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 50W or 75W.
What is the reason for 50W or 75W?
The most common value is 50W, the reason for this value goes back
to the early days of transmitter design and there are several theories
as to why 50W was chosen.
The most likely dates back to the 1920s, when mathematicians
determined that a coaxial transmission line impedance of 77W
provides the lowest loss, while 30W provides the highest power
handling for a given cable size, both assuming air dielectric and
identical inner and outer conductor material.
Dummy load resistance value
50W is a good compromise between lowest
loss and highest power handling for a given
cable size.
30W is the optimum value for handling power
with respect to the ratings of the cable
conductors and insulators. 77W provides the
lowest signal attenuation.
50W simply caught on and managed to
become the de facto standard as
interconnection became common between
circuit sections, not just to the antennas.
(75W became the standard for video signal
transmissions, which explains why TV
antenna systems also use 75W )
Commercially available dummy loads
Home made dummy loads
Relatively 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 50W value.
50mm of #20 wire in free space has an inductance of ~50nH and
presents an inductive impedance of ~32W at 100MHz (this usually
appears in series with the load, making it ~59W)
10pF at 100MHz presents a capacitive impedance of ~160W (this
usually appears in parallel with the load, making it ~47.4W)
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 100W resistors provide a 50W, 2W rated dummy load
3 150W resistors provide a 50W, 3W rated dummy load
20 1kW resistors provide a 50W, 20W rated dummy load
30 1.5kW resistors provide a 50W, 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 50W).
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 50W resistor, it dissipates
(10Vdc x10Vdc)/ 50W = 2W
and
if we apply 10Vac rms to a 50W resistor, it dissipates
10Vac rms x10Vac rms/ 50W = 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.
Note that the diode has a forward voltage
drop and this affects the peak voltage
accuracy, in particular at low input power
levels.
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
Power measurement – more maths
Power = V2/R
so V = √PR
So, at
1 Watt, V = √1W x 50Ω= √50
=7.071Vrms = 10V peak
2Watts, V = √2W x 50Ω = √100 =10Vrms
= 14.14V peak
3Watts, V = √3W x 50Ω = √150 =12.25Vrms = 17.32V peak
4Watts, V = √4W x 50Ω = √200 =14.14Vrms = 20V peak
5Watts, V = √5W x 50Ω = √250 =15.81rms
= 22.36V peak
10Watts, V = √10W x 50Ω = √500 =22.36Vrms = 31.62V peak
20Watts, V = √20W x 50Ω = √1000 =31.62Vrms = 44.71V peak
30Watts, V = √30W x 50Ω = √1500 =38.73Vrms = 54.76V peak
40Watts, V = √40W x 50Ω = √2000 =44.72Vrms = 63.24V peak
50Watts, V = √50W x 50Ω = √2500 =50Vrms
= 70.71V peak
100Watts, 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 50W load and the peak rectifier circuit
1 Watt, V actual = 10V peak - 0.5V = 9.5V, i.e. 5% error
10Watts, actual = V 31.62V peak - 0.5V = 31.12V, i.e. 1.6% error
100Watts, 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 variable resistor is
adjusted during the set
up of the Wattmeter
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.
Power measurement – diode characteristics
Some germanium diode characteristics
OA47, 25V reverse voltage peak,110mA
OA70, 22.5V reverse voltage peak,150mA
OA79, 45V reverse voltage peak,100mA
OA81, 115V reverse voltage peak,150mA <– a good choice
OA90, 30V reverse voltage peak,45mA
OA91, 115V reverse voltage peak,150mA <– a good choice
OA95, 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.