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Is it practically possible to reach 1 A current AC in the metal conductor with frequency 2 GHz?

Or in other words, if I have plain metal wire, or maybe a thin tube/foil to reduce skin effect, what is the maximum current I can reach with 2 GHz frequency (How many Amperes (amps))?

If there is some limit for the current, what is the reason for this limit?

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3 Answers 3

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There is no fundamental theoretical limit on the current. From the formula for the radiated power of a single wire derived in my answer here, the radiation resistance is given by: $$R_{rad}=\frac {2P_{rad}}{I_0^2}=\frac { \eta \ }{4 \pi} \biggl[\frac 1{kl} \biggr(2kl(-Ci(2kl)+\ln(kl)+\gamma-1+\ln 2)+\sin(2kl)\biggr)\biggl]$$

From this you can see that this resistance increases monotonically with frequency $f\sim kl$ (check the plot of my linked answer above). So, in higher and higher frequencies, the voltage needed to supply a specific current increases. But, as long as you have the necessary supply, you can always create such currents. However, there is some point at which practically supplying such a voltage becomes impossible because of the technological limitations on such high power voltage supplies at such high frequencies. At some ridiculously high voltages, even the air surrounding the circuit experiences dielectric breakdown and everything falls apart (the actual limitations are far below such a ridiculous voltage). This is all assuming that you have no skin effect and no frequency dependence in the actual copper resistance of your wire, which introduces further limitations on the current. In an actual wire, the copper resistance increases with frequency, which means that the $E_d=\frac12RI_0^2$ heat created completely melts the wire.

Nonetheless, I don't have enough experience with such high voltage sources at those frequencies to give you any numbers on when the limitations actually become practically impossible to overcome.

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  • $\begingroup$ Thanks a lot for your answer. Maybe you can tell any number from your mind what is really possible? Is it possible to reach practically 1 Amper or it is too high for 1-2 GHz? $\endgroup$
    – Zlelik
    Commented Dec 4, 2017 at 10:11
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For a given wire, there is a limit. This is due to its finite thickness, which won't support currents higher than that which would destroy the wire via melting or internal stressed due to ponderomotive forces.

If a wire is too thin to transport 1A at 2GHz, then there is a reason to use a different wire, with greater diameter. If the frequency is high enough, due to skin effect the current will be mainly near the wire surface, so it is a good idea to use a hollow conductor (a pipe). The greater the diameter, the greater the surface of the wire available for current distribution.

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1 A is only 50 watts down a 50 Ω cable. Most of the cables in this table can handle 50 W at 2 GHz.

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  • $\begingroup$ I think 50Ω is the intrinsic impedance of the coaxial cable, i.e. the ratio of the electric field strength of the dielectric medium to the magnetic field H strength in the dielectric medium, but I wonder if the relationship I=V/R holds ? $\endgroup$
    – HEMMI
    Commented Jan 25, 2023 at 11:01
  • $\begingroup$ @HEMMI $I=V/Z_0$ holds if the standing wave ratio is one (you have a pure traveling wave). For 50Ω cable, this happens when the termination resistance is 50Ω. $\endgroup$
    – John Doty
    Commented Jan 25, 2023 at 14:09

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