
    41303 0.500 0.312 0.125 0.0765 0.074 3.124 0.609 1.65E-005
    40907 0.375 0.220 0.250 0.0380 0.137 2.266 0.774 2.22E-005
    41506 0.520 0.290 0.156 0.0661 0.112 3.055 0.697 2.85E-005
    41407 0.500 0.281 0.188 0.0620 0.129 2.950 0.746 3.33E-005
    41206 0.500 0.203 0.250 0.0324 0.224 2.459 0.900 4.33E-005
    41306 0.500 0.312 0.250 0.0765 0.149 3.124 0.859 4.75E-005
    41406 0.500 0.281 0.250 0.0620 0.172 2.951 0.870 5.07E-005
    41605 0.625 0.350 0.185 0.0962 0.158 3.672 0.833 6.93E-005
    42206 0.870 0.540 0.250 0.2290 0.261 5.419 1.125 3.33E-004
    42207 0.870 0.540 0.312 0.2290 0.326 5.419 1.249 4.68E-004
    42507 1.000 0.610 0.312 0.2922 0.385 6.170 1.346 7.73E-004
    42212 0.870 0.540 0.500 0.2290 0.522 5.419 1.625 9.23E-004
    42908 1.142 0.748 0.295 0.4394 0.369 7.320 1.397 1.03E-003
    43806 1.500 0.750 0.250 0.4418 0.581 8.300 1.644 2.18E-003
    42915 1.142 0.748 0.600 0.4394 0.750 7.320 2.007 2.96E-003
    43813 1.500 0.750 0.500 0.4418 1.110 8.300 2.144 6.10E-003
    43615 1.417 0.905 0.590 0.6433 0.977 9.200 2.190 6.74E-003
    43825 1.500 0.750 1.000 0.4418 2.220 8.300 3.144 1.66E-002
    44920 1.932 1.252 0.625 1.2311 1.220 12.310 2.620 1.68E-002
    44916 1.932 1.332 0.625 1.3935 1.200 12.730 2.594 1.86E-002
    44925 1.932 1.252 0.750 1.2311 1.490 12.310 2.870 2.29E-002
    46013 2.400 1.645 0.500 2.1253 1.200 15.800 2.672 2.75E-002
    46113 2.400 1.400 0.500 1.5394 1.600 14.400 2.756 3.44E-002
    44932 1.932 1.332 1.250 1.3935 2.400 12.730 3.844 5.02E-002
    47213 2.900 1.800 0.500 2.5447 1.760 18.370 3.084 6.14E-002
    47313 2.900 1.530 0.500 1.8385 2.140 16.520 3.182 6.36E-002
    48613 3.375 2.187 0.500 3.7565 1.890 21.510 3.394 9.50E-002

    4 Design Examples

    4.1 MPP Toroid Example

    Design Inductor on MPP Core

    DCR  <  .020 Ohms, L = near 25E-6 Henry at 6.6 Amps.  The DC resistance and size is critical.  The inductance is  desired  to  swing  from some  higher  value  to  approximately  25E-6  Henries.   A MPP is desired, so set UrH = 2800 gauss for a 50% swing in inductance.

    Step 1. Calculate inches ^5 and pick core

    (LI)^2 / DCR = (25E-6*6.6)^2 / .020 = 1.36E-6

    Toroids should avoid fill high fill factors, it leads to issues with repeatability both electrically and mechanically.


               Ac^2Aw        pd^2 * (LI)^2 / R                      0.013 * 1.36E-6
               ----------- =   ---------------------------------- = --------------------------- = 1.1E-4
                 MTL           K1  * (Ur*H)^2 * Ap/Aw          50E-12 *2800^2 *.4

               0.68 OD toroid in^5 = 8.97E-5

               0.83 OD toroid in^5 = 1.81E-4

               The 0.68 OD toroid is chosen because of size considerations.

                0.680 OD, 0.375 ID, 0.280 HT

                 AC = .0298 sq in, AW = .11 sq in, lm = 1.618 in , MTL = 1.09 in

     Step 2. Determine N and Ur

             (17) N    = K2 * (LI) / ({Ur H} {Ac})
                   N    = 15.5E6 * 25E-6 * 6.6/(2800 * .0298) = 30 turns

            You could mix the units you are using if you want.
            Just use the value of K2 for Ac in cm^2 if Ac is in cm^2.
            You'll get the same answer.

    ...

                                     L lm                    25E-6 * 1.61
             (19) Ur   = K3 * ----------- = 31.3E6 ------------------ = 46 u at bias
                                     Ac N^2                .030*30^2

                    oersteds = 0.4*PI*N*I/lm (cm) = 0.4*PI*30*6.6/4.11 = 60 oe

            Because the design is for 50% left, actual permeability of the core should be:

                    46 / (50%) =  92 u the nearest perm is 125u.

             Wire diameter = (0.4 * 0.11/30)].5] = .038 diameter = #19 AWG

             #19 winding diameter = (.043 max + .039 min )/2 = .0414 inches average

          Maximum turns for 1 layer in ID = PI*(.375 - .0414)/.0414 = 25.3

          Single layer windings usually give the "shortest" part height, but have a poor Ap/Aw factor.

    Step 3. Calculate final parameters using 24 turns #19 on 55120-A2 core.

             24T * 1.09"/turn * 1ft/12" * 8.046 Ohms/1000 ft = .0175 Ohms Nom

             72mH is what the data sheet gives for 1000 turns.

             L at zero DC = 72mH * ( 24/1000 )^2 = 41.5 uH
             The core tolerance is + 8%.
             The tolerance that can be tolerated is (- 10%  + 20%)

    The inductance will be lower than desired, so the ripple current will be higher. So we will test the inductance with DC at 7.0 Amps. Some MPP's inductances will increase at high ripple currents. We are not going to take advantage of that and we don't have the test equipment to run that high power of an inductance test.

    oe = .4 PI * 24 * 7/4.11 = 51.4 oersteds = 46% left

    L min at 7A = 41.5 uH * 90% * 46% = 17 uH min

    At this point we call the "customer" and ask if this is acceptable, and the response is "Yes."

    Step 4. Build and Measure it.

      Results:
                     L at 10kHz zero DC = 43.5 uH  Q = 84  SRF  = 17.5 MHz
                     L at 10kHz 7.0A DC = 22.8 uH
                     DCR  = .016 Ohms at 1.5 inch leads

    .....

    4.2 C-Core Design Example

    Design 1 C-Core inductor 1.3mH at 15A DC

             An inductor is wanted with the following design parameters:

             L = 1.3mH @ 15 Amps DC, 5V 4000 Hertz square wave with a .75 volts
             DC voltage drop. The customer states these numbers are flexible.

             A.   .75 volts/15 Amps = 50 milliOhms DCR

             B.   AC current = 5 V / ( 2 PI 4000 * 1.3E-3) = .153 Amps

                  Design Current =  15 + .153 = 15 Amps

                  L^2I^2     (1.3E-3)^2 * 15^2
             C.  ---------  = ------------------------ = 7.6E-3
                   DCR              .050

    Step 1. Calculate inches ^5 and pick core

    Since the AC component is small, design at 12 Kgauss (Ur H) on a 4 mil  C-core. Assume  a  0.5  (50%)  window utilization because of mounting constraints and a .011 wire pd^2. Then  the "Inches to the Fifth" is:

    $ {A_c^2 A_w} /{MTL} = {{pd^2 * {(LI)^2} / {DCR}} /{ K1 * (U_r * H)^2 * {A_p / A_w}} $

      Ac^2Aw
      ------------ = 7.6E-3 (from "C" above) / ( 49.95E-12 * 12E3^2 * 0.5 / .011 )
        MTL

                Ac^2Aw
           D.  ------------ = 23.3E-3 "Inches to the Fifth"
                   MTL
    An Arnold Engineering AH-177 C-core has the following dimensions:

           D = .625 E = .500  F = .500 G = 1.5626 Stacking factor of .9
                 A 20 mil thick bobbin will be used (BT = .020)

                 Ac = .9 * D * E =  .9  * .625 * .500          = .281 Square inches
                 Aw = G * F      = 1.56 * .625                  = .781 Square inches
                 MTL ~ 2*(E + D) + (Ap/Aw)*PI*F + 16BT = 3.34 inches
                 lm = 2F + 2G + 2.9E                             = 5.57 inches

    ...

                      Ac^2Aw
                      ------------ = 18.5E-3
                       MTL

    Note:  Aw  is  already corrected for the window fill factor in the "Inches to the fifth" equation, but the  MTL  is  not.   Since  we originally  designed for a 50% fill factor; if we design for a 70% fill factor the required "Inches to the fifth" becomes:

    23.3E-3 * .5/.7 = 16.6E-3 Thus the core will work.

    Step 2. Determine N and Ur

                   N = 15.5E6 * LI / (Ur H) / Ac

             E.   N = 15.5E6 * 1.3E-3 * 15 / ( 12E3 * .316) = 89.6 = 90 turns

                  Wire area       = (Ap/Aw) * Aw/N
                  Wire Area      = .7 * .781/90 = 6.07E-3 square inches
                  Wire Diameter = .077 inches
                     #14 is .0701 inches flat to flat 
                     #13 is .0780 inches flat to flat

    ---> #14 square wire is available, #13 has a long lead time.

    For producibility, I usually allow a diameter that is half a wire gauge bigger than the actual gauge for easier production; therefore, the diameter of #14 is .074 inches.

              The winding length is G - 2*margin - tolerances:

                     1.562 - 2*.125 -.016 = 1.296 inches.

              The actual winding height is F - bobbin thickness - tolerances

                     0.500 - .020 - .016 = .464 inches.

             The number of layers than can be wound are

                  0.464 / ( .074 + .010 ) = 5.5 = 5 layers

             The wires per layer are

                  1.296/.074 = 17.5 = 17  wires per layer

    ...

    Because of the stiffness of the wire, assume that only 16 turns (17 wires) can be used for a total  of 80 turns.  

          Ur = 31.6E6 * L * lm /( Ac N^2 )

          Ur = 31.6E6 * 1.3E-3 * 5.57" / (.281 * 80^2 )

          Ur = 127

          leff ~ lm / Ur

          leff ~ 5.57/127 = .04378 inches = 43.8 mils = 22 mils each leg

                           1             2*lg           2 S
      F = L'/   = 1 + ----------  * ---------- * Ln -----
                          Kstack      Ac^0.5          lg

          lg = .022, Ac^0.5  = .312^0.5  = .559, S = G = 1.56

                          1       2 * .022         2*1.56
          L'/L  = 1 + ----- * ------------- * ln ----------   = 1.43
                          0.9       .559            .022

      The corrected gap length is now 1.43*.022 = .03

    Using the second approximation (not as accurate):

          F    = (.625 + .044)*(.500 + .044)/(.281) = 1.295

    The corrected gap length method 2 is 1.293*.022 = .028 (Use method 1)

      Use 30 mil air gap in each leg (60 mils total)

    Step 3. Calculate final parameters

    The actual fill factor is (.07*.07*80)/.781 => 50%.

                  #14 Square ~ 2.1 milliOhms/ft

                  DCR = MTL * N * Ohms/Inch

                  MTL = 2(D + E) + 16BT + pi*Wire buildup
                  MTL = 2(.625 + .500) + 16*(.020 + .016) + PI*(5*{.074+.010})
                           = 4.145"

                  DCR = 4.15" * 80T * 1 ft/12" * 2.1mOhm/ft = 58 milliOhms

    ...

                  Power loss = I^2R = 15^2*.058 = 13.1 Watts Copper
                  B gauss = 5 * 1E8/(4.00 * 80 * .281 * 4000) = 1,390 gauss

                  From core loss curves:   
                            loss = 4 Watts/pound

                            4 W/lb * .434 lb = 1.736 watts in the core

                  Total loss = 14.8 watts.

                  In air surface temperature rise is approximately

                       80 * (square inch surface area)^-0.7 * (watts loss)^+0.85

                       80 * (12.8 sq in)^-0.7 * (14.8)^+0.85 = 132 C RISE

    Convection cooling is not adequate.  The  part  will  have  to  be mounted on a cold plate.  The project states the previous part had 76  watts  loss and was running only slightly too hot when mounted on the cold plate so 15 watts is OK.

     Step 4. Build and Measure it.
            
        Results:
            
             The  air  gap  needed for 1.31 mH was 32 mils in each leg (64 mils total).
             The DC resistance with 6 inch leads was 58 milliOhm.
             The inductance at 15 Amps DC 400 Hz was 1.22 mH  (8%  drop).  
             The SRF tested at 1 MHz.  Below is the inductance curve for the core.

             DC current      10     13     14     15     18     19       19.7    20     Amps
             Inductance  1.3   1.33   1.29  1.26   1.22   0.87   0.745    0.65   0.619  mH .

    ...


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