# Tag Info

0

Building is educational and fun or you could just bye the W1 from Electraft (disclosure: I own one and love it) https://elecraft.com/collections/test-equipment/products/w1-100w-wattmeter-kit-1

2

How to make an RF probe for accurate power measurement at QRP power levels?                                (emphasis mine) Now, accurate measurement of RF power is actually pretty hard once you cross ...

2

A basic probe looks something like this: simulate this circuit – Schematic created using CircuitLab Ideally the diode has a low voltage drop, like a Schottky diode. But any ordinary silicon diode will work in a pinch. The capacitor and resistor values aren't really critical as long as the time constant (the resistance times the capacitance) is much ...

0

Most homebrew RF probes use a diode with low voltage drop to rectify the RF. You then measure the resulting pulsed DC voltage to determine the power. Most people use a germanium diode like a 1N34, but these can be hard to find. You can use a schottky diode instead. You want the voltage drop across the diode to be as low as possible. A diode with a high ...

1

Especially for narrowband (by modern standards, I'd call that anything < 50 kHz bandwidth) signals, the oversampling that you can do with even the cheapest SDR devices (e.g. RTL-SDR dongles) gives you excellent detectability, down to thermal noise. (Make sure you turn of any AGC, though.) So, assuming you just want to know what the power of that single ...

6

If the frequency counter is measuring -5.3 dBm, then before 20 dB of attenuation the power was 20 dB more than that, so 14.7 dBm. "dBm" means decibels relative to 1 milliwatt. 14.7 decibels can be converted to a ratio like so: $$10^{14.7/10} = 29.5$$ So 14.7 dBm is 29.5 milliwatts.

Top 50 recent answers are included