**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 = V^{2}/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 = I^{2} 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
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 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 |

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 |

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

There