Coax Cable Tester COMPLETED COAX CABLE TESTER

This is a simple circuit that can be used to test all those coaxial cables that we have or make. It only uses a few components and will test if the cable is OK, has any shorts between screen and inner and if both screen and inner are continuous. The circuit is shown below along with components required. Note: switch shown in “SHORT TEST” position.

Designation Value
R1, R2 680R, 0.25W
D1, D2 LED, Green
D3 LED, Red
SW1 DPDT Switch
CONN1, CONN2 To suit

CONSTRUCTION NOTES:

1. Item MUST be built in plastic box. (See NOTE below)
2. Other connectors can be placed in parallel with CONN1 and CONN2 to give more universal DC testing of cables.
3. LEDs can be mounted in holders to ease build.
4. Designed to work from PP3 battery or other 9V source. Resistors R1 and R2 can be re-calculated for other supply voltages.
5. Additional switch can be placed in +9V line to provide an ON/OFF function.

NOTE: Box must be of an insulating material otherwise the connector screens are shorted together and braid test will give a false reading.

METHOD OF OPERATION:

1. Plug cable to be tested between CONN1 and CONN2.
2. With switch in position shown it is in “SHORT TEST”. If there is a short the red LED D3 will be illuminated showing a cable fault.
3. Switch to other position. This is “CONTINUITY TEST”. If centre core is intact then D2 will illuminate green, if screening is intact then D1 will illuminate green at the same time. If only one LED illuminates then there is a cable fault denoted by which LED is not illuminated. INTERNAL CONSTRUCTION OF COAX CABLE TESTER

Basic AC Theory

During studies for the old RAE or newer Full Licence the concepts or resistance and reactance have been taught and the following equations will have been given:

\text{Inductive reactance:}\quad X_L = 2\pi f L ~\Omega

\text{Capacitive reactance:}\quad X_C = \frac{1}{2 \pi f C} ~\Omega

(Note: at least X is in Ohms and we have a chance of combining it with resistance R, L and C themselves are not in Ohms). Figure 1: Impedance in a Series Circuit

You will also have been taught that inductance and capacitance introduce a phase shift in the circuit between the applied voltage and the current flowing. A circuit has impedance rather than resistance when inductance and capacitance are also involved in a circuit carrying an alternating current.

Again, referring to what has been taught, impedance can be represented by a triangle such as shown in Fig. 1 for a series circuit. It is not correct to write that Z = R + X_L or Z = R + X_C as this has not taken into account the phase shift introduced by the reactive element. Rather, you must use the formulae given in Fig 1.

It would, however, be very convenient it there was a method whereby R and X could be combined in some form without the use of square roots and trigonometric functions. It would allow a consistent set of units — Ohms — instead of dealing with \text{pF}, \text{μF}, \text{μH}, \text{mH} and so forth and also be convenient if reactances could just be added and subtracted. This would help us in, for example, antenna calculations, where we need to get a series reactance that will make an antenna look purely resistive. The next section explain a method for attaining this with some examples.

The “j” Operator

There is a mathematical tool which uses the j operator (it is often called i in mathematical books but engineers use j). This allows us to write Z=R+jX_L or Z= R-jX_C — note the minus sign for capacitive reactance, it is important. The R and the j terms cannot be further simplified, i.e. if Z = 65 + j40 this is its simplest form. The j term implies a quantity that is at 90\degree (or quadrature) to the resistive term.

Two practical examples – see Fig, 2. Using the formulae for reactance given earlier and the frequencies quoted in the examples, then the series circuits can be specified as Z = 220 + j 628.3 and Z=100-j15.9 respectively (note these figures have been rounded off). This gives phase angles of approximately 72\degree (lagging) and 9\degree (leading) respectively. The minus sign indicates the reactance is capacitive and the plus sign denotes inductive. Figure 2: Two Practical Examples

If the series circuit as shown in Fig. 3 is used, the combined impedance is given by:

Z = R_1 + R_2 + j X_L – j X_C

The non j and the j terms can he collected together which gives:

Z = (R_1 + R_2) + j (X_L – X_C) Figure 3: Combined Circuits

Thus series resistance can be added together (something that should be known) as also can series reactance — but taking into account the sign. The reactances can only be added together provided that they are quoted at the same frequency. Taking the examples from Fig. 2 and combining them in series gives:

Z = 100 +330 + j(628.3 – 15.9) \\ Z = 430 + j612.4

This denotes that the combined circuit at 100\text{kHz} has a resistive part of 430\Omega and an inductive reactance of 612.4\Omega (because the j term is positive). This is equivalent to 0.975\text{mH} (or 975\text{μH}). The resulting phase angle of 55\degree is obtained from:

\tan \text{\o} = 612.4/430

A well-known condition is achieved when the resultant j terms equals zero, i.e. when X_L = X_C. From earlier then:

2 \pi fL = 1/(2 \pi fC)

When rearranging this one obtains:

f=\frac{1}{2\pi \sqrt{LC}}

This is the well-known resonant frequency formula.

You are then left with Z = the resistive term only, i.e. a series circuit at resonance is purely resistive — something one learnt for the exams? Figure 4: Antenna System Impedance

A Practical Use

You could well ask what is the use of this, is it just a mathematical exercise? No, it is not: a practical use was hinted at earlier. The following example is just one application.

The impedance at an antenna system is measured at 3.7\text{MHz} using an antenna analyser and it is found that the resistive part is 38\Omega and the reactive part is -j100\Omega (ie Z = 38 – j100). To get maximum power into the antenna it is desirable to eliminate the reactive part so that, from terminals AB, the impedance is purely resistive. Assuming the antenna analyser gives an equivalent series circuit, then a reactance must be added in series to cancel the -j100 term. This is obviously +j100 and the value of the inductance can now be calculated as 4.3\text{μH} at 3.7\text{MHz}.

Conclusion

It is hoped that this short article has provided an insight into the use of the operator j, but the article really only touches the surface regarding the use or this operator.

Sufficient information is given for converting between physical values (ie. farads and henrys and sub-multiples) and equivalent reactances which are expressed in one single unit — the Ohm. The practical examples given will hopefully allow you to use the operator for other applications.

To ease the maths, you can write a simple spreadsheet to do it for you.