Insulation materials.

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For insulation materials used in crystal receivers the following electrical properties are important: insulation resistance, dielectric constant and dissipation factor.

Insulation resistance. 

This is the resistance between two conductors having a insulation material in between.
In most cases the insulation resistance is high enough for use in crystal receivers.

Dielectric constant.

Two conductors with insulation in between form a capacitor.
If the conductors have the shape of parallel plates then we can calculate the capacitance as follows:

C=0.0885 εr A / d

C= capacitance in pF (picofarad)
εr=dielectric constant of the insulator    (ε = the Greek letter Epsilon)
A= area of the plates in square cm.
d= thickness of the insulation in cm

The value of εr indicates the increase of capacitance compared to air insulation.

Dissipation factor.

In the ideal case a capacitor has no loss, if we send a AC current through the capacitor there is no loss of energy.
In practice there will always be losses, if we send a current through the capacitor a part of the energy will get lost by heating up the capacitor.

These losses are caused by:

a- The resistance of the conductors (plates), these losses are left out of consideration here.
b- The dielectric losses in the insulation.
One insulator gives more dielectric losses then the other, this is indicated by the dissipation factor DF.
The dissipation factor "DF" is sometimes also called: "tangens delta".

The lower the value of DF, the better the quality of the insulator.

In the following table some values of εr and DF for different insulation materials.

A capacitor wil have at certain frequency a impedance of:

Zc=1/(2.pi.f.C)

Zc= impedance of the capacitor (Ohm)
pi=3,14
f= frequency (Hertz)
C= capacitance of the capacitor (farad).

Because of the losses caused by the insulator, it looks like there is a resistor connected in series with the capacitor.
This is the series resistance (Rs) of the capacitor.

If the insulation of the capacitor has a certain DF value, the series resistance wil have a value of:

Rs=Zc . DF

If we connect this capacitor to a coil, then we have a LC circuit.
If the coil has no loss, the maximum Q of the LC circuit will be:

Q= Zc / Rs = Zc / (Zc . DF) = 1 / DF

So, if we use a capacitor with nylon insulation, the Q of the LC circuit will have a maximum value of:
Q= 1 / 0.026 = 38.
This is a very low value.

So, a LC circuit with airspaced tunercapacitor has a infinite Q???
No, because there are more losses, like the resistance of the coilwire, the resistance of the capacitor plates, the dielectric losses in the coil, etc.

It is also not possible to use a tunercapacitor with only air insulation.
The rotor (turnable part) and stator (non turnable part) must also be connected to each other by insulation material.
If for instance 4/5th of the capacitance is caused by air insulation, and 1/5th by nylon insulation blocks, then the Q will increase by a factor 5 compared to a full nylon insulation.
So the Q can then be 5x38=190.

The higher the frequency of the circuit, the higher the percentage of capacitance caused by the insulationblocks, and the lower the Q.
Instead of nylon it is better to use a insulator with a lower DF value, and preferable also a low εr value.

For a high Q the following is also important:

Use the insulation material only on places where the distance between rotor and stator is high, this helps to reduce the capacitance and losses caused by the insulator.
Don't use more insulation material then necessary.

Dielectric losses in coils.

When a coil is wounded on a coilformer, there will be dielectric losses in the former.
Also in the insulation of the wires there will be dielectric losses, especially when windings are touching each other.

For the coilformer it is important to use a material with low DF value, and to use as few as possible of this material between the windings so the capacitance and losses are kept to a minimum.

Other dielectric losses.

Also in the wires in the receiver dielectric losses can occur.
For instance if a wire carrying radio frequencies is placed very close to a other conductor.
The two conductors form a capacitor, with the wire insulation as insulator, in this insulation losses can occur.

We can reduce these losses by placing the wire at some distance of the other conductor (some cm. or more).

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