Homebrew Project: Build RF Voltage and Current probes
 RF current probe, voltage probe + dummy load   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.
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.
Voltage probe, current probe schematic
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 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).

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 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 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.