1) Type of magnetic, electric distributions are available in EM simulation tool. Which is the proper explanation of current distribution?
2) theoretically, patches are have high magnitude in the edges and maximum current distribution on center of the patch at the resonant frequency. The distribution strongly depends, height of substrate which is presented for antenna size.
The patch is a short resonant length (nearly half a wavelength) of low-impedance microstrip with an open circuit at each end. At each open circuit, the electric field is close to an anti-node, so alternates between high positive and negative values. The magnetic field is close to a node, so is very small. Under the centre of the patch there is an electric field node, and a magnetic field anti-node, so there is a line with zero electric field and alternating high magnetic field. The electric surface current density at each position on the lower surface is equal to the magnetic field under the patch (Js=H). The radiation from the patch is usually treated as radiation from two magnetic dipoles, one in the dielectric gap at each end of the patch, where the strength of the magnetic dipole field is given by the displacement current edE/dt. The magnetic dipoles run along the length of each slot, and are in phase with each other. In contrast to the gaps at the ends, the sides of the patch do not radiate much. Most of the electric currents are on the underside of the patch and on the ground plane directly below them, as they are on microstrip. The patch and ground currents are close together and in opposite directions, and hidden by the patch, which is why these currents don't radiate.
You can think of the patch as a series resonant circuit with a capacitor to earth at each end (the gaps) and an inductor in the middle. The higher the substrate (and the narrower the patch) the smaller the capacitors and the bigger the inductance, and thus the higher the impedance.
@Malcolm White,
Dear Sir, thanking you for your valuable time and the brief explanation will help for researchers.
Make sure you understand how a half-wave patch radiates. The current must be zero at the end of the patch, and the voltage is out of phase with the current. Hence the voltage is at a peak at the end of the patch, and a half-wavelength away at the start of the patch, the voltage has equal magnitude but out of phase. It is this voltage, out of phase, which produces fringing fields that coherently add in phase and produce radiation. Using this argument, you should understand that if this patch radiates at f0, it will also radiate at 3*f0. Can you think why it would not radiate well at 2*f0?
If you understand this, then you understand a single frequency. If you observe that a patch radiates at some frequency f1, then find the corresponding wavelength, and try to think how the current would flow on the patch (for your configuration) such that the voltage produces fringing fields that add up in phase to produce radiation.
The current distribution must be zero at the ends of the patch; current can't flow "off" the patch. The voltage is out of phase with the current. You have to think about how the current might flow to produce radiation; this takes thought and creativity. You need to think how the various "modes" or how the current will flow at each frequency; the current distribution will be different at different frequencies.
Respected to @ Muhammad Usman Memon ,
Thanking you for sharing discussion, the valuable suggestions are noted down and discussion started. The fringing effects of transmission line model and cavity model are studying very carefully and the dominant modes for patches are also taking notes very seriously.
EM simulation tool (HFSS) helping lot for studying of current distribution, so your comments are very helping for my research. Once again thanking you for sharing.
Yours faithfully,
Mohan K N
Dear @Chow Feng @ Muhammad Usman Memon @Malcolm White,
After the simulation from EM tool,
I observed that thickness of dielectric substrate is small (h
To create radiation there must be charge acceleration and deacceleration or we can say time varying current. Charges constitutes a current. So on a patch how the charges get accumulated and where they are accumulating is explained by the current distribution of the patch.
Charge distribution on the patch surface ,it gives the maximum accommodation of charges at the centre of pach and co called node at resonance frequencies within a BW.Other nodes are found at the (2n+1)*lembda/4 distance from the centre node .If several frequencies with low return loss are available in a given band or multi band antenna,the it becomes decide to consider central frequency and its value of wavelength.
Transmission model analysis will help in this context as the radiation from the patch is usually thought as radiation from two magnetic dipoles. Taking half wavelength patch, You can explain with the charge distribution concept
Thank you all for your guidance, the question had valuable answers and discussions.
Thank you once again all
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