Homebrew Project: Build RF Voltage and Current
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 mulitmeter'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.
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
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
From my NORCAL 40, 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).
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 acutally 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
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.