Air Gap - Graphical Solution for Optimizing

when you run DC into a transformer winding, you polarize the core, the molecules get rotated a bit into a biased position, which makes it harder for the AC to swing these same molecules as they are already under tension,

if too many volt-amps are applied to the core, it will saturate, the molecules will be bent over so far that AC will not even move them,

to limit the force on these molecules due to DC current, we can increase the resistance of the core to this polarizing current,

lets pick an example to get into this from the git go,

if we have an inductor 10,000 turns wound on a square stack of 4%  Grain Oriented 75 EI steel core of 0.014 mil lams, and we want to run 10 ma DC through it for a plate choke, how big should the gap be?

Step 1) Calculate the m.m.f, or magneto-motive force, due to the turns and current-

F= 1.2566 N I-dc, where N = turns and F is in Gilberts, NI is also called Ampere-Turns,

for our inductor F = 1.2566 x (10,000 Turns) x (0.010 Amps-DC)

so F = 125.7 Gilberts.

Step 2) Find the Magnetic Path Length, MPL, or more commonly notated l, for the core in use:

75 EI has an MPL of 11.43 cm. this is the average length of one lap around the core.

Step 3) Find H-dc,

if we divide the total magnetic force, mmf, by the magnetic path length,  we get a new number that represents H-dc, or a magnetic force per centimeter in the core.

remember, this is DC force, not AC. So we need to allow some room in the core for that AC force that is our audio signal after we figure out how to drop down the DC force.

so H-dc = F/l, (using l for MPL), then H=125.7 Gilberts divided by the EI 75 spec,

H-dc = 125.7/11.43 cm  = 11 Gilberts/cm. we can call Gilberts/cm Oersteds.

so H-dc=10.8 Oersteds

this is a measurement of the force per unit of length inside the core. you can think of it as voltage across a resistor, if you have 100 volts across a 100 K resistor, you have 1 volt per k ohm, so Oersteds are the voltage across the core per cm. length.

an Oersted is defined as  80 turns of wire per meter at  1 amp current. for a magnetic feild strength of 1 Oersted. 80 Amp/Turns/Meter, or 0.8 Amp/Turns per cm.

you will see this H-dc number used in graphs that show the difference that DC flux has on the permeability of a core, the higher the H-dc value, the Lower the perm,

so what we are doing with the air gap is adding more resistance to the core so the DC current does not push as much flux thru it,

there are many formulas for computing the air gap, but there are so many variables in the equation such as fringing and flux levels that the answer for how big to make the air gap is best solved graphically. we have everything we need to come up with an optimum gap for our 10,000 turn choke with 10 ma DC on it,

we will plot H-dc, our magnetic force field due to amp-turns, along the X axis,

we will plot the flux density due to DC along the Y axis.

by picking 16 kilo-gauss as our saturation level that we want to avoid, we can draw a graph that will relate this flux to our air gap,

we want to draw a B-H curve, only for DC.

this is the first part of a BH loop, where you see the core being energized for the first time, it is called the magnetizing curve, not the BH curve, the normal BH curve is generated by AC current, the Magnetizing curve is generated by DC current.

we want to pick an air gap which will maximize the inductance of the core.

maximum inductance occurs at maximum permeability.

maximum permeability occurs at the "knee" of out magnetizing curve. so we need to find the gap which will put us on the knee of the mag curve.

all you do is draw a line from the H-dc axis to the B-dc axis which passes through the knee. then you read the flux off the B-dc axis and compute your gap frpm there.

Step 4) Find the gap length -

for a gapped inductor, B-dc  = F/l-gap.

we can isolate the l-gap term with algebra so that it is on the left side of the equation:

l-gap = F/B-dc

l-gap = 125.7 Gilberts /12,000 gauss = 0.0104 cm = 0.0104/2.54 = 0.004 inches,

or in transformer lingo,

l-gap-optimum = 4 mil,
(Kraft Paper used for gaps is sold in 1,2,3,5,10,15,20 and 30 mil weights)

you can see that B-dc will go up as the gap is made smaller, as F stays the same no matter what the gap,
a more stable inductor will be one with a larger gap and thus a smaller inductance value do to a decrease in perm,
so if we drop the loadline down with a bigger gap, the relative permeability drops down to an effective permeability B-dc will go way down, and there will be very little H-dc across the core,
this stabilizes the perm vs frequency and will give you a flatter Inductance vs. Hertz graph.  there is a ratio called the Stability Factor, take your relative perm and divide it by the gapped perm and you get a number like 8000/800=10 for 4% Si,
if you drop your perm by a factor of 10, then you stabilize variations of perm due to flux level and temperature by a factor of 10, so if you had a 10% drift in inductance, you now have 1% drift due to the stabilizing factor of 10.

our graph shows that the amp turns and core length set up the H-dc which sets up the B-dc,
by setting a limit to B-dc and drawing in a mag curve, we can graphically find the gap length.

see that the bigger the gap, the lower the slope of the air load line.
since this is a series circuit, gauss will be the same everywhere, inside the core and inside the air gap. so the left side of the graph with the flux readings are for flux inside the core and gap, H-dc, the magnetic force per centimeter, will be divided by the air gap and core, you can see from the graph that most of the H-dc is across the air gap, and not across the core, H-fe,

think of these H  values as voltage drops across a voltage divider. one resistor would be the air gap, and another resistor would be the core, both of which have a resistance to magnetic flux, but instead of calling it resistance, we call it reluctance, as the core is reluctant to be magnetized,

FYI: Total Flux = F-Gilberts/R-Reluctance, and B-dc=Total Flux/A, where A is the core cross section area in cm.

also see that the optimum gap is Not at a point where H-fe and H-air are equal, watch out for this same presentation in RDH4 which might give you that false impression,

it looks like a 4 mil gap will put us at point P. our operating point for max inductance without saturation, looks like 12 kg is our DC  gauss, leaves 4 k-gauss for AC, maybe 5 mils will allow the core to handle more than 4 kg AC flux which is super-imposed on the DC biased core,

see from the graph that if the air gap could approach a value of 1/infinity, or get close to zero, that B-dc would be close to infinity, lim f(x)= 1/x as x>0= infinity,

so what gives?

the core is not perfect. it has an air gap even when stacked 1 x 1 instead of with a butt joint and Kraft Paper air gap. so there is a limit also to the length of the air gap that is built into the core, we call this leakage flux.

here is the graph, if you had a nickel core, you would just re-label the Y axis for maybe 5 kio-gauss for 80% Ni and 12 k-gauss for 50/50% Ni for B-sat, so if you keep the same mmf, then your gap must increase to satisfy l-gap=F/B-sat,

here is a chart showing how inductance/perm can go up for certain gaps at certain Amp-Turns for silicon steel cores:

and a repeat post of the perm vs H-dc graph,

so the more DC bias you have on a core, the more stable the perm with level,

this translates into a flatter frequency response as inductance will not vary as much with frequency,

so you want the choke to have a linear frequency response like a resistor, and you want to take advantage of the lower voltage drop due to DCR, having the dc bias helps to make this happen by making the inductor more linear,

since the knee of the magnetizing curve is gap-sensitive, it is normal to error on the larger side when picking gap material, you want to be down lower on the  more linear part of the curve, as this will stabilize the inductor.

as you increase the air gap, you tilt the B-H curve over to a more horizontal position, this means that the AC current from the incoming signal will not traverse the y axis as much which means less flux movement, as it is traveling along the B-H curve almost parallel to the x axis,

a useful formula for adjusting the permeability of a certain core after an air gap has been added to the magnetic path is given by the formula

u-effective=u-relative/(1+(U-relative x l-g/MPL)
.
l-g/MPL is also called the gap ratio and can be denoted by b

so if you have 50/50 Ni core stamped on a 625 EI , with a perm of 20,000, and you gap it with a 0.010 gap, what is the new effective perm?

u-e=20,000/(1+(20,000 * (0.0254 cm/9.53 cm)

lets get our gap ratio first to simplify this, b= 0.0254/9.53=0.00267

so u-e=20,000/(1+(20,000 * 0.00267) > u-e=20,000/(1+53.4) = 20,000/54.4= 368

so with a 0.010 inch gap, we go from a perm of 20,000 to 368

the same gap used on the same core made from 4% Si grain orient would be

u-e=8,000/(1+(8,000*b) = 8,000/(!+(8,000 * 0.00267) = 8,000/(1+21.36) = 358.

so by using Nickel, you only get to 268 instead of 358 with the Si, this is why you never see an air gap used on a nickel cores, it is a waste of money in the form of unneeded Nickel lams,


just a note on graph construction,

since u(perm)=B/H, then H=B/u,

so Silicon Steel, with a B-sat of 16,000 kg and a perm of 8,000 will need a field strength of

16,000/8,000 = 2 Oersted = H to reach saturation,

but Supermally with a perm of 80,000 and a b-sat of 5 kg, will only need an H field of

B/u = 5,000/80,000 = 0.0625 Oerteds = H

so that is a difference of 2/0.0625=32.

so to saturate Supermalloy, you only need 1/32 the Amp Turns as you would with 29ga.M6.

this is why you do not want to run too much DC into a high nickel core, it gets polarized very easily.

so adjust your x axis accordingly. here is a chart comparing the various alloys, Magnesil =  4% Si, note the air gap built into the ferrite cores(tilted B-H)>

No more questions on Air gap, Please..................

Perhaps another way to look at it:

Find/guess your AT/in. Divide by 20,000. That's a good gap-ratio for high inductance with DC.

Logic: from CJ's (Langford-Smith's) table, find the maximum incremential permiability for each AT/in column, look over at the gap-ratio for that max-perm, look for a trend. High AT/in wants high gap-ratio (makes sense). Now find a trend in that trend. Optimum (AT/in) / (GR) tends to be 20,000.

This persists even to Zero gap-ratio, _if_ we take the wild guess that EI lams can never be dead-nuts ZERO gap; that a big interleave is still a small gap OTOO 0.000,1. While this can be argued, at the bench any very-very-small gap may as well be interleaved (although in the marginal case CJ knows several partial interleaves which approximate a perfect-flat gap with impossibly thin paper).

I suspect the "20,000" fudge factor varies with different types of iron; and since the data here is 70 years old we could re-test on modern stuff.

here is a spreadsheet on perm vs air gap for various lams,

this sheet does not take into account the natural air gap built into the core,

so it looks like most audio iron will be using a gap in the 1 to 5 mil range,

here we are simply dividing the magnetic path length of the lam in cm. by the gap in cm and finding the resultant perm, the smaller the lam, the bigger hit the perm takes due to the smaller path,
steel perm lams on top u = 8,000,, Supermalloy,  u=80,000 on bottom,>