SADARS

Dummy loads & RF power measurement

What is a dummy load?

A dummy load is a device used to simulate an electrical load, usually for testing purposes.

For RF purposes, especially transmitters, dummy loads are used in place of an antenna, to allow testing without radiating significant energy.

The dummy load should be a pure resistor whose resistance is the same as the antenna (load) as is used with the transmitter.

The energy that is absorbed by the dummy load is converted into heat.

The dummy load has to be chosen or designed to tolerate the amount of power that can be delivered by the transmitter.

Dummy load resistance value

The resistance value of the dummy load has to match the impedance for which the transmitter is designed, usually 50 or 75.

What is the reason for 50 or 75?

The most common value is 50, the reason for this value goes back to the early days of transmitter design and there are several theories as to why 50 was chosen.

The most likely dates back to the 1920s, when mathematicians determined that a coaxial transmission line impedance of 77 provides the lowest loss, while 30provides the highest power handling for a given cable size, both assuming air dielectric and identical inner and outer conductor material.

Dummy load resistance value50 is a good compromise between lowest loss and highest power handling for a given cable size.

30is the optimum value for handling power with respect to the ratings of the cable conductors and insulators. 77provides the lowest signal attenuation.

50 simply caught on and managed to become the de facto standard as interconnection became common between circuit sections, not just to the antennas.

(75became the standard for video signal transmissions, which explains why TV antenna systems also use 75)

Commercially available dummy loads

Home made dummy loadsRelatively easy for HF frequencies (up to 30MHz), but gets progressively more difficult as frequency increases.

This is due to stray capacitance and inductance within the construction of the dummy load becoming more dominant as frequency increases, so causing the impedance value to change from the ideal 50 value.

50mm of #20 wire in free space has an inductance of ~50nH and presents an inductive impedance of ~32 at 100MHz (this usually appears in series with the load, making it ~59

10pF at 100MHz presents a capacitive impedance of ~160(this usually appears in parallel with the load, making it ~47.4)

Home made dummy loads

Most usual to use carbon resistors because these have low inductance.

Do not use wire-wound resistors

If we assume 1W carbon resistors which are connected in parallel, then;

2 100 resistors provide a 50, 2W rated dummy load

3 150 resistors provide a 50, 3W rated dummy load

20 1k resistors provide a 50, 20W rated dummy load

30 1.5k resistors provide a 50, 30W rated dummy load

Home made dummy loads

Power measurement (Wattmeter)

There are several ways to determine the RF power being delivered to the dummy load, the principal two being a thermocouple meter and a peak voltage meter.

In all cases, the measurement relies upon knowing the resistance of the dummy load (usually 50).

Professional measurement equipment usually employ thermo-couple based meters – these use the principal that the application of heat (resulting from power being dissipated in the dummy load) to a junction of dissimilar metals. This junction then generates an EMF (voltage), which can be converted into a current by a resistance. Such meters are usually delicate and relatively expensive.

Amateur measurement equipment most often measures the voltage developed across the dummy load, which is the method now discussed

Power measurement (Wattmeter)

Measuring power with in line equipment

Measuring power with Wattmeter within dummy load

Power measurement (Wattmeter)

For dc, power (watts) = Volts x Amps = (Volts dc)2 / resistance

For ac, power = Volts rms x Amps rms = (Volts ac rms)2 / resistance

The rms value is the equivalent ‘dc’ value for an ac waveform with respect to the ‘work’ that can be achieved from that waveform.

So, if we apply 10Vdc to a 50 resistor, it dissipates

(10Vdc x10Vdc)/ 50 = 2W

and

if we apply 10Vac rms to a 50 resistor, it dissipates

10Vac rms x10Vac rms/ 50 = 2W

Power measurement – the maths

For a sin wave (which is fundamentally a single frequency ac waveform – the harmonic content is very low – then there is a defined relationship between the rms value and the peak to peak value of that waveform.

The peak to peak value of a rms waveform is the rms value x √2 x 2

so 10Vrms = 10 x 1.414 x 2 = 28.28V peak to peak

and the peak value is the rms value x √2

so 10Vrms equates to 10V x √2 = 14.14V peak

If we can measure the peak to peak voltage or the peak voltage developed across our known resistance dummy load, then that voltage directly relates to the power input to the load.

Power measurement - method

Problem 1 – meters, whether they be analogue or digital, are difficult to make respond to a wide range of frequencies.

Answer 1 – convert the RF voltage into a dc voltage.

Problem 2 – converting an ac voltage into a dc voltage requires some form of rectification.

Answer 2 – use either a full wave rectifier or a half wave rectifier to convert the ac voltage into a dc voltage.

Problem 3 – full wave rectification gives requires at least two rectifiers connected in series unless some form of centre tapped signal is available.

Answer 3 – use half wave rectification, which requires only 1 rectifier.

Circuits

The 50Ω dummy load resistor is simply connected between the ‘inner’ and ‘outer’ of the co-axial connector.

The best designs minimise stray inductance and capacitance and, depending on the shape of the dummy load resistor, sometimes fit a tapered shield around the resistor to maintain a uniform falling characteristic impedance over the length of the resistor (as the impedance falls from 50Ω at the ‘hot’ end towards 0Ω at the ‘earth’ end)

Circuits

VSWR meters sometimes include a POWER indication within their circuits.

However, adding a peak detector circuit across the dummy load can also provide a power indication.

Here, the RF voltage developed across the dummy load resistor is peak rectified by the diode and the resultant dc voltage is ‘stored’ on the capacitor. This peak voltage is then applied to a meter and some form of calibration data applied

Note that the diode has a forward voltage drop and this affects the peak voltage accuracy, in particular at low input power levels.

Power measurement – more maths

Power = V2/R so V = √PR

So, at

1 Watt, V = √1W x 50Ω= √50 =7.071Vrms = 10V peak2Watts, V = √2W x 50Ω= √100 =10Vrms = 14.14V peak3Watts, V = √3W x 50Ω= √150 =12.25Vrms = 17.32V peak4Watts, V = √4W x 50Ω= √200 =14.14Vrms = 20V peak5Watts, V = √5W x 50Ω= √250 =15.81rms = 22.36V peak10Watts, V = √10W x 50Ω = √500 =22.36Vrms = 31.62V peak20Watts, V = √20W x 50Ω = √1000 =31.62Vrms = 44.71V peak30Watts, V = √30W x 50Ω = √1500 =38.73Vrms = 54.76V peak40Watts, V = √40W x 50Ω = √2000 =44.72Vrms = 63.24V peak50Watts, V = √50W x 50Ω = √2500 =50Vrms = 70.71V peak100Watts, V = √100W x 50Ω = √5000 =70.71Vrms = 100V peak

Power measurement – accuracy

The rectifier diode is not perfect – it has a forward voltage drop when conducting – let us say 0.5V, which reduces the resultant peak voltage

So, with a 50 load and the peak rectifier circuit

1 Watt, V actual = 10V peak - 0.5V = 9.5V, i.e. 5% error10Watts, actual = V 31.62V peak - 0.5V = 31.12V, i.e. 1.6% error100Watts, actual = V 100V peak - 0.5V = 99.5V, i.e. 0.5% error

It can be seen that as the power level increases, so the error introduced by the diode’s forward voltage drop reduces.

This error can be partially compensated for at the meter’s full scale deflection ‘point’ but the error will increase at lower points on the meter’s scale, so be aware! (particularly at low power measurement readings)

Power measurement – practical circuits

The diode’s reverse voltage rating has to be at least twice the peak voltage value, so for a 20W meter, a 100V diode is needed (just under 90V reverse voltage will be applied to it), but the 0.7V forward voltage of a fast silicon diode is less of a problem at such power levels than the 0.4V forward voltage drop from a germanium or Shottkey diode more suited to lower power measurements. The capacitor needs to be a low impedance at the lowest measurement frequency (0.01uF is 16 Ω at 1MHz). The meter has to be high impedance compared to 50Ω, for example a 50uA movement and >1kΩ.

The variable resistor is adjusted during the set up of the Wattmeter

Power measurement – diode characteristics

Some germanium diode characteristics

OA47, 25V reverse voltage peak,110mAOA70, 22.5V reverse voltage peak,150mAOA79, 45V reverse voltage peak,100mAOA81, 115V reverse voltage peak,150mA <– a good choiceOA90, 30V reverse voltage peak,45mAOA91, 115V reverse voltage peak,150mA <– a good choiceOA95, 115V reverse voltage peak,150mA <– a good choice

For higher power Wattmeters, a good choice would be a UF4004, which is an ultra fast version of the 1N4004 silicon diode. 400V, 1A, 50ns, which should be OK for Wattmeters measuring power up to HF (30MHz) frequencies.

Practical home made unit - circuit

Practical home made unit - description

Combined dummy load, Wattmeter and Field Strength Meter (FSM)

Maximum allowed input power (on 100W range) is 30W, due to detector diode reverse voltage rating.

Uses a 50uA, 100mV meter (so 2k Ω resistance). If a different meter is used then the ‘range’ resistors will need re-calculating.

Uses a 3 pole, 6 way switch. If FSM not included then a 2 pole, 5 way sw.

Four calibrated power measurement ranges

One variable power measurement range (allows comparisons and ‘tune for maximum’ signal type tests)

One FSM range – since we are using a sensitive meter for the power measurement, then make best use of it by adding a FSM.

Wattmeter calibration

Use a variable dc power supply to calibrate the Wattmeter.

Thermocouple type instrument, use ‘rms’ figure

Peak detector type instrument (when dc supply polarity is also important because the RF input signal is half wave rectified) use ‘peak’ figure.

For example, earlier slide shows that

10Watts =22.36Vrms = 31.62V peak

So if checking a thermocouple type unit, then if dc input is set to 22.36Vdc the meter should indicate 10Watts.

If checking a peak detector type unit, then if dc input is set to 31.62Vdc the meter should indicate 10Watts.