Under Construction…..
I was surprised by a lack of information about building a simple detector
probe.
Let’s consider a source of 100 watts and look at how to design an HF probe to
rectify and detect the amplitude of an RF envelope. First we have:
100 volts would be the peak RF voltage of a 100watt PEP transmitter into 50
ohms. RMS voltage would be 70.7 volts.
If we sampled through a diode directly we would have this:
The diode charges capacitor C1 to 100 volts minus the diode voltage drop.
The diode reverse voltage is roughly twice RF peak voltage, or 200 volts. That is a
high voltage to find in a low capacitance diode suitable for detecting high
frequencies! A typical small signal fast diode like a
1N4148
or
1N916 is only around 75 to 100 volts PRV.
Also note we have a conduction threshold region where the detector is very
nonlinear. That would be down around 1 volt, depending on the diode and the
current drawn from the diode. To minimize distortion we want a very high
impedance load and the highest possible voltage. Otherwise we would need a more
complex detector with active bias on the diode.
We also have to be sure there is a low impedance dc path through the source
and load to ground for diode current. We can’t always depend on that. Making a
detector like this, if we want more than relative output, is not as easy as it
might seem.
This is a 10:1 divider, so with 70.7 volts RMS we have .0637 amperes RMS
divider current. This is 4.5 watts total dissipation. There is 7.07 volts across
R3, and 63.63 volts across R2.
R3 dissipates about .45 watts.
Dissipation (use RMS for heat) in R2 E^2/R = 63.63^2/ 1000 = 4 watts. That is
a big resistor and a good deal of heat for a noninductive resistor. It also
upsets load impedance a little bit, making SWR 1.05:1 or so.
10 kHz response is good with a 2000 ohm load and .001uF filtering. We lost
about 1 volt off the top, partially from diode current loading the divider and
partially from diode voltage drop.
