Is it possible to generate 10 GHz frequency RFs at 400 V amplitudes while only using 10-100 W order power? I tried looking up how 10s of GHz frequencies are generated online and found next to nothing except communication products that had no spec sheets available.

The main requirements I have are that

  • I want to generate 10 GHz,
  • have the waves transmit over a distance of 10 cm,
  • while having at least 200 V amplitude at the receiver point
  • and only use about 10-100 Watts but ideally fractions of Watts.

Is such an experimental/instrument set up even physically possible? I know that the medium the waves will transmit through, also affect the decay envelop of the wave function and therefore, the decay of the wave's power, so let's just assume that these waves will propagate through an atmosphere in the 100s of Pascals.

Lastly, the emitter's material is non-specific (Because I do not know what they should be for such high frequencies), but the geometry of the emitter should be a ring. If someone can give me a rough order magnitude of power requirements for these waves and/or the resources so I can start reading into RF/communication technologies and do these calculations on my own with the formulas used in that field I would really appreciate it.


1 Answer 1


UPDATE From clarifying the specs in comments: the poster wants a region with a high electric field generated by as low an input power as possible.

You want standing waves, not propagating waves; therefore you want a resonator, not an antenna. Either make the device you have into a resonator, put a resonator in it, or put it as a whole into a resonator.

Since there seem to be a bunch of specific constraints on geometry which are not described in the question, I can only offer very general guidance:

(a) build a resonant structure

(b) figure out all of your geometry requirements, in the following format: volume where you need your field; volumes of space where you can't have anything; where you could have metal, and where you have to have metal

Now the problem becomes simple: you need to make a resonant cavity which fits that geometry. For a rectangular metal box, basically two walls have to be spaced some multiple of wavelength/2 apart. But it also doesn't have to be rectangular, or metal, or a closed structure - many more ideas here: https://en.wikipedia.org/wiki/Microwave_cavity

Easy geometries are box, parallel plates (which may be shorted together at one edge or two opposite edges), and cylinder. If possible, it may be easier/more efficient to use whatever the device you're working with is made of as part of the resonator (especially if it's already metal), but if not, just put the whole resonator inside it.

A section of transmission line or waveguide with shorted or open ends is also a resonator (and may also have open sides, eg in air stripline). A box can be thought of as a waveguide section with shorted ends. Open resonators will leak through the open sides - two facing parabolic reflectors is probably the least-leaky open resonator possible (like parallel plates but curved).

Resonators can be built out of dielectrics: dielectric resonator, these tend to be much more compact but more lossy than ones built of a conductive material.

Anything you put in your resonant structure (eg the probe) will detune it somewhat, and perturb the field (could be a lot, if it's metal) so you probably need to design something that works both with and without it.

The "ring emitter" described in the question would only work well to the extent you can build a resonant structure that incorporates it (eg: the two ends of a cylinder)

Original answer For any kind of electromagnetic radiation (plane waves) in vacuum, the formula connecting the peak electric field strength $E_0$ and the average energy flux per unit area (intensity) $I$ is:

$E_0 = \sqrt{ \frac{2 I}{c \epsilon_0} }$

where $c$ is the speed of light and $\epsilon_0$ is the dielectric permeability of vacuum (source)

There are several missing pieces in the specs you describe.

To sum up the specs: frequency 10GHz, input power 10-100W, amplitude 200-400V, distance 10cm

The missing parts:

  • the area (cross-section) of the beam
  • the size of the receiver (ie antenna)
  • what is in between the transmitter and receiver?
  • are you trying to create a electrical field, or are you trying to transmit power?

The cross sectional area is unspecified in your case, but let's assume 10x10cm = 100 cm^2.

$I = 100 \ \mathrm{W} \ / \ 100 \ \mathrm{cm}^2 = 1 \ \mathrm{W}/\mathrm{cm}^2$

$E_0 = 2745 \ \mathrm{V}/\mathrm{m} = 274.5 \ \mathrm{V}\ / \ 10 \ \mathrm{cm}$

You can see why the area and the size of the receiver matter: (a) the total power is not important, just the power transmitted per unit area, and (b) the electric field is volts/meter, not just "volts" (so a larger receiver would pick up a larger potential difference across it). However, as long as the receiver is the same size as the width of the beam, the resulting voltage across the receiver ends up being the same - for a narrower beam with the same power the intensity is larger, $E_0$ in V/m is larger, but it is multiplied by the smaller size of the receiver.

This makes all kinds of assumptions, eg the transmission happens through free space in vacuum, is roughly a plane wave of uniform intensity with a certain beam cross-section (*of course it can't be exactly a plane wave if it has a finite cross-section), and the receiver is perfectly absorbing and the same size as the beam. The receiver in this case would normally be called an "antenna" - there is a huge variety of designs that work at 10GHz, from patch antennas to parabolic to dielectric lens.

In practice, unless you really need the transmission to be through free space, you would probably use a waveguide, and the other end of the waveguide may be reflecting or absorbing or somewhere in between. Looking up some waveguide equations would help you calculate the electric field in a waveguide.

A lot of both waveguides and coax cables (and the things that they connect to) are impedance matched to 50 ohms. The (rms) voltage expression is very simple in that case, sqrt(100W * 50 ohm) = 70.7 V.

If you are just trying to create a large field, you want standing waves, ie waves going back and forth inside a waveguide. This is also similar to what happens in a microwave oven. You can think of this as the equivalent of shining a flashlight through a small hole into a tin can or between parallel mirrors; the light inside would bounce off the walls multiple times, and be much brighter than if the walls were black.

It would be easier to answer the question if you describe what you're actually trying to do: transmit information? transmit energy? create a field? something else?

Lastly, a safety note: 100W at 10GHz is enough to cause serious injury; it can focus down to a small spot (including by accident) which would heat up very fast. The way to think about it is somewhere in between radio and a laser beam, with some of the same safety issues as a 100W laser. You don't sound like you're familiar with this, so probably the first piece of advice is to find someone who knows how to work with high power microwaves to help you, and make sure anything you do is safe. Read about maximum permissible exposure limits, get a power meter, shield your setup, test at low powers first, and generally treat it with a healthy amount of caution.

See also:

  • $\begingroup$ Hi! Thank you for the awesome, thorough response! That is correct, I have zero experience, but luckily, I am not interested in doing an experiment involving this, but rather trying to do physics calculations to predict the power requirements given a theoretical setup! $\endgroup$
    – Sophia
    Nov 8, 2023 at 1:50
  • $\begingroup$ Here is the info you also requested in your answer: I just want to create an electric field with a certain magnitude specification a distance d away from the emitter ring, where the emitter must be a ring geometry, but its diameter can range from [1,3] cm. The cross section of the beam should be at least 1 mm diameter at that distance d and there doesn't need to be a receiver at all. The medium the emitter ring will sit in is helium at around 100 Pascals. $\endgroup$
    – Sophia
    Nov 8, 2023 at 1:50
  • $\begingroup$ @Sophia Great! If you want to create a field with as little power input as possible, you probably want standing waves inside a resonator of some type, for example a waveguide, stripline, transmission line, TEM cell or even a length of coax, with reflecting shorts on both sides. The only losses would come from the resistance of the conductive elements forming the resonator. The spot you get the field in would be inside the resonator. $\endgroup$
    – Alex I
    Nov 8, 2023 at 3:23
  • $\begingroup$ Is it a requirement that there not be anything in between the emitter and the area where you need the field? (in other words, could your "field spot" be enclosed, or does it have to have empty space around it?) Enclosed inside a resonator is easiest. If you need empty space, probably the best setup is two parabolic antennas facing each other forming a free space resonator - the beam would reflect back and forth, similar to between optical mirrors $\endgroup$
    – Alex I
    Nov 8, 2023 at 3:33
  • $\begingroup$ In the case of a closed resonator, the only losses would be in the conductor forming the walls (resistive heating), in whatever you have inside the field (dielectric loss, very small if you have only helium gas), and leakage back through the structure you're feeding power in through. I think in practice you can get away with 20-50x less power for a certain field strength in a resonator than in a non-resonant structure. In a "free space resonator" you also have to consider loss due to leakage through the open sides (since the beam is not perfectly directional) $\endgroup$
    – Alex I
    Nov 8, 2023 at 3:50

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