Open circuit/ Short circuit behavior of an antenna

What does physically happen when a receiver antenna in Open circuit/ and short circuit mode is exposed to RF signal? Is the power reflected back?

A dissipative short circuit will absorb the power and the reflection will be less (compared to an open circuit). As an open circuit, the antenna will act as a resonator and reflect some power depending on the configuration.

I think the antenna becomes a generator. Passing photons induce AC voltage in the antenna. The amplitude is greatest when no load is attached. The antenna produces a waveform which is a jumble of frequencies. We can feed it to a bandpass filter (tuning it to one frequency), or to an amplifier. Depending on the type of device, it may load the antenna, drawing current, and reducing voltage.

How does current flow in the antenna? Do the photons induce true AC, or is it AC with a DC component? Etc. The above is merely my non-expert thought experiment, and I could be wrong.

Wonder what a dissipative short circuit is? An ideal short has zero ohms and thus can't dissipate power.

A non-ideal antenna might have internal losses which can show with any load, also open circuit. We would need to analyze the antenna design details.

An unloaded antenna without internal losses doesn't absorb any power from an electromagnetic field, but diffracts it to some extent.

FvM said:

Wonder what a dissipative short circuit is? An ideal short has zero ohms and thus can't dissipate power.

A non-ideal antenna might have internal losses which can show with any load, also open circuit. We would need to analyze the antenna design details.

An unloaded antenna without internal losses doesn't absorb any power from an electromagnetic field, but diffracts it to some extent.

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I was trying to come up with the tentative current waveforms inside a receiver dipole in open circuit configuration, but I faced a contradiction...I assume the current needs to be zero at the both ends of each petal...But how can that be possible without keeping the wavelength unchanged?

but I faced a contradiction...I assume the current needs to be zero at the both ends of each petal...But how can that be possible without keeping the wavelength unchanged?

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Yes, there can't be a current at an open wire antenna end. But contradiction to what? It's not said that the open circuit dipole halves must act as lambda/4 resonators. There can be still some current induced in the wire center.

BradtheRad said:

How does current flow in the antenna? Do the photons induce true AC, or is it AC with a DC component

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Consider, to fix ideas, a single photon passing through a circular antenna (it can be any shape). The electric field induces an AC in the antenna and there will be no DC unless externally applied to the antenna. To see the induced voltage, just cut the circle (antenna) at any point and you will see the AC potential that can be used for your project...

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FvM said:

Wonder what a dissipative short circuit is? An ideal short has zero ohms and thus can't dissipate power.

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An interesting observation: but...

In terms of the current: power P = (current ^2) * resistance. So if the resistance is zero the power will be zero.

In terms of the voltage: power P = (voltage ^2) / resistance. So if the resistance is zero the power will be infinite.

It can be written independent of the resistance: power P = voltage * current. So, if both voltage and current are finite, there will be some power produced. I agree that it is difficult to understand how a circuit with zero resistance can support a finite voltage.

An interesting observation: but...

In terms of the current: power P = (current ^2) * resistance. So if the resistance is zero the power will be zero.

In terms of the voltage: power P = (voltage ^2) / resistance. So if the resistance is zero the power will be infinite.

Click to expand...

Antenna's don't work like this. The antenna is always driven by a finite source impedance (transformed free space impedance https://en.wikipedia.org/wiki/Impedance_of_free_space), respectively current and voltage are always finite.