Measure Antenna Impedance
By CHARLES T RAUCH, JR w8ji
2) An RF bridge can be used with a relatively low-power RF signal source. The signal source can be a sine wave from a simple oscillator (however, it must be stable in both amplitude and frequency over time) or it can be from a simple noise generator (however, its RMS output level must be stable over time). A sensitive un-tuned RF voltmeter can be used to detect bridge balance if a sine wave signal source is used. If a noise generator is used, a selective RF voltmeter or selective radio receiver with an S-Meter can be used to indicate bridge balance instead. However, the S-Meter must be able to show small changes in signal level to enable accurate RF bridge adjustments (most digital S-Meters do not have the required small signal level indication abilities needed for accurate antenna impedance measurements). This method is most appropriate in the HF portion of the RF spectrum. Accurate impedance measurements with this method become increasingly difficult with increasing frequency, because very small distributed inductances and capacitances in bridge circuit component interconnections result in erroneous antenna impedance measurement indications. This is an example of a simple RF noise bridge intended for use in the HF spectrum
QRP-Measuring RF Voltage, RF Current, RF Power, SWR
This tutorial starts with simple detector circuits, then combines them into an SWR meter. At the end, you should be able to understand and homebrew your own calibrated dummy load, measure RF current and build a SWR-Power meter, all tailored for QRP use.
Measuring RF Voltage
My meter measures AC volts - why can't I simply measure the output voltage of my rig this way? You can, as long as the frequency is low, perhaps below a few hundred kilohertz. Look at the detailed specs on your AC meter. You'll find that as frequency increases, accuracy goes down. Here's the spec for AC scales on a good quality Fluke 85 handheld multimeter:
60Hz - 0.5%
45Hz to 1KHz - 1.0%
1KHz to 5 Khz - 2.0%
5KHz to 20KHz - 4.0%
Above 20KHz., accuracy is not specified, but you can bet it gets worse. We're going to make a simple diode detector circuit that accepts a radio frequency sinewave. The detector will convert it to a DC voltage, which our multimeter can read with good accuracy.
What RF voltages would you find at a QRP rig output? Let's take the maximum case of a five-watt QRP rig. This rig will deliver five watts of average power to a 50-ohm resistive dummy load. This five watts is root-mean-square (RMS) power, and we can find the equivalent RMS voltage from P = V2/R. Re-arranging, V(rms) = sqrt(P x R). In our case, V(rms) = sqrt( 5 x 50) = 15.8 volts.
Let's find the current too, from P = I2 x R I(rms) = sqrt(5/50) = 0.316 amps(rms).
As a check, the voltage divided by the current should be equal to the 50 ohm load resistor, i.e.: 15.8/0.316 = 50
The AC waveforms at the dummy load are sine waves. Since our multimeter can't measure with any accuracy the amplitude of these waveforms, we'll have to use some electronic circuits to transform AC amplitude to DC amplitude. Once converted to DC, our multimeter can measure amplitude with good accuracy.
Now we have a problem, since we usually require power measurements to be RMS measurements. The electronic circuits required to give DC output proportional to true RMS voltages (or currents) are complex. We have one easy way out...since we're always dealing with sine waves, we can use a simple peak detector circuit, and scale its output (with a resistor divider, or amplifier) to look like RMS.
We won't go through the math (integrating sine waves isn't fun), but the relationship between the RMS amplitude and peak amplitude of a sine wave is simply this: V(rms) = V(peak) / sqrt(2). Here's a table showing the relationship between RF power, voltages, and currents for a 50 ohm resistive load:
5.0W / 15.8 / 0.316 / 22.3 / 0.447
2.0W / 10 / 0.20 / 14.4 / 0.283
1.0 W / 7.07 / 0.141 / 10.0 / 0.200
0.5W / 5.0 / 0.100 / 7.07 / 0.141
0.2W / 3.16 / 0.0632 / 4.47 / 0.0894
0.1W / 2.24 / 0.0447 / 3.17 / 0.0632
The electronic circuit to detect peak voltage is quite simple - it involves a diode and capacitor. The diode charges the capacitor up to the peak voltage of the waveform at its input. The capacitor remains charged throughout the sine waves cycle, so that the voltage across the capacitor is almost entirely a DC voltage. Once the capacitor is charged up to the peak voltage, the diode only needs to conduct enough to keep the capacitor charged. Since our DC multimeter loads this capacitor very little, the diode conducts very little as well, and the RF source sees very little loading due to the diode. So our DC meter should read 22.3 volts(DC) across the peak-detector's capacitor when our 5W transmitter puts out 5W.
There is one small error that creeps in...the diode "eats up" some voltage, so that it doesn't quite charge the capacitor as high as it should. Some diodes are better than others in this respect. A Schottky diode (or barrier diode) is better than a silicon diode, and a germanium diode is better too. Since these diodes drop almost a constant voltage, then error will be worst for smaller RF voltages. For milliwatters, some effort should be made to compensate for the diode drop.
The diode we choose must be a fast switcher. If not, it remains conducting for a bit after the peak, and drags down the capacitor voltage. Diodes meant for 60Hz power supply use are not built for speed, and shouldn't be used in an RF peak detector. Small signal diodes are more appropriate, and almost any germanium small signal diode will do. IN4148, 1N914 are two common silicon types that are fast enough up to 30MHz (at least). 1N34A, 1N270, 1N191 are some common germanium diodes. 1N5711 is a common small-signal Schottky diode.
Measuring RF Current
Once again, our multimeter on AC scales fails at measuring radio-frequency sinewaves. Current measurement is a bit more difficult than voltage measurement shown above. We shall use the same peak-detector circuit, measuring the voltage drop across a small sampling resistor. This resistor will drop a small voltage as RF current flows through it. The peak detector will measure this voltage drop. Ideally, a very small sampling resistor is best: it must be much smaller than the 50 ohm load. If not, the combination of 50 ohm load and series sampling resistor will increase the load seen by the transmitter.
If we were to use a one-ohm sampling resistor (1st circuit), the voltage available (for a 5W output) would be 0.447 peak volts. Not very much: errors due to diode-drop will be high, and even worse at lower power.
Let's use a transformer to step up the available voltage going into the peak detector (2nd circuit). We'll use a 1:10 turns ratio on our transformer, and use a ferrite core to make sure all the flux links to every winding. If the ferrite has high enough permeability, the primary winding can be one turn, while the secondary winding can be ten turns. It needn't be a big core, since very little power is going into the peak detector.
Rather than place a one-ohm resistor at the primary side, its better to add a 100-ohm resistor on the secondary side (3rd circuit). Since impedance is transformed by the turns-ratio-squared, the primary will still see a one ohm impedance. The diode measures the peak voltage across this 100-ohm resistor. Now instead of 0.447 volts peak, we'll get 4.47 volts peak - a substantial improvement in sensitivity.
Measuring RF Power
We can use either the voltage peak detector, or the current peak detector to find RF power delivered to a known resistive load. In fact, if we combine the two instruments, we can find the RF power delivered to any (unknown) load resistance. All we have to do is take a voltage measurement, and a current measurement, convert both to RMS, and multiply them together. We can also find load resistance by dividing voltage by current.
But there's one requirement: the load must be purely resistive. If our load had been a 50 ohm reactance, (say, a capacitor of 455 pf) our measurements would come out the same, but the power we calculate would be all wet, because a reactance doesn't dissipate any power at all. Sure, there'll be 10.4 volts(rms) across this reactance, and 0.205 amps(rms) flowing through this reactance, but the voltage and current are out of phase, by 90 degrees. Thus there's no real power dissipated. Our measurements fail because we're measuring voltage magnitude, and current magnitude. The net result is that our measurements can tell us the magnitude of the load impedance, but can't tell if the load is resistive, inductive or capacitive or somewhere in between. And so we can get fooled into thinking that there's real power going to the load
There are proper solutions to this impedance problem, but the circuitry is usually complex. Most simple SWR/power meters solve the problem this way - first you get the SWR down close to 1:1, then you can read the power from the meter's scale. If the SWR is not 1:1, the power scale is meaningless. This is the approach we'll take. The SWR circuit incorporates some of the peak detectors we've already discussed. After we come up with a circuit to measure SWR, we'll take a look at how it can measure power too.
Homebrew Project: Build RF Voltage and Current probes
The voltage probe includes a 50-ohm dummy load suitable for QRP rigs (bottom). It supplies a DC voltage that is proportional to the RF amplitude across the dummy load. Output can be measured with a DC multimeter, and (with some simple calculations) calibrated in terms of output watts. It will operate on any HF band up to 50MHz with reasonably good accuracy. The multimeter's probes are simply held across the 1000pf capacitor, while reading the meter.
The current probe (top) is also meant for QRP use. Since it must be wired in series with a load, it includes two BNC connectors (my connector of choice). One connector would be wired with coax into the load (antenna or dummy load) and the other connector would be wired to the rig's output, all with 50-ohm coax cables. Two output jacks have been added so that the DC multimeter's probes can be pushed in for no-hands monitoring. Once again, the DC output voltage can be scaled (with some simple calculations) to find RF current flowing from the rig to load.
Schematics
Dummy load, with voltage probe on left, current probe on right.
The diode was a surplus germanium type scrounged from a very old logic board, 1N617. Substitute a 1N34A. The 1000pf capacitor is silver mica, but a ceramic disk type will do as well. The ferrite bead is an old one from Philips, type VK21029/3B. It is 5mm long, 3.7mm O.D. and 1mm I.D. A Panasonic replacement with the same dimensions, Digi-Key part # P9823-ND should be very similar. Panasonic part # is EXC-CL3225U. The 220 ohm, 2W resistors are carbon film, from Philips. Be careful not to employ wire-wound resistors here.
Calibration scaling The voltage probe's output voltage as measured by the DC multimeter is simply equal to the peak RF voltage across the dummy load. Neglecting diode voltage drop, output power is simply V2/100 watts.
This equation combines the peak-to-RMS conversion with P=V2/R. This voltage probe will give you a meter reading that gives Vpeak between 0.05 to 0.1 volt lower than true peak voltage. You may wish to correct Vpeak by this amount before calculating power.
For the current probe, the output voltage measured by the DC multimeter is the peak voltage across the 100-ohm resistor. This voltage is in turn ten times the current (in amps) flowing on the center conductor of the coax. I = 0.07071 x V. Once again, for better accuracy, you may wish to add between 0.05v - 0.1v to the meter reading before calculating actual current.
From my NORCAL40, the voltage probe gave 14.7 volts DC out. 14.7 x 14.7 / 100 gives 2.16 watts as output power.
With the 50-ohm dummy load connected to one of the ports of the current probe, and the same NORCAL 40 driving the other port of the current probe, measured output voltage is 2.9 volts DC. RF current is 0.0707 x 2.9 = 0.205 amps(rms). Knowing that the load is a pure 50-ohm resistance, let's use P=I2xR to find output power: P = 0.205 x 0.205 x 50 = 2.1 watts. This compares fairly well with output power calculated from the voltage probe (2.16 watts).
Building tips
Stray inductance can hurt accuracy with either of these probes: keep wiring leads as short as possible, especially for use at high frequencies. For the 50-ohm dummy load, I bunched four two-watt resistors together, and actually cut off all their leads (on one end). These resistors had brass end-caps which were all directly soldered to a brass washer. From this washer, an insulated wire connected to the center pin of the RF connector. This wire is hidden in the photo above, snaking down between all four resistors. The ground end of the resistors have very short leads that connect directly to the ground shell of the RF connector.
The current probe was also constructed with short leads in mind. The primary winding of the transformer is actually a straight, insulated wire connecting one BNC jack to the other. It should be as short as possible. You might not see it as "one turn", because it is threaded straight through the bead's center. Don't think that "one turn" is a loop around the toroid: that's two turns! You can't see the small diameter #38 wire around the bead, nor its connections to the adjacent 100 ohm resistor in the photo. But you can see that the 100 ohm resistor is very close to the ferrite bead. You could easily substitute a small ferrite toroid instead of the bead: an FT47-23 would work well.
An op-amp circuit to compensate for diode losses
Our diode peak detector suffers from inaccuracies at low RF input levels. This arises because the diode requires a small amount of voltage across its terminals before it conducts much current. The result is that DC output is lower than the peak RF input voltage. At very low input voltages, the error is great.
One simple solution is to use a similar (matched) germanium diode to add the lost DC voltage on top of the detector's output. A simple DC op-amp circuit can do this easily. R1 drags some DC current through diode D1 to simulate the peak rectifying current in the RF-detecting diode. The op-amp will force its output to a DC voltage so that the voltage across R1 is equal to the input DC voltage. Since D1's forward conducting voltage is in series with R1, the output will be Vin + diode voltage. Thus the diode voltage is added to the input voltage.
The value of R1 depends on the diode. A germanium diode will require R1 to be about 15K. A silicon diode and a Schottky diode require a much higher value, perhaps in excess of 100K ohms.
The op-amp requires a power supply - if a single supply is chosen (say, +9v) then the op-amp's negative supply will be connected to gnd. In this case, the op-amp's "+" and "-" inputs must have a common-mode input range right down to the ground potential. Some bipolar op-amps can do it, but not many. A LM324 quad op amp will work. Many more CMOS op-amps will work well. An excellent example is National Semiconductor's LMC6042: it requires VERYlittle supply current, a battery will last for nearly shelf life. Its a dual op-amp: the other (free) op-amp can amplify this compensator's output to drive a meter.
A Clamp-On RF Current Probe
by: Lyle Koehler, KØLR
The original version of this article appeared in the April 1994 LOWDOWN
Here's a sensor that lets you measure currents in your LowFER, MedFER,or ham antenna and ground system without breaking the circuit. You simply clamp the pliers-like probe around the wire you want to measure. An advantage of a current probe over a simple relative field strength meter is that the measurements are highly repeatable and can provide a fairly good indication of how your antenna system compares to others.
A similar current probe is described in the October, 1992 issue of RADIO COMMUNICATION. However, it is intended for use in the 2 to 30 MHz range, and is made from a large toroid core that must be carefully broken or sawed in half. The probe described here uses a split rectangular ferrite core and has a useful frequency range from 100 kHz to more than 10 MHz. All of the electronic components (including the ferrite core) and most of the hardware such as screws, nuts, spacers and sticky-backed hook-and-loop ("Velcro") tape are available from Radio Shack.
Like the clamp-on ammeters used for 60-Hz circuits, this probe is based on the current transformer principle. When you clamp the probe around a wire, the wire becomes the single- turn primary winding of a transformer. If the secondary winding is loaded with a low resistance R, the RF voltage across R is equal to R times the primary current divided by the number of turns on the secondary winding. A diode detector circuit converts the RF voltage into a DC output which can be read on a standard multimeter.
Figure 1 shows suggested construction details for the clamp-on probe. The pliers can be constructed from 1-inch wide pieces of thin (1/8 to 1/4 inch) plywood or plexiglas. Attach the ferrite core to the pliers assembly with 4 small strips of sticky-backed Velcro or double-sided adhesive foam tape. Two pegs made from 1/2-inch spacers and 6-32 screws press against the outside of the core halves or, to be exact, against the winding. There is enough flexibility in the Velcro fastener so the core halves can align themselves when they are squeezed together. It's also easy to remove the core and reposition it if the alignment is incorrect. If you use double- sided foam tape, it will require considerably more care when you initially stick the core halves in place. The core halves do not have to be perfectly aligned side to side and front to back, as long as there is no air gap between them. There are lots of spring arrangements that can be used to provide the clamping action -- a fat rubber band does the job quite well. The kind of rubber band used in supermarkets to hold bunches of broccoli together is just about right for this application. If you don't like broccoli you'll have to improvise.