Hardie@MIT-MULTICS.ARPA (Maj. Douglas Hardie) (09/23/85)
I was reading the Hints and Kinks in May 85 QST, article on Match Your RF Probe to Your Meter. There were several things I believe misleading and possibly wng in the article and the associated Editor's comments. The calculations given for a 11M or 10M meter are fine, except that the meters I am familiar with have a 1M resistor in the DC probe. Since these RF probes replace the DC probe, the meter no longer has a 11M impedance. Hence it seems that one calculates the proper resistance for the dividor using the 11M for the meter, but then must add 1M to compensate for the missing probe resistor. When I do that for a 11M meter, I get the probe resistor of 5.556M for which I believe there is a standar 5.6M resistor that should be close enough. The Heathkit probe I have that was "designed" for my meter ignores the 1M resistor and uses a 4.7M divider resistor also. Also the orientation of the diode in fig 1 seems to imply that the probe will put out a positive DC voltage. However, I find that it puts out a negative voltage. My Heathkit probe has the diode reversed and a quite differnet explination of its workings. It works, and no amount of playing with fig 1 or 2 made it put out a positive voltage. The voltage divider (fig 3) in the Editor's comments I have tried and found very unsatisfactory. The capacitor C is effectively accross the bottom resistor in the dividor. Hence the division ratio is complex and frequency dependent. I tried to make that scheme work with two fixed resistors to give a 1/10 ratio and found that the same power output into a dummy load on 80 meters gave a very different probe voltage than on 10 meters. If I remember right, it was about a 50% variation. I never found a completely acceptable way to solve the divider problem, but ran across an article which recommended making a wire twist capacitor across the top resistors in the dividor to compensate.
jhs@mitre-bedford.ARPA (09/23/85)
People trying to make voltage dividers etc might make something of the following observation: If you have an R shunted with a C as input impedance of some device like a meter or amplifier, you can make it into a voltage divider / lowpass filter by putting an R shunted by an L in series with the original RC combination. If the R values are equal then choose L so that sqrt(L/C) = R, and the whole thing will look like EXACTLY R at ALL frequencies as seen from the input, while giving a 2:1 voltage division ratio at frequencies well below cutoff. Often the cutoff frequency comes out acceptable. Cutoff frequency is of course 1/(2*pi*sqrt(LC)). The other approach is the capacitive compensation approach, which avoids the cutoff problem but which presents a capacitive input impedance. 73, de W3IKG