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 sinewaves. 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 a DC
amplitude. Once converted to DC, our multimeter can measure amplitude with
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 sinewaves, 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 sinewaves isn't fun), but the relationship between the RMS amplitude and peak amplitude of a sinewave 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:
|Power out||AC Volts rms||AC Amps rms||AC Volts peak||AC Amps peak|
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.