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# Self-Resonance in Inductors

Jared Dwarshuis & Lawrence Morris

## Introduction

In this paper we will be describing standing wave self resonance in an inductor using ideal inductor arguments. We will develop a generalized equation to describe the formation of any number of nodes within an inductor. We will show that under self resonance the frequency of each quarter wave region will be in agreement with the frequency of the inductor as a whole. We will also show how the velocity factor (which others have measured) is describing a relativistic observation.

## Part 1 - B field and inductance

We will start with the classic B field of an inductor with the reminder that this equation assumes that we will be measuring the B field in the middle of an infinitely long solenoid where the lines of magnetic force entirely line up with the long axis of the solenoid. With a short solenoid, lines of magnetic force are not entirely aligned along the axis of the solenoid and we do not get ideal results.

### B-Field

B = μ0 i n (where N = nh)

Then

B = (μ0 i N)/ h

We will multiply the numerator and denominator by (2 π r)

Then

B = (μ0 i (2 π r N)) / (2 π r h)

This expression tells us that the B field of a solenoid can be written as:

B = (μ0 i (wire length))/ (surface area of the solenoid).

This formulation for a solenoid clearly shows the geometric concerns. For any solenoid you will always have a given length of wire which covers a particular surface area.

### Inductance

Now we will use the previous formulation for the B field of a solenoid and put it in the inductance formulation.

Again we need to remind ourselves that the ideal equation of inductance assumes that we have an infinitely long solenoid where all of the lines of magnetic force align with the long axis of the solenoid. With a short solenoid towards the ends, your lines of magnetic force are not aligned with the long axis of the solenoid and you do not get ideal results. Interestingly, approximation equations for inductance will typically converge on the ideal for long solenoids.

L = [N][B][A] / [i]

Substituting our previous expression for B, and the cross-sectional area of the solenoid, we find:

L = ( [N] [μ0 i (2π r N)) / (2π r H)] [π (r)2] ) / [i]

Notice that our current term has canceled out at this point. We are left with only geometric variables which are not time varying.

Multiplying numerator and denominator by 4π then reducing we observe that the wire length can be written as a square:

L = μ0 (2π r N)2 / (4π H)

This formulation emphasizes the geometric aspects of inductance.

We also know by definition that:

μ0 = 1 / (C2 ε0)

So:

L = ((2π r N / C)2) / (4π ε0 H)

This formulation for inductance works nicely and is in full agreement with the classic formulation. This formulation is the ratio of the square of a period (determined by the traversal of the wire length at the speed of light) divided by a capacitance (determined by the height of the solenoid). A circuit traversal description of inductance shows that we will run into problems if we attempt to outflank the speed of light with an RF signal.

### Properties of Inductance

We think that there is confusion about the properties of inductance. Some think that frequency or current and voltage distributions can alter the properties of inductance; but that’s not quite right. Inductance in an ideal inductor is always a constant. With standing waves we can partition an inductor into regions defined by voltage maxima and minima where the inductance of an individual region is less than the inductance of the entire inductor. But when we sum these regions in series, the overall inductance will remain a constant.

Inductance is not ruled by current. We can place a small current or a large current through an inductor and it will have exactly the same inductance. Having current bunched up in a particular place at a given instant does not change the inductance.

## Part 2 - Single quarter wave region

We will now examine a single quarter wave region of an ideal inductor.

### Frequency of Travelling E-M Waves

The equation for describing the frequency of traveling electromagnetic waves is:

C / wave length = frequency

### Frequency of Standing E-M Waves

For standing waves however, we are considering the situation where wire length = wave length / 4.

With standing waves:

Frequency = C / (4 (2π r N))

### Frequency of an L-C System

For an LC system: Frequency = 1 / (2π sqrt(LC))

Substituting L from above into the equation for LC frequency we get:

Frequency = 1/ (2π sqrt (((2π r N/ C)2 / (4π ε0 H)) capacitance))

(The capacitance in the above equation is the self capacitance of the inductor at resonance.)

### Self-resonance

Now we have two expressions for the same frequency; we will set them to be equal.

[C / (4 (2π r N))] = [C / (2π r N)] [1 / (2π sqrt(capacitance/ (4π ε0 H)))]

### Self Capacitance

At resonance in the above equation, the self capacitance must satisfy the equation such that:

[1/ (2π sqrt(capacitance/ (4π ε0 H)))] = 1/4.

We can now calculate the self capacitance of our quarter wave inductor at resonance as:

Capacitance = 16 e H / π

### Self-resonant Frequency

The frequency equation reduces to:

[C / (4 (2π r N))] = [C / (2π r N)] [1/4].

The left side of the equation describes frequency as 1 / (4 times the traversal time of the wire length). The right side of the equation describes frequency as [1 / (the traversal time of the wire length)] multiplied by a unit-less correction factor. The parts of the equation that have units of frequency are related by a factor of 4.

### A Relativistic Interpretation

We consider this factor of 4 to be a relativistic gamma factor because we are making an observation of the relativistic interference pattern of a standing electromagnetic wave.

By applying the Lorentz transformation (with Gamma = 4) we find that:

4 = 1/ sqrt(1- ((v/C)2)).

This gives us an observed wave velocity of sqrt(15/16)C. This velocity is the velocity that others have been measuring.

## Part 3 - Multiple quarter wave regions

Now we will describe multiple node formation within an inductor at resonance. Regardless of the number of voltage nodes we put in our inductor we must adhere to the same velocity factor of:

velocity = sqrt(15/16)C

First we will demonstrate that our accounting procedures must be based on quarter wave regions. We know that for an inductor; voltage from point A to point B = L di/dt. Consider points A to B as being between two current nodes. Now we have a problem since there is no difference in voltage between current nodes, but clearly there must be inductance within these regions. We must consider quarter wave regions since our voltage A to B must be between a minimum and a maximum for our regional inductance to have meaning.

We will need to add features to the frequency equation for a quarter wave region to account for node formation.

Frequency = (n/2) C/(2π r N) = 1/ (2π sqrt( ((2π r N) / (2 n C))2 (2n/(4π ε0 H)) (self capacitance) ))

Where n = 1/2, 2/2, 3/2, 4/2 ... (the number of 1/2 waves)

The features added allow us to divide the wire length (2π r N) and solenoid height (H) when we go to the higher harmonics. The right side of the equation will only tally the inductance of a single quarter wave region found within the inductor.

## A Concrete Example

Let's pick an inductor with easy dimensions to work with. Let the wire length = 1000 meters and let the height (H) of the solenoid = 1 meter.

### A Quarter Wave

We will start with the quarter wave region where n = 1/2 (the fundamental quarter wave frequency).

Then:

Frequency = C/ 4 (2πr N) = 1/ (2π sqrt (((1000/C)2 / (4π ε0 )) self capacitance))

From this we can calculate the self capacitance as:

capacitance = 16 ε0 / π

and the resonant frequency as:

frequency = 75,000 hz

### A Half Wave

Now let's see what happens when we take the very same inductor and double the frequency to get a half wave, where: n = 1

Frequency = (n/2) C/(2π r N) = 1/ [ 2π sqrt( ((2π r N) / (2nC))2 2n/(4π ε0 H) (self capacitance)) ]

Substituting in: n = 1 (Because it is a half wave)

F = C / [ 2 (2π r N) ] = 1/ [ 2π sqrt( (2π r N /(2C))2 2/4π ε0 H (self capacitance)) ]

Then: Frequency = C/ 2 (2πr N) = 1/ [ 2π sqrt( ((500/C)2 2 / (4π ε0) ) (self capacitance)) ]

From this we can calculate the self capacitance of the region as:

Cregion = 8 ε0 / π

Our resonant frequency is now:

Frequency = 150,000 hz

When we examine the LC frequency of the region we have half the inductance of the previous quarter wave example (which we will call L*/2) and half the capacitance (which we will call C*/2). We can simplify this by showing the frequency of the region as:

Fregion = 1/ [ 2π sqrt( (L*/2) (C*/2) ) ]

or twice the frequency of the quarter wave. Now we can examine the two regions in series. The rules say that inductors in series add. So

Ltotal = L*/2 + L*/2

As we can see our total inductance has not changed it is exactly the same as our first example with just a quarter wave. Now let's examine the capacitance in series which add by the reciprocal of the sum of the reciprocals.

Ctotal = 1/ [ 1/(C*/2) + 1/(C*/2) ]

This gives us a series capacitance that now totals C*/4. When we examine the LC frequency of the entire inductor we get:

Ftotal = 1/ [ 2π sqrt( (L*)( C*/4) ) ]

or twice the frequency of our original quarter wave.

So now we see that for self resonance it is a requirement that both the quarter wave region and the entire structure, resonate at the same frequency. This pattern is true of all of the harmonics. We will show a last example

### A Full Wave

For a full wave where n = 2 the outcomes follow the exact same pattern where the frequency is now 4 times that of the original quarter wave and the inductance of the quarter wave region is one fourth of the inductance of the entire inductor. The capacitance of the region is one fourth of C*. The LC frequency for the quarter wave region is:

Fregion = 1/ [ 2π sqrt( (L*/4) (C*/4) ) ]

or four times the frequency of our original quarter wave inductor. Examination of the regions in series shows that our total inductance is

Ltotal = L*/4 + L*/4 + L*/4 + L*/4.

Our capacitance in series now sums to C*/16 and the LC frequency for the entire structure is:

Ftotal = 1/ [ 2π sqrt( (L*) (C*/16) ) ]

or four times the frequency of the original quarter wave.