I want to measure voltage across a T-line to visualize standing waves on the line.

something like this video min 6:53 and this experiment Measuring Standing Waves on a Microstrip Transmission Line

I have purchased this VCO to generate RF source signal.It's 5mW.

I am using 12V 5A power supply.

This is my circuit (T-line is open as I want to meassure open, short, and, matched) My setup

Then I used a voltage multiplier circuit to detect the RF signal(also tried half wave rectifier) and connected it's output to my analog voltameter with 0.5V, 0.25V, 100uA scale, but I didn't get any reading.

I used 1n34a diodes with ceramic 10nF caps.


  • Why doesn't voltmeter read any value?
  • What is the dominant factor that would cause that?
  • How to diagnose the circuit without expensive equipments?
  • Is there any easy to find device to get high power RF(from cellphone, TV, ...)?


  • I used solder on the microstrip, because I am planing to put a stepper and slide the probe through the line(similiar to slotted line device).
  • I don't use lecher line as it would require more powerful RF source.
  • I have designed a circuit to generate around 2Ghz, but I delayed the fabrication to verify the concept with the above VCO.
  • I know that it is more practical to use microstrip wave guide with ground plane, but it would be hard to get these small vias connecting upper plane with the lower one and It would take a long time to arrive.

I want the minimum setup(just working).

Thanks in advance.


Here is detector circuit: RF_probe_schematic

I have tried it and it's working as expected"I havn't soldered buffer port yet". rf_probe

My short circuit is not perfect and it is lowering the maximum amplitude.

@The Photon: thanks a bunch.

  • 1
    \$\begingroup\$ You say you "know it is more practical to use microstrip wave guide with ground plane"...but you call your design a "microstrip".... So do you mean your transmission line does not have a ground plane? If that's what you did, you didn't make a transmission line, you made an antenna. \$\endgroup\$
    – The Photon
    Apr 28, 2018 at 15:34
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    \$\begingroup\$ And no matter how simple, show a schematic of your detector circuit, or a photo of it and how you contacted the "transmission line", so we understand exactly what you did. \$\endgroup\$
    – The Photon
    Apr 28, 2018 at 15:38
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    \$\begingroup\$ @mkeith hope you are not annoyed. you know it would be better to have a small device to do that and get closer to RF area. \$\endgroup\$
    – I.Omar
    Apr 28, 2018 at 17:04
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    \$\begingroup\$ BTW solder tinning is wrong - copper has less skin resistance. If you look at microwave pcbs, they don't tin the tracks. \$\endgroup\$
    – Henry Crun
    Apr 30, 2018 at 5:39
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    \$\begingroup\$ It is easy to make microstrip, by cutting a sliver of pcb 3mm wide, and gluing it onto the copper side of the pcb. This makes the ground accessible. \$\endgroup\$
    – Henry Crun
    Apr 30, 2018 at 5:52

1 Answer 1


This is a fun thing to do, and quite easy, but you need to learn something about radio for it to work.- from what you say, you are failing at every level. The thing you are trying to do is called a "slotted line" Read Radio Amateurs VHF handbook to get some ideas of how to assemble UHF/microwave detector.

A few milliwatts doesn't do it - try 1 Watt 100pF is more like the capacitance at 500MHz

The problem is you might get 400mV at the output of the 5mW VCO. You want to capacitively probe it (otherwise your probe introduces its own relection), so lets be optimistic, and say you get 50mV at your probe. You can't do a voltage doubler - 50mV is not turning the diode on/off. You need to use a single diode. It is acting as a square law detector. So perhaps you get 10mV of DC signal out. This is the kind of voltage you are looking for mV or uV. You can try replacing R2 with your meter on its most sensitive ranges - try both current and voltage ranges. This would work with my multimeter - but it has 0.1uA resolution.

Detectors are made sensitive by chopping the RF power at audio ( eg 1kHz) then use an audio (microphone) amplifier after the diode. This is amplitude modulation. (headphones or speaker makes you "feel" the whole thing more too. This might get it working at 5mW power range.

So here 74HC14 uses 1 as oscillator - tune to 1kHz. 5 invertors in parrallel to power your vco.

Use 100pF capacitors for RF at 500MHz.

Ac couple the audio out to a microphone input of something. You might want to use a simple audio preamp. (Q1)

R3/4 lets you put a small DC current through the diode, this increases is sensitivity as a detector.

You pcb is OK, just replace your D1 with R1. I assume there is a ground plane on the other side, and you drill through to make ground connections. Wrapping foil around the edge of the pcb also makes a good ground.


simulate this circuit – Schematic created using CircuitLab

Here is another approach. Instead of switching the power to the VCO, you chop the RF using D2 as an RF switch, driven on/off by the 1kHz oscillator. When th diode is on, it shorts out the RF, reducing the level. This technique was used to make radar detectors once.


simulate this circuit

OR Find you local radio ham, and get help - they will have a radio that is suitable, enthusiasm and knoledge. "70cm" wavelength handheld is perfect for this. Or use UHF-CB radio, also the same wavelength.

  • \$\begingroup\$ I am using 10 nf \$\endgroup\$
    – I.Omar
    Apr 28, 2018 at 18:19
  • \$\begingroup\$ well that will be wrong by 1000x then... Calculate reactance XC of the cap at your frequency. You want 1-10 ohms \$\endgroup\$
    – Henry Crun
    Apr 28, 2018 at 18:25
  • \$\begingroup\$ It's around 31.5:63 "500Meg:1Ghz", but why should it be from 1-10 ohms? \$\endgroup\$
    – I.Omar
    Apr 28, 2018 at 20:34
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    \$\begingroup\$ As you have more C, it becomes resonant, then inductive, and Z goes up. Look here: i.stack.imgur.com/zmMpK.gif. See that at 500MHz, your lowest Z is a 100pF capacitor, but you can only get down to a couple of ohms. At 2GHz, you can only get down to 10 ohms. \$\endgroup\$
    – Henry Crun
    Apr 30, 2018 at 5:07
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    \$\begingroup\$ It will never be a voltage doubler. You only have a few millivolts of signal. The diodes are not acting as diodes - they don't turn off. They are acting as what is called "square law detectors" The bias current keeps them turned slightly on, which increases sensitivity \$\endgroup\$
    – Henry Crun
    Apr 30, 2018 at 22:42

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